PROBLEMS AND SOLUTIONS INCOMMUTATIVE ALGEBRA

Mahir Bilen Canmcan@tulane.edu

Disclaimer: This file contains some problems and solutions in commutative algebra aswell as in field theory. About first hundred problems are those that we encountered at somepoint probably between years 2003 and 2005. We do not claim correctness of those solutions(neither of the other solutions). Read them at your own risk. However, we do appreciate ifyou send us corrections and suggest new problems and solutions.1

Notation: Unless otherwise stated all rings are assumed to be commutative with unity.

1. Find two ideals I and J in a ring R such that I · J 6= I ∩ J.Solution.

Let I = J be the ideal generated by x in the polynomial ring R = k[x]. ThenI ∩ J = I = (x). Since the product of two ideals consists of finite sum of products ofelements of I and J , the ideal product I ·J is equal to (x2) which is different from (x).

2. Definition: ideals I and J from R are called co-prime, if their sum I+J is equal to R.

Show that if I and J are two co-prime ideals in a ring R, then I · J = I ∩ J.Solution.

For all ideals I and J the inclusion I · J ⊆ I ∩ J is clear. To prove the other inclusionwe observe the simple fact that, for any ideal K of R, the following equality KR = Kis true. Hence, assuming I and J are co-prime, on the one hand we have I ∩ J =(I ∩ J)(I + J) = (I ∩ J) · I + (I ∩ J) · J . On the other hand, (I ∩ J) · I ⊆ J · I and(I ∩ J) · J ⊆ I · J . Equality is now obvious.

3. Definition: A multiplicative subset S of a ring R is a multiplicative submonoid of R.

Let S be a multiplicative subset in a ring R and I be an ideal.1We thank Professor Lex Renner for his comments and critical eye on some of the problems with faulty

solutions. We thank Şafak Özden, also.

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(a) Show that S−1I := {a/s : a ∈ I, s ∈ S} is an ideal in the localized ringS−1R = {r/s : r ∈ R, s ∈ S}.

(b) Show that the localization commutes with quotients: S−1R/S−1I ∼=S−1(R/I). Here we are abusing the notation on the right: of courseR/I is localized at the image of S in R/I.

Solution.

(a) Let a1/s1 and a2/s2 be two elements from S−1I. And let r/s ∈ S−1R . Thenr/s · a1/s1 + a2/s2 = (s2ra1 − s1sa2)/s1s2 ∈ S−1I. Therefore S−1I is an ideal.

(b) Elements of S−1(R/I) are of the form r/s where bar denotes the images of elementsof S in R/I. If we start with an element r/s of S−1R, then r/s makes sense. So wecan define the homomorphism φ : S−1R −→ S−1(R/I) by φ(r/s) = r/s. By itsconstruction φ is surjective. How about its kernel? Suppose r/s = 0 in S−1(R/I).Then there exists s′ in the image of S in R/I such that r · s′ = 0 that is rs′ ∈ I. Butit is always the case that rs′/ss′ = r/s in S−1R. Therefore r/s ∈ S−1I. This showsthat the kernel of φ is S−1I, hence we get the desired isomorphism.

Remark 0.1. Let S be the complement of a prime ideal P in a ring R and I be anideal contained in P so that I ∩ S = ∅. By the above proven fact, RP/IP ∼= (R/I)P .Some authors write IRP for what we are calling IP ; the ideal generated by the imageof I in RP .

4. Let φ : A −→ B be a ring homomorphism by which B has a finitely gen-erated A-module structure. It is easy to verify that for any multiplicativesubmonoid S ⊂ A the image φ(S) ⊂ B is a multiplicative submonoid also.Show that the induced homomorphism φS : AS −→ Bφ(S) gives Bφ(S) a finitelygenerated AS-module structure. (Here, without loss of generality we as-sume that 0 /∈ φ(S). Otherwise, Bφ(S) = 0, which is a finitely generatedAS-module.)

Solution.

Let {b1, . . . , br} be a generating set for B as an A-module. Let a/b be an element ofthe localized ring AS. For c/φ(d) ∈ Bφ(S) with c ∈ B, d ∈ S the action of AS on Bφ(S)

is defined bya

b· c

φ(d)=φ(a) · cφ(bd)

.

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Since B is generated by {b1, . . . , br} as an A-module, c is of the form a1 ·b1 + · · · ar ·br =φ(a1)b1 + ·+ φ(ar)br for some a1, . . . , ar ∈ A. Therefore,

c

φ(d)=φ(a1)b1 + ·+ φ(ar)br

φ(d)=

r∑i=1

aid· bi

1

proving that {b1/1, b2/1, . . . , br/1} is a generating set for Bφ(S) as an AS-module.

5. Let P1, . . . , Pm be a finite set points from Cn such that P1 6= Pj for all j ∈{2, . . . ,m}. Find an explicit polynomial F (x1, . . . , xn) ∈ C[x1, . . . , xn] whichtakes constant value 1 on P1 and 0 on Pj for all j ∈ {2, . . . ,m}.Solution.Let (ak1, . . . , akn) ∈ Cn denote the coordinates of Pk for k = 1, . . . , n. Since P1 6= Pj,there exists smallest index rj ∈ {1, . . . , n} such that ajrj 6= a1rj . Define gj(x1, . . . , xn)by

gj(x1, . . . , xn) :=xrj − ajrja1rj − ajrj

It is clear that gj(Pj) = 0 and gj(P1) = 1. The product of gj’s for j = 2, . . . ,m is thedesired polynomial F .

6. Definitions: Given two ideals I and J , the ideal quotient I : J is defined to be the ideal{h ∈ R : hJ ⊆ I}. The radical of an ideal I, denoted by rad(I), is the ideal consistingof elements r ∈ R such that some power rn, n ∈ N of r lies in I. Logically, we call anideal I radical if rad(I) = I.

Notation: Given an ideal I of a polynomial ring k[x1, . . . , xn], V (I) ⊆ kn denotes theset of points a ∈ kn such that all polynomials from I vanishes on a.

Let I and J be two radical ideals. Show that the ideal of the Zariski closureV (I) \ V (J) coincides with the quotient ideal I : J.

Remark 0.2. (a) Hilbert’s Nullstellensatz in commutative algebra says that for analgebraically closed field k, and for any finitely generated polynomial ideal Jthe ideal of the vanishing locus of J is equal to radical of J . In other words,I(V (J)) = rad(J).

(b) The complement V (I) \ V (J) of V (J) in V (I) need not to be an algebraic set (itis an open subset of V (I)). It doesn’t make sense to talk about the ideal of anopen subset. We must take its closure so that we can talk about the ideal of theclosed set.

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(c) The quotient I : J need not be a radical ideal in general. If I is a radical ideal,then so is I : J ; suppose fn ∈ I : J for some n. Then fnhn ∈ I for any h ∈ J .But I being radical, fh ∈ I hence f ∈ I : J .

(d) f ∈ R is not a zero divisor in R/I if and only if I = I : f . In this case, the varietyof the ideal generated by f and I has dimension one less than V (I).

For obvious reasons we assume that V (I) is not equal to V (J) (otherwise there isnothing to prove).

Let α ∈ V (I) \ V (J) be a point, hence there exists a polynomial f in J such thatf(α) 6= 0.

We claim that each element h of I : J vanishes on α, that is to say h(α) = 0. Indeed,hf ∈ I. But f(α) 6= 0, so every element h of I : J vanishes on α ∈ V (I) \ V (J).It follows that I : J lies in the ideal of the closure of the complement. Conversely,if we take a polynomial f vanishing on the closure of V (I) \ V (J), then obviously itvanishes on V (I) \ V (J). Then for any h ∈ J , fh vanishes on all of V (I); f vanisheson the complement and h vanishes on V (J). Thus fh ∈ I since I is radical. But thenf belongs to the quotient ideal I : J .

7. Let M be a finitely generated R-module and a ⊂ R an ideal. Supposeψ : M → M is an R-module map such that ψ(M) ⊆ aM . Find a monicpolynomial p(t) ∈ R[t] with coefficients from a such that p(ψ) = 0. Here, aMis the module consisting of all finite sums of elements of the form bm, whereb ∈ a and m ∈M .

Solution.

The solution technique is important here. Let {x1, . . . , xm} be a generating set for Mas an R-module. By hypothesis, for each i = 1, . . . ,m we have

ψ(xi) =∑

ai,jxj, (1)

for some ai,j ∈ a. We define Ai,j to be the operator δi,jψ−ai,je, where e is the identityendomorphism of M and δi,j is the Kronecker’s delta function.

It is clear from (1) that∑

j=1Ai,jxj = 0 for all i = 1, . . . ,m. In other words, the matrixof operators A := (Ai,j)

mi,j=1 annihilates the column vector v = (xj)

mj=1. Notice that

we can consider M as an R[ψ]-module, and that Ai,j ∈ R[ψ]. Thus A is a matrix overR[ψ]. Therefore, its adjugate makes sense and multiplying Av = 0 on the left by theadjugate gives us (detA)xj = 0 for all j = 1, . . . ,m. Consequently, detA annihilatesall ofM . Expanding the determinant we obtain a monic polynomial p in ψ with entriesfrom a. Furthermore, p(ψ) = 0 on M .

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8. If M is a finitely generated R-module such that aM = M for some ideal a,then there exists x ∈ R such that 1− x ∈ a and xM = 0.

Solution.

By the previous problem we observe that the identity operator id on M satisfies amonoic polynomial: p(id) = idr + a1id

r−1 + · · ·+ arid = 0 for some aj ∈ a. Therefore,if we define x = 1 + a1 + · · · + ar, then x − 1 ∈ a and furthermore xm = 0 for allm ∈M .

9. If a ⊂ R is an ideal such that every element of 1 + a is invertible, M is afinitely generated R-module, and aM = M , then M = {0}.Solution.

Let x ∈ R be as in the previous problem; 1 − x ∈ a and xM = {0}. In particularx− 1 ∈ a, hence x = 1 + x− 1 is invertible. It follows from xM = {0} that M = {0}.

10. Let Jac(R) denote the intersection of all maximal ideals in R. (Jac(R) iscalled the Jacobson radical of R.) Show that x ∈ Jac(R) then 1 − xy isinvertible for any y ∈ R. Conversely, if 1− xy is invertible for all y ∈ R, thenx belongs to all maximal ideals.

Solution.

Suppose x is from Jac(R). If 1−xy is not invertible, then it is contained in a maximalidealm of R. In particular, since x is fromm, we see that 1 ∈ m which is a contradiction.

Conversely, suppose that 1−xy is invertible for all y ∈ R. If x does not lie in a maximalideal m, then the ideal generated by x and m is equal to R. Hence, xy + m = 1 forsome y ∈ R and m ∈ m. In this case, m = 1−xy ∈ m, so it is not a unit, contradictingwith our initial assumption. Therefore, x ∈ m.

Definition: Have a taste of Zorn’s lemma: A Noetherian ring is a ring in which everynon-empty set of ideals has a maximal element. For other definitions and properties ofNoetherian rings see 0.60.

Fact 0.3. Artin-Rees Lemma: Let a be an ideal in a Noetherian ring R andletM be a finitely generated R-module. If N ⊂M is a submodule, then thereexists a positive integer k such that for all n ≥ k, anM ∩N = an−k((akM)∩N).

11. Let R be a Noetherian ring and a be an ideal such that every element of1 + a is invertible in R. Show that ∩n>0a

n = (0).

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Solution.

Let M denote ∩n>0an. Obviously, M is a R-submodule of a. By Artin-Rees lemma,

there exists n such that an+i ∩M = ai(M ∩ an) for all i ≥ 0. Since set theoreticallywe have M ∩ an+i = M for any n+ i, we get M = aiM for any i ≥ 0. By Problem 9above, we get M = 0.

12. If a is an ideal such that every element of 1 + a is invertible, M a fininitelygenerated R-module and M ′ ⊆ M a submodule, then M ′ + aM = M impliesthat M ′ = M .

Solution.

We consider the R-module M/M ′. Our assumption M ′ + aM = M implies thataM/M ′ = M/M ′. Hence, by Problem 9 M/M ′ = {0}, or equivalently, M = M ′.

13. If a is an ideal such that every element of 1 + a is invertible, M a fininitelygenerated R-module. Show that the elements u1, . . . , un generate M if andonly if the images u1, . . . , un generate M/aM as an R-module

Solution.

The implication (⇒) is obvious. We prove the converse. Suppose u1, . . . , un generateM/aM as an R-module. Let {u1, . . . , un} denote a set of preimages of ui’s, and letM ′ denote the submodule generated by ui’s. It is clear that M ′ + aM = M , hence byProblem 12 it follows that M ′ = M .

Definition: A local ring is a ring with unique maximal ideal.

14. Let (R,m) be a Noetherian local ring and suppose that the images of theelements a1, . . . , an ∈ m generate m/m2 as a vector space. Show that a1, . . . , angenerate m as an ideal.

Solution.

We denote by M the maximal ideal m viewed as an R-module, and denote by a themaximal ideal m viewed as an ideal. The solution is now an application of Problem 13.

15. In the notation of the previous problem, a1, . . . , an generates m/m2 as a vector

space, then a1, . . . , an generates m minimally, that is to say none of ai’s isredundant.

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Solution.

Towards a contradiction, without loss of generality, assume that a1 redundant; a1 =r2a2 + · · ·+ rnan for some ri ∈ R. Then, modulo m2, a1, . . . , an are linearly dependentwhich is a contradiction.

Definition. An ideal I 6= (1) is primary if fg ∈ I implies either f ∈ I, or gm ∈ I forsome m ∈ N.

16. Prove that if Q is primary, then rad(Q) is a prime ideal. Furthermore, inthis case, rad(Q) is the smallest prime ideal containing Q.

Notation: If P denotes the prime ideal rad(Q), then Q is called P -primary.

Solution.

Let fg ∈ rad(Q), hence (fg)n ∈ Q for some n ∈ N. Since Q is primary, either fn ∈ Q,or gnm ∈ Q for some m ∈ N. In other words, either f ∈ rad(Q), or g ∈ rad(Q)implying that rad(Q) is a prime ideal.

If M is a prime ideal such that Q ⊆ M , then rad(Q) ⊆ M because of the followingtwo things: First, for any two ideals I and J , I ⊆ J implies rad(I) ⊆ rad(J). Indeed,f ∈ rad(I), then fn ∈ I for some n ∈ N, hence fn ∈ J . In particular, f ∈ rad(J).Secondly, if an ideal J is prime, then J is equal to its own radical. To see this it isenough to show that rad(J) ⊂ J whenever J is prime. Let f ∈ rad(J), hence fn ∈ Jfor some n ∈ N. It follows primeness that f ∈ J .We apply this observation to our original problem. If M is a prime ideal containingQ, then rad(Q) ⊂ M . Therefore, we conclude that rad(Q) is the smallest prime idealcontaining Q whenever Q is primary.

17. LetM be a finitely generated R-module and S ⊂ R be a multiplicative subsetof R. Show that MS = 0 if and only if an element s of S annihilates M , thatis to say, sM = 0.

Solution.

If s annihilates M , then for any m/r ∈ MS, we have m/r = sm/sr = 0/sr = 0, thusMS = 0. Conversely, assume that MS = 0, that is m/r = 0 for every m ∈ M andr ∈ S. By definition this holds if there is an s ∈ R such that s(1 · m − 0 · r) = 0.Thus s · m = 0. Now, this s is specific to m. Since M is finitely generated, thereexists a finite generating set {m1, . . . ,mn} of M , and there exists a corresponding set{s1, . . . , sn} of annihilators. Since R is commutative, the product s1 · · · sn annihilatesall mi’s, hence it annihilates whole of M .

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Definition: Support of an R-module M is the set of all prime ideal p such that Mp

is non-trivial. Here, Mp is the localization of the module at the multiplicative subsetR− p.

18. Let M be a finitely generated R-module. Show that a prime ideal p is in thesupport of M if and only if the annihilator ideal ann(M) of M is containedin p.

Solution.

By Problem 17 we see that Mp = 0 if and only if there exists s ∈ R − p such thatsM = 0. But then ann(M) cannot be contained in p. This proves one implication.For the converse, suppose that p is in the support ofM . Then by the previous problemagain, there cannot be any s ∈ R \ p such that sM = 0. Hence ann(M) must becontained in p.

19. Show that a short exact sequence of R-modules:

0 −→ A1α1−→ A2

α2−→ A3 −→ 0 (2)

gives rise to a left exact sequence

0 −→ HomR(A3, N)α∗2−→ HomR(A2, N)

α∗1−→ HomR(A1, N) (3)

Solution.

Given a homomorphism f : A3 −→ N , we pull it back to a homomorphism fromA2 to N by fα2 : A2 −→ N . Hence we get a homomorphism HomR(A3, N)

α∗2−→HomR(A2, N).

Next, let us see that α∗2 is injective: suppose α∗2f = α∗2g for some f and g fromHomR(A3, N). Then, α2f = α2g. But α2 is surjective by exactness (1). Thus, f andg agree on every point of A3 showing that they are the same functions. Therefore α∗2is injective.

Next, let us see that im(α∗2) ⊆ ker(α∗1), namely α∗1α∗2 = 0. Let f ∈ HomR(A3, N). Thenα∗1α

∗2f = fα2α1 = f · 0 = 0 again by exactness of (1). Therefore im(α∗2) ⊆ ker(α∗1).

Finally, let us see that im(α∗2) ⊇ ker(α∗1): let f : A2 −→ N be in the kernel of α∗1,namely fα1 = 0. Define g : A3 −→ N as follows, let g(a3) be the value f(a2) forany a2 ∈ A2 such that a2 and g is well-defined. Notice that gα2 = f . Thereforeim(α∗2) = ker(α∗1) and (2) is left exact.

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20. Give an example of a module N and a short exact sequence such thatHomR(−, N) does not give a short exact sequence.

Solution.

Let p ∈ Z be a prime and consider the following exact sequence of Z-modules:

0 −→ Z p−→ Z −→ Z/p −→ 0

Apply HomZ(−,Z) and check that the result is not an exact sequence.

21. Let G be an abelian group and write G ' Zn⊕Gtorsion. Show that HomZ(G,Z) ∼=Zn.Solution.

Let f ∈ HomZ(G,Z), then f is determined by the images to the generators of thecopies of Z in G. The reason for not affected by the torsion part of G is the following:if a ∈ Gtorsion, then n · a = 0 for some n ∈ Z. Then 0 = f(0) = f(n · a) = nf(a) sincef is a Z-module homomorphism. Therefore effect of the torsion part of G is 0 showingthat an f is determined by the images of the generators of the copies of Z in G. So,the result follows.

22. In the category of R = k[x1, . . . , xn]-modules show that HomR(R(−a), R) 'R(a).

Solution.

Note that 1 in R(−a) is an element of degree a. Therefore, a homomorphism f ∈HomR(R(−a), R) ' R(a) of graded R-modules maps 1 to a degree a element in R thatis f(1) is of degree a. Now, since an R-module homomorphism R is determined by itsvalue on 1; we have an isomorphism of R-modules f ∈ HomR(R(−a), R) −→ R viaf 7−→ f(1). Thus if we declare 1 in R to be of degree a, namely, if we regard the imageas the graded module R(−a), then we get an isomorphism of graded R-modules.

23. What is Z/2⊗Z Z/3?Solution.

Let a⊗ b ∈ Z/2⊗Z Z/3. If b is 0 or 2, then a⊗ b = 0⊗ 0. If b = 1, then we can replaceit by 4 (since 4 ≡ 1 mod 3) and get a⊗ b = 0⊗ 0 again. Therefore Z/2⊗Z Z/3 = 0.

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24. More generally, show that Z/a⊗Z Z/b ' Z/ gcd(a, b).

Solution.

First of all remember that gcd(a, b) of two integers is the largest (hence unique) integerthat divides both a and b. Furthermore, there are integers x and y such that gcd(a, b) =ax+ by.

Now, that said, define φ : Z/a⊗Z Z/b −→ Z/ gcd(a, b) by r⊗ s 7−→ rs mod gcd(a, b).This map is a well-defined homomorphism of Z-modules. It is injective because: ifrs = 0 mod gcd(a, b), then rs = n(ax + by) for some n ∈ Z. Thus r ⊗ s = 1 ⊗ rs =1⊗ n(ax+ by) = 1⊗ nax = nax⊗ 1 = 0⊗ 1 = 0.

Surjectivity is clear by 1 ⊗ r 7−→ r mod gcd(a, b). Therefore φ is an isomorphism ofZ-modules.

25. Show that − ⊗R M is a right exact functor on the category of R-modules,but it need not be an exact functor.

Solution.

Let0 −→ A1

α1−→ A2α2−→ A3 −→ 0

be a short exact sequence of R-modules and let M be some other R-module. We wantto show that the following is a right exact sequence:

A1 ⊗M −→ A2 ⊗M −→ A3 ⊗M −→ 0

Since the maps are defined by ai⊗m 7−→ αi(ai)⊗m, right exactness is straightforwardto check.

A counter example to exactness of tensoring is the following:

0 −→ Z p−→ Z −→ Z/p −→ 0

This is an exact sequence of Z-modules. Assume that − ⊗Z Z/p gives us an exactsequence:

0 −→ Z⊗Z Z/pp−→ Z⊗Z Z/p −→ Z/p⊗Z Z/p −→ 0

But Z ⊗Z Z/p ∼= Z/p p−→ Z ⊗Z Z/p ∼= Z/p is the 0 map which, on the contrary, wassupposed to be an injection. Contradiction.

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26. For R-modules M , N , and P , prove that

HomR(M ⊗R N,P ) ' HomR(M,HomR(N,P )).

Solution.

Consider the map f ∈ HomR(M ⊗R N,P )φ−→ HomR(M,HomR(N,P )) 3 φ(f),

defined by φ(f)(m) = f(m⊗−) ∈ HomR(N,P ). Here, it should be clear that f(m⊗−)is a homomorphism from N to P for every m ∈M .

Let us prove that φ is an isomorphism of R-modules: If φ(f) = 0, then f(m⊗ n) = 0for every m⊗n ∈M ⊗RN , hence f is the zero homomorphism in HomR(M ⊗RN,P ).So, φ is injective.

If g ∈ HomR(M,HomR(N,P )), then for every m ∈ M , g(m) is an R-homomorphismfrom N to P . So, define f : M ⊗ N −→ P by f(m ⊗ n) = g(m)(n). It is clear thatwe get a homomorphism. Also, by definition, φ(f) = g. So, φ is surjective, hence anisomorphism of R-modules

27. It is easy to see that every direct sum of modules gives an exact sequence.Prove that the converse is not true; there is an exact sequence

0 −→ A −→ B −→ C −→ 0

such that B � C⊕

A

Solution.

Consider0 −→ Z/2 −→ Z/4 −→ Z/2 −→ 0

with obvious maps. This is an exact sequence. However, Z/4 � Z/2⊕Z/2.

28. For a short exact sequence of R-modules

0 −→ A1α1−−→ A2

α2−−→ A3 −→ 0

show that A2∼= A1 ⊕ A3 if and only if there is a homomorphism β2 (or a

homomorphism β1) with α2β2 = idA3 (with α1β1 = idA1).

Solution.

Let A2φ−−→ A1⊕A3 be an isomorphism. We have the following commutative diagram:

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0 A1 A2 A3 0

0 A1 A1 ⊕ A3 A3 0

id

α1

i

φ

α2

π

id

An element of A1 ⊕ A3 is of the form (a1, a3) for unique a1 ∈ A1 and a3 ∈ A3. Defineβ2 : A3 −→ A2 as follows: given a3 ∈ A3 there exists unique a2 ∈ A2 such that φ(a2)= (0, a3) since φ is an isomorphism. Set β2(a3) = a2. Then by the commutativity ofthe diagram it is clear that α2β2(a3) = a3.

Conversely, suppose that we have a homomorphism β2 : A3 −→ A2 such that α2β2 =idA3 . Then define ψ : A1 ⊕ A3 −→ A2 by ψ((a1, a3)) = α1(a1) + β2(a3). This is a welldefined R-module homomorphism. Let us see why it is an isomorphism.

Suppose ψ((a1, a3)) = α1(a1) + β2(a3) = 0. Then α1(−a1) = β2(a3). Compose thisequality with α2 to get 0 = a3. Then α1(−a1) = 0. But α1 is injective, so a1 = 0.Therefore ψ is injective.

Now, let a2 ∈ A2. Then α2(a2 − β2α2(a2)) = α2(a2) − α2(a2) = 0, therefore a2 −β2α2(a2) ∈ im(α1), say, α1(a1) = a2 − β2α2(a2). Then ψ((a1, α2(a2)) = α1(a1) +

β2(α2(a2)) = a2, therefore A1 ⊕ A3ψ−→ A2 is surjective and hence an isomorphism.

The case of β1 is proven similarly.

Definition 0.4. We recall some very basic definitions from homological algebra. Achain complex is a sequence of abelian groups with homomorphisms between them:

· · · ∂4−→ C3∂3−→ C2

∂2−→ C1∂1−→ C0

∂0−→ 0

satisfying∂n+1∂n = 0 for all n ≥ 0.

Last condition ensures that the image of ∂n+1 lies in the kernel of ∂n, hence nextdefinition makes sense: ith Homology group of the chain complex (C, ∂) is defined by

Hi(C) = ker ∂i/im ∂i+1.

These groups measure how far the chain complex from being exact. Note that thechain complex (C, ∂) does not need to be of infinite length, however, a finite lengthchain complex can be extended by adding trivial groups and trivial connecting homo-morphisms in between them.

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29. Compute the homology of the complex 0 −→ V1φ−→ V0 −→ 0 where V1 = V0 = k3

and φ is: 1 0 −1−1 1 00 −1 1

Solution.

Kernel of φ is:

abc

∈ V1 :a− c = 0−a+ b = 0−b+ c = 0

which is spanned by

111

. So, H1 is iso-

morphic to k. For H0, note that the image of φ is spanned by the image of the basis1

00

,0

10

,0

01

under φ. It is straightforward to see that the image of this basis

is nothing but the same basis. Thus, H0 is trivial.

30. Prove that for a complex V : · · · −→ Vn −→ . . . −→ V0 −→ 0 of finite dimen-sional vector spaces, the following equality is always true:∑

i=0

(−1)i dimVi =∑i=0

(−1)i dimHi(V )

This alternating sum is called the Euler characteristic of the complex. It isclear that the complex is exact then its Euler characteristic is 0.

Solution.

This follows from the basic linear algebra fact that if Vi+1 and Vi are two finite dimen-sional vector spaces and φi+1 is a linear map between them, then

dim(kerφi+1) + dim(imφi+1) = dim(Vi+1).

Alternating sum of these equations (for i ≥ 0) gives us the desired equality.

31. Let k be a field. Suppose k[x0, x1, . . . , xn] is graded by degree. Show that ithgraded component has vector space dimension

(n+ii

).

Solution.

The vector space basis for the ith graded piece is {xp00 xp11 . . . xpnn : t

∑pj = i}. So, the

problem is equivalent to finding the number of ways of distributing i candies to n+ 1children named x0, x1, . . . , xn. Of course, this is done in

(n+1i

)different ways.

13

Definition 0.5. Let M = ⊕n∈ZMn be a Z-graded R-module. The numerical functionHM : Z −→ Z defined as

HM(n) = dimk(Mn)

is called the Hilbert function of M .

The power seriesPM(z) =

∑HM(t)zt

is called the Hilbert series of M.

32. What is the Hilbert series of k[x1, . . . , xn]?

Solution.

Using Problem 31, we see that it is nothing but 1/(1− t)n.

33. Find the Hilbert series of k[x1, x22, . . . , x

nn].

Solution.

The set of monomials contained in the ith graded piece is

{xi11 x2i22 . . . xninn : i1 + 2i2 + · · ·+ nin = i}

Therefore, the Hilbert series is given by∏

r≥11

1−tr .

34. Let P be a function P : N −→ Z such that the associated difference function∆P (i) := P (i) − P (i − 1) is a polynomial with rational coefficients (for suffi-ciently large i). Show that P itself is a polynomial with rational coefficientsand has degree one greater than ∆P .

Hint: Use induction on the degree s of the difference polynomial. The base case istrivial. If the leading coefficient of ∆P is as, then define h = ass!

(i

s+1

), and compute

∆h. By construction, ∆P −∆h will have degree s− 1.

Solution.

As given in the hint, obviously, h is of degree s− 1, and ∆h(i) = ass!(

is+1

)− ass!

(i−1s+1

).

Computing this we find ∆h(i) = ass!(i−1s

). Clearly ∆h is of degree s. At the same

time, the leading coefficient of ∆h is as. Thus, the degree of ∆h−∆P = ∆(P − h) iss− 1. By the induction hypothesis, this says that P − h is a polynomial with rationalcoefficients and of degree s. But h is a polynomial with rational coefficients of degrees+ 1, hence P is has the same type. This finishes the induction.

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35. Let M be a finitely generated, Z-graded module. Show that there exists apolynomial f(x) ∈ Q[x] such that for i� 0, HM(i) = f(i).

Definition 0.6. The polynomial f(i) is called the Hilbert polynomial of M , writtenHP (M, i).

Solution.

Applying induction on the number of variables in the ring over whichM is defined, thebase case is trivial. So, suppose it is true for n− 1 variables {x1, . . . , xn−1}. Using thehomomorphism, multiplication by xn, M(−1)

xn−→M , we can build an exact sequence:

0 −→ ker(xn) −→M(−1)xn−→M −→ coker(xn) −→ 0

Since multiplication by xn kills both ker and coker, they can be regarded as modulesover the polynomial ring of n − 1 variables. (Of course, they are finitely generated).By exactness, HM(i)−HM(i− 1) ∈ Q[i]. We are done by the previous exercise.

Definition 0.7. Let I be a homogenous ideal in R = k[x0, . . . , xn], hence it correspondsto a projective variety Y = V (I) in Pn with ring of global sections R/I. Let us writethe Hilbert polynomial of Y (or that of I) as the Hilbert polynomial of the coordinatering R/I:

HP (R/I, i) =amm!im +

am−1

(m− 1)!im−1 + . . .

Then we define the dimension of the projective variety Y ⊂ Pn as m, and its degree asam.

36. Compute the Hilbert polynomial of Pn and find its degree.

Solution.

The coordinate ring of Pn is k[x0, . . . , xn]. Thus, for large i’s we have HP (Pn, i) =HR(i) = dimk(k[x0, . . . , xn]i) =

(n+ii

)=(n+in

). Obviously this is a polynomial in i with

leading term 1n!in. Recall that for a Hilbert Polynomial as

s!is + . . . the degree is the

dimension of the variety and as is the degree of the variety. Thus, the degree of Pn is1.

Definition 0.8. Given n, d > 0, let M0, . . . ,MN be all the monomials of degree d inthe n+1 variables x0, . . . , xn where N =

(n+dn

)−1. We define a mapping ρd : Pn −→ PN

by sending the point P = (a0, . . . , ad) to the point ρd(P ) = (M0(a), . . . ,MN(a)). Thisis called d-uple embedding of Pn in PN .

15

37. Let θ : k[y0, . . . , yN ] −→ k[x0, . . . , xn] be the homomorphism defined by sendingyi to Mi (in the notation of the above definition). Let a be the kernel ofθ. Show that a is a homogenous prime ideal, and V (a) is an irreducibleprojective variety in PN .Solution.

Recall that an ideal is called homogenous if it is generated by homogenous polynomials.A relation among Mi’s is zero if and only if that relation is a homogenous polynomial.For example, ifM0 = x2

0,M1 = x0x1,M2 = x22, thenM0M2−M2

1 = 0 which is of coursey0y2 − y2

1. This is because each Mi has the fixed degree d. Therefore, the kernel ofθ is generated by homogenous elements. The image of θ is a subring of k[x0, . . . , xn]generated by Mi’s. Obviously being an integral domain is preserved for the subringsof integral domains. Therefore the quotient k[y0, . . . , yN ]/ ker θ is an integral domainshowing that ker θ is a prime ideal.

38. Show that the variety, ρd(Pn) defined in Problem 37 has the ring of globalsections equal to k[y1, . . . , yN ]/ ker θ, namely it is given as the locus of ker θ.

Solution.

The kernel of θ is generated by the polynomials that are zero when evaluated in themonomials M0, . . . ,MN in place of y0, . . . , yN . Therefore these are exactly the genera-tors of the vanishing ideal on the set of points (M0(a); . . . ;MN(a)) in PN . Thereforethe variety ρd(Pn) is cut out by the kernel of θ.

39. Show that ρd is a homeomorphism of Pn onto V (ker θ).

Solution

First, let us see that ρd is an injection. Suppose (M0(a); . . . ;MN(a)) = (M0(b); . . . ;MN(b)),namely ρd(a) = ρd(b) for some a, b ∈ Pn. Observe that ai’s cannot be all zero (oth-erwise Mi(a) = 0 for all i). Say ai 6= 0. Among monomials of degree d we haveadi and ad−1

i aj for j = 0, . . . , n. Since (M0(a); . . . ;MN(a)) = (M0(b); . . . ;MN(b)) inPN , we have (M0(a)/adi , . . . ;MN(a)/adi ) = (M0(b)/bdi ; . . . ;MN(b)/bdi ) in AN . There-fore we have ad−1

i aj/adi = bd−1

i bj/bdi for j = 0, . . . , n. But then aj/ai = bj/bi hence

(a0; . . . ; an) = (b0; . . . ; bn) in Pn. Thus ρd is injective. Since a map is surjective ontoits image. Therefore ρd is a bijection. It is clear that it is continuous (on each openpiece yi 6= 0 in PN it is given by continuous maps). It is enough to show that it has acontinuous inverse. The inverse map ρ−1

d maps (M0(a); . . . ;MN(a)) to a. Since at eachaffine open xi 6= 0, this reduces to ρ−1

d ((M0(a)/adi , . . . ,MN(a)/adi )) = (a0/ai, . . . , an/ai)and amongMj(a)/adi we have aj/ai for j = 0, . . . , n, we see that the inverse is basicallya projection. Therefore, it is continuous hence ρd is a homeomorphism.

16

40. Find the degree of d-uple embedding of Pn in PN .Solution

By definition, we want to compute the Hilbert polynomial of the k-vector space

k[y0, . . . , yN ]/ ker θ,

where θ(yi) = Mi(x). This ring is isomorphic to the subring k[M0, . . . ,MN ] of k[x0, . . . , xn]generated by all the monomials of degree d. Therefore it suffices to compute the Hilbertpolynomial of the graded (by degree) vector space k[M0, . . . ,MN ] =

⊕Si. Obviously

Si = 0 if i is not a multiple of d. Let i = rd for some nonzero r, then Si is equal to theith piece of k[x0, . . . , xn]. Thus the Hilbert polynomial is equal to

(n+rdrd

)=(n+rdn

)as a

function of r = 0, 1, . . . . This is a polynomial in r;(n+ rd

n

)=

(n+ rd)(n+ rd− 1) · · · (rd+ 1)

n!.

