Algebraic number theory - Lecture Notes

Nivedita

2009

Lecture 1

Introduction

Arithmetic was done in the ring Z, which we know has several nice properties,one among them being that Z is a unique factorization domain. However whenlarger rings such asR = Z[

√−5] were investigated, it was found that, for instance

9 could be factorized as follows:

9 = 3.3 = (2 +√−5)(2−

√−5)

Proposition 0.1. 3, 2 +√−5, 2−

√−5 are irreducibles in R

Proof. Look at the norm mapN : Z[√−5]→ Z whereN(a+b

√−5) = a2 +5b2.

One can easily verify that N(a)N(b) = N(ab).Therefore if 3 = (a+ b√−5)(c+

d√−5), then N(3) = 9 = (a2 + 5b2)(c2 + 5d2) where a, b, c, d ∈ Z.

If a2 + 5b2 = 1, this means (a + b√−5)(a − b

√−5) = 1 and so a + b

√−5 is a

unit. So, if 3 is not irreducible, then N(a + b√−5) must be 31. So a2 + 5b2 =

3 =⇒ a2 ∼= 3 mod (5) which is not possible. So 3 is irreducible. Similarly, asN(2 +

√−5) = N(2−

√−5) = 9, we conclude that they are also irreducible

19=3.3 and Z is a U.F.D

1

All this goes to show that R is not a U.F.D. In algebraic number theory, we wouldlike to study how to resolve this issue. Dedekind resolved it by introducing theconcept of factorization of ideals.

Let P1 denote the ideal generated by 3, 2 +√−5 in R and P2, the ideal generated

by 3, 2−√−5.

First note that P1, P2 are prime ideals (in fact, they are maximal)

To show P1 is maximal, let a+b√−5 6∈ P1. Note that 3−(2+

√−5) = 1−

√−5 ∈

P1.

a+ b√−5 + b(1−

√−5)− 3ba+ b

3c = z ∈ R(a+ b

√−5) + P1 = X

Note that z can be 0, 1 or 2. If z 6= 0, then X = R. 2. If z = 0, this means thata+ b ∼= 0 mod 3. Therefore, write a = 3n− b. Then a+ b

√−5 = 3(n) + b(1−√

−5) ∈ P1. So we are done. P2’s maximality follows similarly.

Now if A,B are two comaximal (ie - if A+B = R) ideals in a commutative ring,then A ∩ B = AB3. Therefore P1P2 = P1 ∩ P2 ⊇ (3). Also any x ∈ P1P2 lookslike

∑(3x+(2+

√−5)y)(3z+(2−

√−5)w) =

∑3d ∈ (3). And so P1P2 = (3).

Also, P 21 = (2 +

√−5) because 3(2 +

√−5) + (3 − (2 +

√−5))2 = 2 +

√−5.

Similarly 3(2 −√−5) + (3 − (2 −

√−5))2 = 2 −

√−5 and therefore P 2

2 =(2−

√−5)

Thus, at an ideal level, we have

(9) = (3)2 = (P1P2)2 = (2 +

√−5)(2−

√−5) = P 2

1P22

with no loss to unique factorization.

Number fields

K ⊆ C is an algebraic number field if it is a finite field extension of Q.

Hereafter K will denote a number field unless otherwise mentioned.23 ∈ P1, so 3− 2 = 1 ∈ X . So X = R3Clearly AB ⊆ A ∩ B for any A,B. And since A + B = R, a + b = 1 for a, b ∈ A,B

respectively. Then any x ∈ A ∩B = x(a+ b) = xa+ xb ∈ BA+AB = AB

2

Integral extensions

α ∈ K is said to be integral over A, a sub-ring of K if ∃p(x) ∈ A[x] , a monicpolynomial such that p(α) = 0. And B ⊇ A is said to be integral over A if each bin B is integral over A.

It is generally quite hard to give an explicit monic polynomial satisfied by an ele-ment, which is why the following theorem is very useful in establishing integrality.

Theorem 0.2. The following are equivalent

1. α ∈ K is integral over A where A is a sub-ring of K

2. A[α] is a finitely generated A module

3. ∃B, a sub-ring ofK which containsA and α and is also a finitely generatedmodule over A.

Proof.(1) =⇒ (2)

Let αn + an−1αn−1 + . . . + a0 = 0 with ai ∈ A∀i. Therefore αn is an A linear

combination of 1, α, . . . , αn−1. Similarly αn+1 = α(αn) is again an A linearcombination of 1, . . . , αn−1. Therefore as A[α] is generated by {αi|i ∈ N∪ {0}},1, α, . . . , αn−1 form a set of generators for A[α].

(2) =⇒ (3)

Set B = A[α]

(3) =⇒ (1)

Let u1, u2, . . . uk be a set of generators for B. Now since α ∈ B, we have αui =∑kj=1 aijuj with aijs in A. In matrix language , this translates to

α− a11 a12 . a1n

a21 α− a22 . a2n

. . . .an1 an2 . α− ann

u1

u2

. . .un

=

00.0

3

Now use Kramer’s rule 4 and conclude that det(αI − (aij)) = 0, which tells usthat α satisfies a monic polynomial with coefficients in A.

Corollary 0.3. 1. If A is a sub-ring of K and α1, α2, . . . , αn integral over A,then A[α1, α2, . . . αn] is a finitely generated module over A

2. The set of integral elements over A form a sub-ring of K

3. C ⊇ B ⊇ A . Then C integral over A iff C is integral over B and Bintegral over A

Proof. • Proof for (1) is by induction. Note that the case n = 1 has alreadybeen settled before. Let us assume X = A[α1, . . . , αn−1] is a finitely gener-ated module over A and let {v1, . . . vn} be a set of generators.

Now αn is integral over A, therefore it is also integral over X ⊇ A. There-fore X[αn] is a finitely generated module over X . Let a set of generators be{u1, . . . uk}Therefore {viuj} is a set of generators for X[αn] as a module over A

• If α, β are integral over A, then from (1), it follows that A[α, β] is a finitelygenerated module over A which contains A[αβ]. Therefore by applying the3rd equivalent condition of theorem , we conclude that αβ is integral overA. We can do a similar thing for α + β etc. 5

• If C is integral over A, then clearly, any element of B is integral over A asB ⊆ C. Also since B ⊇ A, any c which is integral over A is also integralover B.

For the other direction, pick c ∈ C. We are given that C is integral over B.Therefore let cn + bn−1c

n−1 + . . . b0 = 0 where bi ∈ B∀i, which means thatc is integral over A[b0, . . . bn−1] = X . Therefore X[c] is a finitely generatedmodule over X , with generators {u1, . . . uk}, say. Now since b0, . . . bn−1 ∈B which is integral overA, by (1), we haveX is a finitely generated moduleover A with generators say v1, . . . vn. Then {uivj} generates X[c] as an Amodule, which proves c is integral over A.

4Consider the entries as sitting inside K5Note that this method is much simpler than explicitly giving a monic polynomial which α+β

satisfies

4

Note that if a satisfies a monic polynomial in Z[x], then it satisfies an irreduciblemonic polynomial in Z[x]. For a proof, pick f its monic irreducible polynomialover Q. f = ch where c ∈ Q and h in Z[x]. Let g denote the monic polynomial inZ[x] which a satisfies. Then f |g in Q[x]. That is fj = g where j ∈ Q[x]. Writej = dk where k ∈ Z[x] and d ∈ Q. So we have g = dchk. By gauss’s lemma,we have hk is primitive. Since g is monic, it is primitive, and therefore dc = 1 .This implies that h is also monic. But f = ch is monic also, so c = 1. Thus f isin Z[x] and irreducible in Q[x].

Integral closure

As proved in the corollary above, the set of all integral elements over A form asub-ring, which is called the integral closure of A in K or A.

