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PMATH 348 LECTURES

ROSS WILLARDUNIVERSITY OF WATERLOO

WINTER 2014

Contents

1. January 6 – Sylow’s 1st Theorem 32. January 8 – Sylow’s 2nd Theorem 53. January 10 – Sylow’s 3rd Theorem 74. January 13 – Irreducibility tests 95. January 15 – Fields of characteristic p 116. January 17 – Fundamental construction 137. January 20 – Multiple roots, derivative test 158. January 22 – Applications of derivative test 179. January 24 – Minimal polynomial, degree 1910. January 27 – Finite simple extensions 2111. January 29 – Product Theorem 2312. January 31 – Applications of Product Theorem 2513. February 3 – Cyclotomic polynomials 2714. February 5 – Splitting fields 3015. February 7 – Newfoundland Theorems 3216. February 10 – Uniqueness of splitting fields 3417. February 12 – Existence of finite fields; automorphisms 3618. Feburary 14 – Continuation of example 3819. February 24 – Counting automorphisms 4020. February 26 – Counting continued 4221. March 3 – Characterizing Aut(K/F ) = [K : F ] 4522. March 5 – The Galois correspondence: example 4823. March 7 – Artin’s Theorem 5024. March 10 – Fundamental Theorem, part 1 5225. March 12 – Fundamental Theorem, part 2 5426. March 14 – Finite fields revisited 5627. March 17 – Galois groups; discriminant test 5828. March 19 – Formula for the discriminant 6129. March 21 – Sick-Day Propositions 64

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2 R. WILLARD

30. March 24 – Solving cubic equations by radicals 6731. March 26 – Radical extensions, solvable groups 6932. March 28 – Strong radical extensions; Galois’ 1st Theorem 7133. March 31 – Expressing primitive roots 7434. April 2 – Galois’ Main Theorem 7635. April 4 – Fundamental Theorem of Algebra 80

PMATH 348 LECTURES 3

1. January 6 – Sylow’s 1st Theorem

Galois theory is the study of:

• polynomials and their roots

viewed through

• the fields generated by the roots• field automorphisms.

The theory uses both ring theory and group theory. The course builds on PMATH 347.

Review of group theoryLet G be a finite group. Recall:G acts on itself by conjugation: each g ∈ G “acts on” a ∈ G by g · a := gag−1, and

this satisfies

(1) g · (h · a) = (gh) · a for all g, h ∈ G and a ∈ G.(2) 1 · a = a for all a ∈ G.

Intuitively, each g “moves” the elements of G around. (Picture, with arrows.)Given a ∈ G, the stabilizer of a is the set {g ∈ G : g · a = a} of elements which do

not move a. This = {g : gag−1 = a} = {g : ga = ag} = CG(a), the centralizer of a.Left cosets of CG(a) move a to distinct conjugates of a.The orbit of a is the set O(a) of distinct conjugates of a; O(a) is also called the

conjugacy class of a.

• |O(a)| = [G : CG(a)]• Hence |O(a)| = 1 iff CG(a) = G iff a ∈ Z(G).

Draw picture of partition; group the 1-element orbits together; get the class equation:

|G| = |Z(G)|+r∑i=1

[G : CG(ai)]

where a1, . . . , ar represent the nontrivial conjugacy classes.Other facts you should know:

(1) Cauchy’s theorem (ask someone to state it): we only need it for abeliangroups.

(2) Fact: if H ≤ Z(G) then H CG.

§4.5 – Sylow’s Theorems

Definition. Let p be prime.

(1) A p-group is any group of order pβ for some β ≥ 0.(2) A p-subgroup of G is the obvious thing.(3) Suppose p divides |G|; say |G| = pαm where p6 |m. A p-subgroup of order pα

is called a Sylow p-subgroup.

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Sylow’s 1st Theorem (Thm. 4.18(1)). Suppose G is a finite group. For everyprime p dividing |G|, G has a Sylow p-subgroup.

Proof. By induction on |G|. If |G| = 1 there is nothing to prove. Suppose |G| > 1and the theorem is true for all smaller groups.