The leading coefficient of this polynomial is dn/n!. Therefore, the degree (leading termtimes n!) of ρd(Pn) is dn.

Fact 0.9. If partially ordered set S has the property that every chain has an upperbound in S, then the set S contains at least one maximal element.

Definition 0.10. The nilradical of a ring is the set of all nilpotent elements of thering.

41. Show that a nilradical is an ideal. Furthermore, show that the nilradical isequal to the intersection of all prime ideals of the ring.

Solution.

Let n denote the nilradical of the ring R. Let x and y be two elements from n,and suppose n,m ∈ N are such that xn = ym = 0. Binomial theorem implies that(x+ y)n+m = 0, hence n is closed under addition. Let a ∈ R. Since R is commutative,(ax)n = 0, hence ax ∈ n. Therefore, n is an ideal. To prove that n is the intersectionof all prime ideals, we first make the following simple observation: If P is prime idealP ⊂ R, then the quotient ring R/P (which is an integral domain) does have anynilpotent elements. Therefore, n ⊆ P . This proves that n ⊆ ∩P : primeP .

For the opposite inclusion, it suffices to show that for any non-nilpotent element x,there is some prime ideal that does not contain x. Towards this contradiction weassume that ∩P : primeP − n is non-empty. Let S be the set of all ideals that does notcontain any power of x. Since n ∈ S, we know that S is non-empty. Furthermore,if I1 ⊂ I2 ⊂ · · · is a nested sequence of elements from S, then ∪Ii ∈ S. Hence, byZorn’s lemma (Fact 0.9) S has a maximal element M ∈ S. Notice that if we prove M

17

is a prime ideal, then we succeed in our goal that there is a prime ideal that does notcontain x.

Assuming M is not prime, we take two elements a, b ∈ R − M such that ab ∈ M .The ideals Ma generated by a and M , and Mb, generated by b and M both properlycontain M , hence both of them contains a power of x: ay+m1 = xp and bz+m2 = xq.Since (ay + m1)(bz + m2) = abyz + aym2 + bzm1 + m1m2 lies in M we conclude thatxpxq = xp+q is contained in M also. This contradiction shows that M is a prime ideal,hence our proof is complete.

Fact 0.11. Recall that the Jacobson radical Jac(R) of a ring R is the intersection ofall maximal ideals of R. Given an ideal I ⊂ Jac(R) and a finitely generated R-moduleM , Nakayama’s Lemma says that

IM = M implies M = 0.

An R-module P is projective if there exists a module K such that P⊕

K ' F forsome free module F . Equivalently, every short exact sequence

0 −→ N −→Mf−→ P −→ 0

splits; there exists P h−→M such that f ◦ h = idP .

Remark 0.12. Being projective is transitive in the following sense: ifM is a projectiveB-module (henceM is a direct summon of a free B-module F ) and if B is an A-algebrawith free A-module structure, then M is a projective A-module as well. The reasonfor this transitivity is that F is a free A-module, hence M is a direct summand of afree A-module.

42. Show that the following are equivalent for a finitely generated Z-module M :

(a) M is projective;

(b) M is torsion free;

(c) M is free.

Solution. (c) =⇒ (b) and (c) =⇒ (b) are obvious. Suppose thatM is torsion free andlet {m1, . . . ,mr} be a generating set for M . Define ψ : M −→ Zr by ψ(mi) = ei andextend by linearity, where ei is the ith standard vector in Zr. Since none of the mi’s aretorsion, this is a well defined map of Z-modules. Furthermore, it is an isomorphism.So (b) =⇒ (c). Finally, since M is a direct summand of a free module, it is torsionfree; (a) =⇒ (b). So, we are done.

18

Definition 0.13. For an integral domain R, a fractional ideal is an R-submodule Aof the fraction field K of R such that for some nonzero element d ∈ R the followinginclusion holds:

d · A =

{da

b:a

b∈ A

}⊂ R.

Equivalently, A = d−1I for some ideal I ⊂ R and a nonzero d ∈ R. In particular,any ideal of R is a fractional ideal. The product of two fractional ideals A = d−1I,B = r−1J is another fractional ideal AB = (dr)−1IJ . This operation defines a monoidstructure on the set of all fractional ideals of R. Indeed, the identity fractional ideal isthe ring R itself.

The unit group of this monoid contains all principal ideals. Obviously, the set of allfractional principal ideals forms a subgroup of the unit group of the monoid. Invertiblefractional ideals modulo its subgroup of principal fractional ideals is called the classgroup of R. Thus the class group of an integral domain R measures how far is R frombeing a PID.

43. Show that in an integral domain R with fraction field K, if a fractional idealA is invertible, then it is a projective R-module.

Solution.

Assume that A is an invertible fractional ideal. Let A−1 be its inverse. Then a1a′1 +

· · · + ana′n = 1 for some ai ∈ A and a′i ∈ A−1 since AA−1 = R. Let S be a free R

module of rank n say generated by y1, . . . , yn. Define φ : S −→ A by φ(yi) = ai andextend it by linearity. Define also ψ : A −→ S by ψ(c) = c(a′1y1 + · · · + a′nyn). Thismakes sense because ca′i ∈ R. Obviously, φψ = idA, so A is a direct summand of F .In other words, A is a projective module.

Remark 0.14. The converse of this problem is true also: if a fractional ideal A isprojective, then it is invertible

44. Show that the ideal (2, 1+√−5) of Z[

√−5] is a projective Z[

√−5]-module but

not free.

Solution.

Note that Z[√−5] = 1 · Z⊕

√−5 · Z, therefore Z[

√−5] is a free Z-module of rank 2.

If I := (2, 1 +√−5) were free Z[

√−5]-submodule of Z[

√−5], then it would be of rank

one. Thus there would be a single generator; I ' αZ[√−5] for some α ∈ I. But it

is easy to see that 2 and 1 +√−5 are Z[

√−5]-linearly independent; αZ[

√−5] cannot

generate I. Therefore I cannot be a free Z[√−5]-module. However, I2 is the ideal (2).

19

Since (2) is an invertible ideal with inverse (1/2)Z[√−5], I is an invertible ideal with

inverse I · (1/2)Z[√−5], hence I is a projective module.

45. Use Nakayama’s Lemma to prove that a finitely generated projective mod-ule over a local ring is free.

Solution.

Let {w1, . . . , wn} be a minimal set of generators for a finite projective module M andlet F be a free module of rank n. Consider the following surjection φ : F →M :

φ(∑n

i=1aiei) =

∑n

i=1aiwi,

where ei’s are the generators for F . Since we have declared wi’s to be the minimalgenerating set for M , if

∑ni=1 aiwi = 0, then we should have ai’s in the maximal ideal

m of the local ring. Otherwise, they would be units and that would contradict theminimality of wi’s. Thus kerφ ⊂ mF . Now, for M being a projective module andhaving a surjection φ on it, we get a splitting of F via an injection ϕ : M −→ Fsuch that φϕ = idM . Therefore we can write F = ϕ(M) ⊕ kerφ. Since we havethe containments mϕ(M) ⊂ ϕ(M) and kerφ ⊂ mF and since kerφ is disjoint fromϕ(M), we see that kerφ ⊂ m kerφ. So, kerφ = m kerφ. Now, by Nakayama’s Lemma,kerφ = 0, hence φ is an isomorphism.

Definition 0.15. Left Derived Functors: Let C denote the category of R-modulesand suppose F is a right exact, covariant, additive (preserving addition of homomor-phisms) functor from C to C. The left derived functors LiF (N) of an R module N aredefined as follows: take a projective resolution of N (for instance, a free resolution)and apply the functor F to the exact sequence omitting N . The new sequence is acomplex and LiF (N) is defined as the ith homology of this complex.

46. Let C denote the category of graded C[x]-modules and let F stand for thefunctor − ⊗C[x] (C/x) on C. Compute the left derived functors LiF of N =C[x]/x2 for all i.

Solution.

Here is a projective resolution for N :

0→ (x2)·x2−→ C[x]→ N → 0

Applying F to it, we get the following complex, say P :

0⊗C[x] C[x]/x0⊗1−−→ (x2)⊗C[x] C[x]/x

·x2⊗1−−−→ C[x]⊗C[x] C[x]/x→ 0

20

Then

L0F (N) = H0(P ) = (C[x]⊗C[x] C[x]/x)/x2 · C[x]⊗ C[x]/x

= (C[x]/x)/(x2 · C[x]/x)∼= (C[x]/x)

because (x2 · C[x]/x) = 0. Therefore L0F (N) = (C[x]/x).

Next,

L1F (N) = H1(P ) = ker(·x2 ⊗ 1)/im(0 · 1)

= ((x2)⊗C[x] C[x]/x)/0

= (x2C[x]/x)/0 = 0.

Of course, LiF (N) = Hi(P ) = 0 for all i > 1.

Definition 0.16. Tor. Left derived functors of the tensor products are called Tor.What we have computed in the previous problems are ToriC[x](N,C[x]/x).

47. Let M be an R-Module and let r ∈ R. Compute the R-modules ToriR(R/r,M).

Solution.

Consider the free resolution

0→ R·r−→ R→ R/r → 0

Apply −⊗RM to this by omitting R/r to get

0→ R⊗RM·⊗1−−→ R⊗RM → 0

This is equal to 0 → M·r−→ M → 0. Therefore ToriR(R/r,M) = M/r · M and

Tor1R(R/r,M) = {m ∈M : r ·m = 0}.

48. Let M be an R-module and let r ∈ R. Compute the R-modules ToriR(R/r,M).

Solution.

Consider the free resolution

0→ R·r−→ R→ R/r → 0

Apply −⊗RM to this by omitting R/r to get

0→ R⊗RM·r×1−−→ R⊗RM → 0

21

This is equal to 0 → M·r−→→ M → 0. Therefore Tor0

R(R/r,M) = M/r · M andTor1

R(R/r,M) = {M ∈M : r ·m = 0}.

49. Prove that for a homogenous polynomial f of degree d and a homogenousideal I ⊂ R, there is a graded exact sequence:

0→ R(−d)/I : f → R/I → R/〈I, f〉 → 0.

Hint: Clearly, 0 → 〈I, f〉/I → R/I → R/〈I, f〉 is exact. How can you get a gradedmap from R to 〈I, f〉/I?Solution.Obviously the exact sequence given in the hint is graded. Consider the graded mapR(−d)

·f−→ 〈I, f〉/I. The kernel is {g ∈ R : fg ∈ I} : f after a glance at the definitionof I : J. Thus, R(−d)/I : f ∼= 〈I, f〉/I. Thus, replacing 〈I, f〉 by R(−d)/I : f in theexact sequence given in the hint, we get the graded exact sequence:

0→ R(−d)/I : f → R/I → R/〈I, f〉 → 0

50. Show that a given f ∈ R is a nonzero divisor on R/I if and only if I : f = I.

Solution.Recall that I : J = {h ∈ R : hJ ⊆ I}. Suppose now that I : f = I, therefore if forsome h ∈ R it happens that h〈f〉 ⊆ I, then h ∈ I. So, f · (h mod I) = fh mod I ≡ 0mod I implies that h ∈ I. Thus, f is a nonzero divisor. Conversely, if f is a nonzerodivisor on R/I then for any h ∈ R with the property that h〈f〉 ⊆ I we must haveh ∈ I.

Remark 0.17. We defined the Hilbert polynomial of a ring in Problem 35. Supposethe initial term of the Hilbert polynomial of a quotient ring R/I is given by

HP (R/I, i) =amm!im + · · ·

If f is a homogenous linear form which is a nonzero divisor on R/I, then we have

HP (R/〈I, f〉, i) = HP (R/I, i)−HP (R/I, i− 1)

=am

(m− 1)!im−1 + . . .

We deduce that, by slicing with the hyperplane defined by f the dimension drops byone while preserving the degree.

22

Definition 0.18. Let M be an R module. An element x of R is called M -regular if itis not a zero divisor on M . An ordered sequence of elements x = x1, . . . , xn from R iscalled M -regular if the following two conditions hold:

(a) xi is M/(xi, . . . , xi−1)-regular for i = 1, · · · , n.(b) M/xM 6= 0

A weak M -regular sequence is defined by requiring only the first condition.

51. Show that given an R-module M over a local ring (R,m), a weak M-regularsequence x ⊂ m is always M-regular.

Solution.

Let x ⊂ m be a weak M -regular sequence and assume M/xM = 0 that is M = xM .But then by Nakayama’s Lemma M must be trivial.

52. Show that x, y(1− x), z(1− x) is an R-sequence, but y(1− x), z(1− x), x is notwhere R = k[x, y, z] a polynomial ring. Here, by an R-sequence we mean anR-regular sequence for R regarded as a module over itself.

Solution.

Multiplication by y(1 − x) on R/xR is the same as multiplication by y on k[y, z](∼=R/xR). So y(1 − x) is R/xR regular. And multiplication by z(1 − x) on R/(x, y(1 −x))R is the same as multiplication by z on k[z](∼= R/(x, y(1 − x))R. So, z(1 − x) isR/(x, y(1− x))R-regular. Furthermore, the ideal I = (x, y(1− x), z(1− x)) is equal to(x, y, z). Therefore R/IR ∼= k 6= 0. Thus x, y(1− x), z(1− x) is an R-sequence.

Since (y(1 − x), z(1 − x), x) is the same ideal with I above, we must show that thissequence fails to be a weak R sequence. Multiplication by z(1−x) on R/(y(1−x))R isnot regular: z(1− x) · y = 0 in R/(y(1− x))R. Therefore (y(1− x), z(1− x), x) cannotbe an R-sequence.

53. Let R be a Noetherian local ring, M a finite R-module, and let x be anM-sequence. Show that every permutation of x is an M-sequence.

Solution.

Since every permutation is the product of adjacent transpositions, it suffices to showthat x1, . . . , xi−1, xi, · · · , xn isM -regular. Note that for the moduleM/(x1, · · · , xi−1)M ,(xi, . . . xn) is a regular sequence. Therefore, by induction, it will suffice to handle thecase of n = 2. We want to show that x2, x1 is M -regular assuming X1, x2 is M -regular.Of course, we may assume that xi is not a unit, otherwise x1M = M , hence we would

23

be done. Now, look at the kernel of multiplication by x2 on M . Since x2 must beregular on M/xiM , ker(·x2) ⊂ xiM . If z ∈ ker(·x2), then z − x1m for some m ∈ M.Since xi(x2m) = x2(x1m) = 0, and x1 is regular on M, (x2m) must be 0. But thenm ∈ ker(·x2). Therefore ker(·x2) = xiker(·x2). Since x1 is a non unit, it is in themaxima ideal, hence any element 1 + (x1) is regular on M. Finally we have to showthat x1 is regular on M/x2M . Assume otherwise; there exists m ∈ M/x2M such thatxim ∈ x2M hence xim = x2m

′ for some m′ ∈M . Here m′ cannot be in xiM otherwisethe equation would imply m ∈ x2M. So, x2 is not regular on M/x1M, contradiction.Thus, x2, x1 is a regular sequence too.

Definition 0.19. Let A be a subring of a ring B and b ∈ B. Then b is call integralover A if b is a root of a monic polynomial with coefficients from A, that is if there isa relation of the form bn + a1b

n − 1 + · · ·+ an = 0 with ai ∈ A.

54. Let A ⊆ B be rings. Show that an element b ∈ B is integral over A if andonly if there exists a ring C between A and B such that b ∈ C and C isfinitely generated as an A-module

Solution.

(⇒) Let b ∈ B be integral over A. The let C be the ring generated by A and b, that isC := A[b]. Obviously A ⊂ CsubsetB. Let us see that C is indeed a finitely generatedmodule over A. Obviously any element of C is a polynomial in b with coefficients ofA. Since bn = −(a1b

n − 1 + · · ·+ an), any element of C can be written as an A-linearcombination of 1, b, b2, . . . , bn−1 hence C is finitely generated A-module.

(⇐) Suppose there exists an intermediate ring C which is a finitely generated Amodule.Then C = Awi + . . . Awn for some wi ∈ C. Let b ∈ C. Then wib =

∑j aijwj for

i = 1, . . . , n. But then we get a relation bn + a1bn−1 + · · · + an = 0 (by the Cayley’s

theorem; expanding det(bδij − aij)).

55. Let B be an integral domain and A ⊂ B a subring such that B is integralover A. Then A is a field if and only if B is a field.

Solution.

(⇒) Suppose B is an integral domain which is integral over a field A. Then any elementb in B satisfies a monic polynomial over A i.e, bn + a1b

n−1 + · · · + an = 0 for someai ∈ A. Since an is invertible, from this equation we see that b is invertible. Therefore,every element of B is invertible, hence it is a field.

(⇐) Suppose B is a field and A ⊂ B a subring such that B is integral over A. It isenough to show that for a ∈ A, the inverse a−1 of a also lies in A. Assume otherwise;b := a−1 6∈ A. But then it satisfies a monic polynomial over A, that is bn + a1b

n−1 +

24

· · · + an = 0 for some ai ∈ A. Multiplying this relation by an−1, we see that b ∈ Awhich is a contradiction. Therefore A is a field, also.

56. Let B be an extension ring of A which is integral over A. Let P ⊂ B be aprime ideal. Then P is maximal if and only if P ∩ A is maximal in A.

Solution.

P ∩A is obviously a prime ideal. The composition A ↪→ B → B/P gives the injectionA/A∩ P → B/P. Note that B/P is integral over A/A∩ P. Therefore, by the previousproblem that A/A∩B is a field if and only if B/P is a field. Hence A∩ P is maximalin A if and only if P is maximal in B.

57. Let A ⊂ B be rings and suppose B is integral over A. Let m be a maximalideal in A. Show that there exists a prime P in B lying over m, that isP ∩ A = m. Furthermore, any such P is maximal ideal of B.

Solution.

Let us first see that mB 6= B. Assume contrary that mB 6= B, then there exists bi ∈ Band πi ∈ m such that

∑ni=1 biπi = 1. Set C := A[b1, ...bn], then C is finite over A and

mC = C. Let C = Au1 +· · ·+Aur for some uj ∈ C. Then, we get ui =∑πijuj for some

πij ∈ m. Therefore ∆ := det(δij − πij) satisfies ∆uj = 0 for every j. Hence ∆C = 0.But since 1 ∈ C,∆ = 0. On the other hand ∆ ≡ mod m simply by its expansion.Hence 1 ∈ m, a contradiction. Now, since mB 6= B, it is contained in a prime ideal,say P. By the previous problem we know that P must be a maximal ideal.

58. Let A ⊂ B be rings ans suppose B is integral over A. Let p be a prime idealof A. Show that there exists a prime p ⊂ B such that P ∩A = p. Furthermore,there is not inclusion between primes in B that lie of p.

Localizing the exact sequence 0→ A→ B at p, we get an exact sequence

0→ Ap → Bp = B ⊗A Ap

in which Bp is an extension ring of Ap and integral over Ap. Using the following com-mutative diagram:

Ap Bp

A B

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we see that the prime ideals of B lying over p corresponds bijectively with the maximalideals of Bp lying over the maximal ideal pAp ⊂ Ap. Hence by the previous problemwe are done.

59. Let A ⊂ B be rings and suppose B is a finitely generated A-module( henceintegral over A.) Show that for a prime ideal of A, there are only finitenumber of prime ideals in B that lie over p.

Solution.

First we assume A to be a local ring with maximal ideal m. Since B is finitely generatedmodule over A,B = Awi + · · · + Awr for some wi ∈ B. Then, we the quotient ringB/mB becomes a vector space over A/m with a generating set {wi}. Every primeP ⊂ B containing mB gives a vector space P/mB ↪→ B/mB. Also note that by theprevious problems we know that if any prime lying over a maximal must be maximaltoo. Therefore these primes are coprime to each others. But then by the ChinesRemainder Theorem we see that B/mB ∼=

∏i Pi/mB. Since each Pi/mB contributes

to the dimension of the vector space, there must be finitely many Pi′s. So, in the localcase we are done. For the general case we make use of the diagram:

Ap Bp

A B

Since primes containing p ⊂ A correspond to the maximal ideals of Bp over pAp in thelocal ring Ap, we reduce to the local case. Hence we are done.

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Note: Next two problems have been added to the file on January 2015.

Definition 0.20. Let p be a prime number and consider the ring of p-adic integersZp, which is constructed as the inverse limit lim←−Z/p

iZ. Let Ai denote Z/pi+1Z forsimplicity. By definition, this is the ring of sequences (ai)i≥0 ∈

∏∞i=0Ai which satisfy

the compatibility condition with respect projections µji : Aj → Ai defined by

µji(aj mod pj+1) = ai mod pi+1 for all j > i.

60. Prove that there exists a bijection between Zp and the set of all formalpower series of the form

α0 + α1p+ α2p2 + · · · , where αi ∈ {0, 1, . . . , p− 1}.

Solution.

Let (ai)∞i=0 be an element from Zp. As ai ∈ Ai = Z/pi+1Z, i = 0, 1, . . . , we assume that

ai is a positive integer less than pi+1. Set α0 = a0 ∈ {0, . . . , p− 1}.Since a1 = a0 mod p, a1 − a0 is divisible by p. Since a1 < p2 and a0 < p, there existsunique α1 ∈ {0, . . . , p− 1} such that a1 = a0 + pα1.

Similarly, by using projections µji, we have that a2 = a1 mod p2, or equivalentlythat a2 − a1 = α2p

2 for a unique integer α2. Since 0 ≤ a2 < p3, we see that α2

has to be less than p. Therefore, we see that a2 = a1 + pα2 = α0 + pα1 + p2α2.Continuing in this manner, we conclude that the nth term of the sequence (ai)

∞i=0 is

equal to∑n

i=0 αipi for some non-negative integers 0 ≤ αj < p, uniquely determined

by a0, a1, . . . , an−1. Therefore, the data of the element (ai) ∈ Zp is represented by theinfinite series α0 + α1p+ α2p

2 + · · · in a unique way.

Let us revert this process to compute αi’s in terms of ai’s. We already know thata0 = α0. α1 = (a1 − a0)/p. More generally

αn =an − (α0 + · · ·αn−1p

n−1)

pn.

61. What are the ring operations on the series representation of p-adic integers?

Solution.

Suppose (ai), (bi) are two p-adic integers represented by the series∑αip

i and∑βip

i,respectively. We first analyze what happens to addition in the series notation. Let(ci) = (ai) + (bi) = (ai + bi) be represented by the summation

∑γip

i.

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Since α0 = a0, β0 = b0, we see that the ‘constant term’ γ0 of the series representationof (ci) has to be equal to α0 + β0 mod p. Note that α0 + β0 = γ0 + pδ0 for a uniqueδ0 ∈ {0, 1}.Next we determine γ1. Since c1 = a1 + b1 = (α0 + β0) + p(α1 + β1) ∈ Z/p2Z is equalto γ0 + γ1p, by the uniqueness of the power series coefficients we see that γ1 has to beequal to δ0 +α1 +β1 mod p. Since δ0 +α1 +β1 < 2p, we see that δ0 +α1 +β1 = γ1 +pδ1

for a unique δ1 ∈ {0, 1}. Thus we write α1 + β1 = γ1 − δ0 + pδ1.

Similarly,

a2 + b2 = α0 + β0 + (α1 + β1)p+ (α2 + β2)p2 mod p3

= γ0 + (δ0 + α1 + β1)p+ (α2 + β2)p2 mod p3

= γ0 + (δ0 + γ1 − δ0 + δ1p)p+ (α2 + β2)p2 mod p3

= γ0 + γ1p+ (δ1 + α2 + β2)p2 mod p3

Similar to the previous case, we write γ2 for δ1 + α2 + β2 mod p, hence the equalityγ2 + δ2p = δ1 + α2 + β2 implies that δ2 ∈ {0, 1}. Thus

a2 + b2 = γ0 + γ1p+ γ2p2 mod p3,

where γ2 is equal to δ1 +α2 + β2 mod p, and δ1 is found from δ0 +α1 + β1 = γ1 + pδ1.

More generally, if an =∑n

i=0 αipi and bn =

∑ni=0 βip

i, then an + bn =∑n

i=0(αi + βi)pi,

and γn is equal to δn−1 + αn−1 + βn−1 mod p, where δn−1 is found (inductively) fromδn−2 + αn−2 + βn−2 = γn−1 + pδn−1.

Next we look at what happens to the nth term of the product (ai)(bi) = (aibi). Suppose∑γip

i corresponds to this product. Clearly γ0 = a0b0 mod p = α0β0 mod p. For γ1

we look at a1b1 modulo pZ, which has to be equal to a0b0 modulo pZ. Since a1b1 =α0β0 + (α1β0 +α0β1)p in Z/p2, the equality a1b1 = a0b0 mod p is straightforward. Onthe other hand, to compute γ1 we need to look at the carry over from α0β0. Indeed,writing α0β0 as γ0 +pu1, we see that γ1 must be α0β1 +α1β0 +u1 mod p. Similarly, γ2

must be u2 +∑2

i=0 αiβ2−i mod p, where u2 is the carry-over from the previous parts.Indeed,

a2b2 = α0β0 + (α0β1 + α1β0)p+ (α0β2 + α1β1 + α2β0)p2 mod p3

= γ0 + (α0β1 + α1β0 + u1)p+ (α0β2 + α1β1 + α2β0)p2 mod p3

= γ0 + γ1p+ (u2 + α0β2 + α1β1 + α2β0)p2 mod p3,

where u2 is found from the equation α0β1 + α1β0 + u1 = u2p + γ1. Let γ2 denoteu2 +α0β2 +α1β1 +α2β0 mod p2, hence we write γ2 = u2 +α0β2 +α1β1 +α2β0 + u3p

2.Iterating this process we see how to multiply power series representations of p-adicintegers by carry-overs.

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Gröbner Bases:

The monomials xa: = xa11 · · ·xann in a polynomial rings k[x1, . . . , xn] over any field kcan be identified with the lattice points (a1, . . . an) ∈ Nn. A total order ≺ on Nn isa term order if the zero vector 0 is the unique minimal element, and a ≺ b impliesa+c ≺ b+c for all a, b, and c in Nn. Given a term order ≺, every nonzero polynomialf ∈ k[x] has a unique initial monomial, denoted in≺(f). If I is an ideal in k[x], then itsinitial ideal is the monomial ideal generated by the initial monomials of the elementsof I. The monomials which do not lie in in≺(I) is a Grobner Basis for I with respectto ≺, if in≺(I) is generated by {in≺(g) : g ∈ G}. Given any polynomial f ∈ k[x], bydivision algorithm, we can write f = h + r uniquely for h ∈ I and for an r ∈ k[x]with no term of r is divisble by any in≺(g) for g ∈ G. Therefore, any polynomial fcan be uniquely written as a linear combination of the standard monomials. It is clearthat the set of standard monomials makes a k-vector space basis for the quotient ringk[x]/in≺(I). Two important term orderings are

• Pure Lexicographic Order: α > β if the leftmost nozero entry of α−β is positive.

• Graded Lexicographic Order: α > β if∑α >

∑β or

∑α =

∑β and the

leftmost nonzero entry of α− β is positive.

62. Fix a term order ≺ . Show that a finite subset G = {g1, . . . gn} ⊂ I is a Grobnerbasis for I if and only if S(gi, gj) reduces 0 mod G where

S(f, g) =l.c.m(in≺(f), in≺(h))

in≺(f)· f − l.c.m(in≺(f), in≺(h))

in≺(h)· h

Solution. Type it later..

63. Fix a term ordering ≺ on R := k[x1 . . . xn]. Let I be an ideal in R. Show thatthere exists a basis Bµ of the vector space R/I consisting of the images ofthe standard monomials,the monomials that do not lie in in≺(I).

Solution.

Let G be a Grobner for I, so, by definition in≺(I) is generated by the monomials n≺(I)for g ∈ G. We know by the division algorithm that any polynomial f is congruentmodulo I to a polynomial r which does not posses any term divisible by in≺(I) forg ∈ G (hence with none of in≺(I) for f ∈ (I)). Note that this implies that r is in thespan of the standard monomials (monomials that do not lie in in≺(I)). therefore anypolynomial f ∈ R, the image R/I is a linear combination of the images of the standardmonomials. Assume for a second that a linear combination of standard monomials liein I; s :=

∑α aαx

α ∈ I. Then in≺(I) would be a standard monomial inside in≺(I)which is absurd. Therefore we must have aα = 0 for all α ∈ Nn. This shows that

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the standard monomials are linearly independent mod I. So we have shown that theimages of the standard monomials is a basis for the vector space R/I.

64. Show that a Grobner Basis G of an ideal I generates the ideal.

Solution.

Recall that a finite subset G ∈ I is called a Grobner basis for I if the initial ideal of Iis generated by the initial terms of G. Denote by the ideal generated by G. We wantto show that G = I. Observe that (in≺(g) : g ∈ G) = in≺(G) = in≺(I). Now, by theprevious problem we know that R/I is a vector space spanned by the monomials thatdo not lie in in≺(I) = in≺(G). Therefore R/I = R/G. Since G ⊂ I, this shows thatI = G.

65. Let I = 〈x2 + y, x+ xy〉. Find a Grobner basis for the ideal I.

Solution.

This is an application of Buchberger’s Algorithm. I will use the Pure Lex with x > y.Let fi = x2 +y and f2 = xy+y. And let G0 = {f1, f2}. Using pure lex, in(f1) = x2 andin(f2) = xy. Then S(f1, f2) = y · f1 − x · f2 = y2 − x2. Reduce y2 − x2 mod G0 : sincey2−x2 + f1 = y2 + y and in(y2 + y) = y2 is not divisible by the initial monomials of f1

and f2, the result is f3 = y2 − y. Now, let G1 = {f1, f2, f3}. Then S(f1, f3) = y3 − yx2

and modulo G1 this is 0. and S(f2, f3) = 0. Therefore algorithm stops here and we geta Grobner basis {f1, f2, f3}.

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USEFUL GEOMETRIC INTERPRETATIONS

We repeat; all rings are assumed to be commutative unless otherwise stated.

Given a ring R we denote by Spec(R) the set of all prime ideals of R. Note that weexclude the whole ring itself from this set, but if R is an integral domain, the zero-idealis assumed to be an element of Spec(R). There is a useful topology on Spec(R), calledthe Zariski topology defined as follows:

For each ideal I ⊂ R, let V (I) ⊂ Spec(R) denote the set of all prime ideals containingI. A closed set in Spec(R), by definition, is either

• intersection of arbitrarily many sets of the form V (I), or

• union of finitely many sets of the form V (I).

66. Show that (a) V (∑Ii) = ∩V (Ii); (b) V (I ∩ J) = V (I) ∪ V (J); (c) V (IJ) =

V (I) ∪ V (J) .

Solution.

We start with the first claim. If P contains the sum of the ideals Ii, then it contains eachideal Ii as well. Therefore, P ∈ V (Ii) for all i, hence V (

∑Ii) ⊆ ∩V (Ii). Conversely,

if a prime ideal contains all Ii’s, then it contains their sum as well, hence the equalityV (∑Ii) = ∩V (Ii) follows.

For (b), let P ∈ V (I ∩ J) be a prime ideal containing I ∩ J . Assume that P does notcontain neither I nor J . Let a ∈ I and b ∈ J be two elements such that a ∈ I −P andb ∈ J − P . Since ab belongs to both of I and J , it lies in I ∩ J , hence it lies in P . Pis a prime ideal, therefore, either a ∈ P , or b ∈ b, both of which gives a contradiction.Therefore, either P ∈ V (I), or P ∈ V (J). Conversely, if P lies in V (I), then it ofcourse lies V (I ∩ J). This finishes the proof.

The proof of (c) is similar to that of (b).

Observe that a prime ideal is pulled back to a prime ideal by ring homomorphisms:Let φ : A→ B be a ring homomorphism, P ⊆ B be a prime ideal, and let a, b be twoelements from A. If ab ∈ ψ−1(P ), then ψ(ab) ∈ P , hence either ψ(a) ∈ P , or ψ(b) ∈ P .It follows that either a ∈ ψ−1(P ) or b ∈ ψ−1(P ). Therefore, there exists an inducedmap in the opposite direction: ψ∗ : Spec(B)→ Spec(A), defined by ψ∗(P ) = ψ−1(P ).

Definition 0.21. A basic open set in a spectrum Spec(R) is, by definition, the com-plement D(f) := Spec(R) − V ((f)), where (f) is the principal ideal generated by an

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element f ∈ R. Thus D(f) is the set of all primes ideals which do not contain f . Thecollection {D(f)}f∈R forms a basis for the Zariski topology.

67. Show that the induced map ψ∗ : Spec(B) → Spec(A) of the ring homomor-phism ψ : A → B pullbacks distinguished open sets to distinguished opensets.

Solution.

Let D(a) ⊂ Spec(A) be the distinguished open set associated with an element a ofA. The pre-image ψ∗−1(D(a)) consists of prime ideals Q in Spec(B) that are mappedinto D(a). If Q ∈ ψ∗−1(D(a)), then ψ∗(Q) = ψ−1(Q) is an element of D(a). Hence,a /∈ φ−1(Q), or φ(a) /∈ Q. In other words, ψ∗−1(D(a)) = D(φ(a)) and this is what wewanted to prove to begin with.

Remark 0.22. Our conclusion from Problem 67 is that the induced map ψ∗ : Spec(B)→Spec(A) is continuous with respect to Zariski topology.

Definition 0.23. Let X be a topological space. A presheaf F (of rings, groups,modules, or anything) on X is an assignment (of rings, groups, modules, or anything)to each open set and satisfying certain compatibility conditions with respect to restric-tions: For each pair of open sets U1, U2 with U2 ⊂ U1, there exists a ‘restriction’ mapr1,2 = rU1,U2 : F (U1)→ F (U2) that satisfies

(a) If U1 = U2, then r1,2 is the identity map;

(b) If U3 ⊂ U2 ⊂ U1, then the following diagram of restriction maps commute:

F (U1) F (U2)

F (U3)

r1,2

r1,3 r2,3

A sheaf is a presheaf F such that for every collection of open sets {Ui} with U =⋃Ui,

the following two conditions are satisfied:

(a) If x1, x2 are two elements from F (U) and rU,Ui(x1) = rU,Ui(x2) for all Ui’s, thenx1 = x2,

(b) If a collection elements xi ∈ F (Ui) satisfies rUi,Ui∩Uj(xi) = rUj ,Ui∩Uj(xj) for alli, j, then there exists x ∈ F (U) such that xi = rU,Ui(x) for all i.