As as expected, A = A, because if b is an integral over A, as A is integral overA, we get A[b] is integral over A by the third part of the corollary, which meansb ∈ A.

A is said to be integrally closed in K if A = A. The following theorem gives usmany examples of integrally closed rings.

Theorem 0.4. Any U.F.D R is integrally closed in its quotient field S

Proof. Let s ∈ S = pq

where p, q ∈ R and H.C.F of p, q is 16. If s is integral

over R, then pn

qn+ rn−1

pn−1

qn−1 . . . + r0 = 0. Clearing the denominators, we getpn + qx = 0 for some x ∈ R, which implies an irreducible factor of q divides p,which is a contradiction, unless q is a unit, which means s ∈ R.

Therefore Z is integrally closed in Q.

Considering we have quite a few examples of integrally closed domains in theirquotient fields, one would like to know an example of a ring which is not closed.

Take R = Z[√d] where d ∼= 1 mod 4, say d = 4a + 1 . It is not closed in Q(

√d)

because we will exhibit an element, namely 1+√d

2, which is integral over Z.

6Since R is a U.F.D, you can talk of the highest common factor

5

1 +√d

2satisfies p(x) = x2 − x− a

Well behaved integral closure

Number theory started out as the study of diaphontine equations over Z. Algebraicnumber theory studies diaphontine equations over other rings of integers

OK = {k|k ∈ K, k is integral over Z} is called the ring of integers of K.

K < ⊃ OK

L

∧.........< ⊃ OL

∧........

Q

∧.........< ⊃ Z

∧........

Let L be a finite extension of Q and K be a finite extension of L.The abovediagram motivates the question : Is the integral closure of OL in K = OK ? Andthe answer is yes!

Proof. Clearly any element ofOK , being integral over Z is also integral overOL.Now if a ∈ K such that a is integral over OL, then as OL is integral over Z, thismeans a is integral over Z and hence belongs to OK .

Note that K is the quotient field of OK because if you pick any s in K, it willsatisfy a polynomial in Q[x], then by clearing denominators , it will satisfy apolynomial in Z[x], say ansn + an−2s

n−1 + . . . a0 = 0. Multiplying by an−1n , we

get (ans) satisfies a polynomial in Z[x] and so lies in OK . an being in Z already,implies s lies in the fraction field of OK .

Also OK is integrally closed in K since if s ∈ K is integral over OK , the latterbeing integral over Z would imply that s is integral over Z which would means ∈ OK .

6

Lecture 2

We would like to describe OK for finite extensions K of Q. And to start off withthe simplest situation, let us investigate quadratic extensions.

First, note that if [K : Q] = 2, then K = Q(√d) where d is square free. K

is also a galois extension over Q. The only Q automorphisms of K are identityand the map σ sending

√d → −

√d. Also note that σ2 is identity and therefore

α + σ(α) and ασ(α) lie in the fixed field. K being a galois extension, this meansthey actually lie in Q for any α ∈ K.

Proposition 0.5. α ∈ OK iff α + σ(α) and ασ(α) belong to Z

Proof. Clearly if α ∈ OK , so is σ(α) 7. Therefore σ(α) + α, ασ(α) ∈ OK aka,they are integral over Z. They are also in Q as shown earlier, which implies theyactually belong to Z as Z is integrally closed in Q.

For the other direction, let α + σ(α) = a1 and ασ(α) = a0 be in Z. Thenp(x) = (x− α)(x− σ(α)) = x2 − a1x+ a0 shows that α is integral over Z.

Therefore if α ∈ OK = a + b√d where a, b ∈ Q, then σ(α) + α = 2a ∈ Z and

σ(α)α = a2− db2 ∈ Z. This means (2a)2− (2b)2d ∈ 4Z. Since 2a ∈ Z, we have(2b)2d ∈ Z. Note that d is square free. Therefore 2b ∈ Z.

So let u = 2a and v = 2b. Note that they are integers. We know that 4a2−4b2d =u2−dv2 ∼= 0mod 4. If u is even, since d is square free, then v is even. Any squareis either 0 or 1 mod 4. Therefore if u is odd, then dv2 ∼= 1 mod 4. So v has to beodd and this implies d = 1 mod 4.It is easy to check that if u and v are integersof the same parity then u

2+√dv

2are indeed algebraic. Thus we get the following

result.

Theorem 0.6. If d ∼= 1 mod 4, algebraic integers form the ring Z + Zω whereω = 1+

√d

2and if d 6∼= 1 mod 4, OK = Z + Zω where ω =

√d.

However describing OK when degree of extension is higher is more subtle

7a0 + a1α+ . . . αn = 0 =⇒ a0 + a1σ(α) + . . . σ(α)n = 0

7

Trace and Norm

Let α ∈ K which sits inside C. And let f be its irreducible polynomial over Q.Note that because rationals have characteristic 0, f(x) has distinct roots, ie - Q(α)is a separable extension. We define the algebraic conjugates of α to be the rootsof f .

Now if K = Q(θ), and θ1, . . . θn denote the n distinct algebraic conjugates of it(extension is of degree n). Then there exist exactly n embeddings of K into C.Namely σi : K → C which takes θ θi.8. This prompts the definition of fieldconjugates of α ∈ K as σ1(α), σ2(α) . . . σn(α). Note that these can be repeated.In fact we can say how many times.

Every finite separable extension is simple, that is, generated by one element. Andsince Q(α) has char 0, K over Q(α) is separable. So, let K = Q(θ) = Q(α)(β) .Let m = [K : Q(α)] and r = [Q(α) : Q]. Therefore n = mr where n = [K : Q].Then there are r embeddings of Q(α) into C. And given any embedding of Q(α)into complex numbers, it can be extended to the whole of K in m ways 9. Sowe have found mr embeddings of K. However mr = n and we know there areexactly n embeddings of K from the first paragraph. Thus, we have found all ofthe n embeddings. And α takes values of its algebraic conjugates , repeated mtimes each. We shall denote the field conjugates of α by α(1), α(2), . . .

We now define the polynomial fα(x) =∏n

i=1(x − σi(α)) where σis denote then embeddings of K into C. If g is the irreducible polynomial of α with respectto Q, then from the above discussion, it must be clear that gm = fα. And sinceg ∈ Q[x], so is fα.

Norm is defined by NKQ

(α) =∏n

i=1 σi(α) and Trace by TrKQ

(α) =∑n

i=1 σi(α).Since these are coefficients of fα, Norm and Trace are rational valued functions.

Lemma 0.7. If α ∈ OK , then fα ∈ Z[x]. In particular, norm and trace areintegers.

Proof. We already know fα ∈ Q[x]. Note that since α ∈ OK , αn + rn−1αn−1 +

. . . r0 = 0. Applying σi to it, we find that σi(α) ∈ OK also.Thus fα ∈ OK [x].This implies it must be in Z[x] as Z is integrally closed in Q.

81 goes to 1. So any embedding is identity on Q. And σi(θ) must also be a root of f . Andonce you know where to send θ, the rest of the map is determined

9send β to the m roots of the irreducible polynomial of β over Q(α).

8

So, NKQ

(α) and TrKQ

(α) (in fact fα(x)) can be thought of as the determinant andtrace (the characteristic polynomial) of the matrix

Given a basis B = {b1, b2, . . . bn}, define matrix M(B) as follows:

M(B) =

b(1)1 b

(1)2 . b

(1)n

b(2)1 b

(2)2 . b

(2)n

. . . .

b(n)1 b

(n)2 . b

(n)n

Let B and C be two bases of K over Q. If X is matrix such that

X

b1.bn

=

c1.cn

Then XM(B) = M(C). Note that X is nonsingular . Therefore either for allbases B, M(B) is singular or for all B, M(B) is nonsingular.