Case 1: p divides |Z(G)|.Z(G) is abelian, so by Cauchy’s theorem, Z(G) has an element a of order p. Let

N = 〈a〉. Then N CG, so we can form G := G/N of order pα−1m. By induction, Ghas a subgroup P of order pα−1; it consists of pα−1 cosets of N . The union of thesecosets is a subgroup of G (exercise) of order pα.

Case 2: p6 | |Z(G)|.Consider the class equation:

|G| = |Z(G)|+r∑i=1

[G : CG(ai)]

where a1, . . . , ar represent the nontrivial conjugacy classes. If p divided each [G :CG(ai)], this would imply p | |Z(G)|, which is false. Hence ∃ ai with p6 | [G : CG(ai)].This forces pα | |CG(ai)|. As [G : CG(ai)] > 1, we have CG(ai) < G, so by inductionCG(ai) has a subgroup of order pα, which is also a subgroup of G. �

Example. |S4| = 24 = 233. A Sylow 2-subgroup must have order 8. By Sylow’s 1sttheorem, S4 has a subgroup of order 8. In fact it has 3 such subgroups: one is

{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3 4), (1 4 3 2), (1 3), (2 4)}

Corollary (Cauchy’s theorem). If G is a finite group, p is prime, and p divides |G|,then G has an element of order p.

Proof. Say |G| = pαm. Let P be a Sylow p-subgroup. Pick any a ∈ P \ {1}. Then

|a| = pk for some 1 ≤ k ≤ α. Let b = apk−1

; then b 6= 1, but bp = (apk−1

)p = apk

= 1,so |b| = p. �

If time: in fact, if |G| = pαm with p6 |m, then G has subgroups of order pβ for all0 ≤ β ≤ α. (You’ll prove this on the first assigment.)


2. January 8 – Sylow’s 2nd Theorem

Conjugate subgroups.

• Suppose H ≤ G and g ∈ G. Then gHg−1 ≤ G and H ∼= gHg−1. gHg−1 iscalled a conjugate subgroup of H.• How many conjugates does H have?

Let S(G) be the set of all subgroups of G. G acts on S(G) by conjugation:g ·H = gHg−1. The orbits of this action are the conjugacy classes. So I am askingfor a formula for |O(H)|.

Given any action of a finite group G on a set A, the size of the orbit containinga ∈ A is equal to the index of the stabilizer [G : Ga].

Thus in our example, the number of conjugates of H is equal to [G : GH ] where

GH = {g ∈ G : g ·H = H}= {g ∈ G : gHg−1 = H}= NG(H).

I.e.,

Proposition (DF Prop. 4.6). For any H ≤ G, the number of distinct conjugates ofH in G is [G : NG(H)].

Example. In S4 let H = 〈(1 2)(3 4), (1 2 3 4)〉, the Sylow 2-subgroup from Monday.Then H ≤ NS4(H) ≤ S4, so NS4(H) = H or S4. Randomly picking (1 2) ∈ S4 and(1 2 3 4) ∈ H,

(1 2)(1 2 3 4)(1 2)−1 = (2 1 3 4) 6∈ H.Thus NS4(H) 6= S4, so H has [S4 : NS4(H)] = [S4 : H] = 3 conjugates.

I said on Monday that S4 has three Sylow 2-subgroups. By the above calculation,they are all conjugate to each other. This is true in general:

Sylow’s 2nd Theorem (DF Thm. 4.18(2)). Let G be finite, p prime, p | |G|. Anytwo Sylow p-subgroups of G are conjugate.

Proof. The following proof is not in the text. Let |G| = pαm with p6 |m. Let P,Q betwo Sylow p-subgroups, so |P | = |Q| = pα. Let A be the set of left cosets of P .

Consider the action of Q on A by left multiplication: i.e., if h ∈ Q and gP ∈ A(g ∈ G), then h · gP := hgP . This is a group action.

Thus A is partitioned into orbits by this action (draw picture). Let’s group the1-element orbits together in a set X; let O1, . . . ,Ok be the nontrivial orbits. Thenobviously

|A| = |X|+k∑i=1

|Oi|.