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As an example, we define a sheaf of rings OR on Spec(R). For each prime ideal P , letRP denote the localization of R at the multiplicative set (submonoid) R−P . For eachopen set U of Spec(R), we define OR(U) to be the set of functions

s : U →⊔P∈U

RP (4)

such that

• s(P ) ∈ RP for all P ∈ U ;• for each P ∈ U there exists an open neighborhood P ∈ U ′ ⊂ U and two elementsa, b in R satisfying 1) b is not contained in any prime Q ∈ U ′, hence a/b lives inRQ, 2) a/b from 1) is equal to s(Q) for all Q ∈ U ′.

It is straightforward to verify that OR is a sheaf on Spec(R).

Finally, let us define the notion of the ‘stalk at a point.’ Let F be a presheaf (ofgroups, rings, ..) on a topological space X and let x ∈ X be a point. The stalk of Fat x, denoted by Fx is the set of germs of sections of F at x. In other words, eachelement of Fx is an equivalence class pairs (U, s), where U is an open set containingx and s ∈ F (U) a section; the pair (V, t) is equivalent to (U, s) if there exists an openset W ⊆ U ∩ V such that s|W = t|W .

A locally ringed space is a ringed space for which stalks of the sheaf are local rings.

68. Show that (Spec(R),OR) is a locally ringed space and furthermore the stalkof OR at P is nothing but the local ring RP .

Solution.

Let (OR)P denote the stalk ofOR at P and define ψ : (OR)P → RP by ψ((U, s)) = s(P ).We claim that ψ is an isomorphism. To prove the surjectivity, let a/b be an elementof the local ring RP , hence, a, b ∈ R and b /∈ P . Then the distinguished open setU = D(b) contains P . We define a section s ∈ OR(U) by s(Q) = a/b for all Q ∈ D(f).In particular s(P ) = a/b. Therefore, ψ is surjective.

Next we prove that ψ is injective. Let (U, s) and (V, t) be two germs with the sameimage a/b at P . Then we need to show that (U, s) and (V, t) represent the sameequivalent class. By definition given in (4) we know that for both section s and t,there are open sets U ′ and V ′ around P and elements a1, b1 and a2, b2 from R suchthat for all Q1 ∈ U ′ and Q2 ∈ V ′ we have b1 /∈ Q1, b2 /∈ Q2 with s(Q1) = a1/b1 andt(Q2) = a2/b2. In particular, a1/b1 = s(P ) = t(P ) = a2/b2 in RP which is true if andonly if c(a1b2 − a2b1) = 0 for some unit c from R − P . This equality is true for allprime ideals Q such that b2, b1, c /∈ Q, or equivalently, for all Q ∈ D(b1)∩D(b2)∩D(c)which is an open set. Therefore (U, s) and (V, t) represents the same germ, hence ψ isinjective.

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69. Given (Spec(R),OR) let OD(f) denote the restriction of the sheaf OR to theopen set D(f), f ∈ R. Show that the locally ringed space (D(f),OD(f)) isisomorphic to (SpecRf ,ORf ).Solution.

Let φ : R → Rf denote the homomorphism defined by φ(a) = a/1. A prime ideal ofRf is of the form PRf where P is a prime ideal of R such that f /∈ P . It is clearthat φ−1(PRf ) = P . The image of the associated map φ∗ : SpecRf → Spec(R). Theimage of φ∗ is the set of all primes in R that does not contain f . In other words,φ∗(SpecRf ) = D(f) in Spec(R). The induced map on the stalks is given by thelocalized homomorphism

φP : RP → (Rf )PRf .

Since P does not contain f , the localization (Rf )PRf is isomorphic to RP , hence thecanonical map φP is an isomorphism. Since we have isomorphisms between their stalks,the isomorphism SpecRf → D(f) is an isomorphism between locally ringed spaces(SpecRf ,ORf ) and (D(f),OD(f)).

Definition 0.24. An affine scheme is a locally ringed space (X,F ) which is isomorphicto a pair (Spec(R),OR) for some ring R. The isomorphism here is a local isomorphism.A scheme is a locally rings space such that around every point x ∈ X there exists aneighborhood x ∈ U for which the pair (U,F |U) is an affine scheme. Here, F |U denotesthe restriction of the sheaf F to U .

Definition 0.25. A scheme (X,OX) is reduced if for every open U ⊆ X the ringOX(U) has no nilpotent elements.

70. Show that reducedness is a local property: (X,OX) is reduced if and only iffor every p ∈ X the local ring (OX)p has no nilpotent element.

Solution.

Let (U, s) and (V, t) be two germs from the stalk at a point P ∈ X. Multiplicationand/or addition of these germs are done in in O(W ), where W ⊆ U ∩ V is a neighbor-hood of P . Therefore, a germ (U, s) is nilpotent if and only if sn = 0 in a sufficientlysmall neighborhood W of P . It follows that OX(U) has nilpotent elements if and onlyif the stalk (OX)P has nilpotent elements.

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71. Let (X,O) denote an affine scheme and Ored denote the sheaf associatedwith the presheaf U 7→ (O(U))red. Here, the subscript red means we aretaking the quotient of the ring O(U) by the ideal of nilpotent elements.Show that (X,Ored) is a scheme. Let us denote it by Xred. Show that thereis a morphism π : Xred → X which a homeomorphism of the underlyingtopological spaces.

Solution.

It is straightforward to verify that U 7→ (O(U))red is a sheaf. The second claimthat Xred is homeomorphic to X boils down to the fact that the nilradical (idealof nilpotent elements) of a ring is the intersection of all prime ideals in the ring.See Problem 41. Indeed, let n denote the nilradical of R. The canonical quotienthomomorphism p : R −→ R/n gives a homeomorphism between topological spaces

π : Spec(R/n)∼−→ Spec(R). (5)

72. Let R be a reduced ring and suppose we have a homomorphism f : R′ → Rfrom another ring R′ into R. Show that there exists a unique ring homomor-phism g : R′red → R such that f = g ◦ p, where p : R′ → R′/n is the canonicalprojection into quotient of R′ by its nilradical.

Solution.

We claim that the map g : R′/n → R defined by g(p(a)) = f(a) is a well-definedhomomorphism. if p(a) = p(b), then a − b is a nilpotent element in R′. Since R hasno nilpotent elements, a− b is mapped to 0 ∈ R via f . Therefore, g(p(a))− g(p(b)) =f(a) − f(b) = f(a − b) = 0. In other words, g is well defined. g is a homomorphismbecause

• g(p(a) + p(b)) = g(p(a + b)) = f(a + b) = f(a) + f(b) = g(p(a)) + g(p(b)) for alla, b ∈ R′;• g(p(a))g(p(b)) = f(a)f(b) = f(ab) = g(p(ab)) = g(p(a)p(b)) for all a, b ∈ R.

Uniqueness is clear from the definition.

73. Let f : Xred → Y be a map from a reduced scheme into another scheme. Showthat there exists unique a scheme map g : Xred → Yred such that f = πg.

Solution.

Reducedness is a local property. The solution follows from Problem 72.

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74. Let x be an element of a ring R. Show that the distinguished basic openset D(x) ⊂ Spec(R) is empty if and only if x is nilpotent.

Solution.

Recall that D(x) is the set of all prime ideals that do not contain x. Therefore,D(x) = ∅, then x lies in all prime ideals of R, hence it belongs to the nilradical.Conversely, if x is nilpotent, it lies in all of the prime ideals, hence D(x) = ∅.

Definition 0.26. A topological space X is called irreducible if it is not a union of twoproper closed subsets.

75. Show that a scheme is irreducible if and only if every open subset is dense.

Solution.

Assume that X is irreducible and let U ⊆ X be an open subset. Since U∪(X−U) = Xand since X − U 6= X, we must have U = X. In other words, U is dense.

Conversely, suppose that every open subset in X is dense. If X = A∪B for two properclosed subsets, then X = (X − A) ∪ (X − B). In this case (X − A) ∩ (X − B) 6= ∅since open sets are dense. But taking complement once again we see that A ∪ B 6= Xwhich is absurd.

76. An affine scheme X = Spec(R) is reduced and irreducible if and only if R isan integral domain.

Solution.

(⇒) First, assume that X is irreducible. Let a, b ∈ R be two non-zero elements suchthat ab = 0. Since any prime ideal contains 0, we have D(a)∩D(b) = D(ab) = D(0) =∅. Since D(a) and D(b) are open sets, we have a contradiction. Therefore, R is anintegral domain.

(⇐) If R is an integral domain it cannot have any nilpotent elements. Hence, Spec(R)is a reduced affine scheme.

77. Spec(R) is irreducible if and only if R has a unique minimal prime ideal.

Solution.

Suppose X = Spec(R) is irreducible. If S is the set of all minimal prime ideals ofR, then ∪P∈SV (P ) = Spec(R). This contradicts with the irreducibility of X unlessS is a singleton. Conversely, if there exists unique minimal prime ideal P , then it is

36

necessarily the nilradical of R. In this case R/n is an integral domain. Since Spec(R/n)is homeomorphic to Spec(R), the irreducibility of the latter scheme follows from Prob-lem 76.

Definition 0.27. Let φ : F → G be a map (morphism) of sheaves. The presheafdefined by U 7→ kerφ|U is a sheaf and denoted by kerφ. The morphism φ is calledinjective if the associated kernel sheaf kerφ is the constant sheaf 0. Equivalently, φ isinjective if and only if for all open sets U the homomorphisms φ(U) : F (U) → G (U)are injective.

78. Let ψ : A → B be a ring homomorphism and let f : Spec(B) → Spec(A)denote the associated morphism between affine schemes. Show that if Bis an integral domain, then ψ is injective if and only if the correspondingsheaf map f ] : OA → f∗OB is injective.

Solution.

(⇐) Suppose f ] : OA → f∗OB is injective. By definition, f ](U) is a ring map (in-duced by φ) from OA(U) to OB(f−1(U)). If U = Spec(A), then OA(U) = A andOB(f−1(U)) = OB(Spec(B)) = B, and the corresponding ring map ψ is injective.

(⇒) Suppose that ψ : A → B is an injective ring homomorphism. We are going toshow that any homomorphism f ](U) : OA(U) → OB(f−1(U)), where U ⊆ Spec(A)is injective. Recall that the elements of the ring OA(U) are the s : U →

⊔P∈U AP

satisfying for each P ∈ U , there exists a neighborhood D(g) and f ∈ A such thats(P ) = f/g ∈ AP . After this refreshing, we observe that the value of f ](U) on asection s ∈ OA(U) is defined by the compositions:

f−1(U)f−→ U

s−→⊔P∈U

APψU−→

⊔Q∈f−1(U)

BQ.

The last map is defined as follows: For each Q ∈ f−1(U) ⊂ Spec(B) there existsunique prime ideal P ∈ U ⊂ Spec(A) such that P = f(Q) = ψ−1(Q), or equivalentlyψ(P ) = Q. Since g /∈ P , and ψ is injective, we see that ψ(g) /∈ ψ(P ). ThereforeψU(f/g) = ψ(f)/ψ(g) ∈ BQ. Now we are ready to check that f ](U)(s) = f ](U)(t)implies that s = t. Indeed, if ψ(f)/ψ(g) = ψ(f ′)/ψ(g′), then cψ(fg′ − f ′g) = 0 forsome c ∈ B − Q. Since B is an integral domain, and since ψ is injective f/g = f ′/g′,hence f ](U) is injective.

Definition 0.28. A morphism of schemes is called dominant if its image is dense.

79. Let φ : A → B be an injective ring homomorphism. Show that the inducedmap f : Spec(B)→ Spec(A) is dominant.

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Solution.We are going to show that every non-empty distinguished open set D(x) in Spec(A)intersects f(Spec(B)). Assume contrary that there exists x ∈ A such that D(x) ∩f(Spec(B)) = ∅. Let Q ∈ Spec(B) be a prime ideal in B. Then f(Q) is the primeideal P = φ−1(Q). Notice that x must be contained in P , otherwise, P ∈ D(x). Itfollows that φ(x) is contained in every prime ideal in B, therefore, φ(x) is nilpotent.Since φ is injective we x must be nilpotent. Then D(x) = ∅ a contradiction to ourinitial assumption.

80. Let φ : A→ B be a surjective ring homomorphism. Show that f : Spec(B)→Spec(A) is a homeomorphism onto a closed subset of Spec(A) and further-more f ] : OSpec(A) → f∗OSpec(B) is surjective.

Solution.The image of a prime ideal of B under f is a prime ideal in A that contains the kernelof φ. Therefore, the image of φ is nothing but the closed set V (kerφ) ⊆ Spec(A). It isclear that f is homeomorphic onto its image.To prove the surjectivity of the sheaf map we look at the corresponding claim on stalks.For P ∈ Spec(A) let Q ∈ Spec(B) be the prime ideal such that φ−1(Q) = P . Since thestalk of OA at P is AP and the stalk of f∗OSpec(B) at Q is BQ, for each a/b ∈ BQ withb /∈ Q, it is enough to find c/d ∈ AP such that φ(c/d) = φ(c)/φ(d) = a/b. To this end,let d be an element from φ−1(b). It is clear that d /∈ P . Similarly, let c ∈ φ−1(a). Thenc/d lies in AP , and furthermore, φ(c/d) = a/b.

81. Show that the converse of the above problem is true: if f : Spec(B) →Spec(A) is a homeomorphism onto a closed subset of Spec(A) and if f ] :OSpec(A) → f∗OSpec(B) is surjective, then φ : A→ B is surjective.

Solution.Type later.

82. Show that for any OX-module F , HomOX (E ,F) ∼= E ⊗OX F .Solution.Consider it locally; on the left hand side we have HomOXx (Ex,Fx) and on the rightHomOXx (Ex,OXx)⊗Fx. Thus

HomOXx (Ex,OXx)⊗Fx ∼= HomOXx (Ex,OXx ⊗Fx)∼= HomOXx (Ex,Fx),

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so we are done.

83. Show that for any OX-modules F , G, and E,

HomOX (E ⊗ F ,G) ∼= HomOX (F , HomOX (E ,G))

Solution. Think local once again. It from Problem 26.

84. If f : (X,OX) −→ (Y,OY ) is a morphism of ringed spaces, and if F is anOX-module and E is a locally free OY -module of finite rank, then there is anatural isomorphism

f∗(F ⊗OX f ∗E) ∼= f∗(F)⊗OY E .

Solution.

Type later.

Definition 0.29. A morphism f : X −→ Y is called "locally of finite type" if thereexists a covering of Y by open affine subsets Vi = Spec(Bi), such that for each i,f−1(Vi) can be covered by open affine subsets Uij = Spec(Aij), where each Aij is afinitely generated Bi-algebra. The morphism f is of finite type if in addition eachf−1(Vi) can be covered by a finite number of the Uij.

85. Show that a morphism f : X −→ Y is locally of finite type if and only if forevery open affine subset V = Spec(B) of Y , f−1(V ) can be covered by openaffine subsets Uj = Spec(Aj), where each Aj is a finitely generated B-algebra.

Solution.

Suppose that f is locally of finite type and V = Spec(B) is an affine open subsetof Y . Then it is clear from definition that there exists an open covering of the formVj = Spec(Aj) for which Aj is a finitely generated B-algebra.

For the converse first recall the definition of a scheme: a locally ringed space (X,OX)in which every point has an open neighborhood U such that the topological spaceU , together with the restricted sheaf OX |U is an affine scheme, that is isomorphic to(Spec(B),OSpec(B)) for some ring B.

39

Thus by definition of a scheme we get an open covering consisting of affine open sub-schemes Vj = Spec(Bj). By the hypothesis we have a covering of f−1(Vj) with affineopen subsets Uij = Spec(Aij), where Aij is a finitely generated Bj-algebra. But this isexactly the definition of being locally of finite type. We are done.

86. Let f and g be two elements from a ring A. Show that the basic opensubsets D(fg) = D(f) ∩D(g).

Solution.

By definition D(fg) is the set of all primes in A that do not contain fg. But thennone of those primes can contain neither f nor g. Therefore D(fg) ⊆ D(f) ∩D(g).

Conversely, a prime that does not contain both f and g cannot contain fg. ThereforeD(f) ∩D(g) ⊆ D(fg) hence we get the equality.

87. Show that {D(fα)}, a family of distinguished open subsets of Spec(R), is acovering if and only if 1 is in the ideal generated by fα’s.

Solution.

Suppose Spec(R) =⋃αD(fα). Then

Spec(R) =⋃α

D(fα)

=⋃α

(V (fα))c

=

(⋂α

(V (fα))

)c

= (V (〈fα〉))c

Therefore ∅ = V (〈fα〉). But then 〈fα〉 = R.

88. Show that Spec(R) is quasi compact, that is any cover has a finite subcover.

Solution.

Suppose {Uα} is an open cover for Spec(R). Each Uα is covered by basic opens Uα =⋃βD(fαβ). Therefore {D(fαβ)} is an open covering for Spec(R). By the previous

problem, 1 is in the ideal geberated by fαβ that is 1 = a1fα1β1 + . . .+ arfαrβr for someai ∈ R. Then {D(fαiβi)}i=1...r hence {Uαi}i=1...r is an open covering by the previousproblem. Therefore Spec(R) is quasi compact.

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89. Let A be a ring. Show that the following conditions are equivalent:

(a) Spec(A) is disconnected.

(b) There exists nonzero elements e1 and e2 in A such that e1e2 = 0, e21 = e1,

e22 = e2, and e1 + e2 = 1 (these elements are called orthogonal idempo-tents).

(c) A is isomorphic to a direct product A1 × A2 of two nonzero rings.

Solution.

(a) ⇒ (b) Suppose Spec(A) is disconnected that there are two disjoint open (henceclosed) proper subsets U1 and U2 with U1 ∪ U2 = Spec(A). Therefore there are dis-tinguished opens D(fα) that cover U1 and distinguished opens D(gα) that cover U2.Since Spec(A) is quasi-compact, and totality of those basic opens is a covering for A,we can choose a finite subcover: D(f1), . . . , D(fr), D(g1), . . . , D(gs). We can safelyassume that none of the fi’s or gj’s are nilpotent or unit. Otherwise correspondingbasic open would be the whole space or empty. One remark here is that for any ringand for any element a of it, D(an) = D(a). This is because a prime containing somepower of a will also contain it. Thus we may replace each basic open D(fi) by D(fni )if needed. Now since these basic opens cover Spec(A), 1 is in the ideal generatedby f1, . . . , fr, g1, . . . , gs. Say 1 = a1f1 + . . . + arfr + b1g1 + . . . + bsgs. Before go-ing further, we observe that D(fi) and D(gj) are disjoint for any i and j. ThereforeD(fi) ∪D(gj) = D(figj) = ∅. Thus figj is nilpotent. Thus (figj)

mij = 0 for some mij.Let m be the maximum of those natural numbers for i = 1, . . . , r and j = 1, . . . , s.By the previous remarks we can replace each fi and gj by fmi and gmj and still get acovering of Spec(A). Now let e1 = a1f1 + . . .+ arfr and e2 = b1g1 + . . .+ bsgs so thate1 + e2 = 1 and e1 · e2 = 0. Furthermore, e1e2 = e1(1 − e1) implies that e1 = e2

1 ande2 = e2

2.

(b) ⇒ (c) Since we can write ae1 + ae2 = a, we have the following homomorphismof rings A φ−→ A/(e1) × A/(e2) via a 7−→ (ae2, ae1). This homomorphism is injectivebecause (ae2, ae1) = (be2, be1) implies that (a − b)e2 + (a − b)e1 = 0 thus a = b. Thishomomorphism is surjective since for any (x, y) ∈ A/(e1)×A/(e2), we have the equality(x, y) = (e2x, e1y) from e1 + e2 = 1 and hence φ(e2x + e1y) = (e2(e2x + e1y), e1(e2x +e1y)) = (e2x, e1y). Therefore A is the direct sum of two nonzero rings A1 = A/(e1)and A2 = A/(e2).

(c) =⇒ (a) Suppose A = A1 × A2, and let e1 = (1, 0) and e2 = (0, 1). Thereforee1e2 = 0, e1 + e2 = 1, e2

1 = e1 and e22 = e2. Now let V1 = V (e1) and V2 = V (e2) that is

the closed subsets of Spec(A) defined by e1 and e2 respectively. If a prime p containse1 then it cannot contains e2 otherwise it would contain e1 + e2 = 1. Therefore V1 andV2 are disjoint proper subsets of Spec(A). Thus Spec(A) is disconnected.

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90. A morphism of schemes f : X −→ Y is quasi compact if there is a cover of Yby open affines Vi such that f−1(Vi) is quasi compact for each i. Show thatf is quasi compact if and only if for every open affine subset V ⊂ Y , f−1(V )is quasi compact.

Solution.

⇐ Suppose that for every affine open subset V ⊂ Y , the preimage f−1(V ) is quasicompact. Since Y is a scheme, every point has an affine open neighborhood which isisomorphic to a spectrum of a ring. These affine neighborhoods give us a covering thatis required.

⇒ Suppose that f is quasi compact so that we have an open affine cover {Vi =Spec(Ri)} of Y such that f−1(Vi) is quasi compact. Let V ⊂ Y be an arbitraryaffine open subset. Since affine schemes are quasi compact, we can choose finitelymany {V1, . . . , Vn} affine opens such that

⋃ni=1 Vi = V . Now if we can show that

each open f−1(Vi ∩ V ) is quasi compact, then we are done because a finite union ofquasi compact sets is quasi compact. Let Z := f−1(Spec(Ri)) be the preimage ofVi in X and let f |f−1(Z) = g. Now we have a morphism g : Z −→ Spec(Ri) withg−1(Spec(Ri)) = f−1(Spec(Ri)) = Z is quasi compact, and we would like to showthat any open subset V ′ ⊂ Spec(Ri) has quasi compact preimage. Since Z is quasicompact we can cover it by finitely many open affines, say Z =

⋃kj=1 Spec(Bj). So,

if we can show that the preimage of V ′ in Spec(Bj) is quasi compact, then we aredone. Repeating it one more time we want to show that the preimage of any openV ′ ⊂ Spec(Ri) under morphism h = g|Spec(Bi) : Spec(Bi) −→ Spec(Ri) is quasi com-pact. Any open subset V ′ of Spec(Ri) is covered by finitely many basic open D(fα).Therefore, we would like to show that these have quasi compact preimages in Spec(Bj).Note that for any two rings A and B, the preimage of a basic open D(f) ⊂ Spec(A)under a morphism h : Spec(B) −→ Spec(A) is D(φ(f)) where φ : A −→ B is theinduced homomorphism by h on the rings. But we know from problems above thatD(φ(f)) = SpecBφ(f). Therefore the preimage of a basic open in an affine scheme isaffine hence quasi compact and we are done.

Definition 0.30. Recall that a morphism f : X −→ Y is called "locally of finitetype" if there exists a covering of Y by open affine subsets Vi = Spec(Bi), such thatfor each i, f−1(Vi) can be covered by open affine subsets Uij = SpecAij, where eachAij is a finitely generated Bi-algebra. The morphism f is of finite type if in additioneach f−1(Vi) can be covered by a finite number of the Uij.

91. Show that a morphism f : X −→ Y is of finite type if and only if it is locallyof finite type and quasi compact.

Solution.

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If a morphism f : X −→ Y is of finite type then it is necessarily of locally finite type.Let us see why it is quasi compact. By definition of being of finite type, there exists acovering of Y by open affine subsets Vi = Spec(Bi), such that for each i, f−1(Vi) canbe covered by a finite number of open affine subsets Uij = SpecAij. Since affine setsare quasi compact and finite number of quasi compact subsets is still quasi compactwe see that the preimages f−1(Vi) are quasi compact hence one direction is done.Conversely suppose we have the locally of finite type and quasi compact situation.So, we have a covering of Y by open affine subsets Vi = Spec(Bi), such that for eachi, f−1(Vi) can be covered by open affine subsets Uij = SpecAij, where each Aij is afinitely generated Bi-algebra. Since f is quasi compact, by the previous question thepreimages f−1(Vi) are quasi compact hence can be covered by finite number of openaffine subsets Uij = SpecAij, hence we are done.

92. Conclude from the above question that f is of finite type if and only if forevery open affine subset V = Spec(B) of Y , f−1(V ) can be covered by a finitenumber of open affines Uj = Spec(Aj), where each Uj is a finitely generatedB-algebra.

Solution.

If for every open affine subset V = Spec(B) of Y , f−1(V ) can be covered by a finitenumber of open affines Uj = Spec(Aj), where each Uj is a finitely generated B-algebra,then by the previous two questions we know that f is of locally finite type and quasicompact therefore it is of finite type. Conversely if it is of finite type, then it is of locallyfinite type and hence by the first question about locally finiteness above, for every openaffine subset V = Spec(B) of Y , f−1(V ) can be covered by open affines Uj = Spec(Aj),where each Uj is a finitely generated B-algebra. Since f is quasi compact too, we havefinite number of Uj = Spec(Aj)’s. Hence we are done.

93. Show that if f is of finite type, then for every open affine subset V =Spec(B) ⊂ Y , and for every open affine subset U = Spec(A) ⊂ f−1(V ), A is afinitely generated B-algebra.

Solution.

This is rather long. Let⋃ni=1 Spec(Ai) be a finite cover of f−1(V ) by affine open

Ui := Spec(Ai) where each Ai is a finitely generated B-algebra. Since V ∩Ui is an opensubset of V , it is covered by the basic open subset D(gV ) for gV ∈ V . Each of thosebasic open is open in Ui so they are unions of basic opens of Ui. But D(gV ) ∼= SpecBgV

is quasi compact as being affine. Therefore, each D(gV ) is a finite union of the basicopen from Ui, say D(gV ) = D(gVi1) ∪ · · · ∪ D(gVir). But this last union is equal toD(gVi1 · · · gVir). This last one is in turn isomorphic to Spec(Ai)gVi1···gVir hence a finitely

43

geerated B-algebra. So is SpecAgV . Now note that we can cover Spec(A) by thoseSpecAgV . Since Spec(A) is quasi compact, we can cover it by finitely many SpecAgi ,where each Agi is a finitely generated B-algebra. Since they cover Spec(A), we havethe relation g1 + · · · + gr = 1. Therefore we have a purely algebraic question; givenfinitely many gi ∈ A generating the unit ideal, and each Agi is a finitely generatedB-algebra, we would like to conclude A is a finitely generated B-algebra. To showthat we will construct a finitely generated B-algebra A and show it is equal to A. Tostart with, first put gi’s in it. Then, for each i put a finitely many elements from Asuch that adding 1/gi to the algebra gives Agi . Since each Agi is a finitely generatedB-algebra, we need finitely many elements altogether. Therefore, this finite set A,when 1/gi added, generates Agi . Let us see why A generates A as a B-algebra. Letα ∈ A be an element of A. Then, in Agi it is of the form ai/g

mαi for some ai in A and

integer mα. Therefore glαi (gmαi α − ai) = 0 for some integer lα. Since gi and ai are inA, it follows gmα+lα

i α ∈ A. Now choose a sufficiently large power m of gi for each i sothat gmi α will be in A for each i. Now since g1 + . . .+ gn = 1, we can take sufficientlylarge power r of 1, so that in each term we get one of powers of the gi’s gets largerthan m. Therefore α = α · 1 ∈ A hence A = A.

Definition 0.31. Recall that a morphism is finite morphism if there exists a coveringof Y by opening affine subsets Vi = Spec(Bi), such that for each i, f−1(Vi) is affine,equal to Spec(Ai) where Ai is a Bi-algebra which is a finitely generated Bi-module, inother words, Ai is finite over Bi.

94. Show that a morphism f : X −→ Y is finite if and only if for every openaffine subset V = Spec(R) of Y , f−1(Y ) is affine, equal to Spec(A), where Ais a finite B-module.

Solution.

See Mumford’s (red) book page 145.

Definition 0.32. A morphism f : X −→ Y between schemes is called quasi-finite ifpreimage f−1(y) of a point y ∈ Y is a finite subset of X.

95. Show that a finite morphism is quasi-finite.

Solution.

Suppose that f : X −→ Y is a finite morphism. Let y ∈ Y be a point and U = Spec(B)is an affine neighborhood of x such that f−1(U) = Spec(A) is affine open with theproperty that A is a finite B-algebra (that is finitely generated as a B-module). Wecan restrict f to Spec(A) and loose nothing about the problem. Therefore we may

44

assume that we have a surjective morphism f : Spec(A) −→ Spec(B) where A is afinitely generated B-module. Since f is onto, any prime p of B is a preimage φ−1(q)of a prime q in A where φ : B −→ A is the ring homomorphism inducing f . Thereforewe may say that every prime ideal of B contains the kernel of φ. After this remark wemay well assume that φ is an injection; because Spec(B) is identical to Spec(B/kerφ)for our purpose. Furthermore, we may identify B with its image in A, that is B ⊂ A.Now what we exactly want to show here is that for each prime p ⊂ B, there are onlyfinitely many prime ideals of A that lie above p. Well, this is already done as one ofthe exercises above; we are done.

Definition 0.33. A sheaf of OX-modules F on a scheme X is called quasi-coherent ifX can be covered by open affine subsets Ui = Spec(Ai), such that for each i there isan Ai-module Mi with F|Ui ∼= Mi. It is called coherent if furthermore each Mi can betaken to be a finitely generated Ai-module.

96. Show that a sheaf of OX-,modules F on a scheme X is quasi coherent if andonly if every point of X has a neighborhood U such that F|U is isomorphicto a cokernel of a morphism of free sheaves on U . If X is Noetherian, thenF is coherent if and only if it is locally a cokernel of a morphism of freesheaves of finite rank.

Solution.

This solution is taken from Mumford’s (red)book at page 139. Let us name someproperties that will be equivalent at the end of the proof.

(a) For all x ∈ X, there exists a nbhd U of x and an exact sequence of OX |U -modules

OX |U (I) −→ OX |U (J) −→ F|U −→ 0

Note that this is exactly the statement that we want to prove.(b) There is an open affine cover {Ui} of X such that for all i, F|Ui ∼= Mi for some

Γ(Ui,OX)-module Mi.This is exactly the definition of a quasi coherent sheaf.

(c) For all U ⊂ X affine open, F|U ∼= M for some Γ(U,OX)-module M .(d) For all V ⊂ U open affines, the canonical map

Γ(U,F)⊗Γ(U,OX) Γ(X,OX) −→ Γ(X,F)

is an isomorphism.Strategy is to prove (d) ⇒ (a) ⇒ (b) ⇒ (c) ⇒ (d). Assume (d) and let x ∈ X.Take any open affine nbhd U of x and let R = Γ(U,OX). The R-module Γ(U,F)has a presentation

R(I) −→ R(J) −→ Γ(U,F) −→ 0

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Since R is nothing but OX |U and also R(I) = OX |U (I), we get the exact sequenceof sheaves

OX |U (I) −→ OX |U (J) −→ Γ(U,F) −→ 0

Therefore we must show the isomorphism F|U ∼= Γ(U,F). It is enough to dothat around basic opens. Since Γ(Uf , Γ(U,F)) = Γ(U,F) ⊗R Rf and we havethe obvious map Γ(Uf , Γ(U,F)) −→ Γ(Uf ,F), this obvious map becomes anisomorphism by (d). This proves (a).Next, we show (a)⇒ (b). Cover X by affine opens Ui = Spec(Ri) on which thereare exact sequences

OX |Ui(Ii)φi−−→ OX |Ui(Ji) −→ F|Ui −→ 0

Since R(Ii)i = OX |Ui

(Ii), and R(Ji)i = OX |Ui

(Ji), the cokernel of φi is a module oftype K because of the previous problem. But, then we are done for (b). Next(b)⇒ (c) then (c)⇒ (d). I will show those later.

Regular functions on Pnk :

97. Let A be a commutative ring with unity and S = A[x0, . . . , xn] be a gradedpolynomial ring over A, graded by degree. Show that (Sxi)0 = S[x−1

i ]0 =A[x′0, . . . , x

′n], the polynomial ring with generators x′j = xj/xi. Of course

x′i = 1, hence we can omit it.

Solution.

Note that the first equation is definition of localization at an element xi, so there isnothing to prove; it is the definition. For the second, let f/xmi ∈ S[x−1

i ]0. Then fmust be a homogenous (this is crucial for f/xmi to be in the zeroth piece) polynomialof degree m. That is f =

∑α x

α with deg(xα) = m. Then f/xmi becomes a polynomialin xj/xi. Converse is clear because xj/xi is of degree 0.

Fact 0.34. Let R = k[x0, . . . , xn] and let Uxi ⊂ Pnk be the open subset defined by thenon-vanishing of the i’th coordinate (or the complement of the closed subset V (xi) inPnk). Then, regular functions on Uxi are OPnk (Uxi) = (Rxi)0 the 0’th graded piece of thelocalized ring k[x0, . . . , xn]xi = k[x0, . . . xn, 1/xi] at the element xi. Therefore, a regularfunction on Uxi is of the form g/xdi where g is of degree d. By the above discussionOPnk (Uxi) can be identified with the polynomial ring k[x′0, . . . , x

′n] where x′j = xj/xi on

the affine open Uxi = kn (= Spec(k[x0, . . . , xn, 1/xi])).

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98. Justify the above description of regular functions on Pnk in the case of k = Cand show that global section of Pnk are constants.

Solution.

Let’s assume k = C and let us work on P1k. On Ux0 6=0 regular functions are of the form

g/xd0 because, by definition a regular function is a quotient g/h of two polynomials witha non-vanishing condition on h. Of course on any subset of Pn, these two polynomialsmust be homogenous and of the same degree in order for making a well defined function.Since we want h to be non-vanishing on Uxi , it cannot contain any variable other thanxi because otherwise it would always have a root as a polynomial over C. Therefore,it must be a power of xi only. This justifies that regular functions on Ux0 6=0 ⊂ P1 areof the prescribed form g/xd0 with degree(g) = d. Now let us check if there is a non-vanishing global regular function on P1 (idea will generalize to all n automatically).So, we are looking for a function f which is like g/xd0 on Ux0 and like g′/xl1 on Ux1 .Consider the overlap Ux0 ∩Ux1 . Points (1;x1) for x1 6= 0 all lie in the overlap. Hence asx1 becomes unbounded, the value of g/xd0 at such points becomes unbounded (unlessg is constant). But the value of g′/xl1 is either constant or goes to 0. Similarly lookingat the points of the form (x0; 1) for x0 6= 0 we see that f must be constant function.

Remark 0.35. Recall that the construction of the sheaf of regular functions OPnk onPnk depends on the graded structure of the polynomial ring. What if we use shiftedgrading on the polynomial ring. Then we get a sheaf of OPnk -modules in a very specialway. This is very close to the sheaf of regular functions itself and its notation OPnk (m) ifwe shift it by m. Explicitly, the construction is as follows: on the open Uf , OPnk (m)(Uf )is the OPnk (Uf )-module of elements in the 0’th piece of the graded k[x0, . . . , xn]f -modulek[x0, . . . , xn](m)f , that is (k[x0, . . . , xn](m)f )0.