Let us choose a special B , namely {1, θ, θ2, . . . θn−1} where K = Q(θ). ThenM(B) is the vandermonde matrix for θ which has determinant nonzero as θi 6= θj

when 0 ≤ i < j ≤ n− 1. Thus for all B, M(B) is nonsingular.

D =

α(1)

α(2)

.α(n)

Now look at the map T : K → K which sends x αx. This is a Q linear map(it is a K linear map in fact). However, let W = {w1, w2, . . . wn} be a basis forK as a vector space over Q. And let A = (aij) be the matrix of T with respect tothis basis.

a11 a12 . a1n

a21 a22 . a2n

. . . .an1 . . ann

w1

w2

.wn

=

αw1

αw2

.αwn

So a11w1 +a12w2 + . . . a1nwn = αw1 . Acting σi on it, we get a11w

(i)1 +a12w

(i)2 +

. . .+a1nw(i)n = α(i)w

(i)1 etc . Thus you can check thatAM(W ) = M(W )D. Since

M(W ) is nonsingular, A and D are conjugates and the characteristic polynomialof A is also fα

9

Lecture 3

In this lecture, we examine an important invariant of an extension named the dis-criminant. Note that TrK

Qis a Q linear map.

If {x1, x2, . . . xn} is a basis forK over Q, then we define discriminant with respectto this basis as follows:

D(x1, x2, . . . xn) = det(TrKQ

(xixj))

Lemma 0.8. D(x1, x2, . . . xn) = det(σi(xj))2 6= 0

Proof. Since TrKQ

(xixj) =∑n

k=1 σk(xixj) =∑σk(xi)σk(xj),

D(x1, x2, . . . xn) = det(n∑k=1

σk(xi)σk(xj)) = det(σi(xj))2

If this was 0, that is det(σi(xj)) = 0, this means that for some u1, u2 . . . uk not allzero, u1σ1 + . . . unσn = 0 (as x1, . . . xn is a basis). We recall Dedekind’s lemmadone in field theory, which states that if G is a group and L a field , and σ1, . . . σn, n distinct homomorphisms from G into L∗, then they are linearly independentover L. Here G = K∗ and L = C. This shows that discriminant can’t be 0.

Corollary 0.9. The bilinear form B : K × K → Q which sends (x, y) TrK

Q(xy) is nondegenerate.

Proof. Observe that Tr(x, x−1) = n.

A lengthier proof which will work for any characteristic of field is:

Fix a basis for K, say {x1, . . . xn}. If B(x, y) = 0 for all y, write x = a1x1 +. . . anxn with ais in Q. Then xxj =

∑aixixj . Since B(x, y) = 0∀y, we have∑

aiTr(xixj) = 0 for all i, j. But we know det(Tr(xixj)) 6= 0 by above lemmawhich forces ai = 0∀i, which means x = 0.

10

Given any nondegenerate bilinear form B and a basis U = {u1, u2, . . . un} for K,then there exists another basis V = {v1, v2, . . . vn} such that B(ui, vj) = δij . V issaid to be a dual basis for U . For a proof consider the map B̂ : K → Hom(K,C)which sends x [y → B(x, y)]. Since B is non-degenerate, B̂ is injective.Dimension of K is same as dimension of Hom(K,C), which therefore meansthat B̂ is an isomorphism. Therefore Ti : K → C where Ti(uj) = δij lies inthe image of B̂. Let B̂(vi) = Ti. Since B̂ is injective, we can find vis uniquely.Now we just have to show V is basis. To see that, note that Tis form a basis forHom(K,C) and vi is the inverse image of Ti.

Thus, given a basis U, we can find another basis V, such that Tr(uivj) = δij .

OK as a Z module

We call a ∈ Q a denominator of b ∈ K if ab ∈ OK .

Lemma 0.10. Every b ∈ K has a denominator.

Proof. Since b is algebraic over Q, it satisfies a polynomial in Q[x] and thereforea polynomial in Z[x], say anbn +an−1b

n−1 + . . . a0 = 0. Multiply by an−1n and we

can see that anb satisfies a monic polynomial in Z[x].

Theorem 0.11. If [K : Q] = n, then OK is a free Z module of rank n.

Proof. Let X = {x1, x2 . . . xn} be a basis for K. By the above lemma, we knowthat there exist integers a1, a2, . . . an such that a1x1, . . . anxn lie in OK . And theystill remain a basis ofK. Let us call this new basis consisting of algebraic integersto be w1, w2, . . . wn.

Fix a dual basis Y = {yi} such that Tr(wiyj) = δij

Now pick any w ∈ OK . Since wi ∈ OK , wwi ∈ OK . Since Y is a basis, we canwrite w = c1y1 + c2y2 + . . . cnyn where ci ∈ Q∀i.

But Tr(wwi) = Tr(c1wiy1) + Tr(c2wiy2) + . . . T r(cnwiyn) = ci and we knowthat w,wi ∈ OK , so wwi is an algebraic integer also, and thus trace is an integer.

So w ∈ Zy1 + Zy2 + . . .Zyn = M . Thus OK is a submodule of the free moduleM .

11

Lemma 0.12. If A is a Z module with generators a1, a2, . . . an and B is a sub-module of A, then we can find generators b1, b2, . . . bk for B.In fact we can find anupper triangular matrix X of dimension k × n such that

X

a1

.

.an

=

b1.bk

(Rough sketch of the proof: Let H1 = {z1|b ∈ B, b = z1a1 + z2a2 + . . . znan}.10

This is an ideal in Z , say (x1). Now pick an element b1 ∈ B such that b1 =x1a1 + z2a2 + . . . znan for some z2, . . . zn. The first row of the matrix X isx1, z2, . . . zn. Now let G = {b − nb1|b ∈ B, n chosen such that b − nb1 =s2a2 + . . . snan for some si}. And let H2 be the set of all the coefficients of a2

when you write each element of G in terms of ais. This is an ideal again in Z, say(x2). Pick a b2 in G such that b2 = x2a2 + s3a3 + . . . sna − n. The second rowof X is 0, x2, s3, . . . sn and so on. Note that you can only carry this process foratmost n steps.)

If A was free, since X is upper triangular, this would imply that b1, b2, . . . bk wereZ linearly independent. Hence as M is free, OK is a free Z sub-module and letZ = {z1, z2, . . . zm} be a Z basis for OK . Note that therefore these are Q linearlyindependent as well and so m ≤ n.

Now any k ∈ K has a denominator in Q, say d.Thus dk ∈ OK = s1z1 + s2z2 +. . . smzm where si ∈ Z. Therefore k = s1

dz1 + . . . sm

dzm, and so Z spans K as

well. Thus it is a basis and so m = n.

Note that | MOK | is finite, in fact, equal to determinant X , as {e1a1 + e2a2 +

. . . enan|0 ≤ ei < xi} is all of MOK

.

A Z basis for the free module OK is called an integral basis. 11

So far, the discriminant was defined in a manner that it depended on a basis. Wenow show that upto a square, it is the same.

Proposition 0.13. If W = {w1, . . . wn} is an integral basis for OK and Y ={y1, . . . yn} be any basis forK such that yi ∈ OK for all i, thenD(Y ) = a2D(W )for some integer a.

10There may be many different ways of writing b in terms of ais. Put all the first coefficients inH1

11Any integral basis is also a Q basis for K as rank of OK = dimension of K over Q and Zindependence implies Q independence

12

Proof. Since yi is in OK we have yi =∑aijwj where aij ∈ Z.(ie - if A = (aij)

then AW = Y )

The required a will be det(A).

To see this, recall the definition D(Y ) = det(Tr(yiyj)).

Tr(yiyj) =∑n

k=1

∑nh=1 aikajhTr(wkwh) and thus (Tr(yiyj)) = ATr(wkwh)A

T ,which shows that D(Y ) = det(A)2D(W )

Corollary 0.14. OK = Zy1 + . . .+ Zyn iff D(y1, . . . yn) = D(w1, . . . wn) whereyis and wis have the same meaning as in the proposition above.