We have:

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• |A| = [G : P ] = m.• For each i, if we pick giP ∈ Oi then |Oi| = [Q : QgiP ], where QgiP is the

stabilizer of giP under this action.– Thus |Oi| = pβi for some 1 ≤ βi ≤ α.

• Hence p divides∑

i |Oi|.• Hence p6 | |X|.• In particular, X 6= ∅.

This proves that there exists a 1-element orbit {gP}. Hence

• ∀h ∈ Q, hgP = gP• So ∀h ∈ Q, hg ∈ gP .• So Qg ⊆ gP .• So Q ⊆ gPg−1.• So Q = gPg−1 as |Q| = |gPg−1| = pα.

�

Corollary. G finite, p prime, p | |G|. Any two Sylow p-subgroups are isomorphic.

Combining Sylow’s 2nd theorem with our formula for the number of conjugatesubgroups gives:

Corollary. G finite, p prime. Let P be a Sylow p-subgroup. Then the number np ofdistinct Sylow p-subgroups is [G : NG(P )], so is a divisor of |G|.

The last result is a technical lemma needed on Friday.

Corollary (cf. DF Lm. 4.19). G finite, p prime, p | |G|. Suppose P,Q are Sylowp-subgroups of G. Then P ⊆ NG(Q) iff P = Q.

Proof. (⇐) Clearly P ⊆ NG(P ).(⇒). First observe that if P ≤ H ≤ G, then P is also a Sylow p-subgroup of H.Let H := NG(Q). Obviously Q ⊆ H. Now assume that P ⊆ H. By the above

observation, P and Q are both Sylow p-subgroups of H. But QCH, so Q has onlyone conjugate as a subgroup of H. Thus P = Q by Sylow’s 2nd theorem. �


3. January 10 – Sylow’s 3rd Theorem

Sylow’s 3rd Theorem (DF Thm. 4.18(3)). Suppose G is finite, p prime, p | |G|.Recall that np is the number of Sylow p-subgroups. Then np ≡ 1 (mod p).

Proof. Let A be the set of all Sylow p-subgroups of G, and pick P ∈ A.P acts on A by conjugation. As usual, A is partitioned into orbits. Let X be the

union of the 1-element orbits. Let O(Q1), . . . ,O(Qk) be the nontrivial orbits. Thus

|A| = |X|+k∑i=1

|O(Qi)|.

We have

• |A| = np.• Each |O(Qi)| = [P : PQi

], so is a power of p.• Thus p |

∑i |O(Qi)|.

• Thus np ≡ |X| (mod p).

What are the 1-element orbits? Given Q ∈ A, first note that

PQ = {h ∈ P : hQh−1 = Q}= P ∩NG(Q).

Hence

{Q} is a 1-element orbit ⇐⇒ |O(Q)| = 1

⇐⇒ [P : PQ] = 1

⇐⇒ PQ = P

⇐⇒ P ∩NG(Q) = P

⇐⇒ P ⊆ NG(Q)

⇐⇒ P = Q by Weds Corollary.

I.e., there is exactly one 1-element orbit, {P}. �

Summary: Let |G| = pαm with p6 |m. Let P be a Sylow p-subgroup and let np = thenumber of Sylow p-subgroups.

• np = [G : NG(P )], so np | |G|.• np ≡ 1 (mod p).• Also P CG iff NG(P ) = G iff np = 1.

Examples

(1) Consider S5. |S5| = 5! = 120 = 5 · 24. What is n5?• I.e., how many subgroups of order 5 does S5 have?• n5|120 = 5 · 24 and n5 ≡ 1 (mod 5); hence n5 | 24; hence n5 = 1 or 6.

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• Each Sylow 5-subgroup contains 4 5-cycles. There are more than 4 5-cycles; hence n5 = 6.• This also shows that S5 has 6 · 4 = 24 5-cycles.