99. Describe the space global sections of OPnk (m) explicitly.

Solution. They are homogenous polynomials of degree m in k[x0, . . . , xn].

100. Prove that a nonzero section of the line bundleOP1C(m) has total sum oforders of vanishing m over P1

C. Hence show that no two such bundles areisomorphic.

Solution.

Let R = k[x0, . . . , xn] be the graded (by degree) polynomial ring over the field k.Recall that OPnC(m)(Uf ) for a basic open Uf in PnC is defined to be ((Rf )0)(m); them’th piece of the graded ring (Rf )0. In particular, on Uxi ⊂ P1, OPnC(m)(Uxi) is theOPnC(Uxi) = k[x0, . . . , xn, x

−1i ]-module (or algebra) k[x0, . . . , xn, x

−1i ]0(m). Therefore, a

section s ∈ OPnC(m) on Uxi is an element of k[x0/xi, . . . , xn/xi](m) hence deg(f) = m+r

47

for some nonnegative integer r. In particular , on Ux0 ⊂ P1, it is a polynomial of degreem + r in the variable x1/x0. Hence, its order of vanishing is m + r on Ux+0. On Ux1this section is of the form (x0/x1)m · f . But this has a pole of order r. Therefore, sumof the order of vanishing of s over P1 is m+ r − r = m.

101. Let X be the variety P2 minus an irreducible conic. Let OX(m) be therestriction of O2

P to X. Show that OX(2) is trivial, but that OX(1) is not.Hence O2

P is a 2-torsion element of PicX.

Fact 0.36. Suppose L is an invertible sheaf on a variety X, and that there is a globalsection s of L vanishing no where. Then L is isomorphic to OX , the trivial sheaf.

Solution.

Since X is missing a conic in P2, the equation of that conic is a non-vanishing 2 form onX. This is an element of O2

P(2) and hence when restricted to X an element of OX(2).By the fact above, OX(2) is trivial. How about OX(1)? Elements (sections) of OX(1)are 1-forms. Of necessity they vanish at a point of X (because they must intersectnontrivially). Therefore OX(1) is not a trivial sheaf. Since OX(2) = (OX(1))⊗2 we aredone.

Remark 0.37. OX(1) is called hyperplane bundle; it attaches a hyperplane (a linein P2) to every point of X. Since a hyperplane is a linear space in Pn, elements ofOX(1) are 1-forms. And the intersection of these hyperplanes with the variety givesthe vanishing of section on X. Since varieties of proper dimension must intersect inPn, the hyperplane bundle OX(1) has no non-vanishing section in it.

Interlude on Čech cohomology of sheaves.

Let us assume that we have a sheaf of modules F at hand. The Čech n-cochains ofF over an open cover U is the module

Cn(U,F) =∏

{i0,...,in}

F(Ui0 ∩ · · · ∩ Uin)

where product runs over all indices of the opens of U. The coboundary operatorsd : Cn(U,F) −→ Cn+1(C,F) are defines as follows:

d((fi0,...,in)) = (gi0,...,in+1)

where

gi0,...,in+1 =n+1∑k=0

(−1)kfi0,...,in|Ui0,...,ik,...,in

48

Thus, using previous cochain data, d gives a new cochain on the smaller opens (furtherintersections). It is easy to check that d2 = 0. Thus we get a chain complex. Thecohomology modules are called the Čech cohomology modules. Let us work those ona specific example.

The constant sheaf on a space X is defined by giving Z the discrete topology, andthen setting Z(U) to be the continuous functions from U to Z. (note: there is an easyextension of this definition from Z to any abelian group).

102. Is it true that Z(U) ∼= Z for any open set U? If not, what additional assump-tion on U would make this true?

Solution.

It is not true unless U is a connected open set. And otherwise it is⊕

n Z with as manycopies of Z as the connected components of U .

103. Compute the Čech cohomology of the constant sheaf Z on S2 using thecover U consisting of U1, the top hemisphere, U2 and U3 bottom two quarter-spheres (all open are slightly enlarged to have non empty overlap).

Solution.

The Čech complex for the open cover U and the constant sheaf Z is

0 −→ Z(U1)⊕ Z(U2)⊕ Z(U3)d1−→ Z(U1 ∩ U2)⊕ Z(U2 ∩ U3)d2−→ Z(U1 ∩ U2 ∩ U3) −→ 0

Since each Ui is connected, Z(Ui) ' Z with the generators say f1, f2 and f3 forU1, U2 and U3 respectively. In fact, it is obvious that fi = 1 for each i. So, anarbitrary Čech cochain looks like (hi)i=1..3 where hi = nifi for some ni ∈ Z. Letfij denote the restriction of fi on the open Ui ∩ Uj. Let us look at the kernel ofd : C0(U,Z) −→ C1(U,Z). Since d((fi)) = (fj − fk), we see that kernel of d isgenerated by (f1, f2, f3) because the differences fi − fj is 0 on the overlaps of thedomain. Therefore kerd is a rank 1 free submodule of

∏Z(Ui). So, H0(Z,U) ∼= Z.

Next, let f12 = (f12, 0, 0), f13 = (0, f13, 0) and f23 = (0, 0, f23) be the generators ofC1(U,Z) = Z(U1 ∩ U2) ⊕ Z(U1 ∩ U3) ⊕ Z(U2 ∩ U3): what is the kernel of the nextcoboundary map d then? By definition d(fij) = ∓fij|U1∩U2∩U3 . Therefore images arethe restrictions of the constant functions, fij on a space of two connected componentU1 ∩U2 ∩U3 (check that U1 ∩U2 ∩U3 has two connected component and hence Z(U1 ∩U2∩U3) ∼= Z⊕Z). Thus the rank of the image of d will be 1 in Z⊕Z. Thus, the kernelmust be of rank 2. But observe also that the image of d : C0(U,Z) −→ C1(U,Z) mustbe of rank 2. And since d2 = 0, that is the image of d : C0(U,Z) −→ C1(U,Z) must liein the kernel of d : C1(U,Z) −→ C2(U,Z), they must be the same. Thus H1(Z,U) ∼= 0.

49

Finally, since the image of the second d is of rank 1 in C1(U,Z), we get Z2, H2(Z,U) ∼=Z.Given a short exact sequence of sheaves (short exact on the stalks)

0 −→ F1 −→ F2 −→ F3 −→ 0

there is a long exact sequence in sheaf cohomology

. . .→ H i−1(F3)→ H i(F1)→ H i(F2)→ H i(F3)→ H i+1(F1)→ . . .

Proof of this fact along the others is in Ch.III.4 of Hartshorne’s book. An importantinstance of this short exact sequence is the ideal sheaf sequence; let X = V (I) be theprojective variety in Pn defined by the ideal I. Then we have the exact sequence ofmodules 0 −→ I −→ R −→ R/I −→ 0. Exactness on the stalks gives us the exactsequence on the sheaves on Pn, 0 −→ IX −→ OPn −→ OX −→ 0.

104. Prove that for any m, H i(OPn(m)) = 0 unless i = 0 or i = n. Use the definitionof Čech cohomology and the fact that the standard open cover of Pn is Leray.Solution. See Hartshorne’s book Chapter 3, Section 5, Theorem 1.

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Highlights

(a) Localization is an exact functor.

(b) Localization commutes with quotients.

(c) HomR(−, N) is a left exact functor and it need not be exact.

(d) −⊗RM is a right exact functor and it need not be exact.

(e) Direct limit commutes with tensor products.

(f) Let s ∈ OP1C(m) be a global section. Then, on the affine open Ux0 , s is in the 0’thpiece of k[x0, x1](m)x0 . 0’th graded piece of k[x0, x1](m)x0 is {f/xd0 : deg(f) =m + d}. Therefore, s|U0 = f/xd0 for some homogenous polynomial f of degreem + d. On U1, this section will be of the form g/xr1 for some g ∈ k[x0, x1] ofdegree m+ r. Note that on the overlap they must be divisible by xd0 and hence sis a polynomial of degree m.

(g) PicPn = Z. In general regular sequences are dependent on the ordering of thesequence.

(h) Let f : Spec(B) −→ Spec(A) be a morphism of affine schemes. Then f is a con-tinuous map between topological spaces as well as a map of sheaves OSpec(A) −→f∗OSpec(B) on Spec(A). Let U be an open subset of Spec(A). Then f∗OSpec(B)(U) =OSpec(B)(f

−1(U)). Therefore at every open U in Spec(A) we have a ring homo-morphism OSpec(A)(U) −→ OSpec(B)(f

−1(U)).

51

INTEGER THEORY

Next problems are added in January, 2015.

105. Suppose R is a unique factorization domain and K is its fraction field. Letα ∈ K be a root of a monic polynomial p(x) ∈ R[x]. Show that α lies in R.

Solution.

Suppose p(x) = xn + an−1xn−1 + · · ·+ a1x+ a0 for some a0, . . . , an−1 ∈ R and α = a/b

with relatively prime a, b ∈ R. Since

0 = p(α) =an

bn+ an−1

an−1

bn−1+ · · ·+ a1

a

b+ a0,

the following is immediate:

an = b(−an−1an−1 − · · · − bn−2a1a− bn−1a0).

Therefore b is a unit and a/b ∈ R.

106. Let p(x) = anxn + · · · + a1x + a0 be a polynomial over an integral domain R.

Let K denote the fraction field of R. If a/b ∈ K is a root of p(x), then showthat a|a0 and b|an. Here, we assume that a and b are relatively prime.

Solution.

Since p(a/b) = 0 we have an an

bn+ · · ·+ a1

ab

+ a0 = 0, or

anan = −an−1ba

n−1 + · · · − a1abn−1 − a0b

n = b(−an−1an−1 · · · − a0b

n−1).

Thus b divides an. Similarly,

a(−anan−1 − an−1ban−2 − · · · − a1b

n−1) = bna0

implies that a divides a0.

Notation: Fq denotes the finite field with q elements. In particular, if q is a primenumber, then Fq = Z/qZ.

107. Show that x3 +x+ 1 is irreducible over F2 and let θ be a root. What are thepowers of θ in F2(θ).

Solution.

Checking that p(x) = x3 +x+ 1 is irreducible over F2 can be done in several ways, theeasiest here is to substitute 0 and 1 into p(x) and see if the results is 6= 0.

52

Next, calculate a few values inductively.

θ3 = −θ − 1

θ4 = −θ2 − θθ5 = −θ3 − θ2 = −θ2 + θ + 1

θ6 = −θ3 + θ2 + θ = θ2 + 2θ + 1 = θ2 + 1

θ7 = θ3 + θ = −1 = 1

Since we have reached θ7 = 1 the process (exponentiation) repeats periodically withperiod 7.

108. Show that a finite field F has cardinality pn for some prime number p andn ∈ N.Solution.

Since the characteristic of F is p, F contain the field Z/p. More precisely, the subfieldgenerated by 1 in F is isomorphic to Z/p. Since F is a finite dimensional vector spaceover this field, by choosing a basis e1, . . . , en for F, we see that the cardinality of |F| isa prime power pn.

109. Show that the group of nonzero elements of a finite field F with pn elementsis isomorphic to the cyclic group Zpn−1 with pn − 1 elements.

Solution.

Since non-zero elements of F is a finite abelian group, by the Fundamental Theoremof Finitely Generated Abelian Groups, we have

F∗ ∼= Za1 × Za2 × · · · × Zar (6)

for some integers a1, a2. . . . , ar that satisfy the divisibility condition: ai|ai+1 for i =1, . . . , r − 1. (Here, Zai is the cyclic group of order ai and it is isomorphic to additivegroup Z/aiZ.) It follows that any element x ∈ F∗ satisfies the polynomial equationxar = 1. Since xar − 1 has at most ar roots over a field, |F| ≤ ar + 1. Since |F∗| ≥ arby (6) we see that |F| = ar + 1, thus F∗ ∼= Zar = Zpn−1.

110. Let x ∈ F be an element from a finite field of characteristic p. Prove thatthe map Fr : F → F defined by Fr(x) = xp is an automorphism of F .

53

Solution.

Let x, y ∈ F be two distinct elements. It is clear that xp = yp if and only if xp−yp = 0,or (x − y)p = 0. Since F has no zero divisors, x − y = 0, or x = y. Therefore, weconclude that Fr : x 7→ xp is an injective map, hence it is a bijection. It is easy tocheck that Fr is a homomorphism, hence it is an automorphism of F .

Definition 0.38. Let F ⊂ K be a field extension. The minimal polynomial of anelement α ∈ K over F is a monic polynomial minα,F (x) ∈ F [x] of smallest degree suchthat minα,F (α) = 0.

A useful and simple fact about minimal polynomial of an element α ∈ K is the follow-ing:

Fact 0.39. Let F ⊂ K be a field extension. The degree of the minimal polynomialminα,F (x) ∈ F [x] is the dimension of the F -vector space F (α).

111. Show that minα,F (x) is irreducible and uniquely determined by α and F .

Solution.

Assume that minα,F (x) = a(x)b(x) for some non-constant polynomials a(x), b(x) ∈F [x]. Without loss of generality we assume that a(x) and b(x) are monic. Sincea(α)b(α) = 0, and F is an integral domain, either a(α) = 0, or b(α) = 0. This contra-dicts with the degree of minα, F (x) being minimal. Hence, minα,F (x) is irreducible.Uniqueness follows from the fact that the set of polynomials that vanish at α is aprincipal (in F [x]) hence it has a unique minimal irreducible generator up to invertiblemultiples.

112. Compute the minimal polynomial of 1 + i over Q where i ∈ C is the secondroot of unity.

Solution.

First we manipulate some equations to find a candidate. (i + 1)2 = −1 + 2i + 1 = 2i.So, (i+ 1)2 − 2(i+ 1) = −2. We conclude that f(x) = x2 − 2x+ 2 has i+ 1 as a root.Since f(x) is of degree 2, to show that it is irreducible it is enough to prove that it hasnot rational root. If α = a/b ∈ Q is a root of f(x), then by Problem 106 a divides 2and b divides 1. Possibilities for a/b are then ±2. But f(2) = 2, f(−2) = 10.

54

113. Prove that Q(√

2,√

3) = Q(√

2 +√

3).

Solution.

The inclusion ⊃ is obvious. To prove the opposite inclusion we compute the degreesof these two fields over Q. Let α denote

√2 +√

3. Since α2 = 5 + 2√

6 we see that(α2 − 5)2 = 24, or that α is a root of

p(x) = x4 − 10x2 + 1.

By Problem 106 we see that p(x) has no rational root. Also, by Problem 115 it sufficesto show that p(x) is irreducible over Z. Assume that p(x) is reducible, hence, it is ofthe form p(x) = (a + bx + cx2)(d + ex + fx2) for some a, b, c, d, e, f ∈ Z. Equatingcoefficients we have the following equalities and implications:

ad = 1 =⇒ a = d = ±1

cf = 1 =⇒ c = f = ±1

ae+ bd = 0 =⇒ e = −bbf + ce = 0 =⇒ e = −b

af + be+ cd = −10 =⇒ ±1− e2 ± 1 = −10

The last equality is absurd since there is no square equal to 8 or 12. Therefore, p(x) isirreducible, hence it is the minimal polynomial of α over Q. In other words, the field ex-tension (Q,Q(α)) has degree 4. On the other hand, the field extension (Q,Q(

√2,√

3))has degree at most 4, the degree of both of the field extensions (Q,Q(

√2)) and

(Q,Q(√

3)) is 2. Since the Q(α) is a vector subspace of Q(√

2,√

3), the followinginequalities finishes our argument:

dimQQ(α) = 4 ≤ dimQQ(√

2,√

3) ≤ 4.

Next problem is a nice application of the field theory to the superpositions of functions.

114. Let f(x) be an irreducible polynomial of degree n over F and let g(x) ∈ F [x]be any polynomial. Show that every irreducible factor of the compositionf ◦ g(x) has degree divisible by n.

Solution.

Without loss of generality we assume that f(x) is monic. Since f(x) is irreducible,it is the minimal polynomial of any of its roots. Let α be a root of an irreduciblefactor h(x) ∈ F [x] of f(g(x)). In particular f(g(α)) = 0, hence g(α) is a root of f(x).Let K and K ′ denote respectively the field extensions F (α) and F (g(α)). ClearlyF (g(α)) ⊆ F (α). Without loss of generality we assume that h(x) is monic, hence h(x)

55

is the minimal polynomial of α over F . The degree of the field extension F ⊂ K = F (α)is equal to deg h(x). On the other hand, the following implication is easy to verify:

F ⊂ K ′ ⊂ K = F (α)⇒ dimF K = dimF K′ · dimK′ K.

Therefore, deg h(x) is divisible by deg f(x) = dimF K′.

115. Let R be a UFD (unique factorization domain) and K denote its fractionfield. Show that an element f(x) of R[x] is irreducible in R[x] if and only ifit is irreducible over K[x].

Solution.

If f(x) is irreducible in K[x] but of the form f(x) = a(x)b(x) for some a(x), b(x) ∈R[x] ↪→ K[x], then obviously we have a contradiction. Let us prove the converse thatthe irreducibility in R[x] implies irreducibility in K[x].

First of all, let us recall some basic facts: Since R is a UFD, R[x] is a UFD, also. In aUFD, irreducible elements are prime that is to say, the principal ideal generated by anirreducible element is a prime ideal. Now, towards a contradiction, assume that f(x)factorizes as in f(x) = a(x)b(x) for two polynomials a(x), b(x) from K[x]. Let c1 bethe product of the denominators of coefficients of a(x) and let c2 denote the productof the denominators of coefficients of b(x). Define c = c1c2. Then cf(x) = a′(x)b′(x),where a′(x) = c1(x) ∈ R[x] and b′(x) = c2b(x) ∈ R[x]. Since (f(x)) ⊂ R[x] is a primeideal, and since cf(x) ∈ (f(x)), it follows that either a′(x) ∈ (f(x)) or b′(x) ∈ (f(x)).Without loss of generality we assume that a′(x) = d(x)f(x) for some d(x) ∈ R[x].Then cf(x) = a′(x)b′(x) = d(x)f(x)b′(x) implies c = d(x)b′(x). Hence, we concludethat both d = d(x) and b′ = b′(x) are from R, thus invertible in K. But then b(x) = b′

c2is a unit in K[x], hence f(x) is irreducible in K[x].

116. Let F be a field and F (x) denote the field of rational functions over F forsome variable x. Let t ∈ F (x) be a rational function t = P (x)/Q(x) whereP (x) and Q(x) are two relatively prime polynomials from F [x].

(a) Show that the polynomial P (X) − tQ(X) as an element of F (t)[X] isirreducible and has x as a root.

(b) Show that the dimension of the field extension F (t) ⊂ F (t)(x) = F (t, x)is equal to dimF (t) F (x) = max{degP (x), degQ(x)}.

Solution.

(a) It is clear that P (x)− tQ(x) = 0. Let us prove the irreducibility of P (X)− tQ(X).By Problem 115 it is enough to prove it in R = F [t] for which K = F (t) is the

56

fraction field. To this end, assume that P (X)− tQ(X) is reducible over R that thereexists two polynomials At(X) and Bt(X) from F [t][X] = F [t,X] = F [X][t] such thatP (X)− tQ(X) = At(X)Bt(X). Without loss of generality we assume that the t-degreeof At(X) is 0, and the t-degree of Bt(X) is 1. Thus we write Bt(X) = tU(X)+V (X) forsome U(X), V (X) ∈ F [X]. In particular P (X)− tQ(X) = At(X)V (A)+ tAt(X)U(X),or P (X) = At(X)V (X) and Q(X) = At(X)U(X). Since P (X) and Q(X) are relativelyprime, At(X) must be a unit, which implies that P (X) − tQ(X) is an irreduciblepolynomial.

(b) It is obvious that the X-degree of P (X)− tQ(X) as a polynomial over F (t) is themaximum of the degrees of P (X) and Q(X). By Fact 0.39, it is enough to computethe degree of the minimal polynomial of α = x over F (t). If c(t) ∈ F (t) denotes thecoefficient of the highest X-degree term in any of the polynomials P (X) or tQ(X), thenthe polynomial ft(X) := 1

c(t)(P (X) − tQ(X)) ∈ F (t)[X] is monic, and furthermore,

irreducible by Part (a). Therefore, ft(X) is the minimal polynomial of x over F (t),hence its X-degree, which is max{degX P (X), degX Q(X)}, is the dimension of thefield extension F (t) ⊆ F (t)(x).

Notation and a simple fact: Let K be a field and let K1 and K2 be two subfields. Thesmallest subfield of K that contains both K1 and K2 is denoted by K1K2.

117. Let K1 and K2 be two finite extensions of a field F all contained in a largefield K. Prove that the F -algebra K1 ⊗K2 is a field if and only if

dimF (K1K2) = (dimF K1)(dimF K2).

Solution.

Let a1, . . . , am and b1, . . . , bn denote F -vector space bases for K1 and K2, respectively.Clearly {aibj}i,j spans K1K2 as an F -vector space.

If K1 ⊗F K2 is a field, then its action on K1K2, which is defined by linearly extending

x⊗ y · aibj = (xai)(bjy), where x ∈ K1, y ∈ K2,

gives a K1⊗K2-vector space structure on K1K2. Since K1⊗K2 is an F -vector space,the dimension of K1K2 as an F -vector space is (dimF K1⊗K2)(dimK1⊗K2 K1K2). SincedimF K1K2 ≤ mn, and since dimF K1 ⊗K2 = mn we see that mn = dimF K1K2.

Next we prove the converse. Since dimF K1K2 = mn, then we know that {aibj} is abasis for K1K2. Define f : K1 ⊗F K2 → K1K2 by extending a ⊗ b 7→ ab by linearity.We first show that f is an F -vector space isomorphism. Its linearity is clear. It is alsoclear that f is surjective. To prove the injectivity, assume that f(

∑(i,j)∈I ri,jai⊗ bj) =

57

∑(i,j)∈I ri,jaibj = 0 for some ri,j ∈ F , (i, j) ∈ I ⊂ [m] × [n]. Since {aibj} is a basis of

K1K2 we must have that ri,j = 0 for all (i, j) ∈ I, hence f is an injective linear map.Since F -dimensions of K1 ⊗F K2 and K1K2 are the same, f is a linear isomorphism.Next we show that f is a ring isomorphism. With what we have so far, it suffices toshow that f respects multiplication:

f

∑(i,j)∈I

ri,jai ⊗ bj

∑(i′,j′)∈J

si′,j′ai′ ⊗ bj′

= f

∑(i,j)∈I,(i′,j′)∈J

ri,jsi′,j′aiai′ ⊗ bjbj′

=

∑(i,j)∈I,(i′,j′)∈J

ri,jsi′,j′aiai′bjbj′

=

∑(i,j)∈I

ri,jaibi

∑(i′,j′)∈J

si′,j′ai′bj′

,

which is equal to f(∑

(i,j)∈I ri,jai ⊗ bi)f(∑

(i′,j′)∈J si′,j′ai′ ⊗ bj′). Therefore, f is an

isomorphism between the rings K1 ⊗F K2 and K1K2. Since the latter is a field, weconclude that K1 ⊗F K2 is a field, also.

118. Prove by an example that in general the tensor product of two field exten-sions does not need to be an integral domain.

Solution.

Let K1 and K2 both be equal to Q(√−1), and let x and y be the non-zero elements in

K1 ⊗F K2 defined by x = 1⊗ 1 +√−1⊗

√−1 and y = 1⊗ 1−

√−1⊗

√−1. In this

case,

xy = (1⊗ 1)2 − (√−1⊗

√−1)2

= 1⊗ 1− 1⊗ 1 = 0.

Definition 0.40. Let F be a field and f(x) ∈ F [x] be a polynomial over F . Thesplitting field of f(x) is the smallest field extension K of F such that f(x) splitscompletely into linear factors in K[x] but not over any proper subfield F L K.A splitting field of f(x) is easy to construct in inductive manner by using roots ofthe irreducible factors of f(x), and furthermore, ant two splitting field of f(x) areisomorphic to each other.

An algebraic extension K of F is called a splitting field over F if K is the splittingfield of a collection of polynomials over F .

58

119. Determine the splitting field and its degree over Q of the polynomial f(x) =x4 + x2 + 1.

Solution.

By using Problem 106 we see that f(x) does not have a root in Q. However, thisdoes not mean it is irreducible over Q. It could be equal to a product of two degree 2polynomials. Indeed, let ξ6 denote the ‘primitive 6th root of unity,’ exp(2πi/6). Since

f(x) = x4 + x2 + 1 = (x2)2 + (x2) + 1

=(x2)3 − 1

x2 − 1

=x6 − 1

x2 − 1(7)

=(x3 − 1)(x3 + 1)

(x− 1)(x+ 1)

=x3 − 1

x− 1· x

3 + 1

x+ 1

= (x2 + x+ 1)(x2 − x+ 1). (8)

we see that all 6th root of unities, except 1 and ξ36 (which makes the denominator of

eqn 7 0) are roots for f(x). Therefore, f(x) = (x2 + x+ 1)(x2− x+ 1) = ((x− ξ26)(x−

ξ46))((x−ξ6)(x−ξ5

6)). Clearly Q(ξ6) contains all roots of f(x), and furthermore, it is ofdegree 2 over Q. To see this, it is enough to observe that ξ6 is the root of the irreduciblepolynomial x2 − x + 1. Thus Q(ξ6) is the smallest field in which f(x) = x4 + x2 + 1splits into linear factors.

Definition 0.41. An algebraically closed field is a field L such that every polyno-mial in L[x] has a root in L. An algebraic closure of a field F , denoted by F is analgebraic extension of F that is algebraically closed.

120. Let K be an algebraic extension of F . Suppose we have an injective fieldhomomorphism (embedding) ψ : F ↪→ F . Then there exists an injectivehomomorphism ψ : E → F such that ψ(x) = ψ(x) for all x ∈ F .Solution.

Follows immediately from Zorn’s lemma by considering the poset of all pairs (E ′, τ),where E ′ is an algebraic extension of F and τ : E ′ → F is an injective homomorphism.The partial ordering is given by (E ′, τ) ≤ (E ′′, σ) if E ′ ⊆ E ′′ and σ|E′ = τ .

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121. Let K be an algebraic extension of F . Suppose K is contained in an algebraicclosure F of F . Prove that K is a splitting field over F if and only ifany injective field homomorphism φ : K ↪→ F that is constant on F is anautomorphism of K.

Solution.

(⇒) Let {fi(x)}i∈I be a collection of polynomials over F for which K is a splittingfield over F and let α ∈ K be a root of fi(x) for some i ∈ I. Without loss of generalitywe assume that fi(x) is irreducible over F . On one hand if g(x) denotes the imageof fi(x) under φ (applied to coefficients), then φ(α) is a root of g(x). On the otherhand φ(fi(x)) = g(x). Therefore, the subfield F (α) of K is mapped injectively intoitself in F . Since F (α) is a finite extension (finite dimensional vector space) of F ,φ : F (α) → F (α) is a surjection. In particular it is an isomorphism. We repeatthis process for all roots α, γ, β, . . . of all fi(x), i ∈ I to conclude that φ restricts toisomorphisms on all F (α), F (γ), F (β), . . . , hence it is an isomorphism of F (α, γ, β, . . . )onto itself. Since F (α, γ, β, . . . ) is the splitting field K, we are done.

(⇐) Let α ∈ K be an element and let p(x) denote its irreducible polynomial over F . Ifthere is any other root β of p(x), it lies in F . Let φ′ denote the field map F (α)→ F (β)defined by α 7→ β and c 7→ c for all c ∈ F . By Problem 120 we know that φ′ extendsto an injection φ : K ↪→ F , hence, by our assumption to an automorphism of K. Iffollows that β is an element of K. Repeating this process for all other roots we seethat p(x) splits completely in K. In particular we see that K is the splitting field ofthe set of all such polynomials over F .

122. Let K be an algebraic extension of F . Prove that K is a splitting field overF if and only if every irreducible polynomial over F that has a root in Ksplits completely in K[x].

Solution.

(⇒) Since K is algebraic over F it is contained in an algebraic closure of F . Letf(x) ∈ F [x] be an irreducible polynomial and suppose α ∈ K of f(x). Let β ∈ F beanother root of f(x) and define ψ : F (α) → F (β) by ψ(α) = β and ψ(c) = c for allc ∈ F . By Problems 120 and 121, ψ extends to an automorphism of K. Therefore,β ∈ K, hence f(x) splits in K[x].

(⇐) Conversely, for α ∈ K let pα(x) denote the irreducible polynomial over F of α.Then pα(x) splits completely in K[x]. Hence K is the splitting field of {pα(x)}α∈K .

123. Let F ⊂ K be a field extension and let K1 and K2 be two intermediate fieldswhich are finite extensions of F . Suppose both of K1 and K2 are splitting

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fields of some polynomials over F . Show that both of K1K2 and K1 ∩K2 aresplitting fields over F .

Solution.

Let φ : K1K2 ↪→ F denote an injective morphism, constant on F , into the algebraicclosure of K. Since K1 and K2 are splitting fields over F and since the restrictionsφK1 and φK2 are injective, by Problem 121 these restrictions are automorphisms of K1

and K2. Let x1y1 + · · · + xnyn, xi ∈ K1, yi ∈ K2 denote an element of K1K2. Then,φ(x1y1 + · · · + xnyn) = φ(x1)φ(y1) + · · · + φ(xn)φ(yn) lies in K1K2 also. Therefore, φis an automorphism of K1K2, hence, K1K2 is a splitting field by Problem 121.

To prove that K1 ∩K2 is a splitting field over F we use Problem 122: let α ∈ K1 ∩K2

be a root of an irreducible polynomial p(x) over F . Since both of the fields K1 andK2 are splitting fields, p(x) splits completely over them, hence it splits completely overK1 ∩K2. In particular, K1 ∩K2 is a splitting field.

124. Let d and n be two positive integers. Prove that xd − 1 divides xn − 1 if andonly if d divides n.

Solution.

(⇐) Suppose n = dk for some k ∈ N. If ξd is the primitive dth root of unity, thenξnd = (ξd)

dk = 1k = 1, hence every root of xd− 1 is a root of xn− 1 proving that xd− 1divides xn − 1.

(⇒) Conversely, suppose that xd − 1 divides xn − 1 and suppose r ∈ {0, 1, . . . , d− 1}.We claim that xr− 1 is not divisibly by xd− 1. Indeed, for every m = 1, . . . , d the mthpower of ξd would be a root of xr − 1. In particular ξd and ξr+1

d would be roots. Butthen 0 = (ξd)

r+1 − 1 = (ξd)rξd − 1 = ξd − 1, which is absurd, hence xd − 1 - xr − 1 for

any 0 ≤ r < d. Next we assume that d does not divide n, hence by division we haven = dq+r for some q, r ∈ N and 0 < r < d. Since xd−1 divides xqd+r−xr = xr(xqd−1),by the first part of our solution we obtain a contradiction from the equation

xn − 1 = xqd+r − 1 = (xqd+r − xr) + (xr − 1).

125. Improve the previous problem by showing that for any fixed integer a > 1,d|n ⇐⇒ (ad − 1)|(an − 1).

Solution.

Let us validate our hint: by Problem 124, we know that d|n ⇒ (xd − 1)|(xn − 1), inparticular ad − 1 divides an − 1 by evaluation. Conversely, assume that ad − 1 dividesan − 1 but d - n, that there exists 0 < r < d such that n = qd + r for some non-negative integer q. By the same trick we used in the proof of Problem 124 we write:

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an−1 = aqd+r−1 = (aqd−1)ar−(ar−1). Equivalently, −(an−1)+((ad)q−1)ar = ar−1.Since (ad)q−1 = (ad−1)((ad)q−1 +(ad)q−2 +· · ·+(ad)1 +1) the left hand side is divisibleby ad− 1 but (ar− 1)/(ad− 1) is not an integer because 0 < r < d. This contradictionshow shat d|n.

126. Let Fpd denote the finite field with pd elements. Show that if Fpd ⊆ Fpn, thend divides n.

Solution.

Recall that F∗q the group of non-zero elements of the finite field with q elements isisomorphic to the cyclic group of order q − 1. On the other hand, if Z1 and Z2

are two cyclic groups then Z1 ↪→ Z2 if and only if |Z1| divides |Z2|. Therefore, theFpd ⊆ Fpn , then F∗pd is a subgroup of F∗pn hence pd − 1|pn − 1. Now the result followsfrom Problem 125.

127. Let F be a field of characteristic p. Show that for any polynomial f(x) ∈ F[x]we have fp(x) = f(xp).

Solution. It suffices to observe (a + b)p = ap +(p1

)ap−1b + · · · +

(pp−1

)abp−1 + bp which

is equal to ap + bp in a characteristic p field.

Fact 0.42. Eisenstein’s Criterion: Let P be a prime ideal in an integral domain R.Suppose that a monic polynomial f(x) = xn+an−1x

n−1+· · · a1x+a0 from R[x] satisfiesthe following properties:

• each coefficient ai belongs to P ;

• constant coefficient a0 does not belong to the square of P , that is a0 /∈ P 2.

Then f(x) is irreducible over R (hence over K, the fraction field of R).

128. Let p and ` be two prime numbers, and let [`]x denote the “`th cyclotomicpolynomial” 1 + x+ · · ·+ x`−1.

(a) Show that [`]x is an irreducible element of the polynomial ring Q[x].

(b) Show that [`]x is divisible by x− 1 in the polynomial ring Fp[x] if p = `.Here Fp is the finite field Fp = Z/pZ.

(c) Suppose now that p is different than `, and let a be the order of p inF`, that is to say, a is the smallest positive integer such that pa = 1 inF`. Show that a is the first value of m for which the group of invertible

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m × m matrices with entries from Fp contains an element of order `.Hint: Use the formula for number of elements in GLm(Fp).

(d) Show that [`]x is not divisible by any polynomial of degree smaller thana in Fp[x]. Here, a is as in the previous part. Hint: Consider thecompanion matrix of any such divisor and use the previous part.

(e) Let A denote a matrix of order ` from GLm(Fp). Assume that m issmallest possible number a for which such A exists (see part (c)). IfmA(x) ∈ Fp[x] is the minimal polynomial of A, then show that mA(x) isirreducible of degree a, and it divides [`]x in Fp[x].

(f) Prove that [`]x is irreducible in Fp[x] if and only if ` − 1 is the smallestpower of p which is congruent to 1 modulo ` (in other words, a = `− 1is the smallest positive integer such that pa − 1 = 0 in F`).