Proof. If Y is another integral basis for OK , then we have , by the proposi-tion above, that D(Y ) = det(aij)

2D(W ) and D(W ) = det(bij)2D(Y ). Thus

det(aij)2det(bij)

2 = 1, which implies det(aij)2 = 1 (as aij and bijs are all inte-gers) and thus D(Y ) = D(W ).

For the converse, we have D(Y ) = det(aij)2D(W ) and if D(Y ) = D(W ), then

det(aij) = ±1 where aij ∈ Z. Thus inverse of (aij) is also a Z matrix (inverse= adjoint

determinant) say (bij), and hence (bij)Y = W and thus every wi is a Z linear

combination of yis and wis span OK , therefore Y is an integral basis.

Thus, we define discriminant of OK to be the discriminant with respect to anintegral basis. Note that discriminant is always an integer as it is det(Tr(wiwj))and wiwj ∈ OK .

The following (obvious)corollary is very useful when one wants to identify whethera basis is integral or not.

Corollary 0.15. If Y = {y1, y2 . . . yn} is a basis of K such that yi ∈ OK for all iand D(Y ) is square free, then Y is an integral basis.

13

Lecture 4

We know that Z is a domain, a principal ideal domain at that, and that it is inte-grally closed in Q. Since every ideal is principal, it is noetherian as well. In thislecture we see what properties carry over to the ring of integers of K, OK .

Theorem 0.16. OK is noetherian, integrally closed domain in which every non-zero prime ideal is maximal.

Definition: A domain R which is noetherian, integrally closed in its fractionfield and in which every nonzero prime ideal is maximal is termed as a Dedekinddomain.

Proof. OK is a sub-ring of K , a field. Hence it is a domain.

We have already seen that OK is closed in K. And that OK is a free Z module offinite rank, ie - it is a finitely generated free abelian group. Let I be an ideal ofOK . It is firstly a subgroup of OK and therefore is a free subgroup of rank lessthan rank of OK . So let I be generated by i1, i2, . . . im as a Z module. Clearly,the same generators generate I as an ideal. Hence every ideal is finitely generatedand OK is noetherian.12

Take P to be a non-zero prime ideal in OK and let Q = P ∩ Z. Then Q is primein Z (because if x, y ∈ Z , so xy ∈ Q, therefore xy ∈ P , which implies x ∈ Por y ∈ P , which implies x ∈ Q or y ∈ Q). Because P is non-zero, Q is alsonon-zero. To see this, take p 6= 0 ∈ P . So it satisfies pd + ad−1p

d−1 + . . . a0 = 0where ais are in Z. Since p is in P , it follows that a0 is in P also.

Thus , let Q = qZ 13. Now since Z sits inside OK and P ∩ Z = Q, this inducesa map i : Z

Q→ OK

Pwhich sends z + Q z + P . Note that this is well defined

because if z + Q = w + Q, then z − w ∈ Q ⊆ P and it is clear that i is ahomomorphism and injective as well (i(z+Q) = i(w+Q) =⇒ z−w ∈ P , butz, w ∈ Z, therefore z − w ∈ P ∩ Z = Q).

Claim: OKP

is a finite dimensional vector-space over ZQ

To see it’s a vector-space, define (z+Q)(x+P ) = (z+P )(x+P ) = zx+P . Thisis well defined as z−w ∈ Q, x− y ∈ P , then zx−wy = zx−wx+wx−wy =x(z − w) + w(x− y) ∈ Q+ P = P .

12However sub-rings of Noetherian rings needn’t be noetherian. Take a non-noetherian domainX and its fraction field Y

13Note that Q is maximal in Z

14

SinceOK is a finitely generated Z module, say with generators a1, a2, . . . an, thena1 + P, a2 + P, . . . an + P span OK

P. Therefore, it is a finite-dimensional vector

space.

But ZQ

is a finite field! Therefore OKP

being finite dimensional actually implies, itis a finite domain. And it is well-known that any finite domain is actually a field.Therefore P is maximal in OK .

Having established similarities with Z, we now examine the properties of uniquefactorisation in OK , which is given by the following result

Theorem 0.17. OK is a U.F.D iff it is a principal ideal domain

Proof. Clearly PID implies U.F.D. For the other direction, let a ∈ P , where P isa prime. Writing a = xr11 . . . xrdd as product of irreducibles, xi ∈ P for some i.Now the ideal generated (xi) is prime, because if ab ∈ (xi), then ab = rxi andwriting a and b as product of irreducibles, because of unique factorisation, we findeither a ∈ (xi) or b ∈ (xi). However since prime ideals are maximal, P = (xi).Hence all prime ideals are principal. We shall prove that every ideal of OK is aproduct of prime ideals (unique product in fact) which shows that OK is indeed aprincipal ideal domain.

Unique factorisation of ideals

Lemma 0.18. Every ideal contains a product of non-zero prime ideals

Proof. Let R = OK .

S = {A|A proper ideal which doesn’t contain product of prime ideals}.

If S = ∅ we are done. If not, because OK is noetherian, pick a maximal elementM in S. Since M ∈ S, it is clearly not prime (asM ⊇M ). Therefore ∃a, b ∈ OKsuch that a, b 6∈M but ab ∈M .

Look at M + Ra and M + Rb. Note that (M + Ra)(M + Rb) ⊆ M whichis a proper ideal. Hence one of M + Ra and M + Rb is a proper ideal, sayM + Ra. If M + Rb = OK , then (M + Ra)(M + Rb) = (M + Ra) whichproperly contains M , which is not possible. So M + Rb is also a proper ideal.

15

Also note that M +Ra and M +Rb properly contain M , hence they can’t belongto S. Therefore M + Ra ⊇ P1P2 . . . Pr,M + Rb ⊇ Q1Q2 . . . Qk. ThereforeM ⊇ (M +Ra)(M +Rb) ⊇

∏Pi

∏Qj which contradicts the fact that M ∈ S.

Now we define an inverse of a prime ideal P as follows:

P−1 = {x ∈ K|xP ⊆ OK}

The product of an idealA and P−1 is defined to beAP−1 = {∑

finite sum aixi|ai ∈A, xi ∈ P−1}.

Lemma 0.19. P−1 6= OK for P a non-zero prime ideal.

Proof. First note that OK is contained in P−1. We now produce an element in Kwhich is not OK which is in P−1 as follows:

Choose a 6= 0 ∈ P . Now (a) ⊇ P1P2 . . . Pr where Pis are prime ideals (notneccesarily distinct). Choose the smallest r possible.

So P ⊇ Pi for some i, say i = 1, but primes are maximal, which implies P = P1.Now since r is the minimum possible such number, P2P3 . . . Pr 6⊆ (a). So choosea b in P2 . . . Pr which is not in (a). If a−1b ∈ OK , this means a−1ba = b ∈ (a),which is a contradiction. Hence a−1b 6∈ OK .

Now bP = bP1 ⊆ P1P2 . . . Pr ⊆ (a), therefore a−1bP ⊆ OK (because if a−1bp =s, then bp = as. Now bp ∈ (a), so bp = ra for r ∈ OK . Therefore ra = as whichimplies r = s and hence s ∈ OK . Therefore a−1b ∈ P−1

Remark: This proof can be modified to show that for any ideal A, A−1 6= OK bylooking at A ⊆ M where M is a maximal ideal . Pick x ∈ A and let r be theminimum number such that there exist prime ideals Pi where P1P2 . . . Pr ⊆ (x) ⊆A ⊆ M . Therefore M = P1 , say. Pick b ∈ P2 . . . Pr which doesn’t lie in (x).Therefore x−1b 6∈ OK . However bP1 = bM ⊆ P1P2 . . . Pr ⊆ (x). So bM ⊆ (x)and hence x−1bM ⊆ OK which implies x−1bX ⊆ OK as X ⊆M .

Lemma 0.20. Let A be any nonzero ideal of OK . Then AP−1 6= A

16

Proof. Note that 1 ∈ P−1, hence A ⊆ AP−1. Let a1, a2, . . . an be generatorsof A (as OK is noetherian). If AP−1 = A, let x ∈ P−1. Then aix =

∑bijaj ,

where bij ∈ OK . Then det(xδij − bij )=0, which means x ∈ OK , which impliesP−1 = OK which is a contradiction.

Remark: Similarly you can show that for any ideal X , AX−1 6= A

Lemma 0.21. If AP−1 ⊆ OK , then AP−1 is an ideal

Proof. If r ∈ OK and z =∑aixi ∈ AP−1 (where ai ∈ A, xi ∈ P−1), then

rz =∑raixi and rai ∈ A as A is an ideal. So rz ∈ AP−1. The other properties

are equally trivial to check.

Corollary 0.22. PP−1 = OK

Proof. P ⊆ PP−1 as 1 ∈ P−1. By lemma 0.20 P 6= PP−1. Also PP−1 ⊆ OKby the definition of P−1. Hence by lemma 0.21, we have PP−1 is an ideal whichcontains P . Recall that any prime ideal is maximal, which forces PP−1 to be thewhole ring.

Now we are all set to prove the unique factorisation, which we state again.

Theorem 0.23. Any ideal of OK can be unqiuely factorised (upto order) intoprime ideals.

Proof. Let us first prove that any ideal is product of prime ideals. As usual, wetake S = {A|A is not a product of prime ideals}. If S is empty, we are in goodshape. Else, since OK is noetherian, pick M to be its maximal element. M isclearly not prime (otherwise it can’t belong to S) and hence is contained in aprime ideal, say P .

Now M ( MP−1 ⊆ PP−1 = OK . Hence MP−1 is an ideal strictly biggerthan M and therefore can’t belong to S, which means MP−1 =

∏Qi where

Qi are prime ideals. Multiply both sides by P . Since PP−1 = OK , we haveM = P

∏Qi, which means M is a product of prime ideals, a contradiction to the

fact that M ∈ S.

17

The uniqueness part follows trivially. If A =∏Pi =

∏Qi, then Pi ⊇

∏Pi =∏

Qi and therefore Pi ⊇ Qj for some j. Since primes are maximalQj = Pi. Nowmultiply by Q−1

j to cancel the factor and induct on the number of prime factors.

Observe that any object of the formQPaiiQQbii

where {Pi} and {Qj} are disjoint sets

is in OK iff {Qj} is the empty set! (It is clear that if there is no denominator, theobject is inOK . If

QPaiiQQbii

∈ OK , multiply by∏Qbjj . So we get

∏P aii ⊆

∏Qbjj ⊆

Qj . This means Qj ⊇ Pi for some i. As primes are maximal, Qj = Pi which isnot possible, as we have assumed {Pi} and {Qj} are mutually disjoint).

Now we are in a position to imitate the behaviour in Z.

Lemma 0.24. If A ⊆ B, then there exists an ideal C such that BC = A. And ifBC = A, then A ⊆ B

Proof. If A ⊆ B, writing A and B in terms of prime factors as∏P aii and

∏Qbjj ,

then (∏Qj−bj)

∏P aii ⊆ OK and by the discussion above, this is possible iff each

Q−1j cancels off with some Pi. In other words if B =

∏Qbji , then A =

∏Qaji X

where aj ≥ bj . Hence we can find the ideal C (it is X∏Qaj−bjj ).

The other implication is clear

We say that B|A if A = BC for some ideal C. We can now define the notions ofGreatest Common Divisor and Least common multiple as follows:

• D = GCD(A,B) if D|A, D|B and if C|A,C|B then C|D.

• L = LCM(A,B) if A|L, B|L and if A|X,B|X , then L|X

Theorem 0.25. GCD(A,B) = A+B and LCM(A,B) = A ∩B

Proof. Let D be the GCD. Then D|A which implies A ⊆ D. Similarly B ⊆ D.Therefore A+B ⊆ D. However A+B|A,A+B|B and hence A+B|D whichmeans D ⊆ A+B , which gives us A+B = D.

18

Similarly if L is the LCM, then L ⊆ A, L ⊆ B , therefore L ⊆ A ∩ B. HoweverA|A ∩ B and so does B. Therefore L|A ∩ B, which implies A ∩ B ⊆ L, andtherefore A ∩B = L

Now if A =∏P aii and B =

∏P bii where Pi ranges over all prime ideals and ai

is 0 except for finitely many is and so also for bis. Denote min(ai, bi) by xi andmax(ai, bi) by yi. Then in fact

GCD(A,B) =∏

P xii = S, LCM(A,B) =

∏P yii = T (Check!)

The ideal group

Definition: X ⊂ K is a fractional ideal if it is an OK module such that thereexists m ∈ Z , mX ⊆ OK

Note that an object like P−1 is a fractional ideal as r ∈ OK , x ∈ P−1, thenrxP = x(rP ) ⊆ xP ⊆ OK , so it is an OK module. Also since PP−1 = OK ,P ∩ Z = pZ, then pZP−1 ⊆ OK . In fact, we can exactly say what a fractionalideal looks like as shown by the following:

Lemma 0.26. Every fractional ideal is finitely generated as an OK module

Proof. If M is a fractional ideal, then there exists m ∈ Z such that mM ⊆ OK .Since OK is noetherian , mM is finitely generated, say by x1, x2, . . . xk as a Zmodule and hence as an OK module. Then M is generated by x1

m, . . . xk

m

Theorem 0.27. Any fractional ideal M looks likeQPaiiQQbii

where Pis and Qjs are

prime ideals such that {Pi} ∩ {Qj} = ∅

Proof. Let M be a fractional ideal and m ∈ Z such that mM ⊆ OK . Let mM =∏P aii and (m) =

∏Qbii . Multiplying by (m)−1 on both sides of the expression

for mM , we get M =QPaiiQQbii

and cancelling off the common factors, we get the

disjointedness condition also.

Lemma 0.28. Sum and product of fractional ideals is again a fractional ideal.

19

Proof. If M , N are two fractional ideals such that mM,nN ⊆ OK for somem,n ∈ Z. Then mn(M +N) and mn(MN) sit inside OK and MN,M +N areOK modules. Hence they are fractional ideals.

We would like to now make the set of fractional ideals a group. A candidate for theidentity is clearly OK . The operation is product of fractional ideals and the set offractional ideals , we have shown, is closed under this operation. As our notationsuggests, we take the inverse of a prime ideal P to be P−1 = {x ∈ K|xP ⊆ OK}.If A is any arbitrary ideal, such that A =

∏P aii , then we can define A−1 in two

ways , namely X =∏P−aii or Y = {x ∈ K|xA ⊆ OK}. It turns out that both

are the same. To see this, if x ∈ Y , then xA ⊆ OK , therefore xAX = xOK ⊆ X ,which shows that Y ⊆ X . If x ∈ X , then x =

∑ ∏xi where xi ∈ P−1

i . 14

Therefore xiPi ⊆ OK , which gives us that∏xiA ⊆ OK and therefore X ⊆ Y .

However this is too large a group . So an equivalence relation is introduced so thatthe equivalence classes form a smaller group which is termed as the class group15.The equivalence relation is defined as follows:

M ∼ N if ∃a, b ∈ OK such that aM = bN 16

Let us check that the equivalence classes indeed form a group under the inducedbinary operation.