Solution.(a) First we make an easy observation: a polynomial f(x) is irreducible over Q if andonly if f(x+ 1) is irreducible over Q. Thus, it suffices to prove that

[`]x+1 =(x+ 1)` − 1

(x+ 1)− 1=x` +

(``−1

)x`−1 + · · ·+

(`1

)x1

x

= x`−1 +

(`

`− 1

)x`−2 + · · ·+

(`

2

)x+

(`

1

)= x`−1 + `x`−2 + · · ·+ `(`− 1)

2x+ `

is irreducible. This follows from Eisenstein’s criterion.(b) Suppose p = `. If p = 2 then [2]x = 1 + x which is equal to 1− x in F2[x]. If p > 2,then

[p]x =xp − 1

x− 1=

(x− 1)p

x− 1= (x− 1)p−1,

hence [p]x is divisible by x− 1 in Fp[x].(c) Let us compute the number of elements in GLm(Fq), where q is a prime power: Lete1, . . . , em denote the standard basis for Fmq . If A ∈ GLm(Fq), then Ae1, . . . , Aem is abasis for Fmq . Conversely, any basis for the vector space Fmq put together as the columnsof a matrix gives an invertible m×m matrix over Fq. So, we count the number of basesfor Fmq . The first element of a basis B = {f1, . . . , fm} is any non-zero vector from Fmq soit can be chosen in qm−1 different ways. The second element f2 of B is any vector that isnot a scalar multiple of f1, hence it can be chosen in qm−q different ways. f3 is a vectorthat is not in the span of f1 and f2. In other words, f3 ∈ Fmq −{af1 + bf2 : a, b ∈ Fq}.There are qm − q3 choices for f3. Continuing in this manner we see that fi is one ofthe qm − qi possible vectors from Fmq − {

∑i−1j=0 aifi : ai ∈ Fq}. Therefore;

|GLm(Fq)| = (qm − 1)(qm − q) · · · (qm − qm−1). (9)

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Going back to the original problem, we are going to show that the order a of p in F`is the smallest m such that GLm(Fq) contains an element of order `. The necessarycondition for this to happen is that ` divides |GLm(Fp)| = p(

m2 )∏m

i=1(pi − 1). Here,since ` 6= p, we see that the first value of m that ` divides p(

m2 )∏m

i=1(pi − 1) is when` divides the highest term pm − 1 of the product for the first time. Obviously, thishappens exactly when pa − 1 = 0 mod `.

(d) Suppose [`]x is divisible by a polynomial f(x) of degree s. Without loss of generalitywe assume that f(x) has a non-zero constant term, hence its companion matrix A isinvertible (it belongs to GLs(Fp)). Since the characteristic polynomial of a companionmatrix A of a polynomial f(x) is the polynomial f(x) itself, we see that f(A) = 0.Therefore, the value of the polynomial [`]x = (x`− 1)/(x− 1) on A is 0. It follows thatA` = id. In other words, A ∈ GLs(Fp) is an element of order `. By Part (c) we seethat this is possible only if s ≥ a.

(e) Let A ∈ GLa(Fp) be a matrix of order `, where a is smallest possible index forwhich such A exists. If mA(x) is the minimal polynomial of A over Fp, the irreduciblegenerator of the principal ideal consisting of all polynomials that vanish on A. SinceA is of order `, x` − 1 is divisible by mA(x). It follows that [`]x = (x` − 1)/(x − 1)is divisible by mA(x). The degree of the minimal polynomial of a matrix is less thanor equal to the degree of the characteristic polynomial of a matrix. Since the degreeof the characteristic polynomial is the size of the matrix, we see that degmA(x) ≤ a.But by part (d), we know that a ≤ degmA(x), therefore, we have the equality.

(f) Let A denote the companion matrix of [`]x, and let mA(x) denote the minimalpolynomial of A which divides the characteristic polynomial of A, which is [`]x. SincemA(x) is irreducible we know that the degree of mA(x) is greater than or equal to a.Since A satisfies [`]x we know that mA(x) divides [`]x. Therefore, if [`]x is irreducible,then mA(x) = [`]x and furthermore its degree is `− 1, which is the smallest number asuch that pa = 1 mod `. Conversely, if `−1 is the smallest number such that p`−1 = 1mod `, then m = ` − 1 is the smallest number such that GLm(Fp) has an element oforderm. In other words, A`−1 = id for some A ∈ GLm(Fp). It follows that [`]x vanisheson A, hence it belongs to the ideal generated by the minimal polynomial of A. By part(d), we know that the degree of mA(x) is ≥ a, hence degmA(x) = a = ` − 1. Sincedeg[`]x = `− 1, we see that [`]x = mA(x), hence irreducible.

Definition 0.43. A polynomial f(x) over a field F is called separable if its roots inF are all distinct from each other. Otherwise f(x) is called inseparable.

Let K be an extension of F and let x ∈ K be an element. Multiplication by x isan F -linear operator on K. We denote its minimal polynomial by px(t) ∈ F [t]. Iffor all x ∈ K the minimal polynomial px(t) is a separable polynomial, then K iscalled a separable extension over F . If there exists at least one element x ∈ K with

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inseparable minimal polynomial, thenK is called, logically, an inseparable extensionof F . If minimal polynomial of any element of K is inseparable over F , then K is calledpurely inseparable.

Fact 0.44. A straightforward to verify observation about having a root with multi-plicity is this: a polynomial f(x) has a root α with multiplicity ≥ 2 if and only if α isa root of f ′(x), the derivative of f(x). In other words, a polynomial f(x) is separableif and only if f(x) and f ′(x) are relatively prime.

129. Let f = f(X) ∈ F [X] be an irreducible polynomial. Prove that f is separableover k if and only if f ′ is non-zero.

Solution.

We already know that if f is separable, then f ′ and f are relatively prime, therefore,f ′ cannot be zero. We prove the converse by contrapositive, so, assume that f isnot separable. Then f and f ′ have a non-unit common divisor. On one hand f isirreducible, hence the common divisor of f and f ′ is f . On the other hand, the degreeof f ′ is less than that of f . Therefore, f | f ′ implies that f ′ = 0.

130. For any prime p and nonzero element a ∈ Fp, prove that the polynomialf(x) = xp − x+ a is irreducible and separable over Fp.Solution.

Let A(x) denote xp − x − a. First of all, A(x) is separable because A′(x) = −1 isnon-zero. Second, A(x) has no root in Fp, since bp− b− a = −a for any b ∈ Fp. Noticethat A(x + a) = (x + a)p − (x + a) − a = xp + ap − x − a − a = xp − x − a = A(x).Therefore, by iteration, we see that A(x + ba) = A(x) for all b = 1, 2, . . . . As b runsover all positive integers, ba runs over {0, . . . , p − 1} = Fp. This means that in thesplitting field K of A(x) over Fp, if α is a root, then α + k, k ∈ {0, . . . , p − 1} is aroot, also. Since the degree of A(x) is p, it follows these are all the roots of A(x)that is A(x) =

∏p−1k=0(x − α − k). If A(x) is not irreducible over Fp then there exist

r(x), s(x) ∈ Fp[x] such that A(x) = r(x)s(x) and the 0 < d = deg r(x) < p. On theother hand, being a divisor of A(x), r(x) has the form r(x) =

∏k∈J(x − α − k) for

some subset J ⊆ {0, . . . , p − 1}. The coefficient of xd−1 of r(x) is the negative of thesum of all roots of r(x), hence it is equal to

∑i∈J α+ i = |J |α+

∑i∈J i = dα+

∑i∈J i.

Since this coefficient lies in Fp, and since the degree d of r(x) is non-zero, we obtainthe contradictory statement that α ∈ Fp.

131. Show that for any prime p and a positive integer n there exists a finite fieldwith q = pn elements.

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Solutions.

We know that Z/p, which we denote by Fp, is a finite field with p elements. Considerf(x) = xp

n − x over Fp. Since f ′(x) = pnxpn−1 − 1 = −1, f(x) is separable, hence it

has pn distinct roots. It is easy to verify that all of these roots form a field K; closedunder multiplication, inverses exist, and closed under addition. Therefore, there existsa field K that contains F and has pn elements. Note: Fp is contained in K (since 1 isa root of f(x), and 1 generates Fp). In fact, K is the splitting field of f(x) over Fp.

132. In this problem we ask for an inverse of Problem 126. If d divides n, thenFpd is a subfield of Fpn.Solution.

By Problem 131 we know that both Fpd and Fpn exist. It suffices to show that thesplitting polynomial that defines Fpd is a divisor of Fpn . Indeed, the former’s splittingpolynomial is xpd − x = x(xp

d−1 − 1) and the latter’s is xpn − x = x(xpn−1 − 1). By

Problem 125 we know that pd − 1 divides pn − 1. Then by Problem 124, xpd−1 − 1divides xpn−1 − 1. Therefore, Fpd is a subfield of Fpn .

133. Show that a finite field F does not contain any ring R which is not a field.

Solution. If R ⊆ F is a subring, then the non-zero elements of R is a submonoid of thegroup of invertible elements of F, which is a cyclic group {1, g, . . . , gd}. Any submonoidof a cyclic group is a subgroup, hence it is cyclic. Therefore, R is a subfield of F.

134. Recall (from Problem 110) that in a finite field F of characteristic p anyelement is a pth power. Find a field K of characteristic p which has anelement that is not a pth power.

Solution.

Let X be a variable and consider the rational function field K = Fp(X). ClearlyK is of characteristic p. Let us show that X is not a pth power in K. Assumeotherwise that X = (f(X)/g(X))p, for some polynomials f(X), g(X) and hence Xdivides f(X). This leads to a contradiction as follows. Write f(X) = Xrf1(X), wherer ≥ 1 and X - f1(X). Simplifying the polynomial equality g(X)pX = f(X)p, we obtaing(X)p = Xrp−1f1(X)p. Thus X divides g(X). Write g(X) = Xsg1(X) with X - g1,and s ≥ 1. It follows that Xsp = Xrp−1 which is impossible.

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135. Find an irreducible polynomial f(x) over a field F which is not separable.

Solution.

As we know from Problem 134, a variable X is not a pth power in F = Fp(X), hencethe polynomial f(Y ) defined by f(Y ) = Y p − X ∈ F [Y ] has no root in F . Let Kdenote the splitting field of f(Y ) over F , and let x ∈ K be a root. The characteristicof K is p, therefore, it follows from binomial expansion that (Y − x)p ∈ K[Y ] is equalto f(Y ) = Y p −X ∈ K[Y ]. In other words, the multiplicity of the root x is p, hence,f(Y ) is not separable.

We claim that f(Y ) is irreducible in F [Y ]. First of all, R = Fp[X] is an integral domainand the fraction field of R is F = Fp(X). Since R is a UFD, by Problem 115, to showthat f(Y ) is irreducible over F [Y ], it is enough to prove its irreducibility over R[Y ].Let P ⊂ R[Y ] denote the prime ideal generated by X in R[Y ] = Fp[X, Y ]. Since X ∈ Pbut X /∈ P 2 = (X2), by Eisenstein’s criterion (Fact 0.42), f(Y ) is irreducible in R[Y ],hence in F [Y ] = Fp(X)[Y ].

Definition 0.45. Problem 135 brings us to the notion of a “perfect field.” A field Fis called perfect if every irreducible polynomial is separable.

136. Let F be a characteristic 0 field. If f(X) ∈ F [X] is irreducible, then it isseparable. In other words, F is perfect.

Solution.

If f(X) ∈ F [X] is a non-constant irreducible polynomial, then its derivative f ′ isnon-zero in F [X], hence f is separable by Problem 129.

137. Prove that any finite field is perfect.

Solution

Let K be a finite field with |K| = pn, and let f(x) ∈ K[x] be an irreducible polynomial.Assume towards a contradiction that f(x) is not separable, or equivalently that f ′(x) =0 (by Problem 129). Then each monomial in f(x) is of the form axn where n is divisibleby p. We analyze f(x) further by writing it explicitly:

f(x) = amxmp + am−1x

(m−1)p + · · ·+ a1xp + a0

= bpm(xm)p + bpm−1(xm−1)p + · · ·+ b1xp + bp0 (since every coefficient is a pth power)

= (bmxm + · · ·+ b0)p.

But this means f(x) is not irreducible, a contradiction.

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Fact 0.46. Let M be an n × n matrix with entries from a field F . If F contains allthe eigenvalues of M , then M is similar to a diagonal matrix over F if and only if theminimal polynomial of A has no repeated roots. Here, by a minimal polynomial wemean a polynomial of the smallest degree p(x) ∈ F [x] such that p(A) = 0. (See thesection on RATIONAL CANONICAL FORMS.)

138. Let A ∈ GLn(C) be an invertible matrix with entries from C such that Ak = idfor some k ≥ 1. Show that A is diagonalizable. Show that the matrix

B =

(1 α0 1

)∈ GL2(Fp)

satisfies Bp = id but it is not diagonalizable.

Solution.

Since A ∈ GLn(C) satisfies the polynomial equation xk−1 = 0 the minimal polynomialof A is a divisor of xk − 1, hence it has distinct roots. In particular, by Fact 0.46, A isdiagonalizable.

For the second part observe that, for all k ≥ 1 Bk =

(1 kα0 1

), hence Bp = id.

Therefore, the minimal polynomial is a divisor of xp − 1. Since xp − 1 = (x − 1)p

in Fp[x], and since B 6= id, we see that the minimal polynomial has a repeated root,therefore, by Fact 0.46 B is not diagonalizable.

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MODULES OVER PRINCIPAL IDEAL DOMAINS

We begin with some warm-up exercises.

Definition 0.47. An element m of a module M over an integral domain is calledtorsion if r ·m = 0 for some r ∈ R. The set of all torsion elements of a module M isdenoted by tor(M). It is straightforward to verify that tor(M) is an R-submodule ofM (commutativity plays a role in the proof).

The rank of a moduleM over an integral domain R is the maximal number of R-linearlyindependent elements.

139. Let M be a module over an integral domain R. Show that the rank of M isthe same as that of M/tor(M).

Solution. We begin with showing that the rank of tor(M) is zero. Indeed, there is noR-linearly independent set of elements from tor(M): for any set of elements x1, . . . , xsfrom tor(M) there exists a nonzero r ∈ R such that s·(x1+· · ·+xs) = sx1+· · ·+sxs = 0.

Next, let B = {y1, . . . , yr} ⊂M be a maximally R-linearly independent set of elementsfrom M . None of the yi’s lie in tor(M) otherwise elements of B satisfy

a1y1 + · · ·+ aryr = 0

with not all ai’s are zero, hence B is linearly dependent. Thus, the images of theelements of B in M/tor(M) form a maximally R-linearly independent elements.

Definition 0.48. An R-moduleM is called a torsion R-module, if for a givenm ∈Mthere exists r ∈ R such that rm = 0.

140. Let R be an integral domain and let M be an R-module. (a) Suppose Mis of rank n. Prove that if B = {y1, . . . , yn} ⊂ M is a maximally R-linearlyindependent subset, then the R-submodule N of M that is generated by Bis a free R-module of rank n, and furthermore M/N is a torsion R-module.(b) Prove the converse: if N ⊆ M is a free R-submodule of rank n and ifM/N is a torsion R-module, then the rank of M is equal to the rank of N .

Solution.

(a) Since B is an R-linearly independent subset it is clear that N ∼= Ry1⊕· · ·⊕Ryn. Inother words, N is free of rank n. Let m ∈M be a non-zero element. The set {m} ∪ Bis linearly dependent by the maximality assumption on B. It follows that there existsnot all zero elements r0, r1, . . . , rm from R such that r0m+r1y1 + · · ·+rnyn = 0. Hence,r0m ∈ N , or equivalently, r0 ·m = 0 in M/N . Since m is arbitrary, M/N is a torsionR-module.

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(b) Since M/N is torsion, for an element m ∈ M , there exists r ∈ R such thatrm ∈ N . Since N is free of rank n, it has a basis B = {y1, . . . , yn} ⊂ N . Thusrm = r1y1 + · · · + rnyn for some r1, . . . , rn ∈ R. It follows that we cannot add anyelement of M to B and keep the R-linear independence. Therefore, B is a maximallyR-linearly independent subset of M , hence the rank of M is n, also.

141. Let R be an integral domain and N ⊂M be two R-modules. If the ranks ofM , N , and of M/N are given by, respectively, n, r and s, then prove thatm = r + s.

Solution.

Let x1, . . . , xs be a list of elements fromM such that their images x1, . . . , xs inM/N isa maximally R-linearly independent. If there is an R-linear relation among xi’s, thenit carries into M/N , hence their images would not be linearly independent. Therefore,x1, . . . , xs are linearly independent in M . Let {y1, . . . , yr} be a maximally R-linearlyindependent subset of N . We claim that {x1, . . . , xs, y1, . . . , yr} is maximally R-linearlyindependent in M . Assume otherwise that there exist a1, . . . , as, b1, . . . , br such that∑aixi +

∑bjyj = 0. On the one hand we know that not all of ai’s are simultaneously

zero, and similarly, not all bj’s are simultaneously zero. On the other hand, reducingmod N we see that

∑aixi = 0 in M/N . By our initial assumption on xi’s we then

have ai = 0, a contradiction. Therefore, {x1, . . . , xs, y1, . . . , yr} is maximally R-linearlyindependent set and the rank of M is s+ r.

142. Here is an example of an R-module of rank 1 which is not free: Let R denotethe polynomial ring Z[x] and M ⊂ R denote the ideal generated by 2 and x.Show that M is an R-module of rank 1 but it is not free. More generally,if R is any integral domain and M ⊆ R a non-principal ideal, then M is atorsion-free R-module of rank 1 but not a free R-module.

Solution. Every ideal in R has a natural structure of an R-module. Let y1, . . . , yn bea maximal set of R-linearly independent elements from M . Since y1, y2 belongs to R,we define a1 = −y2, a2 = y1, a3 = · · · = an = 0. Then a1y1 + a2y2 + · · ·+ anyn = 0 eventhough not all of ai’s are zero. We conclude that a maximal set of linearly independentelements in M has only 1 element. In other words, M is of rank 1. Next we provethat M is not a free R-module. Let N ⊂M be the subideal generated by 2. Since anyelement of N is of the form 2f(x) for f(x) ∈ R, N is a free R-module of rank 1. Noticethat M/N is the Z[x]-module consisting of polynomials with odd integer coefficients,hence it is non-zero, proving that M is not free.

This argument generalizes verbatim to the setting of an arbitrary integral domain anda non-principal ideal.

70

143. Let R be an integral domain. Find an example of a torsion R-module Msuch that the annihilator of M is 0.

Solution. The Z[x]-moduleM =⊕∞

i=1 Z[x]/(xi) is a torsion module: If f = (f1, f2, . . . )is an element of M , then only finitely many fi’s are non-zero (meaning that fi /∈ (xi)).If j is sufficiently large positive integer, then xr · f = (xrf1, x

rf2, . . . ) = (0, 0, . . . ). Onthe other hand, for any polynomial g(x) ∈ Z[x] of degree d, the element g(x)f ∈ M ,where f = (f1, f2, . . . ) with fd+1 = 1, is non-zero. Hence, the annihilator of M is {0}.

144. Let R be an integral domain. Prove that if N is a finitely generated torsionR-module, then the annihilator of N is nonzero.

Solution.

Let {y1, . . . , yk} be a generating set for N . Since N is a torsion module, for eachi ∈ {1, . . . , k} there exists ri such that riyi = 0. Then the nonzero element r = r1 · · · rklies in the annihilator of N .

Definition 0.49. Let R be a PID, M a torsion R-module, p ∈ R a prime element.(Recall that an element p is called prime if and only if the principal ideal generatedby p is a prime ideal. In a UFD an element is principal if and only if it is irreducible.)The p-primary component of M is the set of all elements of M that are annihilatedby some positive power of p.

145. Let R, M and p be as in the definition.

(a) Prove that p-primary component of M is a submodule.

(b) Prove that M is the direct sum of its p-primary components as p runsover all prime elements.

Solution.

(a) Let m1 and m2 be two elements from p-primary component of M and let r ∈ R bean arbitrary element. This means pum1 = pvm2 = 0 for some positive integers u and v.Since pu+v(m1 + rm2) = pu+vm1 + rpu+vm2 = 0, the p-primary subset is a submodule.

(b) Given a prime element p ∈ M let us denote by M(p) the p-primary component.We claim that M(p) ∩ M(q) = 0, where M(p) is the p-primary component of M(M(q) is defined similarly). Assume otherwise that there exists a nonzero elementm ∈ M(p) ∩ M(q). Then there exist positive powers pa and qb that belong to theannihilating ideal AnnR(m) of m in R. Since R is a PID, pa and qb are relativelyprime, the ideal is the whole ring, which is absurd (1 ∈ R cannot annihilate anynon-zero element).

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Next, we show that any element m ∈ M is of the form m = m1 + · · · + mk for somemi ∈M(pi), where p1, . . . , pk are distinct prime elements. SinceM is a torsion module,there exist r ∈ R such that rm = 0. Suppose that r = upa11 · · · p

akk is the irreducible

decomposition of r. It is clear that the list r1, . . . , rk, where ri = r/paii (1 ≤ i ≤ k)consists of relatively prime elements. Therefore, there exists f1, . . . , fk ∈ R such that∑firi = 1. Denoting by mi the element firim, we have the following decomposition:

m = m1 + · · ·+mk. Note that mi is annihilated by paii , therefore, m ∈ ⊕ki=1M(pi) andthe proof is complete.

146. Let R be a PID and M be a free R-module. Show that any R-submodule Nof M is free, also.

Solution.

Note that we do not assume M is finitely generated. Let {ei} be a basis for M whichwe assume to be totally ordered (by Axiom of Choice). Let pi : M → R denotethe ith projection homomorphism defined by pi(

∑ajej) = ai. Clearly, the element

fi = riei ∈ M is in N ∩ Rei. We claim that {riei} is a basis for N . To this end,let n ∈ N be an arbitrary element. Clearly, n =

∑siei, where si = pi(n). Since

pi(n) ∈ (ri) we see that si is a multiple of ri, say si = αiri for some αi ∈ R. Thus,n =

∑αiriei showing that n lies in the span of {riei}. Since the set {ei} is R-

linearly independent, the set of scalar multiples {riei} is R-linearly independent, also.Therefore, it is a basis for N .

Remark 0.50. Recall that one of the several equivalent definitions of a projectivemodule is that a module M is a projective R-module if it is a direct summand of afree R-module. Problem 146 shows that over a PID a finitely generated R-module isprojective if and only if it is free.

147. Let R be a PID and let M be a finitely generated R-module. Prove thatM ∼= Rm ⊕ tor(M) for some (unique) m ∈ N.Solution.

Let {ui}ni=1 be a generating set for M . Define φ : Rn →M by φ(∑aiei) = aiui, where

e1, . . . , en is the standard basis for the free R-module Rn = R × · · · × R. It is clearthat φ is a surjective R-module homomorphism. Therefore, Rn/ kerφ ∼= M . Sincekerφ is a submodule of the free module Rn, we know from Problem 146 that it is freeand has a basis of the form {riei}i∈I for some elements ri ∈ R, and I ⊂ {1, . . . , n}.We split this basis into two subsets: A1 = {riei : ri is a unit in R} and A2 = {riei :ri is a non-unit}. Let m1 and m2 denote the cardinalities of A1 and A3, respectively.

72

Thus, kerφ = Rm1⊕(⊕

riei∈A2Rriei

)⊆ Rn, and therefore

M ∼= Rn/ kerφ ∼= (Rn−m1)⊕( ⊕

riei∈A2

R/Rri

). (10)

It is clear that the second component⊕

riei∈A2R/Rri gives the torsion submodule of

M . It is also clear from this argument that m = n − m1 is equal to the rank of M ,hence it is unique.

Definition 0.51. We know from Problem 147, specifically from eqn. (10) that anyfinitely generated module M over a PID R is isomorphic to Rs⊕R/(r1)⊕ · · · ⊕R/(rl)for some nonnegative integer s and some non-invertible elements r1, . . . , rl from R. Byusing Chinese Remainder Theorem we have a further decomposition of each factor:

R/(ri) ∼= R/(pa11 )⊕ · · · ⊕R/(pamm ),

where ri = pa11 · · · pamm is the decomposition of ri into irreducible factors in R. Anelementary divisor of M is one of the prime powers paii appearing in any of thesummands R/(ri).

148. Let R be a PID andM be a torsion R-module. Prove that the decompositioninto elementary divisors of M can be re-organized so that we have

M ∼= R/(s1)⊕ · · · ⊕R/(sk),

for some (unique) elements si ∈ R such that si | si+1 for i = 1, . . . , k − 1.

Solution.

Since R is a PID, it is a UDF. Therefore, each ri is a product of irreducible elements:ri = p

ri,1i,1 p

ri,2i,2 · · · p

ri,dii,di

for some positive exponents. By Chinese Remainder Theorem:

R/(ri) ∼=(R/(p

ri,1i,1 ))⊕(R/(p

ri,2i,2 ))⊕ · · · ⊕

(R/(p

ri,dii,di

)).

Next, we prepare a right justified array of the powers of the irreducible factors appearingin any of the summands R/(ri). Along each row we place the powers of pi (that appearin any of R/(ri)) ordered so that the exponents increase from left to right. Obviously,there are rows of various lengths. If the length of the longest row is k, then we completeour array into an l × k array by placing a 1 in each available entry. For example, forM ∼= R/(q4t8) ⊕ R/(p3t) ⊕ R/(p2t2) ⊕ R/(p3t8) ⊕ R/(t6), the list of prime powers isp2, p3, p3, q4, t, t2, t6, t8, t8, hence, our array is:

1 1 p2 p3 p3

1 1 1 1 q4

t t2 t6 t8 t8

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Next, we multiply the entries along each column of the array and get k elementss1, . . . , sk with the property that si | si+1 for i = 1, . . . , k − 1. Since their componentsare rearrangement of each other, there is an isomorphism between the following directsums:

M ∼=k⊕i=1

R/(si) ∼=l⊕

i=1

R/(ri). (11)

The uniqueness of the factors si are clear from the uniqueness of the array we con-structed above for proving (11).

149. Let a be a nonzero element of a principal ideal domain R. If M = R/(a) andp ∈ R is a prime element, then prove that

pk−1M/pkM ∼=

{R/(p) if k ≤ n

0 if k > n

where n is the power of p dividing a.

Solution.

Let a = pnq be the factorization of a with p and q are relatively prime. There ex-ists f, g ∈ R such that fp + gq = 1. Thus, for any x ∈ M = R/(a) we have thedecomposition x = fpx+ gqx.

If k > n, then k − 1 ≥ n, hence pk−1x = fpkx + gpk−1qx = pkfx. In other words,pk−1x = 0 modulo pkM .

If k ≤ n, then pk−1x = fpkx + gpk−1qx, hence pk−1x = pk−1gqx mod pkM . Defineφ : pk−1M/pkM → R/(p) by φ(pk−1x mod pkM) = x′ mod (p), where x′ ∈ R isany representative of x ∈ R/(a) in R. Lets check that φ is well-defined: if pk−1x1

mod pkM = pk−1x2 mod pkM , then pk−1(x1 − x2) ∈ pkM , hence x1 − x2 is divisibleby p. It follows that x′1 − x′2 = 0 in R/(p). Now we know that φ is well-defined weprove that it is an isomorphism. The kernel of φ is clearly those pk−1x ∈ pk−1M withany representative x′ of x is divisible by p, hence, x ∈ pkM . To prove the subjectivity,let u ∈ R/(p) be a nonzero element. Hence p and u′ (a representative of u in R) arerelatively prime. Then pk−1u′ is not divisible by pk, and furthermore pk−1u′ is not zeroin M = R/(a) = R/(pnq) (since k−1 < n). Thus, pk−1u′ represents a nonzero elementin pk−1M/pkM which is mapped to u mod (p). It is clear that φ is an R-module map,therefore, it is an isomorphism.

150. Let R be a PID, p ∈ R be a prime element. Prove that if M is a finitelygenerated torsion R-module, then show that pk−1M/pkM ∼= F nk , where F is

74

the field F = R/(p) and nk is the number of elementary divisors ps of M suchthat k ≤ s.

Solution.Since M is a finitely generated torsion R-module, we know of various decompositionsof M as in eqn. (11). Write, for example,

M ∼= R/(r1)⊕ · · · ⊕R/(rl)

for some ri ∈ R. Let us denote factor R/(ri) by Mi so that M ∼= M1⊕ · · · ⊕Ml. Sincethe sum is direct, we have

pk−1M/pkM ∼=l⊕

i=1

pk−1Mi/pkMi.

By Problem 149 we know that pk−1Mi/pkMi is isomorphic to R/(p) for all k ≤ n(i),

where n(i) is the power of p that divides ri. In other words, pk−1Mi/pkMi

∼= R/(p)if the elementary divisor pα of Mi satisfy k ≤ α. Thus, pk−1M/pkM has one copy ofR/(p) for each elementary divisor ps of M with s ≥ k.

Remark 0.52. In particular, if M1 and M2 are two isomorphic finitely generatedtorsion R-modules, then for every k ≥ 0 M1 and M2 has the same the same set ofelementary divisors.

151. Let R be an integral domain with quotient field K and letM be an R-module.Prove that the rank of M is equal to the dimension of the K-vector spaceK ⊗RM .

Solution. Let {ui}i∈I be an R-linearly independent maximal set. We claim that {ui ⊗1}i∈I is a linear independent set in K⊗RM . Assume otherwise that there is a relationamong its elements: a1(u1⊗1)+· · ·+ar(ur⊗1) = 0 for some scalers ai ∈ K. Multiplyingby a common multiple of the denominators of ai’s we assume that ai lie in R. On theone hand, since tensor products are linear, we have (

∑ri aiui) ⊗ 1 = 0. On the other

hand, since we work with an integral domain, this primitive tensor is 0 if and only if∑ri aiui = 0, hence ai = 0 for all i = 1, . . . , r. Hence rankRM ≤ dimK K ⊗RM .

Conversely, if B = {ai/bi ⊗ ui}i∈I is a vector space basis for K ⊗R M , then for anyfinite subset {a1/b1 ⊗ u1, · · · , an/bn ⊗ un}, let b =

∏bi. We claim that the set

{ba1/b1u1, . . . , ban/bnun} is an R-linearly independent set. Indeed, otherwise, thereexists not all zero elements c1, . . . , cn ∈ R such that

∑cibai/biui = 0. Then

bc1(a1/b1 ⊗ u1) + · · ·+ bcn(an/bn ⊗ un) = ((bc1a1/b1)⊗ u1) + · · ·+ ((bcnan/bn)⊗ un)

= (1⊗ (bc1a1/b1)u1) + · · ·+ (1⊗ (bcnan/bnun))

= 1⊗ (bc1a1/b1u1 + · · ·+ bcnan/bnun))

= 1⊗ 0 = 0.

75

This contradiction shows that dimF F ⊗RM ≤ rankRM .

76

CANONICAL FORMS OF LINEAR OPERATORS

Let V be a finite dimensional vector space over a field F and T : V → V be an F -linearmap. This data is equivalent to giving an F [x]-module structure on V via the actionp(x) · v = p(T )v, where p(x) =

∑aix

i ∈ F [x] and so p(T )v = anTnv + · · ·+ a0v ∈ V .

Since V is a finitely generated F -module, it is finitely generated over F [x], as well.Also, since any free F [x]-module is an infinite dimensional vector space over F , we seethat V is a finitely generated torsion F [x]-module. Therefore, by Problem 148 we have

V ∼= F [x]/(a1(x))⊕ F [x]/(a2(x))⊕ · · · ⊕ F [x]/(ak(x)) (12)

for some polynomials ai(x) ∈ F [x] such that ai(x) | ai+1(x) for i = 1, . . . , k − 1. Iden-tifying the linear action of T on V by the multiplication by x on the right hand side ofeqn. 12, we represent T in an another matrix form. Since the images of 1, x, . . . , xdi−1

in F [x]/(ai(x)) is a vector space basis, where di = deg ai(x), we see that multiplica-tion by x on this particular component has a pleasant matrix form, namely, by thecompanion matrix of ai(x) = xdi + bdi−1x

di−1 · · ·+ b1x+ b0:

Ci =

0 0 0 · · · 0 0 −b0

1 0 0 · · · 0 0 −b1

0 1 0 · · · 0 0 −b2

0 0 1 · · · 0 0 −b3

......

.... . .

......

...0 0 0 · · · 1 0 −bdi−2

0 0 0 · · · 0 1 −bdi−1

The block diagonal matrix with diagonal blocks C1, . . . , Ck, called the rational canon-ical form of T and denoted by rat(T ), determines the similarity class of T . Note thatthe characteristic polynomial of the operator T is the product of the characteristicpolynomials of Ci’s, which, in turn, are equal to ai(x)’s. The minimal polynomial ofT is the ‘largest invariant factor’ ak(x). Therefore, some power of the minimal poly-nomial is divisible by the characteristic polynomial, and conversely, the characteristicpolynomial is divisible by the minimal polynomial. In particular, this gives us varioususeful information: minimal polynomial and characteristic polynomials have the sameroots.

152. Recall from Problem 110 that the Frobenius map φ : Fpn → Fpn defined byφ(a) = ap is an Fp-linear automorphism of Fpn. Prove that φn is the identitytransformation on Fpn and no smaller power of φ is identity.

Solution. Since φ(a) = ap, a ∈ Fp we see that φ2(a) = φ(ap) = ap2 , and by induction

φi(a) = api for all i = 1, . . . , n. Since Fpn is the splitting field of the polynomial xpn−x

over Fp, we see that φn(a) = apn

= a. To prove that no smaller power of φ gives

77

the identity transformation we need to show that for all i = 1, . . . , n − 1, there existsa ∈ Fpn such that api − a 6= 0. Assume otherwise that there exists 1 ≤ i < n such thatap

i − a = 0 for all a ∈ Fpn . But this is impossible because the polynomial xpi − x is ofdegree pi and Fpn has pn elements.

153. Determine the rational canonical form of the Frobenius map φ : Fpn → Fpn.Solution.

By Problem 152 we know that φn = id and for no i ∈ {1, . . . , n−1} φi = id. Therefore,the minimal polynomial of φ is xn − 1. In particular, its degree is n. Since Fpn is ann-dimensional vector space over Fp, we see that the rational canonical form of φ is thecompanion matrix of the minimal polynomial, which given by

rat(φ) =

0 0 0 · · · 0 0 11 0 0 · · · 0 0 00 1 0 · · · 0 0 00 0 1 · · · 0 0 0...

......

. . ....

......

0 0 0 · · · 1 0 00 0 0 · · · 0 1 0

.

154. Determine the Jordan canonical form of the Frobenius map φ : Fpn → Fpnconsidered as an Fp-linear automorphism. Here we assume that Fp containsall eigenvalues of φ.

Solution.