• Well defined: If M ∼ N and A ∼ B, then MA ∼ NB (as mM =nN, aA = bB, so maMA = nbNB)

• Closure: [A][B] = [AB]

• Associativity: follows from the associativity of the product of fractionalideals

• Identity: [A][OK ] = [AOK ] = [A] = [OK ][A] 17

• Inverse: [A]−1 = [A−1].

Henceforth the class group will be denoted by HK and its cardinality by |HK |.We have said that the group has become smaller. But how small has it become?

14er, Pis may be repeated, So pick an xI ∈ I−1 in each copy of the prime factors of A15The group of equivalence classes is termed a Picard group for other dedekind domains16Transitivity follows by noting that mM=nN, xX = jN, mjM=njN=xnX17as A is an OK submodule and 1 ∈ OK , AOK = A

20

One of the main theorems which shall be proved in subsequent lectures is that theclass group is actually finite!

Let [A] be an equivalence class of a fractional ideal A. Since aA ⊆ OK , thismeans a(A) = 1(aA) and hence A ∼ aA. Thus each equivalence class admitsan integral ideal representative. In fact, let us find that a ∈ Z which makes aAintegral. Let A = B

Cwhere B,C are coprime integral ideals. Let (m) = C ∩ Z.

Since (m) ⊆ C, it means C|(m), so CX = (m). Therefore mA = (m)BC

= BXwhich is integral.

Note that A, an integral ideal, is in the same equivalence class as OK iff A is aprincipal ideal.

( If A = (a), then 1.A = aOK and hence A ∼ OK . If A ∼ OK ,then xA = yOK .Then A = (x)−1yOK . However (x)−1 = (x−1) because if s ∈ (x)−1, thensx ∈ OK and s = (sx)(x−1) which therefore is in (x−1). The otherway is clear.So, we have A = x−1yOK , but 1 ∈ OK , so x−1y ∈ OK which tells us thatA = (x−1y).)

So OK is a PID iff |HK | = 1 (!) which tells us that the cardinality of the classgroup, which shall be called the class number is an important invariant of K.

21

Lecture 5

Since we have begun treating ideals like elements of a group, we now extend thenotion of norm.

Definition: If A is an ideal, then N(A) = |OKA|

And to show that this is really extending the definition, let us prove the following:

Theorem 0.29. If a ∈ OK , then N(aOk) = |N(a)|.

Proof. First let (b) = (a) ∩ Z. Therefore OK(a)

= X is a module over Z(b)

= Y .SinceOK is finitely generated as a Z module, X is a finitely generated Y module.As Y is a finite ring, X is a finite module. So |OK

(a)| does make sense.( In more

bombastic language , OK(a)

is a torsion module of Z. Every element of it is killedby b. Any finitely geneerated abelian group which is a torsion module is finite, asX =

∑Zxi and you can only have combinations of xi , where the coefficients

≤ b.)

Let e1, e2, . . . en be a Z basis of OK . Since X is finite, (a) has the same rank (n)as OK . By lemma 0.12, there exists a basis for (a) of the form xi = ciei + jiwhere j ∈

∑nk=i+1 Zek.

Define map T : OK → (a) which sends ei xi. The matrix will be a lowertriangular one with ci’s on the diagonal. Thus det(T ) = c1c2 . . . cn.

However ae1, ae2, . . . aen is also a basis for (a). Define map S : (a) → (a)which sends xi aei. S is an integer matrix (for {xi} is a Z basis and this mapis invertible as it takes a basis to a basis and the inverse matrix also has integercoefficients as {aei} also is a Z basis. Thus det(S) = ±1.

Now ST (ei) = aei . Thus ST (x) = ax for any x ∈ OK . Therefore det(ST ) =N(a) which means that det(T ) = ±N(a). Thus N(a) = |c1c2 . . . cn|.

|c1c2 . . . cn| is |X| ! In fact X = {[a1e1 + . . . anen]|0 ≤ ai < |ci|} where [x] refersto the equivalence class in which x belongs. (check!) 18

Corollary 0.30. N(A) is finite for any A.

18Use the fact that {xi} is a Z basis of (a).

22

Proof. Pick a ∈ A. We know that N(aOK) = N(a) which is finite. Also A ⊇aOK . So |OK

A| ≤ | OK

aOK|.

23

Lecture 6

Theorem 0.31. Norm is multiplicative.(ie) - If A,B denote ideals in OK , thenN(AB) = N(A)N(B)

Proof. If we prove that N(AB) = N(A)N(B) for A,B coprime ideals andN(P n) = N(P )n for P prime, then because of unique factorisation of idealsinto prime ideals, we are done.

Let A,B be coprime ideals of OK , (ie) A + B = R. Therefore AB = A ∩ B.19.Using chinese remainder theorem, we find that

OKAB' OK

A× OK

B

.

But N(X) = |OKX|. Hence we find that N(AB) = N(A)N(B)

Now, let P denote a prime ideal. N(P n) = |OKPn|. Look at the sequence

OK ⊇ P ⊇ P 2 . . . ⊇ P n.

And by the third isomorphism theorem, OKPn−1 =

OKPn

Pn−1

Pn

. Note that these quantities

are all finite groups as N(P n) is finite. Anyway, from the above we can concludethat

|OKP n| =

n−1∏k=0

| Pk

P k+1|

Claim: |OKP| = | Pk

Pk+1 | for any k ≥ 1

Because of unique factorisation, P k 6= P k+1 for any k ≥ 1. Therefore pick ana ∈ P k which doesn’t belong to P k+1. Therefore P k+1 ⊆ aOK + P k+1 ⊆ P k.Now write aOK + P k+1 in terms of its prime factors. It is not P k+1, 20 and iscontained in P k, so aOK + P k+1 = P k.

We will set up a set map ψ as follows:

19a+ b = 1, if x ∈ A ∩B, then x(1) = xa+ xb ∈ BA+AB = AB20a ∈ aOK + P k+1

24

ψ :OKP→ P k

P k+1which sends x+ P xa+ P k+1

Note that if x + P = y + P , then xa − ya ∈ P.P k = P k+1 and so ψ is welldefined. If ψ(x+ P ) = ψ(y + P ), then xa− ya ∈ P k+1, therefore as a 6∈ P k+1 ,we have x− y ∈ P , so x + P = y + P . So ψ is injective. Since the domain andrange are finite, ψ is surjective as well.

Corollary 0.32. If A 6= (0), is an ideal in OK such that N(A) is prime in Z, thenA is prime in OK .

Theorem 0.33. N(A) = r implies r ∈ A

Proof. N(A) = |OKA| = r. x ∈ OK , then r(x + A) = 0 + A. So rx ∈ A for any

x ∈ OK . Putting x = 1, we get r ∈ A.

Corollary 0.34. If P is non-zero prime ideal of OK , then P contains exactly oneprime integer.

Proof. Let N(P ) = r = pa11 p

a22 . . . pakk . Since r ∈ P and P prime, P contains one

of pis, say p1.

If it contains any other prime, say q, then 1 ∈ P which is a contradiction.

Ramification formula

Let p be a prime in Z. Let us denote by A the ideal pOK . Let n = [K : Q].Therefore OK is a free module of rank n over Z. Now pZ is a maximal ideal in Zand therefore OK

pOKis a vector space over Z

pZ of the same dimension n (Prove! Thisis how you prove rank of free modules is well defined).

ThereforeN(A) = pn. LetA = P e11 P

e22 . . . P en

n be the prime factorisation. There-fore N(A) =

∏N(Pi)

ei . Let fi denote [OKPi

: ZpZ ].

Therefore n =∑eifi.

25

Lecture 7

Theorem 0.35. Fix a real number c > 0. Then there are only finitely many idealsA such that N(A) ≤ c

Proof. Let p1, . . . pr be the finite set of positive prime integers in Z less than orequal to c. If A =

∏P aii , then N(A) =

∏N(Pi)

ai ≤ c. So each of N(Pi) ≤ c.Also N(Pi) ∈ Pi and since Pi is prime, it contains one factor of N(Pi). Since anyprime ideal contains exactly one prime integer, Pi must contain a prime integerwhich is less than c and no other prime.

Claim : There are only finitely many prime ideals containing a prime p

Let (p) = Qa11 . . . Qak

k . If X is a prime ideal which contains p, then X ⊇∏Qaii ,

which means X = Qi for some i.

Thus , there are only finitely many choices for Pis ( pick from set of prime idealswhich contain pi for some i). Also , if X is a prime ideal which contains pi,pi|N(X) and N(X) ≤ c, so there is a maximum power mi so that Xmi+1 doesn’tdivide A21.

So, finally, there are only finitely many ideals of norm ≤ c.

Lemma 0.36. ∃ constantHK called Hurwitz constant, such that given any a ∈ K,there exists a b ∈ OK and an integer t ∈ Z such that |t| ≤ HK and |N(ta−b)| < 1

Proof. Fix an integral basis for OK , say {wi}.

Since {wi} is a basis, let a =∑ciwi where cis are ∈ Q. Let {x} denote the

fractional part of x ∈ R and [x], the greatest integer less than or equal to it. Lookat the following map :

φ : Z→ [0, 1]n, t ({tc1}, {tc2}, . . . , {tcn})

Pick an arbit integer m and divide the interval [0,1] into m equal parts, each oflength 1

m. This divides [0, 1]n into mn cubes. Now if you pick integer t such that

0 ≤ t ≤ mn, then there are mn + 1 choices of t, so by pigeon hole principle,

21if it does, then pmi+1i |N(A) ≤ c which is not possible

26

there exist t1 6= t2 ≤ mn such that φ(t1), φ(t2) belong to the same cube. That is|{t1ci} − {t2ci}| ≤ 1

mfor each i ≤ n.

Put b =∑

([t1ci] − [t2ci])wi. Thus b ∈ OK . Also (t1 − t2)a − b =∑

({t1ci} −{t2ci})wi.

Let t = t1 − t2, so that

|N(ta− b)|= |

∏σi(ta− b)|

≤∏|σi(ta− b)|

=∏|σj(

∑({t1ci} − {t2ci})wi)|

≤∏|∑|σj(({t1ci} − {t2ci})wi)||

≤∏| 1m

∑|σj(wi)||

≤ 1mn

∏|∑|σj(wi)||

Put HK = mn and choose m such that mn >∏|∑|σj(wi)||. Now since |t| =

|t1 − t2| ≤ mn = HK , we are done.

Theorem 0.37. ∃ a constant CK such that for every ideal A ⊆ OK , there is anequivalent ideal B with N(B) ≤ CK .

Proof. Let A be an integral ideal. Pick β ∈ A such that |N(β)| is minimal andnonzero. Given α ∈ A, let a = α

β. So by above lemma, there exists a wα ∈

OK and a tα ∈ Z where |tα| ≤ HK such that |N(tααβ− wα)| < 1. That is

N |( tαα−βwαβ

)| < 1 which implies |N(tαα− wαβ)| < |N(β)|.

Since α, β ∈ A, tαα−wαβ ∈ A, and as β is so picked such that norm is minimal,we have tαα− wαβ = 0

Set m =∏|t|≤HK t. Therefore mA ⊆ (β), which means (β)X = (m)A for some

integral ideal X . This is equivalent to saying that X ∼ A.

Since (β)X = (m)A, N(β)N(X) = N(m)N(A). As β ∈ A, N(β) ≥ N(A),which therefore means N(X) ≤ N(m).

Set CK = N(m) and we are done.

And lo!

27

Corollary 0.38. Class number is finite

Proof. Obvious ! 22

22Since there are only finitely many ideals with norm ≤ CK

28

Lecture 8

Theorem 0.39. Let K = Q(θ) and OK = Z[θ]. Let f be the minimal polynomialof θ over Z[x]. If p is a prime in Z such that f ∼= f e11 f

e22 . . . f ekk (mod p) where

fis are irreducible in Fp[x], then pOK = Qe11 Q

e22 . . . Qek

k where Qi = (p, fi(θ)) isprime and N(Qi) = pxi where xi denotes the degree of fi.

Proof. Z[x](p,fi(x))

= Fp[x]fi(x)

which is a field. Look at ψ : Z[x](p,fi(x))

→ Z[θ](p,fi(θ))

whichsends x θ. (p, fi(x)) ⊆ Kernel ψ. But we know (p, fi(x)) is maximal. Thusit is the kernel. Therefore Z[x]

(p,fi(x))∼= Z[θ]

(p,fi(θ))and as the former is a field, we have

shown that Qi is maximal.

f(x)− f e11 (x) . . . f ekk (x) ∈ pZ[x]. Thus, as f(θ) = 0, f e11 (θ) . . . f ekk (θ) ∈ pOK =pZ[θ]. If Q is any prime ideal which contains pOK , then p ∈ Q and fi ∈ Qfor some i. Thus Qi ⊆ Q for some i. However primes are maximal and henceQ = Qi.

Let pOK = Qd11 . . . Qdk

k and let gi = [OKQi

: ZpZ ]. Thus n =

∑digi which is also

degree of f 23. But we know degree of f =∑eixi. But gi = xi (use OK

Qi= Fp[x]

fi).

Also ei ≤ di for all i as∏Qeii ⊆

∏f eii (θ) + pOK = Qd1

1 . . . Qdkk . Thus, we have

di = ei by comparing n =∑xiei =

∑digi.

23n=f

29

Lecture 9

Let p ∈ Z be a prime and let pOK = P e11 P

e22 . . . P ek

k .

• p is said to ramify in OK if ei > 1 for some 1 ≤ i ≤ k. The integer ei iscalled the ramification index of Pi.

• p is said to split if ei = 1∀i

• p is said to remain inert if pOK is a prime ideal in OK

D−1

D−1 = {x ∈ K|Tr(xOK) ⊆ Z}

Lemma 0.40. D−1 is a fractional ideal.

Proof. If x ∈ OK and y ∈ D−1, then Tr(xyOK) = Tr(yxOK) ⊆ Tr(yOK) ⊆Z. Hence xy ∈ D−1.

Let w1, w2, . . . wn be an integral basis of OK and {w∗i } be the dual basis, soTr(wiw

∗j ) = δij .

Therefore Zw∗1 + Zw∗2 + . . .+ Zw∗n ⊆ D−1

Claim: Zw∗1 + . . .Zw∗n = D−1

Let x ∈ D−1. Since {w∗i } is a dual basis, x =∑aiw

∗i , where ai ∈ Q.Now

Tr(xwj) = Tr((∑aiw

∗i )wj) = aj ∈ Z24. Therefore claim is proved.

Now for each w∗i , ∃bi ∈ Z such that biw∗i ∈ OK . Putting b =∏bi, we find that

bD−1 ⊆ OK

D−1 is termed as the different of OK .

Denote by D , the inverse of D−1. So DD−1 = OK . As 1 ∈ D−1, D ⊆ DD−1 ⊆OK . And as shown earlier, any fractional ideal contained in OK is an integralideal and so D is an integral ideal.

24as Tr(xOK) ⊆ Z

30

Theorem 0.41. N(D) = |∆K |

Proof. SinceD−1 is fractional, let a ∈ Z such that aD−1 ⊆ OK . As shown earlierD−1 = Zw∗1 + . . .Zw∗n. Therefore aD−1 = Zaw∗1 + . . .Zaw∗n. Let

aw∗i =∑

aijwj, aij ∈ Q =⇒ w∗i =∑ aij

awj

wi =∑

bijw∗j , bij ∈ Q

So we have that the matrix (bij)−1 = (

aija

)

Recall that ∆K = det(Tr(wiwj)) and Tr(wiwj) = Tr((∑bikw

∗k)wj) = bij .