We know that the minimal polynomial and the characteristic polynomial of φ both arethe same and equal to xn − 1. Since we assume that all roots of xn − 1 lie in Fp, itdecomposes into linear factors as in

xn − 1 = (x− λ1)s1 · · · (x− λk)sk

for some positive integers si such that s1 + · · ·+ sk = n, and λi ∈ Fp’s are its distinctroots. Thus, the Jordan canonical form of φ over Fp is of the form

J1 0 0 · · · 00 J2 0 · · · 00 0 J3 · · · 0...

......

. . ....

0 0 0 · · · Jk

,

78

where Ji is the si × si Jordan block matrix having λi on its diagonal. We furtheranalyze the situation with respect to divisibility of n by p:

If p does not divide n, then xn − 1 is separable (the derivative of xn − 1 is non-zero inFp), hence the Jordan canonical form is an n× n diagonal matrix with distinct entriesas the roots of xn − 1 in Fp.If p divides n, say n = pem (with e ∈ Z+ and p - m), then xn−1 = xp

em−1 = (xm−1)pe .

Therefore, the set of roots of xn− 1 is precisely the set of roots of xm− 1. In this case,there are m distinct Jordan blocks J1, . . . , Jm, each having the same size pe × pe withan mth root of unity along its diagonal.

Fact 0.53. Let A be an n × n matrix with entries from a field F and assume thatF contains all eigenvalues of A. In this case, A is diagonalizable if and only if theminimal polynomial of A has no repeated roots.

155. Determine the characteristic polynomial of the Frobenius map φ : Fpn → Fpnand prove that φ is diagonalizable over Fp if and only if n divides p− 1, andis diagonalizable over the algebraic closure of Fp if and only if gcd(n, p) = 1.

Solution.

We repeat the determination of the characteristic polynomial: Fpn is n-dimensionalover Fp. By Problem 152 we know that φn = id and for no i ∈ {1, . . . , n− 1} φi = id.Therefore, the minimal polynomial of φ is xn−1, hence it is equal to the characteristicpolynomial.

The operator φ is diagonalizable if and only if its minimal polynomial xn−1 factorizesinto distinct linear factors in Fp[x]. If xn − 1 =

∏α∈S(x− α) in Fp[x], then S forms a

subgroup of invertible elements of Fp. Therefore, n = |S| divides p − 1 = |F∗p|. Overthe algebraic closure of Fp, xn− 1 always splits into linear factors. The only conditionthat remains to be checked is if all roots are non-repeated, or if xn − 1 is separable.This is the case if n 6= 0 in Fp, or if gcd(n, p) = 1.

79

AUTOMORPHISMS OF EXTENSIONS

Notation. Given a ring extension A ⊂ B, we denote by AutA(B) the group of all ringisomorphisms σ : B → B such that σ(a) = a. (Equivalently, AutA(B) is the group ofall A-algebra automorphisms of B.

156. Determine AutF (F [x]), where F is a field, x is a variable.

Solution. Since σ : F [x] → F [x] is constant on F , it is uniquely determined by σ(x).It is clear that x has to be mapped to an element of the form ax + b (it cannot bemapped to a scalar, neither it could be mapped to a power xk with k ≥ 2) with a 6= 0.Therefore,

G = AutF (F [x]) ∼= {(a, b) ∈ F × F : a 6= 0} = F ∗ × F,

where the group structure on the last space is given by (a, b) · (c, d) = (ac, ad+ b).

157. Determine AutQ(R).

Solution.

If x ∈ R is a positive real number, then x = y2 for some y ∈ R. Thus, σ(x) = σ(y2) =(σ(y))2, hence it is a positive number. If x ∈ R is a negative number, then x = −y2

for some y ∈ R, hence σ(x) = −(σ(y))2, hence it is a negative number. Let x, y ∈ Rbe two real numbers such that x < y. Since y − x is positive, σ(y − x) = σ(y)− σ(x)is positive also. Therefore, x < y implies that σ(x) < σ(y) for all x, y ∈ R.Next we prove that σ is a continuous map. To this end, it is enough to prove that aninfinitely small neighborhood of a point x is mapped to an infinitely small neighborhoodof the image σ(x). Equivalently, we prove that −1/m < a − b < 1/m =⇒ −1/m <σ(a) − σ(b) < 1/m for every positive integer m and for any a, b ∈ R. Let x denotea− b. Since σ is order preserving we have σ(−1/m) < σ(x) < σ(1/m), or equivalently,−1/m < σ(x) = σ(a)− σ(b) < 1/m.

Now that we know σ is a continuous map, the function f : R → R defined by f(x) =x−σ(x) is continuous also. If f(y) 6= 0 for some real number y ∈ R, then by continuitythere exists a rational number x near y such that f(x) 6= 0. This contradiction showsthat there are no non-trivial (6= id) continuous functions on R that fixes Q.

Notation. Occasionally we use the abbreviated statement “f is an A-automorphism ofB” to mean that f is an A-algebra automorphism of B.

158. Prove that the F -automorphisms of the field F (x) = {f(x)/g(x) : f(x), g(x) ∈F [x], g(x) 6= 0} are precisely the fractional linear transformations. In otherwords, φ ∈ AutFF (x) if and only if there exits a, b, c, d ∈ F such that bc−ad 6= 0and φ(f(x)) = f

(bx+adx+c

)for all f(x) ∈ F (x).

80

Solution.

Let φ : F (x) → F (x) be an F -automorphism. Since φ is a field isomorphism, itis enough to determine its action on the element x ∈ F (x). Let us define the de-gree of an element f(x)/g(x) ∈ F (x) to be max{deg f(x), deg g(x)} and denote it bydeg(f(x)/g(x)). Here, we assume that f(x) and g(x) are relatively prime. We claimthat deg(φ(x)) = 1. Assume otherwise that the degree m of φ(x) is different than1. First of all, m cannot be 0, otherwise, φ(x) ∈ F hence φ(F (x)) = F . Supposeφ(x) = f(x)/g(x), where f(x) and g(x) are two relatively prime polynomials withg(x) 6= 0, r = deg f(x) and s = deg g(x). Assume that max(r, s) ≥ 2.

Let u(x) =∑n

i=0 aixi and v(x) =

∑mj=0 bjx

j be two relatively prime polynomials suchthat

x = φ

(u(x)

v(x)

)=u(φ(x))

v(φ(x))=

∑ni=0 ai(φ(x))i∑mj=0 bi(φ(x))j

=

∑ni=0 aif(x)ig(x)n−i(x)

g(x)n∑mj=0 bjf(x)jg(x)m−j(x)

g(x)m

=g(x)m (

∑ni=0 aif(x)ig(x)n−i)

g(x)n(∑m

j=0 bjf(x)jg(x)m−j) ,

hence

xg(x)n

(m∑j=0

bjf(x)jg(x)m−j

)= g(x)m

(n∑i=0

aif(x)ig(x)n−i

). (13)

Let α be a root of f(x). Since f(x) and g(x) are relatively prime, g(α) 6= 0. Evaluatingboth side of the eqn. 13 at α, we obtain αg(α)n+mb0 = g(α)n+ma0. Since u(x) andv(x) are relatively prime both of their constant terms, a0 and b0, cannot be zerosimultaneously. This means that f(x) has at most one root; α = a0/b0. It followsfrom our initial assumption that the degree s of g(x) is ≥ 2. Now that we know0 ≤ r ≤ 1 < s, we are able to argue over the leading terms of the polynomials on bothsides of eqn. 13. The degree of the leading monomial of the left hand side is 1+(m+n−j) deg g(x) for some j ∈ {0, . . . ,m−1}, but the leading monomial of the right hand sideis (m+ n− i) deg g(x) for some i ∈ {0, . . . , n− 1}. Modulo deg g(x) these two integersare not equal, hence we have a contradiction. Therefore, 0 ≤ deg g(x), deg f(x) ≤ 1.Since we have eliminated the possibility that deg f(x) = deg g(x) = 0, (if f(x) isconstant, then g(x) is not, and vice versa) we see that f(x) = bx + a, g(x) = dx + cwith different roots. This amounts to b0c0 − a0d0 6= 0.

159. Determine the fixed subfield of the F -automorphism u : F (x)→ F (x) definedby x 7→ x+ 1.

81

Solution.

We analyze the situation according to the characteristic of the underlying field. Firstassume that the characteristic is 0 and let f(x) and g(x) be two relatively primepolynomials such that f(x)/g(x) ∈ F (x) is u-fixed. Since f(x+1)/g(x+1) = f(x)/g(x),replacing x by x + 1, or by x − 1, we see that f(x + a)/g(x + a) = f(x)/g(x) for alla ∈ Z. Define G(x, y) := f(x+ y)g(x)− f(x)g(x+ y) ∈ Q[x, y]. It is clear that G(x, y)vanishes on Z2, therefore, it is identically zero. It follows that f(x)/g(x) is constant.

If F has characteristic p, then the answer is not as simple as in the previous case. Weclaim that f(x)/g(x) is invariant under x 7→ x+ 1 if and only if it is of the form

r(∏p

i=0(x− i))s(∏p

i=0(x− i))

for some polynomials r(x), s(x) ∈ F [x] with s(x) 6= 0. We prove this by inductionon the total degree n = deg f(x) + deg g(x). If n = 0, then there is nothing toprove because f(x)/g(x) is a constant. Assume that the result holds for n < d, andlet f(x)/g(x) be a rational function that is invariant under translation x 7→ x + 1and deg f(x) + deg g(x) = d. Without loss of generality, we assume that deg f(x) ≥deg g(x), since otherwise its reciprocal satisfies the same property as well.

Now, since f(x)/g(x) is translation invariant, so is f(x)/g(x) − f(a)/g(a) for anya ∈ {0, . . . , p − 1}. We fix one such a, and let c = f(a)/g(a). Since f(x)/g(x) − c =(f(x) − cg(x))/g(x), we see that f(x)/g(x) − c is translation invariant if and only iff(x)− cg(x) vanishes at every x = 0, . . . , p− 1, or equivalently, it is divisible by u(x).But then, (f(x)/g(x) − c)/u(x) is a rational function of total degree d − p < d andstill translation invariant. Hence, by induction hypothesis, it is a rational function inu(x), hence so is f(x)/g(x). This finishes the proof of our claim and the answer to thequestion.

160. If K is the splitting field of a polynomial f(x) over a field F , then

|AutFK| ≤ dimF K. (14)

Moreover, eqn. (14) is an equality when f(x) is separable.

Solution.

We start with a simple observation. We fix a root α ∈ K of an irreducible factorf1(x) of f(x). We claim that the degree of f1(x) is greater than or equal to number ofembeddings σ : F (α)→ K such that σ(x) = x for all x ∈ F . Moreover, we claim thesetwo numbers are equal when f1(x) is separable. To prove our claims, first, we extend σto an automorphism σ : K → K (see Problem 121). Since all coefficients of f1(x) arefrom F , σ(f(x)) = f(x), hence σ permutes the roots of f1(x) among themselves. In

82

other words, σ(α) = β for another root β of f1(x). Therefore, the number of injectivehomomorphisms σ : F (α)→ K which are constant on F are bounded by the degree off1(x), and simply equal to the degree when all roots of f1(x) are distinct.

Now, we are ready to prove eqn. (14) by using induction on the degree of the extensionK over F . If the extension degree is 1, then there is nothing to prove. Assume otherwisethat dimF K = n and that our claim is true for all splitting fields of degree d < n.Let α be a root of an irreducible divisor f1(x) ∈ F [x] of f(x) ∈ F [x]. By the aboveobservation we know that any σ ∈ AutFK gives us an embedding (that is to say, aninjective field homomorphism) ι : F (α) ↪→ K and the number of such embeddings is≤ dimF F (α) with equality when f1(x) is separable.

Now, the question is ‘how many different F -automorphisms of K restrict to the samemap ι?’ Let σ1 and σ2 be two elements of AutFK such that σ1|F (α) = σ2|F (α) = ι. Sinceσ−1

2 σ1 fixes F (α) point-wise, we see that the number of distinct F -automorphisms ofK that restricts to the same embedding ι : F (α) ↪→ K is bounded by AutF (α)K.Therefore,

|AutFK| ≤ |AutF (α)K| deg f1(x)

= |AutF (α)K| dimF F (α)

≤ dimF (α) K dimF F (α) (by induction)= dimF K.

161. Prove that if dimF K is finite, then K is the splitting field of a single poly-nomial f(x) over F .

Solution.

Let α1, . . . , αk ∈ K be a basis for K as a vector space over F and let f1(x), . . . , fk(x)denote their minimal polynomials over F of αi’s. We claim that K is the splitting fieldof f(x) :=

∏ki=1 fi(x). Indeed, if L is an intermediate field F ⊂ L ⊂ K such that f(x)

splits completely into linear factors over L, then αi’s are contained in L. Obviously,this implies that L = K.

Fact 0.54. LetG be a subgroup of the automorphism groupAutFK of a field extension.The fixed field L = {x ∈ K : σ(x) = x for all σ ∈ G} of G has the right dimension:

|G| = dimLK.

In particular, if K is a finite extension of F , then |AutFK| ≤ dimF K.

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162. Let F ⊂ K be a finite extension. Any subgroup G of AutFK is equal toAutEK, where E = KG is the fixed subfield of G. Furthermore, the mapG 7→ E which assigns a subgroup G its fixed subfield E = KG is injective.

Solution.

By definition, G ⊂ AutEK. Since |AutEK| ≤ dimEK = |G| by the above fact, we seethat AutFK = G. Of course, finiteness played a role in here. Now, our second claim isobvious: if G1, G2 are two subgroups of AutKF with the same fixed subfield E, thenG1 = AutEK = G2.

163. Prove that Q(√

2) is not isomorphic to Q(√

3).

Solution.

It is clear that any isomorphism from Q(√

2) to Q(√

3) (or from Q(√

3) to Q(√

2))has to map Q to Q. If a + b

√2, a, b ∈ Q is mapped to

√3 by an isomorphism

f : Q(√

2)→ Q(√

3), then f(a2+2b2+2ab√

2) = 3, or f(√

2) = (3−f(a2+2b2))/f(2ab),which is an element of Q. This contradicts with the fact that f−1, the inverse of fmaps the rational number (3− f(a2 + 2b2))/f(2ab) to

√2.

164. Determine the splitting field of f(x) = (x2 − 2)(x2 − 3)(x2 − 5) over Q andcompute its degree over Q.Solution.

Roots of f(x) are ±√

2,±√

3,±√

5. We already know that the splitting field of (x2 −3)(x2− 2) is Q(

√2,√

3). Next we show that√

5 /∈ Q(√

2,√

3). Assume otherwise thatthere exists a, b, c, d ∈ Q such that

√5 = a

√2 + b

√3 + c

√6 + d. Taking the square of

both sides of the equation, we see that

5 = (a√

2 + b√

3 + c√

6 + d)2

= 2a2 + 3b2 + 6c2 + d2 + 2ab√

6 + 4ac√

3 + 2ad√

2 + 6bc√

2 + 2bd√

3 + 2cd√

6

= 2a2 + 3b2 + 6c2 + d2 +√

6(2ab+ 2cd) +√

2(2ad+ 6bc) +√

3(4ac+ 2bd).

Thus, 1) ab+ cd = 0, 2) ad+ 3bc = 0, 3) 2ac+ bd = 0 and 4) 2a2 + 3b2 + 6c2 + d2 = 0.First three equations imply that a2 = 3c2, 2a2 = d2, 2a2 = 3b2. By the 4th equationwe see that 8a2 = 5. Since this equation has no solution in rationals, we see that

√5 /∈

Q(√

2,√

3), and since the minimal polynomial of√

5 is of degree two, the splitting fieldof f(x) is Q(

√2,√

3,√

5). Let K denote Q(√

2,√

3,√

5). Since dimQQ(√

2,√

3) = 4,and K is a degree 2 extension over Q(

√2,√

3), we see that the vector space dimensionof K over Q is 8.

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165. Determine the automorphism group of the splitting field of f(x) = x3−3x+1over Q.Solution.

f(x) is irreducible over Z (hence over Q as well). To see this we reduce mod 2 andevaluate: f(0) = 1, f(1) = 1. Therefore, f(x) is irreducible.

Next, we analyze the derivative of f(x) to obtain information about its roots. f ′(x) =3x2 − 3 is positive outside the interval (−1, 1) and negative in the interval (−1, 1).Since f(−1) = 3 and f(1) = −1 we see that there are three real roots, denoted α1, α2

and α3. Since f(x) is irreducible of degree 3, dimQQ(α1) = 3.

LetK denote the splitting field of f(x). Thus,K is either degree 6 overQ, or it is degree3 over Q. In the latter case the automorphism group is cyclic group of order 3. Let usprove that the former case is not possible. In fact, in the former case the automorphismgroup G = AutQK is S3, the symmetric group on 3 letters. Since the roots of f(x) arepermuted by the elements of G, we let σ ∈ S3 be the automorphism that maps α1 to α3,α3 to α1 and α2 to α2. Write f(x) = (x−α1)(x−α2)(x−α3) = x3−3x+1. By taking thederivative of f(x) we obtain (x−α2)(x−α3)+(x−α1)(x−α3)+(x−α1)(x−α2) = 3x2−3.Substituting α2 into this equation gives

(α1 − α2)(α1 − α3) = 3α21 − 3

(α2 − α1)(α2 − α3) = 3α22 − 3

(α3 − α1)(α3 − α2) = 3α23 − 3

On the one hand, multiplying right hand sides gives: 33∏3

i=1(αi − 1)(αi + 1) which isequal to 27f(1)f(−1) = 27(−1)3 = −81. On the other hand, multiplying left handsides gives −

∏1≤i<j≤3(αi − αj)2. In summary, after simplification (canceling negative

signs and taking square roots) we have∏1≤i<j≤3

(αi − αj) = 9.

The right hand side is sent to its negative by the permutation σ we picked above, but 9is obviously fixed by σ. Therefore, σ cannot be in G. It follows that the automorphismgroup G is not equal to S3, therefore, it must be the cyclic group of order 3. Moreover,Q(α1) = Q(α1, α2, α3).

Definition 0.55. A field extension F ⊆ K is called Galois, if it is

• algebraic,

• normal (splitting field of a family of polynomials over F ), and

• separable.

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The Galois group of a Galois extension is AutFK.

166. It follows from Problem 160 that the splitting field K of a polynomial f(x) ∈F [x] is Galois, then |AutFK| = dimF K. Prove more generally that for a finiteextension F ⊂ K the following are equivalent:

(a) F ⊂ K is Galois;

(b) |AutFK| = dimF K;

(c) K is the splitting field of a single separable polynomial over F .

Solution.

Let F ⊂ K be a finite extension and let α1, · · ·αk be an F -vector space basis for K.Let p1(x), . . . , pk(x) denote the minimal polynomials of αi’s over F . Let S denote theset of linear factors appearing in any of pi(x)’s, and let g(x) denote the polynomial∏

s(x)∈S s(x). In other words, g(x) is obtained from product of pi(x)’s by reducing themultiplicity of each repeated linear factor to 1. It is clear that g(x) is separable. Sinceeach σ ∈ AutFK permutes the roots of pi(x), the polynomial g(x) is invariant undereach σ ∈ AutFK, hence g(x) is defined over E, the fixed field of AutFK.

Now, we are ready to prove our claims.

((b) ⇒ (a)) First, assume that |AutFK| = dimF K. By Fact 0.54 we know thatdimF E = |AutFK|. It follows that dimF E = dimF K, hence, F = E. Therefore, g(x)is defined over F . Since g(x) is separable, and K is the splitting field of g(x) ∈ F [x],K is a Galois extension of F .

((a) ⇒ (b)) Suppose that K is Galois over F . By Problem 160, it is enough to showthat the polynomial g(x), which is defined in the first paragraph, is contained in F [x].To this end, let α ∈ K be a root of p1(x). Since p1(x) is irreducible in F [x], it is theminimal polynomial of α over F . Therefore, for all i = 2, . . . , k, we have pi(α) 6= 0. Inother words, p1(x), . . . , pk(x) do not share any common root. Thus, g(x) =

∏ki=1 pi(x),

hence it lies in F [x].

((c) ⇒ (a)) This follows from definitions.

((a) ⇒ (c)) This follows from the solution of ((a) ⇒ (b)).

Fact 0.56. The fundamental theorem of Galois theory (finite case): Let F ⊂K be a finite Galois extension, and letG denote AutFK. There exists an order reversingisomorphism between the inclusion lattice of all subgroups of G and the inclusion latticeof all intermediate fields F ⊂ E ⊂ K. Explicitly, such an isomorphism is given byE 7→ AutEK, and its inverse is H 7→ fixed subfield of H. Furthermore, a subgroup Hof G is normal in G if and only if the corresponding subfield E is Galois over F . (Notethat K is always Galois over E by Fact 0.54 and Problem 166.)

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167. Determine the automorphism (Galois) group of the splitting field of f(x) =(x2 − 2)(x2 − 3)(x2 − 5) over Q.Solution.

We know from Problem 164 that the splitting field of f(x) overQ isK = Q(√

2,√

3,√

5)and dimQK = 8. Hence, the Galois group G of this extension is a finite group of order8. Let H ⊂ G be the subgroup generated by the automorphisms σ, τ ∈ AutQK definedby σ(

√5) = −

√5, σ(

√3) =

√3, σ(

√2) =

√2, and τ(

√5) =

√5, τ(

√3) = −

√3,

τ(√

2) =√

2. Then H fixes Q(√

2) point-wise, and furthermore H ∼= Z/2 × Z/2.Since H is of index 2 in G, it follows that G is an abelian group. By the fundamentaltheorem of finitely generated abelian groups, the only possibility for G is that G ∼=Z/2× Z/2× Z/2. (G cannot be cyclic because H is not!.)

168. Determine the automorphism (Galois) group of the splitting field of f(x) =x4 + 2x+ 3 over Q.Solution.

Unfortunately f(x) does not split as nicely as x4 + 3x2 + 2, so things are more compli-cated (but not horrible). In any case, we look at

y2 + 2y + 3 =

(y − −2 +

√−8

2

)(y − −2−

√−8

2

)=(y − (−1 +

√2i))(

y − (−1−√

2i))

Therefore, four roots x1, x2, x3, x4 of f(x) are x1 =√−1 +

√2i, x2 = −x1, x3 =√

−1−√

2i, and x4 = −x3. Thus, the splitting field of f(x) over Q is equal toQ(x1, x3).

First, we make some obvious observations:

• Since f(x) is irreducible of degree 4, henceQ(x1) andQ(x3) are degree 4 extensionsof Q;• x2

1 − x23 =√

2i, hence Q(x1) ∩Q(x3) contains Q(√

2i);

• x21x

23 = 3, hence x1 = ±

√3/x3;

• x1 =√−1

2+√

32

+ i√

12

+√

32

(this requires a bit of calculation but it is not sobad).

Next, we claim that Q(x1) 6= Q(x3). To this end, let us assume otherwise that x3 ∈Q(x1) and find a contradiction. Indeed, if x3 ∈ Q(x1), then x1 = ±

√3/x3 implies that√

3 ∈ Q(x1), hence that Q(√

3) ⊆ Q(x1). In this case Q(x1) is a degree 2 extension ofQ(√

3). Therefore, there exists α, β ∈ Q(√

3) such that x21 +αx1 +β = 0. We write α =

87

a+√

3b, β = c+√

3d with a, b, c, d ∈ Q, and set x1 = u+ iv with u =√−1

2+√

32

and

v =√

12

+√

32. Thus, we have the equation −1+

√2i+(a+b

√3)(u+iv)+(c+d

√3) = 0.

A complex number is equal to 0 if and only if the real and imaginary parts are both0. Thus, −1 + u(a + b

√3) + (c + d

√3) = 0 and

√2 + (a + b

√3)v = 0. It suffices

to use the second equation: 2 = (a + b√

3)2v2, or equivalently, 4 = (a2 + 2ab√

3 +3)(1 +

√3). Simplifying the equation gives 0 = (a2 + 6ab− 4) +

√3(a2 + 2ab + 3). A

straightforward calculation shows that there are no rational numbers a and b satisfyingthe last equation.

Now that we know Q(x1) 6= Q(x3), we know that Q(x1) ∩ Q(x3) = Q(√

2i) also.Furthermore, we see that X2 − x2

3 = X2 − (−1 −√

2i) ∈ Q(x1)[X] is the minimalpolynomial of x3 over Q(x1). Therefore, Q(x1, x3) is a degree 2 extension of Q(x1),which is degree 4 extension of Q.Finally we are ready to determine the automorphism group of K = Q(x1, x3). Ob-viously an automorphism of K maps x1 to one of x1,−x1, x3, or −x3. In particular,any permutation of these “symbols” extends to a Q-automorphism of K. Let us chooseσ : K → K to be the one with σ(x1) = x3 (hence it maps −x1 to −x3) and σ(x3) = x1.Also, choose τ : K → K to be the one with τ(x1) = −x3 and τ(x3) = x1. The relationson σ and τ are as follows:

σ2 = id, τ 4 = id, στ = τ 3σ (this needs verification but it is tedious).

These are precisely the relations that are required to generate D8, the dihedral groupof order 8.

169. Let F ⊂ K ⊂ L be three field extensions and assume L is Galois over F . Weknow that there exists a subgroup H ⊂ AutFL such that the fixed subfieldof H is K. Let SH denote the set of all coset representatives of H in AutFLand define the “norm” of an element α ∈ K by

NK/F (α) =∏σ∈SH

σ(α)

(a) Show that NK/F (α) ∈ F .(b) Show that NK/F (αβ) = NK/F (α)NK/F (β).

(c) If K is the quadratic extension Q(√D) for some square-free integer

D ∈ Z, then NK/Q(a+ b√D) = a2 −Db2.

(d) If minα,F (x) =∑d

i=0 aixi ∈ F [x] is the minimal polynomial of α ∈ K over F ,

prove that a) d divides n = dimF K; b) each term β = σ(α) in the product∏σ∈SH σ(α) occurs with multiplicity n/d; c) NK/F (α) = (−1)na

n/d0 .

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Solution.1) If τ ∈ AutFL, then τ(NK/F (α)) is equal to NK/F (α) since τ permutes the cosetrepresentatives among themselves. Therefore, the norm of an element is stable underany F -automorphism of L, hence it lies in F .2) The multiplicativeness is clear from the multiplicativeness of automorphisms σ ∈ SH .3) Consider a number D′ that is relatively prime to D and let L be the field ex-tension Q(

√D,√D′). The automorphism group of L over Q is generated by the Q-

automorphisms σ and τ defined by σ(√D) = −

√D, σ(

√D′) =

√D′, and τ(

√D) =√

D, τ(√D′) = −

√D′. Since |AutQ(L)| = 4 = dimQ(L), L is a Galois extension of Q.

Let H denote the subgroup generated by τ . It is obvious that the fixed subfield of His K, and the minimal coset representatives for H in AutQL are id, σ. Thus

NK/Q(a+ b√D) = id(a+ b

√D)σ(a+ b

√D) = (a+ b

√D)(a− b

√D) = a2 −Db2.

4.a) LetK ′ denote the subfield F (α) ⊂ K. SinceK ′ is isomorphic to F [x]/(minα,F (x)),which is a degree d extension of F , we see that d = dimF K

′ divides dimF K = n.4.b) Since K ⊂ L is a Galois extension, by Problem 166, |H| = dimK L, therefore,dimF L = n · |H|, hence n is the number of coset representatives of H in G = AutFL.Let R = {α1, . . . , αd} denote the set of all roots of minα,F (x) with α = α1. Thefull automorphism group G acts on R transitively, and H ⊂ G acts trivially. Inother words, G · α = R and H ⊆ StabG(α). If σ, τ ∈ G are two elements such thatσ(α) = τ(α) = β, then τ−1σ is in the stabilizer subgroup of α. Therefore, the numberof occurrence of β in the product

∏σ∈SH σ(α) is equal to StabG(α) ∩ SH . This set

is nothing but the coset representatives of H in StabG(α), hence its cardinality isequal to |StabG(α)|/|H|. It remains to compute |StabG(α)|. Observe that K ′ is thefixed field of StabG(α), hence AutK′L = StabG(α). By Problem 166, we know that|AutK′L| = dimK′ L and this is equal to dimF L/ dimF K

′ = n · |H|/d. Therefore,|StabG(α)|/|H| = n · |H|/(d · |H|) = n/d.4.c) Consider the polynomial f(x) =

∏σ∈SH (x− σ(α)). Clearly,

f(0) = (−1)|SH |∏σ∈SH

σ(α) = (−1)nNK/F (α).

From part 4.b) we know that each linear factor in f(x) occurs with multiplicity n/d,and of the form x − αi for some root αi ∈ R of the minimal polynomial minα,F (x).Therefore, f(x) = minα,F (x)n/d. Therefore, f(0) = (

∑di=0 aix

i)n/d|x=0 = an/d0 . Thus,

(−1)nNK/F (α) = an/d0 , or NK/F (α) = (−1)na

n/d0 .

170. We keep the notation from Problem 169. The trace of α ∈ K is defined by

TrK/F (α) =∑

σ∈AutFL

σ(α).

89

Prove that

(a) Show that TrK/F (α) ∈ F .(b) Show that TrK/F (α + β) = TrK/F (α) + TrK/F (β).

(c) If K is the quadratic extension Q(√D) for some square-free integer

D ∈ Z, then TrK/Q(a+ b√D) = 2a.

(d) TrK/F (α) = −ndad−1, where minα,F (x) =

∑di=0 aix

i ∈ F [x] is the minimalpolynomial of α ∈ K over F .

Solution.

1) Since τ(TrK/F (α)) =∑

σ∈AutFL τσ(α) =∑

σ′∈AutFL σ′(α) = TrK/F (α) for all τ ∈

AutFL, we see that TrK/F (α) ∈ F .2) Additivity is clear from the additivity of the automorphisms.

3) The automorphism group AutQQ(√

2) is {id, σ}, where σ(√D) = −

√D is defined

by

TrK/Q(a+ b√D) = id(a+ b

√D) + σ(a+ b

√D) = (a+ b

√D) + (a− b

√D) = 2a.

4) Define f(x) =∏

α∈AutFK(x − σ(α)). On one hand, f(0) = −∑

α∈AutFK σ(α) =−TrK/F (α). On the other hand, as in the solution of Problem 169 4.b) L is a Galoisextension of F (α) with AutF (α)L = StabG(α), and furthermore, since F (α) is Galoisover F , StabG(α) is a normal subgroup of L. Let S denote the set of coset represen-tatives for StabG(α). Then S is isomorphic to AutFF (α) and each τ(α) is a root ofminα,F (x). Therefore,

f(x) =∑

σ∈StabG(α), τ∈S

(x− τσ(α)) =∑τ∈S

|StabG(α)|(x− τ(α)) =n

d

∑τ∈S

(x− τ(α)).

It follows that f(x) = nd

∑αi∈R(x− αi). Therefore,

−TrK/F (α) = f(0) = −nd

∑αi∈R

αi = −nd

(−ad−1).

171. Let F ⊂ K be a field extension of degree n. Show that the norm andtrace of α ∈ K that are defined in Problems 169 and 170 are precisely thedeterminant and the trace of the linear operator Tα : K → K defined byTα(β) = αβ.

Solution.

90

First of all, let us show that the minimal polynomial of the operator Tα : K → Kis equal to the polynomial of minimal degree f(x) ∈ F [x] such that f(Tα) = 0 as anoperator on K: If f(x) =

∑aix

i, then f(Tα)(β) =∑aiT

iα(β) =

∑aiα

iβ for anyβ ∈ K. Therefore, f(Tα) = 0 as an operator on K if and only if f(α) = 0. Since f(x)has the smallest degree, f(x) is the minimal polynomial of the element α ∈ K over F .We recall some facts from linear algebra: the characteristic polynomial of an operatoron an n-dimensional vector space is a polynomial of degree n, and its constant termis the determinant (of the operator) times (−1)n. The minimal polynomial dividesthe characteristic polynomial. Moreover, the roots of the minimal polynomial and theroots of characteristic polynomial are the same, hence, the characteristic polynomialdivides some power of the minimal polynomial. Finally, the sum of all roots of thecharacteristic polynomial is the trace of the operator.In the case of our operator Tα we know that the minimal polynomial minα,F (x) is anirreducible polynomial over F , therefore, the characteristic polynomial charTα(x) is apower of minα,F (x). In particular, if the degree of minα,F (x) is d, then charTα(x) =(minα,F (x))n/d. Therefore, the determinant of Tα is equal to

detTα = (−1)ncharTα(0) = (−1)n(minα,F (0))n/d = (−1)nan/d0 .

For the trace, we use the equality charTα(x) = (minα,F (x))n/d once again: each rootof the minimal polynomial is repeated n/d times in charTα(x). Since the sum of allroots of minα,F (x) is equal to −ad−1, we see that the sum of all roots of charTα(x) isnd(−ad−1). By Problem 170, we know that this quantity is precisely the trace of the

element α ∈ K.

172. Let K be a finite extension of F . For α ∈ K let φα : K → F denote the mapdefined by φα(β) = TrK/F (αβ). Here, TrK/F is as defined in Problem 170.Prove that the assignment α 7→ φα is an injective linear transformation,establishing an isomorphism between K and the space of all F -valued linearfunctionals on K.

Solution.

Let us denote by Φ the map α 7→ φα and let α, γ be two elements from K. For c ∈ F ,we have Φ(α + cγ) = φα+cγ and

φα+cγ(β) = TrK/F ((α + cγ)β) =∑

σ∈AutFK

σ((α + cγ)β) = φα(β) + cφγ(β).

It follows that Φ is linear. To prove its injectivitiy of Φ, we compute its kernel. Sinceφα(α−1) = TrK/F (1) = |AutFK|, the image of Φ is always non-zero, therefore it has nokernel.

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173. Suppose K is a Galois extension of F with AutFK a cyclic group of order ngenerated by σ ∈ AutFK. Suppose α ∈ K satisfies NK/F (α) = 1. Prove that αis of the form β/σ(β) for some non-zero β ∈ K.

Solution.