Thus ∆K = det(bij)

Claim: If A ⊆ B ⊆ OK are ideals, then D(A) = [B : A]2D(B)

Therefore D(aD−1) = D(aw∗1, aw∗2, . . . aw

∗n) = N(aD−1)2∆K , but we already

know the base change matrix for aw∗1, . . . aw∗n tow1, w2, . . . wn. ThusD(aD−1) =

D(aw∗1, aw∗2, . . . aw

∗n) = det(aij)

2∆K .

Equating the two expressions for D(aD−1) , we get |det(aij)| = N(aD−1).

N(aD−1) = N(a)N(D−1) = anN(D−1) = |det(aij)|=⇒ N(D)−1 = |det(aij

a)|

=⇒ N(D) = |det(bij)| = |∆K |

Theorem 0.42. p be a prime in Z and P , prime in OK and D−1 , the different ofK. If P e|(p), then P e−1|D.

Proof. Let pOK = P eA. Let x ∈ PA, so let x =∑piai where pi ∈ P and ai ∈

A. So xm for some suitable power lies in P eA = (p). Therefore Tr(xpm) ∈ pZ.(The reason why we take pm will be clear from discussion below)

Tr(xpm

) ∈ pZ=⇒

∑σi(x

pm) ∈ pZ=⇒ (

∑σi(x))p

m ∈ pZ25

=⇒ Tr(x)pm ∈ pZ

=⇒ Tr(x) ∈ pZ26

25The reason26since Tr(x) is already an integer as x is an algebraic integer

31

=⇒ Tr(p−1PA) ⊆ Z27

=⇒ Tr(p−1PAOK) ⊆ Z=⇒ p−1PA ⊆ D−1

=⇒ D ⊆ pP−1A−1 = P e−1

=⇒ P e−1|D

Corollary 0.43. If p is ramified , then p|∆K

Proof. Since p ramifies, it means ∃P prime, such that P e|(p) where e ≥ 2. So byabove theorem, P e−1|D and therefore P |D.

Let D = PA where A is an integral ideal.

N(D) = |∆K | = N(P )N(A) and N(P ) = pfp . Therefore p|∆K

The converse (a theorem of dedekind) is also true!

Theorem 0.44. Let K = Q(a) and let OK = Z[a]. If f = c0 + c1x1 +

. . . cn−1xn−1 + xn is the minimal polynomial for a over Q, then D = f ′(a).

Proof. Let f(x) = (x − a)g(x) where g(x) = bn−1xn−1 + . . . b0. Note that f

has distinct roots, say a = a1, a2, . . . an. Since a ∈ OK , all ai ∈ OK . Thusg(x) ∈ OK [x]. Also f ′(ai) 6= 0 as f has distinct roots.

Define for 0 ≤ r ≤ n− 1, the polynomial gr(x) =∑n

i=1f(x)x−ai

arif ′(ai)

− xr. Observe

that f(x)x−ai evaluated at aj is δijf ′(aj) and conclude that gr(ai) = 0∀1 ≤ i ≤ n.

Thus gr ∼= 0 as degree of gr ≤ n− 1.

For any polynomial h(x) = h0+h1x+. . . hnxn ∈ K[x], define Tr(h) = Tr(h0)+

Tr(h1)x1 + . . . T r(hn)xn which is a polynomial with rational coefficients.

Tr(f(x)ar

(x− a)f ′(a)) = Tr(

(b0 + . . . bn−1xn−1)(ar)

f ′(a)) =

n−1∑i=0

Tr(bia

r

f ′(a))xi

.

Also note that Tr( f(x)ar

(x−a)f ′(a)) =∑ f(x)ari

(x−ai)f ′(ai)28 = xr − gr(x) = xr.This shows

that Tr( biar

f ′(a)) = δir which means that we have found a dual basis for 1, a, a2 . . . an−1.

27Tr is a Q linear map28Not clear

32

We have already shown that if w1, w2, . . . wn is an integral basis for OK , thenD−1 = Zw∗1 + . . .Zw∗n where {w∗i } is the dual basis for {wi}. Thus D−1 =

1f ′(a)

∑Zbi.

Now, to wrap it up, we will show that∑

Zbi = OK and hence D = f ′(a). Notethat since bi ∈ OK , one way is clear. For the reverse, bn−1 = 1 as f is monic. ThusZ ⊆

∑Zbi. Now cn−1 = bn−2 − abn−1 = bn−2 − a. Therefore a = bn−2 − cn−1.

Note that cn−1 ∈ Z ⊆ Zbn−1. Thus a ∈ Zbn−2 + Zbn−1. Continuing in this vein,we find that ai ∈

∑n−1k=n−i−1 Zbk. ThusOK = Z[a] ∈

∑n−1k=0 Zbk and we are done.

33

Bits and pieces

• The class group can be thought of as a free group generated by prime idealsof OK (where principal ideals become trivial).

• K = Q(√−5) has OK = Z[

√−5]. The class group has size 2 where the

nontrivial element is [P ] where P = {3x+ y(2 +√−5)|x, y ∈ OK}.

• If K = Q(θ), then

D(1, θ, . . . , θn−1) = (−1)n(n−1)

2

∏i 6=j

(σi(θ)− σj(θ)) = (−1)n(n−1)

2 N(f ′(θ))

• IfR is a domain, thenR is a dedekind domain iff every ideal can be uniquelyfactorised into prime ideals.

• Another proof for why OK is noetherian is to use the fact that |OKA| is finite

for all ideals A. So if there is an ascending chain A1 ( A2 . . ., then |OKA1| >

|OKA2| . . . , which has to terminate as |OK

A1| is finite.

• Discriminant is 0 or 1 mod 4. To see that, write det((σi(wj))) = P − Nwhere P refers to the terms with positive sign in front of them and thenegative parts. Note that ∆K = (P − N)2 = (P + N)2 − 4PN . SinceP +N,PN are symmetric functions, they are fixed by all embeddings of Kand hence belong to Q 29. Also P +N,PN ∈ OK . Thus they must belongto Z , and hence ∆K = (P +N)2 − 4PM ∼= 0 or 1 mod (4).

•

Proposition 0.45. A dedekind domain with finitely many prime ideals is aprincipal ideal domain.

Proof. Let the non-zero prime ideals be P1, . . . Pr. Enough to show theseare principal. To show P1 is principal we use chinese remainder theorem.

R

P 21P2 . . . Pr

∼=R

P 21

× R

P2

. . .× R

Pr

Choose c ∈ P1 which is not in P 21 . (This you can do by unique factorisation

of ideals into prime ideals. Thus P 21 6= P1), and by chinese remainder

theorem, find an x ∈ R which maps to (c, 1, 1, . . . 1). x ∈ P1. No Pi fori ≥ 2 divides (x) as x ∼= 1 mod Pi for i ≥ 2. Therefore (x) = P k

1 and notethat x 6∈ P 2

1 . Thus (x) = P1.29why ? Is extension galois ?

34

• The chinese remainder theorem is used frequently to prove things like ifI and J are given ideals, then ∃L such that IL is principal and J, L arecomaximal. The idea is to find an a ∈ I such that a 6∈ IP for any P , aprime ideal which divides J .

Now (a) ⊆ I , therefore IS = (a). If P |S for some P which also divides J ,then S = PX and IS = IPX = (a). Therefore (a) ∈ IP , a contradiction.

This a can be found by writing I = P a11 . . . P ak

k and J = P b11 . . . P br

r Qc11 . . . Qcs

s .By chinese remainder theorem, choose an x such that x ∼= 1 mod Qi s andx ∼= xj mod P aj+1

j where xj ∈ Pajj but not in P aj+1

j .

35