For i = 0, . . . , n − 1, define χi(x) = σi(x), x ∈ K. It is clear that {χ0, . . . , χn−1}is a set of distinct characters, hence, they are linearly independent. In other words,for any list of not-all-zero elements a0, a1, . . . , an−1 ∈ K there exists θ ∈ K such thatχ0(θ)a0 + χ1(θ)a1 + · · · + χn−1(θ)(x) 6= 0. In particular, for the choice of a0 = 1,ai := ασ(α)σ2(α) · · ·σi−1(α), i = 1, . . . , n− 1 we have an element θ ∈ K such that

θ + ασ(θ) + (ασ(α))σ2(θ) + · · ·+ (ασ(α) · · ·σn−2(α))σn−1(θ) = β,

and β is non-zero. Notice that, since σn = id,

σ(β) = σ(θ + ασ(θ) + (ασ(α))σ2(θ) + · · ·+ (ασ(α) · · ·σn−2(α))σn−1(θ))

= σ(θ) + σ(α)σ2(θ) + σ(α)σ2(α)σ3(θ) + · · ·+ σ(α)σ2(α) · · ·σn−1(α)σn(θ)

Multiplying both sides with α, we obtain

ασ(β) = ασ(θ) + · · ·+ ασ(α)σ2(α) · · ·σn−2(α)σn−1(θ) + ασ(α)σ2(α) · · · σn−1(α)σn(θ)

= β − θ + ασ(α)σ2(α) · · ·σn−1(α)σn(θ)

= β − θ + NK/F (α)σn(θ)

= β − θ + σn(θ)

= β

174. Here is a basic problem in group actions: Let G be a group acting transi-tively on a finite set B and let H be a normal subgroup of G. If A1, · · · , Asis the list of orbits of H in B, then prove that

(a) G acts on {A1, . . . , As} transitively;(b) if a is any element from A1, then the cardinality of A1 is equal to

|H|/|H ∩ StabG(a)|,

and the number s of orbits of H is equal to

|G|/|H · StabG(a)|.

Solution.

(a) Let a, b be two elements from A1 and let g ∈ G. We claim that g(a) and g(b)belongs to the same set Ai, for some i ∈ {1, . . . , s}. To see this, let 1 ≤ j, k ≤ s be two

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indices such that g(a) ∈ Aj and g(b) ∈ Ak. Since A1 is an H-orbit, there exists h ∈ Hsuch that a = h(b). Then g(a) = g(h(b)). Since H is normal in G, ghg−1 belongs toH. Therefore, g(a) = g(h(b)) = ghg−1g(b) = h′g(b) for some h′ ∈ H. But h′Ak = Ak,since Ak is an H-orbit. Therefore, g(a) belongs to the orbit to which g(b) is a memberof. Therefore, G acts on {A1, . . . , As}. Since the action of G on B is transitive, itsaction on the set of H-orbits is transitive, also.

(b) Since A1 is an H-orbit, H acts on A1 transitively. Let a ∈ A1. Then StabH(a) =H ∩ StabG(a), and therefore, |A1| = |H|/|H ∩ StabG(a)|. For our last claim, we aregoing to show that

StabG(a) ·H = StabG(A1).

The containment ⊆ is obvious. By definition, if g ∈ StabG(A1), then g(b) ∈ A1

for any b ∈ A1. In particular, g(a) ∈ A1. Let h ∈ H be the element such thath(g(a)) = a. Since H is normal in G, g−1hg is an element h′ of H. Therefore,a = hg(a) = gg−1hg(a) = gh′(a) implies that gh′ ∈ StabG(a), hence g ∈ StabG(a) ·H.

175. Let f(x) ∈ F [x] be an irreducible polynomial of degree n over F , and let Lbe the splitting of f(x) over F . If α ∈ L is a root of f(x), and K is any Galoisextension of F contained in L, then show that the polynomial f(x) splitsinto a product of m irreducible polynomials each of degree d over K, wherem = dimF (F (α) ∩K) and d = dimK K(α).

Solution.

Suppose f1(x) · · · fs(x) is the factorization of f(x) into irreducible factors over K. LetA1, . . . , As denote the sets of roots of the factors f1(x), . . . , fs(x), respectively. We fixan irreducible factor, say f1(x), and let α ∈ L be one of its roots. The degree d off1(x) is the degree of the extension K ⊆ K(α).

We know that both F ⊂ L and K ⊂ L are Galois extensions. Let G and H denoteAutFL and AutKL, respectively. Since K is Galois, H is normal in G. Since H fixesK point-wise, each irreducible factor fi(x) is invariant under H, hence, Ai is invariantunder the action ofH. Furthermore, since fi(x) is irreducible, H acts on Ai transitively,hence Ai is an orbit of H. By Problem 174.b), we see that the cardinality of Al = σj ·Aiis the same as that of Ai. Thus, the degree of (any) irreducible factor fl(x) of f(x) isequal to that of f1(x).

On the other hand, by Problem 174, the number s of orbits of a normal subgroup H isequal to the cardinality of the quotient G/HH1, where H1 = StabG(α), where α ∈ A1.Note that H1 = StabG(α) is the automorphism group of F (α) over F , and F (α) is aGalois extension. Therefore H1 is normal in G. The product of two normal subgroupsis normal as well. It follows that the fixed field K ′ of HH1 in L is a Galois extensionover F . Since K ′ = K ∩F (α), by Problem 166, we see that dimF K ∩F (α) is equal to|G/HH1| = s.

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176. Determine the splitting field of xp − x− a over Fp, where a 6= 0, a ∈ Fp.Solution.

We already know from Problem 130 that A(x) = xp−x−a is irreducible and separableover Fp. Let K denote its splitting field over Fp. Also, we know from the solution ofProblem 130 that if α ∈ K is a root, then A(x) =

∏p−1k=0(x − α − k). Therefore, if

φ ∈ AutFpK is an automorphism, since A(x) is invariant under φ, we see that φ(α) is aroot of A(x), thus, φ(α) = α+l for some l ∈ Fp. Clearly, in this case, φ(α+k) = α+k+lfor all k ∈ Fp. It follows that if σ is the automorphism defined by σ(α) = α + 1, thenφ = σk, hence σ generates AutFpK, as a cyclic group of order p. Since K is Galoisover Fp, we know now that p = |AutFpK| = dimFp K. On the other hand, since A(x) isirreducible, Fp[x]/(A(x)) is a field extension of degree p, and has a root of A(x), hencecontained in the splitting field of A(x). Therefore, K is equal to Fp[x]/(A(x)).

Definition 0.57. Let x1, . . . , xn be algebraically independent variables over Q. Apolynomial f(x1, . . . , xn) ∈ Q[x1, . . . , xn] is called symmetric if for any permutationσ ∈ Sn, the polynomial σ · f(x1, . . . , xn) = f(xσ(1), . . . , xσ(n)) is equal to f(x1, . . . , xn).

Let k be a number from {1, . . . , n}. The kth elementary symmetric polynomial invariables x1, . . . , xn is denoted by ek and defined by

ek = ek(x1, . . . , xn) =∑

1≤i1<···<ik≤n

xi1 · · ·xik . (15)

The kth power sum symmetric polynomial in variables x1, . . . , xn is denoted by pk anddefined by

pk = pk(x1, . . . , xn) = xk1 + · · ·+ xkn. (16)

177. Prove Newton’s identities:

pk − pk−1e1 + · · ·+ (−1)k−1p1ek−1 + (−1)kkek = 0.

Solution.

Let y be a variable. Define E(y) =∏n

i=1(1 − xiy). By expanding the product we seethat

E(y) =n∏i=1

(1− xiy) = 1− e1y + e2y2 + · · ·+ (−1)neny

n. (17)

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Since 1 + t + t2 + · · · = 11−t = − (1−t)′

(1−t) = − ddt

ln(1 − t), we see that − ln(1 − t) =

t+ t2/2 + t3/3 + · · · . Therefore,

lnE(y) =n∑i=1

ln(1− xiy) = −n∑i=1

∞∑j=1

(xiy)j

j

= −∞∑j=1

pjjyj

Taking the derivative of both sides with respect to y gives

E ′(y)

E(y)= −

∞∑j=1

pjyj−1 or E ′(y) = −E(y)

∞∑j=0

pj+1yj.

For good reasons, it is convenient to define e0 = p0 = 1. By comparing the coefficientsof yk−1 on the expansion of E ′(y) and the product on the right hand side of the lastequation, we obtain Newton’s identity

(−1)kkek = −k−1∑j=0

(−1)jejpk−j−1.

178. Prove that over a field F of characteristic 0, an n× n matrix A is nilpotentif and only if the trace of Ak is zero for all k = 1, 2, . . . .

Solution.Assume that A is nilpotent, hence the minimal polynomial of A in F [x] is xk, wherek is the smallest positive integer such that Ak = 0. It follows that the characteristicpolynomial of A is equal to xn. The trace of A is equal to an−1, where −an−1 is thecoefficient of xn−1 in the characteristic polynomial of A. Therefore, we see that thetrace of A is 0. Similarly, since Ar is nilpotent for any power r ≥ 2 of A, the traces ofAr’s are zero, as well.Conversely, assume that trace(Ak) = 0 for all k ≥ 1. If λ is an eigenvalue of A, then λkis an eigenvalue for Ak. Since the trace of an operator is the sum of all eigenvalues, wesee that the sums of powers of all eigenvalues of A are 0. On the other hand, by usingProblem 177 we see that the elementary symmetric polynomials e1, . . . , en evaluatedat the eigenvalues of A are all zero, also. Since the coefficients of the characteristicpolynomial of an operator is precisely the elementary symmetric polynomials evaluatedat the eigenvalues, we see that the characteristic polynomial of A is equal to xn. Sincean operator is annihilated by its characteristic polynomial, we see that An = 0, henceA is nilpotent.

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179. Prove that over a field of characteristic 0 two n× n matrices A and B havethe same characteristic polynomial if and only if trace(Ak) = trace(Bk) forall k ≥ 0.

Solution.

If A and B have the same characteristic polynomial then the elementary symmetricpolynomials in eigenvalues of A are equal to that of B. By Newton’s identity, in-ductively, we express the power sums of eigenvalues of A in terms of the elementarysymmetric polynomials. Therefore, the trace of Ak, which is equal to the value of thekth power sum symmetric polynomial evaluated at the eigenvalues of A is equal to thatof B. Conversely, if trace(Ak) = trace(Bk) for all k ≥ 0, then the elementary symmet-ric polynomials evaluated at the eigenvalues of A is equal to that of B. Therefore, Aand B have the same characteristic polynomial.

Remark 0.58. For the last two problems, it is enough to restrict k to {1, . . . , n}.

180. By using the previous problem, show that the characteristic polynomial ofAB and that of BA are the same.

Solution.

Let us first assume that the trace of AB is the same as that of BA for any two n× nmatrices A and B. It follows that the traces of (AB)m and (BA)m are the same (byletting C = A(BA)m−1, and D = B) for all m ≥ 2. By Problem 179 we see that thecharacteristic polynomials of AB and BA are the same.

Next, we show that trace(AB) = trace(BA). Let Aij and Bij, i, j = 1, . . . , n denotethe entries of A and B, respectively. The trace of AB is the sum of its diagonal entries:

trace(AB) =n∑k=1

n∑j=1

AkjBjk

Interchanging the sums as well as the entries we have the same result:

n∑k=1

n∑j=1

AkjBjk =n∑j=1

n∑k=1

BjkAkj = trace(BA).

Remark 0.59. This result is true over arbitrary field.

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Definition 0.60. A (commutative, of course) ring R is called Artinian if any de-creasing sequence of nested ideals I ⊃ I1 ⊃ I2 ⊃ · · · stops (becomes an equality) afterfinitely many steps.

A (commutative, of course) ring R is called Noetherian if any increasing sequence ofnested ideals I ⊂ I1 ⊂ I2 ⊂ · · · stops (becomes an equality) after finitely many steps.

These two definitions extend to an R-module M , by replacing ideals by submodules.

181. Prove that in a Noetherian ring R every ideal is finitely generated andconversely a commutative ring R in which every ideal is finitely generatedis Noetherian.

Solution.

Suppose R is a Noetherian ring and I ⊂ R is an ideal. Let I0 = (a), a ∈ I be a subideal of I generated by one element. If I0 = I, then we proved our claim. Otherwise,let a1 ∈ I − I0 be another element from I and let I1 be the subideal generated by aand a1. If I1 = I, then we proved our claim. Otherwise, we continue. Since we havean increasing sequence of ideals I0 ⊂ I1 ⊂ I2 ⊂ · · · , this process stops at step n (byour Noetherian assumption). Therefore, In = I.

Conversely, if every ideal in R is finitely generated, and I0 ⊂ I1 ⊂ · · · is an increasingchain of ideals in R, then we let I = ∪Ii the union of all ideals in the sequence. Itstraightforward to verify that I is an ideal. Therefore, I has finitely generated, say bya0, a1, . . . , an. Let Im be one of the ideals of the sequence that contains all of theseelements. Then I = Im = Im+i for all i ≥ 0, hence R is Noetherian.

182. Determine if the ring C([0, 1]) of all real valued continuous function on theunit interval [0, 1] is Noetherian or not.

Solution.

For n ∈ Z+ let An denote the interval [0, 1/n]. Let In denote the set of all continuousreal valued functions on [0, 1] such that f(x) = 0 for all x ∈ In. It is straightforwardto check that a) In is an ideal for all n ∈ Z+ and b) In ⊂ Im for all n ≤ m.

On the other hand, not all functions in In+1 are contained in In. For example, letg(x) : [0, 1]→ R denote the function that takes value 0 on [0, 1/n+ 1] and

g(x) =(n+ 1)(x− 1)

n+ 1 for x ∈ [1/(n+ 1), 1].

It is clear that g(x) is continuous, g(x) ∈ In+1 but g(1/n) 6= 0. Therfore, C([0, 1]) isnot Noetherian.

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183. Show that the ring of all functions from X to Z/2 is not Noetherian if Xhas infinite number of elements.

Solution.

This is similar to the previous problem: let xn, n = 1, . . . be a sequence of distinctelements from X, An denote the set {xn, xn+1, . . . }. Then the set In of all functionsf : X → Z/2 such that f(x) = 0 if x ∈ An is an ideal. It is also clear that In ⊂ In+1.On the other hand, the function defined by f(x) = 1 on {x1, . . . , xn} and f(x) = 0 onAn+1 does not belong to In. Therefore, I1 ⊂ I2 ⊂ · · · does not stabilize.

184. A module M over a field F is Artinian if and only if M is Noetherian if andonly if M is a finite dimensional vector space over F .

Solution.

Let us first prove that Artinian ⇐⇒ finite dimensional. The implication “ ⇐′′ isobvious. Conversely, let M be an Artinian F -module. If B is a basis for M , thenlet M1 be the proper vector subspace spanned by the set B1 = B − {u1} obtainedfrom B by omitting one of its elements u1 ∈ B, M2 be the subpace of M1 spannedby B2 = B1 − {u2}, and so on. We construct Mi from Mi−1 similarly by taking outone of the basis elements of Mi. Since M is Artinian, there exists n ≥ 1 such thatMn = Mn+k for all k ≥ 0 thus Bn = Bn+k = ∅ for all k ≥ 0. In other words, B is finite.

Next, we prove Artinian ⇐⇒ Noetherian.

An increasing sequence of vector subspaces

{0} ⊂ · · · ⊂M1 ⊂ · · · ⊂Mn−1 ⊂Mn ⊂ · · · ⊂M

gives a decreasing sequence of quotients in the other direction

M/{0} ⊃ · · · ⊃M/M1 ⊃ · · · ⊃M/Mn−1 ⊃M/Mn ⊃ · · · ⊃M/M

If M is Artinian, hence finite dimensional, by extending a basis of Mi to a basis of M ,we see that there exists a subspace Wi (spanned by the newly added basis elements) ofM such that Wi

∼= M/Mi. Furthermore, there exists n ≥ 1 such that Wn+k = Wn forall k ≥ 0. It follows that M/Mn+k = M/Mn for all k ≥ 0, or equivalently, Mn+k = Mn

for all k ≥ 0, hence M is Noetherian. Conversely, if M is Noetherian, let B be a basis,and let Mi be the submodule of M generated by the first i elements of B. Clearly,Mi ⊂ Mi+1 for all i, and since M is Noetherian, the sequence stabilizes. Therefore, Bis finite, hence M is a finite dimensional vector space, implying that M is Artinian.

Remark 0.61. Our proof above relies on Axiom of Choice..

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185. Show that if R is Artinian (respectively, Noetherian), I ⊂ R is an ideal, thenR/I is Artinian (respectively, Noetherian).

Solution. Any nested decreasing (respectively, increasing) sequence of ideals in R/Ipulls back to a to nested decreasing (respectively, increasing) sequence of ideals in Rvia the canonical projection π : R → R/I, therefore, it stabilizes after finitely manysteps.

186. Let 0 → M ′′ f−→ Mg−→ M ′ → 0 be a short exact sequence of R-modules.

Show that M is a Noetherian R-module if and only if both M ′′ and M ′ areNoetherian.

Solution.

First, assume that M is Noetherian. Any increasing subsequence of submodules in asubmodule M ′′ of M is an increasing subsequence of submodules in M . Therefore,any R-submodule of M is Noetherian. If 0 → M ′′ → M → M ′ → 0 is short exact,then M ′ ∼= M/M ′′. Any increasing sequence of R-modules in M/M ′′ pulls back to anincreasing sequence of R-submodules containing M ′ in M . Since M is Noetherian, thesequence of preimages stabilizes at finitely steps, so does the images in M ′. Therefore,M ′ is Noetherian, as well.

Conversely, assume that both of the R-modules M ′′ and M ′ are Noetherian, and let{Ni} be an increasing sequence of R-submodules in M . The set of R-submodules{g(Ni)} of M ′ is an increasing sequence as well, hence, there exists a sufficiently largem0 such that g(Nn+m0) = g(Nm0) for all n ≥ 0. Since {f(M ′′) ∩Nm} is an increasingsubmodule of the Noetherian R-module f(M ′′), we see that there exists a sufficientlylarge integer m1 such that f(M ′′) ∩ Nm1+n = f(M ′′) ∩ Nm1 for all n ≥ 0. Set m :=max{m0,m1}. Now, on the one hand we have f(M ′′)∩Ns = ker g∩Ns for all s ≥ 0, thus,ker g ∩Ns+m = ker g ∩Nm for all s ≥ 0. On the other hand we have g(Ns+m) = g(Nm)for all s ≥ 0. Therefore, Ns+m = Nm for all m ≥ 0, hence M is Noetherian.

187. Let 0 → M ′′ f−→ Mg−→ M ′ → 0 be a short exact sequence of R-modules. Show

that M is Artinian if and only if both of the R-modules M ′′ and M ′ areArtinian.

Solution.

The arguments of the proof are the same as in that of Problem 186, where we replaceNoetherian by Artinian.

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188. Let M be a maximal ideal in a ring R and suppose that Mn = 0 for somen ≥ 1. Prove that R is Noetherian if and only if R is Artinian.

Solution. As a convention we set M0 = R. The sequence of R-modules

0→M i+1 →M i →M i/M i+1 → 0 (for all i)

is exact. Therefore M i is Artinian if and only if M i+1 and M i/M i+1 are Artinian.Similarly, M i is Noetherian if and only if M i+1 and M i/M i+1 are Noetherian. In par-ticular, at the last two steps: Mn−2 is Artinian (respectively, Noetherian) iffMn−1 andMn−2/Mn−1 are Artinian (respectively, Noetherian), and Mn−1 is Artinian (Noethe-rian) iff Mn = {0} and Mn−1/Mn are Artinian (Noetherian).

We first assume that R is Noetherian and show that it is Artinian. If R is Noethe-rian, then for all i = 0, . . . , n − 1, the modules M i+1, M i/M i+1 are Noetherian. ButR-module structure on the quotient module M i/M i+1 is the same as R/M -modulestructure on it. Since R/M is a field, by Problem 184, M i/M i+1 is Noetherian iff it isArtinian. In particular, at the last step we have Mn−1 ∼= Mn−1/Mn, therefore, Mn−1

is Noetherian iff it is Artinian. Now, since Mn−1 and Mn−2/Mn−1 are Artinian, byProblem 187, Mn−2 is Artinian. Thus, going down in the indices, we see that M0 isArtinian.

The argument for the converse (R is Artinian =⇒ R is Noetherian) is identical exceptwe use Problem 186 instead of Problem 187 at a suitable place.

Definition 0.62. The Krull dimension of a ring R is the length of the maximumpossible length of chains P0 ⊂ P1 ⊂ · · · ⊂ Pn of distinct prime ideals in R.

189. Prove that if Krull dimension of an integral domain R is 0, then R is a field.

Solution. In an integral domain {0} is a prime ideal. If the Krull dimension of R is 0,then it means there is no maximal ideal other than {0}, therefore, R is a field.

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Definition 0.63. A discrete valuation on a field K is an integer valued functionν : K → Z ∪ {∞} such that

• ν is surjective and ν(x) =∞ ⇐⇒ x = 0;

• ν(x) + ν(y) = ν(xy) for all x, y ∈ K;

• ν(x+ y) ≥ min{ν(x), ν(y)} for all x, y ∈ K.

The subset {x ∈ K∗ : ν(x) ≥ 0} ∪ {0} is called the valuation ring of ν. A discretevaluation ring R is an integral domain such that R is the valuation ring of a valuationof a discrete valuation on its field fractions K. We abbreviate discrete valuation ringby DVR.

190. Show that if R is a DVR, then for each element x from the fraction field Kof R either x ∈ R, or 1/x ∈ R.Solution.

Suppose R is a DVR with valuation ν, and let x ∈ K − R. Then ν(x) < 0. Observethat ν(1) = ν(1 · 1) = ν(1) + ν(1), therefore, ν(1) = 0. It follows from 0 = ν(1) =ν(x · 1/x) = ν(x) + ν(1/x) that ν(1/x) = −ν(x). In other words, 1/x ∈ R.

Remark 0.64. The converse of this problem is also true; if R is an integral domainsuch that for all x ∈ K either x ∈ R, or 1/x ∈ R, then R is a DVR.

191. Let R be an integral domain which is local. Suppose that the maximal idealM of R is principal and of the form M = (t), t ∈ R, with ∩n≥0(tn) = {0}. Provethat R is a DVR.

Solution. Let x ∈ R be an arbitrary element. If x is a unit, then we define ν(x) = 0.We claim that any element x of R is of the form utm for some m ≥ 0. Indeed, givena non-zero element x ∈ R, let m denote the highest power of t that divides x, so thatwe have x = tma for some a that is not divisible by t. Notice that the finiteness ofm is guaranteed by the fact that ∩n≥0(tn) = 0. Since t - a, a does not belong to M ,hence invertible (since there is only one maximal ideal). Thus, it makes sense to defineν(x) = m. In particular, ν(a) = 0 if a is invertible. Next, if x/y is an element of thefraction field K, then there exists b ∈ R a unit, and s ∈ Z such that x/y = bts. Indeed,if x = a1t

m1 , y = a2tm2 withm1,m2 ≥ 0, a1, a2 ∈ R are units, then x/y = a1a

−12 tm1−m2 .

Note that m1−m2 is uniquely determined by x/y. Indeed, if x′, y′ ∈ R is another pairsuch that x/y = x′/y′, then xy′ = x′y (R is an integral domain). It follows that thedifference of the highest powers of t that divide x′ and y′ is equal to the differenceof the highest powers of t that divide x and y, which is m1 −m2. Finally, we defineν(0) =∞. Thus, we have a well-defined function ν from K onto Z ∪ {∞}.

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It is clear that the valuation ring of ν is equal to R. First of all, since ν(ts) = s forany s ∈ Z, subjectiveness is evident. Secondly, if z, w ∈ K, then z = u0t

s, w = v0tr

for some r, s ∈ Z and units u0, v0 ∈ R. Therefore, ν(zw) = ν(u0v0tr+s) = r + s =

ν(z) + ν(w). Next, without loss of generality assume that s ≤ r. In this case, z +w =ts(u0 + v0t

r−s). Since r − s is positive, u0 + v0tr−s is an element of M , therefore,

ν(u0 + v0tr−s) = 0. Thus, ν(z + w) = ν(ts) + ν(u0 + v0t

r−s) = s which is greater thanor equal to min{ν(z), ν(w)} = s.

Definition 0.65. A field extension K over F is called purely inseparable if for eachα ∈ K the minimal polynomial of α over F has only one root (with multiplicitiesallowed).

192. Assume that K is an algebraic extension of F and the characteristic of K isp. Prove that the following are equivalent:

(a) K is purely inseparable over F ;

(b) if α ∈ K is separable over F , then α ∈ F ;(c) if α ∈ K, then αp

n ∈ F for some n = nα, and minα,F (x) = xpn − αpn.

Solution.

(a)⇒ (c): If K is purely inseparable over F and α ∈ K then we have two cases; eitherα ∈ F in which case there is nothing to prove, or α /∈ F . In the latter case, the minimalpolynomial f(x) of α is of the form f(x) = (x − α)m for some m. If α is separable,by definition, m = 1 so our claim follows. If α is not separable, then f ′(x) = 0 byProblem 129. Since f ′(x) = m(x− α)m−1, this is possible if and only if m is power pnof p.

(c) ⇒ (b): Let α ∈ K be an element which is separable. By our assumption, αpn ∈ Ffor some n = nα and the minimal polynomial of α is f(x) = xp

n − αpn . Once again,by Problem 129, we see that f ′(x) must be non-zero. But f ′(x) = pnxp

n−1, which is 0,unless n = 0. In this case, f(x) = x− α, hence α ∈ F .(b) ⇒ (a): Let α ∈ K be an element. If α is separable, α ∈ F , hence its minimalpolynomial f(x) = x−α has a unique root. If α is not separable, and f(x) =

∑mi=0 aix

i

is its minimal polynomial, then since f ′(x) = 0 by Problem 129, the exponent of eachvariable in any monomial in f(x) is divisible by p. Therefore, f(x) = f1(xp) for somepolynomial f1(x) ∈ F [x]. Notice that f1(x) is irreducible, otherwise f(x) would bereducible. If f1(x) is not separable, then its derivative vanishes, and we repeat theabove argument. Since f(x) has finite degree, this process eventually stops, hence wearrive at a smallest possible positive integer k such that fk(x) ∈ F [x] is separable andirreducible with f(x) = fk(x

pk). Notice that αpk is a root of fk(x). Since fk(x) isseparable, by our initial assumption, we have that αpk ∈ F hence that fk(x) = x−αpk .

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Therefore, f(x) = xpk − αpk . Notice also that if β 6= α is another root for f(x), then

fk(βpk) = 0, or (β −α)p

k= βp

k −αpk = 0 implying that β = α. (Our proof proved (b)⇒ (c), also.)

193. Prove that purely inseparable extensions are normal, that is to say, splittingfield of a family of polynomials.

Solution.

By Problem 122 we know an algebraic field extension K over F is a splitting field ifand only if any irreducible polynomial f(x) ∈ F [x] with a root in K splits completelyinto linear factors in K[x].

Now, let us assume that K is purely inseparable over F , and let us let f(x) denotean irreducible polynomial over F with a root α in K. Since the minimal polynomialminα,F (x) divides f(x), they must be equal. By Problem 192, then f(x) = xp

n − αpn ,for some n, and furthermore, αpn ∈ F . Since the characteristic is p, f(x) = (x− α)p

n ,hence f(x) splits completely, hence K is normal.

194. Suppose we have field extensions F ⊂ K ⊂ L, an element θ ∈ L, and f(x) =minα,F (x). Prove that K ⊗F F (θ) ∼= K[x]/(f(x)) as K-algebras.

Solution.

Let k and∑m−1

i=0 aiθi be two elements from K and F (θ), respectively and let φ be the

map

φ(k ⊗m−1∑i=0

aiθi) =

m−1∑i=0

kaixi mod (f(x))

extended by linearity to a K-algebra map φ : K ⊗F F (θ) → K[x]/(f(x)). In otherwords, φ(k ⊗ g(θ)) = kg(x) mod (f(x)). In particular, if g(x) =

∑sj=0 bix

i ∈ K[x]with deg g(x) < deg f(x), then

φ(b0 ⊗ 1 + b1 ⊗ θ + · · ·+ bs ⊗ θs) = b0 + b1x+ · · ·+ bsxs.

Therefore, φ is surjective. Obviously, a polynomial g(x) ∈ K[x] is divisible by f(x),then g(θ) = 0. Therefore, φ is injective, hence it is an isomorphism of K-algebras.

195. Suppose we have field extensions F ⊂ K1 ⊂ L and F ⊂ K2 ⊂ L with K1 andK2 are algebraic over F , and assume that L is separable. Prove that thenilradical of F -algebra K1 ⊗F K2 is trivial.

103

Solution.

Let N denote the nilradical of K1 ⊗F K2. We claim that it is enough to show thatN ∩ (K1 ⊗F F (θ)) = {0} for all nonzero θ ∈ K2.

To see this, suppose x =∑r

i=1 ai ⊗ bi ∈ K1 ⊗F K2 is a nilpotent element. In this case,x is a nilpotent element in K1 ⊗F F (b1, . . . , br). Since K2 is separable, F (b1, . . . , br) isseparable over F . By primitive element theorem, there exists θ ∈ F (b1, . . . , br) suchthat F (b1, . . . , br) = F (θ). In other words, x is a nilpotent element in K1 ⊗F F (θ).

To see that K1 ⊗F F (θ)) is reduced, we use Problem 194: if x ∈ K1 ⊗F F (θ) werenilpotent, then its image φ(x) ∈ K1[x]/(f(x)) is nilpotent also, where f(x) is theminimal polynomial of θ. Since f(x) is irreducible, hence prime in K1[x], K1[x]/(f(x))is an integral domain. Therefore, φ(x), hence x is not nilpotent.

196. Prove that F (x) is not a finitely generated F -algebra.

Solution.

Assume otherwise that the rational functions f1(x)g1(x)

, . . . , fn(x)gn(x)

generates F (x) as an F -algebra. Let f(x) denote f1(x)

g1(x)+a∈ F (x), where a ∈ F is such that gcd(g1+a, gi) = 1 for

all i = 1, . . . , n as well as gcd(g1 +a, f1) = 1. Let P (y1, . . . , yn) be the polynomial withcoefficients from F such that P

(f1(x)g1(x)

, . . . , fn(x)gn(x)

)= f(x). Equating the denominators

on the left hand side gives us the equality

H(f1(x), . . . , fn(x), g1(x), . . . , gn(x))

g1(x)s1 · · · gn(x)sn=

f1(x)

g1(x) + a

for some non-negative integers s1, . . . , sn and for some polynomial H(z1, . . . , z2n) withcoefficients from F . This contradicts with our initial assumption that g1(x) + a isrelatively prime with each of the polynomials f1(x), g1(x), . . . , gn(x).

Definition 0.66. A numerical semigroup is a subset of non-negative integers whichis closed under addition.

197. Show that numerical semigroups are finitely generated.

Solution.

Let S be a numerical semigroup. If all elements of S have a common divisor a ∈ Z+,then S ′ := {b/a : b ∈ S} is a numerical semigroup, also. Furthermore, S ′ is finitelygenerated, then so is S. Therefore, without loss of generality we assume that S containstwo relatively prime numbers a, b ∈ S.

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The fact, which goes back to James Joseph Sylvester, is that the largest integer thatdoes not belong to the numerical semigroup generated by a and b is ab − a − b. Inother words, if n > c = ab− a− b, then n ∈ S. It follows that S is (finitely) generatedby its elements smaller than or equal to c.

198. Show that any subring R of F [x] that contains F is Noetherian, but notnecessarily a UFD.

Solution.

Let I ⊂ R be a non-trivial ideal.

The degrees of elements from I forms a semigroup in Z+, that we denote by SI . Weknow that numerical semigroups are finitely generated by Problem 197. Let n1, . . . , nkdenote the ordered set of generators for SI and let f1(x), . . . , fk(x) be monic elementsfrom I such that deg fi(x) = ni for i = 1, . . . , k. We claim that fi’s generate I. Weuse induction on SI . To this end, let J denote the ideal generated by fi(x)’s, letg(x) ∈ I be a polynomial of degree n1, and let b denote the leading term of g(x). Theng(x)− bf1(x) lies in I and its degree is less than n1, hence g(x) = bf1(x).

Next we assume that our claim is true for all m ∈ SI with m < N and let f(x) ∈ Ibe an element of order N . Since i1n1 + · · ·+ iknk = N for some i1, . . . , ik ∈ Z+, thereexists b ∈ F (the leading coefficient of f(x)) such that f(x)− b

∏Ns=1 fs(x)is has degree

< N , hence it lies in J by our induction hypothesis. Thus f(x) lies in J . Therefore,J = I, hence I is finitely generated. Therefore, by Problem 181 R is Noetherian.

To give a counter example to the UFD’ness of R consider R = F [x2, x3]. Obviously,both x2 and x3 are irreducible elements in R. The factorizations x6 = x2 · x2 · x2 andx6 = x3 · x3 shows that R is not a UFD.

199. Prove that submodules, quotient modules, and finite direct sums of Noethe-rian modules are Noetherian as well.

Solution.

The fact that submodules and quotient modules of a Noetherian module are Noetherianfollows from Problem 186. To prove that a finite direct sum of Noetherian modules isNoetherian, let M1, . . . ,Mn be a finite list of Noetherian R-modules, and let {Sj}j≥1

be an ascending chain of R-modules in M1 ⊕ · · · ⊕Mn. Clearly, for any submoduleS ⊂M1 ⊕ · · · ⊕Mn that is true:

S = (S ∩M1)⊕ · · · ⊕ (S ∩Mn),

and that S ∩Mi is a submodule of Mi for each i = 1, . . . , n is true. Therefore, for eachi = 1, . . . , n the ascending chain {Sj ∩Mi}j≥1 stabilizes, say at step mi. Let m denote

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max{m1, . . . ,mn}. Then

Sm = Sm ∩M1 ⊕ · · · ⊕ Sm ∩Mn = Sm+k ∩M1 ⊕ · · · ⊕ Sm+k ∩Mn = Sm+k

for all k ≥ 1. In other words, M1 ⊕ · · · ⊕Mn is Noetherian.

200. If R is a Noetherian ring, then an R-module M is Noetherian if and only ifM is finitely generated.

Solution.

Suppose M is a Noetherian R-module. Let M1 be a submodule generated by oneelement f1 from M , and let Mi (for each i = 2, . . . ) be the R-module generated byMi−1 and an element fi ∈ M , if there is any, that is not contained in Mi−1. Thus wehave an increasing sequence of R-modules: M1 ⊂ M2 ⊂ · · · . Since M is Noetherian,this chain stabilizes, say at i = n, hence Mn = M . In other words, the elementsf1, . . . , fn+1 generate M .

Conversely, supposeM is a finitely generatedR-module, say by the elements f1, . . . , fn ∈M , and let M1 ⊂ M2 ⊂ · · · be an ascending chain of R-submodules in M . Since Ris a Noetherian ring, it is a Noetherian R-module, also. Thus, by Problem 199, thedirect sum R ⊕ · · · ⊕ R (n copies), denoted by Rn, is a Noetherian R-module aswell. The map φ : Rn → M defined by φ(r1, . . . , rn) =

∑ni=1 rifi is a surjective R-

module map. The pre-images ofMi’s in Rn given an ascending chain of R-submodules:φ−1(M1) ⊂ φ−1(M2) ⊂ φ−1(M3) · · · . Since Rn is Noetherian (Problem 199) this chainstabilizes, say at the kth step. Thus, their images in M stabilize at the kth step aswell. In other words, M is Noetherian.

201. Recall that rad(I), the radical of an ideal I is the set of all elements r ∈ Rsuch that rn ∈ I for some n ∈ N. Show that the rad(I) is equal to theintersection of all prime ideals containing I.

Solution.

Recall that the radical of a ring is the intersection of all prime ideals of the ring and itis equal to the set of all nilpotent elements (Problem 41). If I ⊂ R is an ideal, then theradical of R/I is the set of nilpotent elements in R/I and also equal to the intersectionof all primes of R/I. Since the prime ideals of R/I are in one-to-one correspondencewith the prime ideals of R that contain I in an inclusion preserving manner, we seethat the radical of R/I corresponds to the intersection of all primes containing I. Atthe same time, the set of all nilpotent elements in R/I corresponds to the set of allelements r of R with some power of r contained in I.

106

202. Let I and J be two ideals from R. Prove that the following are equivalent:

(a) V (I) ⊆ V (J)

(b) J ⊂ rad(I)

(c) rad(J) ⊂ rad(I)

Solution.If V (I) ⊂ V (J), then a prime ideal containing I contains J also. Thus intersectionof primes that contain I contains the intersection of all primes that contain J . ByProblem 201, this shows that (a) ⇒ (c).If rad(J) ⊂ rad(I), then since J ⊂ rad(J) the implication (c) ⇒ (b) is obvious.Finally, if J is contained in rad(I), then J is contained in prime ideal that contains I.Therefore, V (I) ⊂ V (J).

203. Show that the nilradical of R is equal to Jacobson radical of R if and onlyif every open set contains a closed point.

Solution.Recall that Jacobson radical is the intersection of all maximal ideals, and the radical(also known as the nilradical) is the intersection of all prime ideals.Suppose that nil(R) = rad(R). It is snough to prove our claim for basic open sets inSpec(R). Let D(f) be a non-empty basic open set. Recall that D(f) is equal to setof all prime ideals in R that does not contain f . Towards a contradiction assume thatD(f) does not contain a closed point. Closed points corresponds to maximal ideals inZariski topology. In other words, we assume that D(f) does not have any maximalideal. This means that all maximal ideals of R contains f . Therefore, f lies in theJacobson radical, hence it lies in the nilradical of R. Therefore, f is nilpotent. Butthen D(f) = ∅ by Problem 74.Conversely, suppose that every open set contains a closed point and assume rad(J) 6=Jac(R). In particular, it follows our hypothesis that all non-empty basic open setscontain closed points. Equivalently, for each f ∈ R, if D(f) 6= ∅, then there exists amaximal ideal Qf such that f /∈ Qf . If f is contained in all maximal ideals but notin a prime ideal, then it means D(f) 6= ∅. At the same time, if f ∈ Jac(R), f iscontained in all maximal ideals, in particular it has to be contained in Qf , which is acontradiction. Therefore, our assumption is wrong.

204. Let φ : R→ S be a ring homomorphism. If I ⊂ R is an ideal, then show thatφ(rad(I)) ⊂ rad(φ(I)). If φ is surjective and I contains the kernel of φ, thenshow that φ(rad(I)) = rad(φ(I)).

107

Solution.

Let x ∈ rad(I). Then xn ∈ I for some n. In this case, φ(x)n = φ(xn) ∈ φ(I), therefore,φ(x) ∈ rad(φ(I)). Assume now that φ is surjective and its kernel is contained in I. Lety ∈ rad(φ(I)), and suppose x ∈ R is such that φ(x) = y. Since ym ∈ φ(I) for some m,we see that φ(xm) ∈ φ(I). Since the kernel of φ is contained in I we see that xm ∈ I.In other words, x ∈ rad(I).

205. Let a be a non-zero element from the unique factorization domain A. Provethat the nilradical of A/(a) is equal to the intersection of finitely many primeideals.

Solution.

Suppose that a = upe11 · · · perr is the factorization of a into its irreducible factors. (uis invertible, pi’s are irreducible.) It is clear that (a) ⊂ (pi) for i = 1, . . . , r. Let Pidenote the prime ideal in A/(a) generated by the image pi of pi in A/(a). Let x ∈ Abe an element such that x ∈ ∩ri=1Pi. It follows that x is divisible by all of pi’s. Thus, asufficiently large power of x is divisible by a. In other words, x is a nilpotent element.This means P1 ∩ · · · ∩ Pr ⊆ nil(A/(a)). Since the nilradical is the intersection of allprimes, ∩ri=1Pi = nil(A/(a)).

Definition 0.67. Let R ⊂ S be a pair of rings. An element α ∈ S that satisfiesf(α) = 0 for some monic f(x) ∈ R[x] is called an integral element of S over R. Theterminologies “integral extension, integrally closed, integral closure” are defined in away that they makes sense. An integral domain R is called normal if R is integrallyclosed in its fraction field S = K.

206. Prove that UFD’s are normal.

Solution.

Let R be a UFD and K be its fraction field. We have to show that any integral elementof K over R lies in R. This is the statement of Problem 105.

207. Let R denote the following quotient ring: R = F [x, y]/(x2− y3) for some fieldF . Show that

(a) R is an integral domain;

(b) if t denotes the element x/y in the fraction field K of R, then K is equalto F (t);

108

(c) F [t] is the integral closure of R in K = F (t).

Solution.

(a) The ring F [x, y] is a UFD. Therefore, if x2 − y3 is irreducible, then the idealgenerated by (x2 − y3) is a prime ideal, hence R is an integral domain. To prove thatx2 − y3 is irreducible, it is enough to prove that it is irreducible in F (y)[x]. Assumetowards a contradiction that x2− y3 is reducible. Since it is a quadratic polynomial inx over F (y), it is reducible if and only if it has a root in F (y). Suppose f(y)

g(y)is a root.

We assume that f(y) and g(y) have no common divisor. But(f(y)g(y)

)2

− y3 = 0 impliesf(y)2 = g(y)2y3. Thus an irreducible factor of g(y) divides f(y)2, hence divides f(y),a contradiction.

(b) Let t denote x/y in the fraction field K of R. Since x2 = y3 in R, we see that y =(x/y)2 = t2 in K. Also, x = yx/y = yt = t2t = t3. Thus, any element f(x, y) ∈ R =F [x, y]/(x2−y3) (viewed inside K) is a polynomial in t. Therefore, f(x, y)/g(x, y) ∈ Kwith g(x, y) 6= 0 lives in F (t). On the other hand, F (t) ⊂ K, hence, the equalityK = F (t) follows.

(c) Let h(t) =∑m

i=0 aiti be an element of F [t] ⊂ F (t). Arguing as in part (b), we

see that h(t) = a1t + g(x, y) for some g(x, y) from R (replace t2 with y and replace t3with x). Since (h(t) − g(x, y))2 = a2

1t2 = a2

1y, we see that h(t) is a root of the monicpolynomial X2 − 2g(x, y)X + g(x, y)2 − a2

1y ∈ R[X]. This shows that F [t] is integralover R. To prove that F [t] is the integral closure of R in F (t), it remains to show thatany integral element of F (t) over R lies in F [t]. To this end, let h(t) ∈ F (t) be anintegral element over R. In particular, since R ⊂ F [t], h(t) is integral over F [t]. SinceF (t) is the fraction field of F [t], and since F [t] is a UFD, F [t] is integrally closed byProblem 206. Therefore, h(t) belongs to F [t].

208. Let R be an integral domain and let i and j be two relatively prime numbers.Show that (xi − yj) is a prime ideal in R[x, y].

Solution.

Let φ : R[x, y]→ R[t] be the ring homomorphism defined by φ(x) = tj and φ(y) = ti. Itis clear that the kernel of φ contains the ideal generated by xi− yj. If f(x, y) ∈ R[x, y]is mapped to 0 by φ we claim that it belongs to (xi − yj). To see this, we considerf(x, y) in R[x, y]/(xi−yj): Replace each occurrence of yj in monomials of f(x, y) by xiand write f(x, y) = f0(x)+yf1(x)+ · · ·+yj−1fj−1(x) for some fi(x) ∈ R[x, y]/(xi−yj).Since φ(f(x, y)) =

∑j−1s=0 φ(yfs(x)) =

∑j−1s=0 t

sfs(tj), and since any monomial of tsfs(tj)

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is distinct from that of ts′fs′(tj) for 0 ≤ s 6= s′ < j, we see that tsfs(tj) = 0 for alls = 0, . . . , j − 1. In other words, fs(x) = 0 for all s = 0, . . . , j − 1, hence f(x, y) = 0in R[x, y]/(xi − jj). It follows that the kernel of φ is equal to the ideal generated byxi − yj. Since R[ti, tj] = im(φ) is an integral domain, (xi − yj) is a prime ideal.

209. Let R denote the polynomial algebra F [x, y]/(xi − yj) for some field F andrelatively prime numbers i and j. We know from Problem 208 that R is anintegral domain. Find the normalization of R.

Solution.

We know from the solution of Problem 208 that the ring R = F [x, y]/(xi − xj) isisomorphic to F [ti, tj]. So we consider everything inside of the fraction field F (t) ofF [t].

By Problem 206, we know that F [t] is normal. We claim that the normalization S ofF [ti, tj] is equal to F [t].

Since F [ti, tj] ⊂ F [t], it is clear that S is contained in F [t].

Conversely, let f(t) be a polynomial from F [t] and decompose it in the form f(t) =h(t) + g(ti, tj), where g(x, y) ∈ F [x, y] and h(t) ∈ F [t] with no monomial of h(t) is ofthe form dtai+bj, a, b ∈ N, d ∈ F . In other words, no monomial of h(t) lies in F [ti, tj].Let N denote ij − i − j + 1. We know that any integer m greater than or equal toN is of the form ai + bj for some a, b ∈ N (see the solution Problem 197). Therefore,any monomial in h(t)N , in particular (f(t)−g(ti, tj))N = h(t)N , belongs to F [ti, tj]. Inconclusion, f(t) is a root of the monic polynomial (X−g(ti, tj))N−h(t)N ∈ F [ti, tj][X],hence it belongs to the normalization S of F [ti, tj].

210. Let R denote the polynomial algebra F [x, y]/(y2 − x3 − x2) for some field F .

(a) Prove that R is an integral domain;

(b) Compute the normalization of R.

Solution.

(a) We claim that y2− x3− x2 is irreducible in F [x, y]. It is enough to prove that it isirreducible in F (x)[y]. Assume towards a contradiction that y2 − x3 − x2 is reducible.Since the y-degree of y2 − x3 − x2 is 2, it is reducible if and only if it has a root inF (x). Suppose f(x)

g(x)is a root. We assume that f(x) and g(x) have no common divisor.

But(f(x)g(x)

)2

− x3 − x2 = 0 implies f(x)2 = g(x)2(x3 − x2) = (g(x)x)2(x − 1). Thus,x − 1 is a divisor of f(x), hence (x − 1)2 divides f(x)2. But since x is not divisibleby x − 1, it follows that g(x) is divisible by x − 1, which is a contradiction. Since

110

y2 − x3 − x2 is irreducible in F [x, y], the ideal that it generates is prime. Therefore,F [x, y]/(y2 − x3 − x2) is an integral domain.

(b) Let t denote y/x in the fraction field K of R = F [x, y]/(y2−x3−x2). In particular,F (t) is contained in K.

On the one hand 0 = y2 − x3 − x2 = x2(y2/x2 − x− 1) = x2(t2 − x− 1). On the otherhand K is an integral domain. Therefore, t2 − x − 1 = 0, or x = t2 − 1. In this case,y = xy/x = (t2 − 1)t. It follows that any element from R is a polynomial in t, henceR ⊂ F [t]. Therefore, K ⊂ F (t). Thus, K is equal to F (t).

Next we compute the integral closure S of R = F [x, y]/(y2 − x3 − x2) in K = F (t).

Let f(t) =∑m

i=0 aiti be an element of F [t] ⊂ F (t). Let s = 2k be an even number.

Then

ts = (t2)k = ((t2 − 1) + 1)k =k∑i=0

(k

i

)(t2 − 1)i =

k∑i=0

(k

i

)xi. (18)

If s = 2k + 1 is an odd number with s ≥ 3, by using eqn 18, we obtain

ts = ts − ts−2 + ts−2 = ts−3(t2 − 1)t+ ts−2 =

(s−3∑i=0

(s− 3

i

)xi

)y − ts−2. (19)

Since s − 2 is an odd integer, repeating this argument for ts−2, by induction we seethat ts is of the form g(x, y) + at for some a ∈ Z and a polynomial g(x, y) ∈ R. Inparticular, combining (18) and (19) we see that f(t) is of the form h(x, y)+ bt for someb ∈ Z and a polynomial h(x, y) ∈ R. Then f(t) is a root of the monic polynomial

(X − h(x, y))2 − b2 − b2x2 ∈ R[X].

It follows that f(t) belongs to S. In other words, F [t] ⊆ S.

But since R is contained in F [t], and F [t] is integrally closed, we know that S ⊆ F [t].Therefore, they are equal.

211. Let R ⊂ S be a pair of rings and let s ∈ S. Prove that the following areequivalent to each other:

(a) s is integral over R,

(b) the polynomial ring R[s] is a finitely generated R-module,

(c) there exists an intermediary ring R ⊂ T ⊂ S such that s ∈ T and T isfinitely generated as an R-module.

111

Solution.

(a) ⇒ (b) If s is integral over R, there exist elements a0, . . . , an−1 ∈ R such thata0 + a1s + · · · + an−1s

n−1 + sn = 0. In particular, if m ≥ n, then sm = −sm−n(a0 +a1s+ · · ·+an−1s

n−1). Therefore, by using this identity recursively we see that sm is anR-linear combination of the elements 1, s, . . . , sn−1. Hence R[s] is a finitely generatedR-module.

(b) ⇒ (c) Since R ⊂ R[s] ⊂ S, this claim is obvious.

(c) ⇒ (a) Let s ∈ S be an element such that there exits an intermediary ring Tcontaining s which is finitely generated as an R-module. Let s1, . . . , sn be a list ofgenerators for T as an R-module. In particular, we have ssi = ai1s1 + · · · + ainsn forsome coefficients ai1, . . . , ain ∈ R. Thus, we have a matrix equation

s

s1

...sn

=

a11 · · · a1n

.... . .

...an1 · · · ann

s1

...sn

. (20)

Let A denote the coefficient matrix A = (aij), and let u denote the column matrix ofsi’s. We re-write the matrix equation 20 in compact form as follows: (sI − A)u = 0,where I is the n×n identity matrix. Notice that if det(sI−A) is non-zero, then sI−Ais invertible, hence u = 0. Since this is absurd, we see that det(sI − A) = 0. But thisshows that there exists a monic polynomial with coefficients from R having s as a root.

212. Let R ⊆ S be a ring extension, and let T ⊂ S denote the set of elements thatare integral over R. Prove that T is a ring.

Solution.

Let r, s ∈ T be two elements. We are going to prove that r − s ∈ T and rs ∈ T .There exists elements s1, . . . , sm and r1, . . . , rm′ such that R[s] = Rs1 + · · ·Rsm andR[r] = Rr1 + · · · + Rrm′ . It is clear that S ′ = R[s, r] is equal to

∑mi=1

∑m′

j=1Rsirj,hence finitely generated. By Problem 211, r±s as well as rs are integral elements overR since both of them belong to S ′.

213. Let R ⊆ S be a ring extension such that S is an integral domain and integralover R. Prove that R is a field if and only if S is.

Solution.

Suppose R is a field. Since S is an integral extension of R, if s ∈ S, there exists afinitely generated R-submodule T of S such that s ∈ T and T is a ring. In particular,T is a finite dimensional vector space over R, hence 1, s, s2, . . . is linearly dependent:

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there exists a0, a1, . . . , am ∈ R such that∑m ais

i = 0. Without loss of generality weassume a0 6= 0. Thus, −a0 = s(a1 + · · ·+ ams

m−1), so s is invertible, hence S is a field.Conversely, if S is a field and r ∈ R, then we are going to show that 1/r belongs toR. Indeed, since 1/r ∈ S and S is integral over R, a0 + 1

ra1 + · · · + 1

rmam + 1

rm+1 = 0for some ai ∈ R, i = 0, . . . ,m. Equivalently, a0r

m + a1rm−1 + · · · + am = −1

r, which

implies that 1/r ∈ R.

214. Prove that if K is a finite extension of Q, then any element of K has aninteger multiple that is integral over Z.Solution.

Let α ∈ K be an element. Since K is algebraic over Q (K being finite dimensionalvector space over Q), there exist a0, . . . , am ∈ Q such that a0 + a1α+ · · ·+ amα

m = 0.Multiplying by the common denominator of ai’s we assume that the coefficients areintegers. Multiplying by the m− 1th power of am we obtain

a0am−1m + a1a

m−2m (amα) + a2a

m−3m (a2

mα2) + · · ·+ ammα

m = 0.

Thus, amα is an algebraic over Z.

215. Let K be a finite extension of Q, and assume that K is contained a fieldL that is Galois over Q. Recall that the trace of an element α ∈ K overF is defined in Problem 170. Prove that if α ∈ K is integral over Z, thenTrK/Q(α) is an integer.

Solution.

We know from Problem 170 that TrK/Q = −ndad−1, where n = dimQK, d is the degree

of the minimal polynomial minα,Q(x) =∑d

i=0 aixi of α over Q, and furthermore, d | n

(by Problem 169). Therefore, it is enough to prove that the minimal polynomial of αis defined over Z.Since α is integral over Z there is a monic polynomial f(x) with integer coefficientssuch that f(α) = 0. Let f(x) be a smallest degree such polynomial, hence irreducibleover Z. But since Z is a UFD, by Problem 115 we know that f(x) is irreducible over Zif and only if it is irreducible over Q. Since the minimal polynomial of α divides f(x),f(x) is not irreducible over Q unless it is equal to the minimal polynomial.

216. Prove that if K is a finite extension of Q, then the set S of integral elementsover Z of K forms:

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(a) a ring;

(b) a free Z-module.

Furthermore, the rank of S as a Z-module is equal to n = dimQK.

Solution.

The fact that S is a ring follows from Problem 212. Let α1, . . . , αn be a basis of Kas a Q-vector space. By Problem 214, there exists d ∈ Z such that dα1, . . . , dαn ∈ S.Obviously, the Z-module generated by dαi’s is a free Z-submodule of S.

We claim that there are n elements that are related to dαi’s (by some sort of duality)and they form a basis for S as a Z-module, hence, S is free and furthermore its rankis n = dimQK.

By Problem 172 we know that K, as a vector space over F = Q, is isomorphic toHomQ(K,Q) the space of Q-valued linear functionals on K via Φ(α) = φα, whereφα(β) = TrK/F (αβ). Let f1, . . . , fn denote the dual of the basis dα1, . . . , dαn. In otherwords, fi = φdαi , and fi(dαj) = 0 if i 6= j. We choose elements α′i from K such that

fj(α′i) =

{1 if i = j,

0 otherwise .

Now, let β be an element from S and we expand it in the α′i-basis: If β = a1α′1 + · · ·+

anα′n for some ai ∈ Q. Multiplying by dαi and then applying fi we see that fi(dαiβ) =

ai. Since fi(dαiβ) is equal to TrK/Q(dαβ) and since dαiβ ∈ S, by Problem 215 we knowthat ai ∈ Z. Therefore, S is contained in the free Z-module spanned by α′1, . . . , α′n.Since S does not have any torsion elements (as it is contained in a field of characteristic0) S itself is a free Z-module. Since S is squeezed between two rank n free Z-modules,its rank is n, also.

217. Suppose R ⊂ S is an integral ring extension. Let D be a multiplicativesubmonoid of R. Prove that

(a) D−1S is integral over D−1R;

(b) If P is a prime ideal of S, then S/P is integral over R/P ∩R.

Solution.

(a) First of all, if D ⊂ R is multiplicatively closed, then it is multiplicatively closed inS as well, therefore, it makes sense to localize S at D.

Let α/d, α ∈ S, d ∈ D be an element of D−1S. Since α is integral over R, there existsa monic polynomial f(x) = xm + am−1x

m−1 + · · · + a0 ∈ R[x] such that f(α) = 0.

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Dividing both sides with dm we obtain

f(α)

dm=(αd

)m+am−1

d

(αd

)m−1

+ · · ·+ a0

dm= 0.

Since the coefficients of the left hand side is from D−1R and this is a monic expression,we see that α/d is an integral element of D−1S over D−1R.

(b) If α ∈ S and f(x) ∈ R[x] is a monoic polynomial such that f(α) = 0, then theimage of α in S/P satisfies the polynomial f(x) ∈ (R/R ∩ P )[x] that is obtained fromf(x) by reducing its coefficients modulo P ∩ R. Since f(x) has a monic leading term,f(x) is not identically zero. Therefore, the image of α in S/P is integral over R/R∩P .

218. Let D be a multiplicative submonoid of R. Prove that there exists one-to-one correspondence between prime ideals not intersecting D in R and primeideals of D−1R.

Solution.

Let φ denote the map sending a prime ideal I of R to the ideal generated by the imageof I in D−1R. It is clear that, if I ∩D 6= ∅, then φ(I) = D−1R. Therefore, we restrictour attention to the prime ideals that do not intersect D.

Let r ∈ φ(J) for some other prime ideal J from R. Thus, r is of the form r1s1/d1 +· · · rks1/dk for some r1, . . . , rk ∈ J , s1, . . . , sk ∈ R and d1, . . . , dk ∈ D. Since s1r1/d1 +· · · skrk/dk = d′1s1r1/d + · · · + d′kskrk/d, where d =

∏di, we see that r is of the form

r′/d′, where r′ ∈ J and d′ ∈ D. Thus, we identify the ideal generated by J in D−1Rwith the set {r′/d′ : r′ ∈ J, d′ ∈ D}.Now, if φ(J) = φ(I), then r ∈ φ(I), hence there exists r′′ ∈ I, d′′ ∈ D such thatr′/d′ = r′′/d′′, which implies that r′ ∈ I. In other words, J ⊂ I. The same argumentshows that I ⊂ J , hence I = J . Therefore, φ is injective.

Conversely, let U ⊂ D−1R be a prime ideal and let I = U ∩D−1R. We claim that Iis prime, I ∩D = ∅, and furthermore, φ(I) = U . If r, r′ ∈ R are two elements from Rsuch that rr′ ∈ I, then in particular rr′ ∈ D−1R. Thus, either r ∈ D−1R or r′ ∈ D−1R.But r and r′ are from R, therefore, at least one of them is contained in I. In otherwords, I is prime. It is clear that I ∩ D = ∅, otherwise U contains a unit. Finally,r/d ∈ U , then d · r/d = r ∈ U . Since r is an element of R, we see that r ∈ I. It followsthat I generates U in D−1R.

219. Let P be a prime ideal in R and let D = R − P denote the correspondingmultiplicative submonoid. Prove that there exists a unique maximal idealin D−1R (that is generated by P in D−1R).

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Solution.

We know from Problem 218 that prime ideals of R not intersecting D are in one-to-onecorrespondence with the prime ideals of D−1R. In particular the correspondence φ isgiven by sending I to the ideal φ(I) generated by I in D−1R. It is obvious that φpreserves the inclusion relation: I ⊂ J implies φ(I) ⊂ φ(J). Since the unique largestprime ideal not intersecting D is P , there is a unique maximal ideal in D−1R.

220. Let R ⊂ S be a pair of rings such that S is integral over R. If P is a primeideal in R, then prove that there exists a prime ideal Q in S such thatP = Q ∩R.Solution.

The idea is to analyze the situation locally. Let D denote the complement of P in R.Then D is a multiplicative monoid. Notice that D is a multiplicative monoid in S aswell.

By Problem 217 we know that D−1S is integral over D−1R and also if M is a primeideal in D−1S then D−1S/M is integral over D−1R/D−1R ∩M . Now, suppose M isa maximal ideal in D−1S. (The existence of M is guaranteed by the Zorn’s lemma.)Since D−1S/M is a field, by Problem 213 we see that D−1R/M ∩D−1R is a field, also.In other words, the ideal M0 := M ∩ D−1R is maximal in D−1R. Since D−1R has aunique maximal ideal (generated by P ) we have M0 ∩R = P . See Problem 219 .

Similarly, M ∩ S is a prime ideal in S and furthermore M does not intersect D.Therefore, its restriction to R, namely M ∩R does not intersect D neither. Thus it iscontained in P = R−D. Combining this with the fact that M0 ∩R = P , we see thatP = M .

221. Suppose P is a maximal ideal of R, and S is ring that contains R. If thereexists a finite number of elements s1, . . . , sn ∈ S that generate S as a ring overR (that is to say, the coefficients of the polynomial expressions in s1, . . . , snare all from R), then prove that there are only finitely many maximal idealsQ in S such that Q ∩R = P .

Solution.

By Problem 220 we know that S has a prime ideal Q such that Q ∩R = P . We claimthat there exists a maximal ideal M in S such that M ∩ R = P . If Q is maximal,then there is nothing to do, so we assume that Q is contained in a maximal ideal M .Clearly M ∩ R contains P . If this is a proper containment, since P is maximal, thenM ∩R has to contain a unit, which is absurd. Therefore, M ∩R = P .

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Next, we claim that there only finitely many maximal idealsM of S such thatM∩R =P . Notice that, since M ∩R = P , the canonical injection R ↪→ S induces an injectionR/P ↪→ S/M . Therefore, S/M is a field extension of R/P . We analyze the generatorsof S/M as a field over R/P . It is clear that the images of si’s generate S/M as field withcoefficients from R/P . Since si’s are integral over R, we have polynomials fi(x) ∈ R[x]such that fi(si) = 0. Reducing the coefficients of fi’s modulo P , we see that theminimal polynomial over R/P of si’s are divisors of fi(x) mod P . We conclude thatthe field S/M is obtained from R/P by adjoining some roots of fi(x) mod P . Sincefi(x) has finitely many roots, there are finitely many different field extensions of theform S/M over R/P . This proves that there are only finitely many different maximalideals of S that restricts to the same maximal ideal P in R.

222. Let S be an integral extension of R. Suppose P1 ⊆ · · · ⊆ Pn is a sequence ofprime ideals of R and suppose that there exists a sequence of prime idealsQ1 ⊆ · · · ⊆ Qm of S such that Qi ∩ R = Pi for i = 1, . . . ,m. Here, 1 ≤ m < n.Prove that there exists prime ideals Qm+1, . . . , Qn in S such that Qj ⊆ Qj+1

and Qj ∩R = Pj for all j = 1, . . . , n.

Solution.

This is a simple application of the Problem 220: We look at the integral extensionS/Qm of R/Pm. Since the image of Pm+1 in R/Pm is prime, there exists a prime idealQm+1 in S, whose image in S/Qm lies over that of Pm+1. It is clear that Qm+1 containsQm.

223. Let k be a positive integer. Prove that the following equalities are true fortwo ideals I and J from a ring R:

(a) Ik ⊆ J =⇒ rad(I) ⊆ rad(J);

(b) Ik ⊆ J ⊆ I =⇒ rad(J) = rad(I);

(c) rad(IJ) = rad(I ∩ J) = rad(I) ∩ rad(J);

(d) rad(rad(I)) = rad(I);

(e) rad(I) + rad(J) ⊆ rad(I + J) = rad(rad(I) + rad(J)).

Solution.

(a) We start with an easy observation: if I1 and I2 are two ideals such that I1I2 ⊂ Pfor some prime ideal P , then either I1 ⊂ P , or I2 ⊂ P . Indeed, assuming otherwisewe find two elements a ∈ P − I1 and b ∈ P − I2 such that ab ∈ P . Since P is prime,this produces a contradiction. This observation implies also that if P contains a powerof an ideal Ik, then I ⊂ P . Since rad(I) is the intersection of all primes containing

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I (Problem 201), we see that rad(I) ⊂ rad(Ik). On the other hand, as Ik ⊂ J , anyelement x ∈ R such that xm ∈ Ik satisfies xm ∈ J . Equivalently, rad(Ik) ⊂ rad(J).Therefore, rad(I) ⊂ rad(J).

(b) By part (a) we know that rad(I) ⊂ rad(J). Since J ⊂ I =⇒ rad(J) ⊂ rad(I)the equality follows.

(c) Since IJ ⊂ I ∩ J , we see that rad(IJ) ⊂ rad(I ∩ J). Conversely, let x ∈ R bean element of rad(I ∩ J) that is to say xm ∈ I ∩ J for some m ≥ 1. Then xm ∈ Iand xm ∈ J . Thus x2m ∈ IJ . It follows that x ∈ rad(IJ). (Almost) the sameargument shows that rad(I) ∩ rad(J) ⊂ rad(I ∩ J). Since I ∩ J ⊂ I, J we see thatrad(I ∩ J) ⊂ rad(I), rad(J). Therefore, rad(I ∩ J) = rad(I) ∩ rad(J).

(d) The inclusion rad(rad(I)) ⊂ rad(I) follows from the definition of radical ideals.Conversely, I ⊂ rad(I) implies that rad(I) ⊂ rad(rad(I)).

(e) Let x ∈ rad(I), y ∈ rad(J) be two elements. Then xm ∈ I and yn for somepositive integers m,n. Then terms of (x + y)mn are of the form

(k+lk

)xkyl with k +

l = mn. In this case, either k ≥ m, or l ≥ n. Therefore, either(k+lk

)xkyl ∈ I

or(k+lk

)xkyl ∈ J . It follows that (x + y)mn ∈ I + J . Finally, we prove the last

equality. Since rad(I)+rad(J) ⊂ rad(I+J), taking radicals of both sides implies thatrad(rad(I) + rad(J)) ⊂ rad(rad(I + J)) = rad(I + J) (by the previous part of theproblem). Conversely, if x ∈ rad(I+J), then there exists m ≥ 1 such that xm ∈ I+J .Since I + J ⊂ rad(I) + rad(J), we have xm ∈ rad(I) + rad(J). In other words, xbelongs to the radical of rad(I) + rad(J). This finishes the proof.

224. Suppose R is a Noetherian local ring. Prove that ifM is the unique maximalideal of R and dimR/M M/M2 is a finite dimensional vector space over R/M .If d = dimR/M M/M2, then prove that any generating set for M has at leastd elements.

Solution.

Let U denote M/M2. For m ∈ M and r ∈ R we denote by m and r the images ofm and r in U and R/M , respectively. By the action r ·m = rm of R/M , U becomesa module over R/M . Since R is Noetherian, every ideal in R is finitely generated.In particular M is finitely generated. The image in U of a generating set for M is agenerating set for U as an R-module. In particular, U is finitely generated R-module.But any element from M acts as a zero on U , thus U is a finitely generated R/M -module, hence a finite dimensional vector space. Let d = dimR/M M/M2 be the vectorspace dimension of M/M2 and let x1, . . . , xn be a list of generators of M as an ideal.Since x1, . . . , xn in U is a spanning set we see that n ≥ d.

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225. Suppose R is a Noetherian integral domain. We assume further that R isa local ring of Krull dimension 1. Prove that if M is the unique maximalideal of R and dimR/M M/M2 = 1, then R is a DVR.

Solution.

Since R/M is a field and since dimR/M M/M2 = 1, we have an element t ∈ M suchthat t is the basis vector for M/M2.

We claim that Mn/Mn+1 is an R/M -vector space spanned by the image of tn inMn/Mn+1. We prove this by induction on n. n = 1 case is given by the hypothe-sis and it is clear that Mn/Mn+1 is a vector space over R/M . Let y ∈ Mn/Mn+1 beany element. Then y = x1y1 + · · · + xryr for some xi ∈ M , yi ∈ Mn−1. Write xi inthe form ait + p where ai ∈ R is a unit and p ∈ M2. Thus, xiyi = aityi + pyi. Byinduction hypothesis yi is of the form cit

n−1 + q where ci ∈ R and q ∈ Mn. Thus,aityi + pyi = aicit

n + aitq + pyi. Since aitq and pyi are from Mn+1, and since aici is aunit, we see that xiyi modulo Mn+1 is equal to aicitn, hence our claim follows.

Next, we show that R is a DVR by exhibiting a valuation for R. By the above discussionwe see that any element x of R is of the form atn modulo Mn+1 for some unit a ∈ R.Furthermore, the exponent n is unique. Thus, the function ν : R → Z≥0 defined byν(x) = n is well-defined. We extend ν to the fraction field K of R by defining it tobe ν(x/y) = ν(x) − ν(y). Well-definedness of the extension is clear. Checking that νsatisfies the properties of a valuation on K is straightforward.

226. Suppose R is a Noetherian integral domain. We assume further that R isa local ring of Krull dimension 1. Prove that if M is the unique maximalideal of R and if every ideal is a power of M , then R is a DVR.

Solution.

Let x1, . . . , xn ∈ M be a minimal list of generators for M . Without loss of generalitywe assume that x1 /∈ M2. The principal ideal I generated by x1 is a power of Mby hypothesis: I = Mk for some k ≥ 1. Since x1 /∈ M2, in particular, x1 /∈ Mk forany k > 1. Therefore, I = M . In particular, M/M2 is spanned by x1. Therefore,dimR/M M/M2 = 1, hence by Problem 225 the problem is finished.

227. Prove that if R is a DVR, then R is a PID. Conversely, if R is a PID witha unique maximal ideal, then R is a DVR.

Solution

(⇒) Let K denote the fraction field of R, and let ν : K → Z denote the valuation of R.Thus R = {x ∈ K : ν(x) ≥ 0}. First of all, we claim that the set {y ∈ R : ν(y) = 0}

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is the set of invertible elements of R. Indeed, 0 = ν(1) = ν(y/y) = ν(y) + ν(1/y)implies that if ν(y) = 0, then ν(1/y) = 0, hence 1/y ∈ R.Next we show that M = {y ∈ R : ν(y) > 0} is the unique maximal ideal and it isgenerated by a single element. The maximality of M is clear. To see that M is anideal let y ∈M and z ∈ R. Then ν(zy) = ν(z) + ν(y) > 0, therefore, zy ∈M .

Finally, we show that R is a PID. Let I ⊂ R be an ideal. Let z ∈ I be an element withν(z) = m. Also, let t ∈M be an element such that t ∈M −M2. In this case ν(t) = 1.Since ν(zt−m) = ν(z) + ν(t−m) = m−m = 0, we see that zt−m is an invertible elementof R. Therefore, z = atm, where a ∈ R is a unit. In other words, I is generated by tn,where n is the smallest power such that tn ∈ I.(⇐) Define a valuation ν on K as follows: if r ∈ R, then ν(r) = n; the unique non-negative integer such that r ∈ Mn − Mn+1. It is straightforward to verify that νsatisfies the axioms of being a valuation.

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