(2) Let |G| = 15. Let P be a Sylow 3-subgroup and Q be a Sylow 5-subgroup;thus |P | = 3 and |Q| = 5. P,Q are both cyclic, say P = 〈a〉 and Q = 〈b〉.• n3 | 15 and n3 ≡ 1 (mod 3), so n3 = 1. Hence P CG.• n5 | 15 and n5 ≡ 1 (mod 5), so n5 = 1. Hence QCG.• Clearly P ∩Q = {1}.• It follows that every element of P commutes with every element of Q

(exercise). In particular, ab = ba.• It follows that G is abelian, so is cyclic.

(3) A group is simple if it has no normal subgroups other than {1} and itself. Iwill prove that no group of order 12 is simple.

Suppose |G| = 12. Let P be a Sylow 2-subgroup and Q a Sylow 3-subgroup,so |P | = 4 and |Q| = 3. I will show that P CG or QCG.• n2 | 12 and n2 ≡ 1 (mod 2); hence n2 = 1 or 3.• n3 | 12 and n3 ≡ 1 (mod 3); hence n3 = 1 of 4.

Suppose Q 6C G. Then n3 = 4, so G has 4 subgroups Q = Q1, Q2, Q3, Q4 oforder 3. Note that Qi ∩Qj = {1} for i 6= j. Thus G has 8 elements of order3. This leaves only 4 elements from which to build subgroups of order 4; son2 = 1, so P CG.

Question: |A4| = 12; which alternative holds for A4?(4) If time: show that every group of order 30 has a subgroup of order 15 (which

must then be normal).• Let P,Q ≤ G with |P | = 3 and |Q| = 5.• P ∩Q = {1}, so |PQ| = 15.• If P CG or QCG, then PQ ≤ G and we’re done.

So assume P,Q 6C Q. Thus n3 = 10 and n5 = 6. This would imply 20elements of order 3 and 24 elements of order 5, impossible.


4. January 13 – Irreducibility tests

Review of polynomial rings

Definition. A field is a commutative ring with 1 in which every a 6= 0 has a multi-plicative inverse. Notation: F,E,K,L,M .

Examples: R, Q, C, Fp = Z/pZ. Any R/I where R is a commutative ring with 1 andI is a maximal ideal (DF Prp. 7.12).

Definition. The characteristic of a field F is the least n > 0 such that

1 + 1 + · · ·+ 1︸︷︷︸n

= 0

if such n exists, and is 0 otherwise.

Subfield. Notation: E ≤ F .

Fact: Every field F has a unique smallest subfield F0, called the prime subfield ofF .

(1) F0 is “generated” by 1.(2) It is isomorphic to Q if F has characteristic 0, and is isomorphic to Fp if F

has characteristic p.

Definition. F [x], F [x, y], etc.

Theorem (DF Cor. 9.4). Let F be a field. Then F [x] is a PID and hence a UFD.

Corollary (DF §§8.2, 8.3). Hence irreducible elements in F [x] are prime and gener-ate maximal ideals.

In general, it is hard to determine when a polynomial f(x) ∈ F [x] is irreducible.For polynomials of degree 2 or 3 it is easy:

Proposition (DF Prp. 9.10). If F is a field, f(x) ∈ F [x], and deg(f(x)) = 2 or 3,then f(x) is irreducible in F [x] iff f(x) has no root in F .

The remainder of today’s lecture gives some techniques that can help to decidewhether polynomials in Q[x] are irreducible.

Rational Roots Test (DF Prp. 9.11)

Lemma. Given a prime p, the map Z[x]→ Fp[x] which “reduces coefficients (mod p)”is a ring homomorphism.

Gauss’s Lemma (DF Prp. 9.5). Suppose f(x) ∈ Z[x], deg(f) ≥ 2. If there ex-ists a nontrivial factorization f(x) = g(x)h(x) in Q[x], then there exists a similar

factorization f(x) = g1(x)h1(x) in Z[x].

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Proof. By multiplying f(x) = g(x)h(x) by a large enough integer d ≥ 1 we getdf(x) = a(x)b(x) with a(x), b(x) ∈ Z[x]. If d = 1 we’re done. Else write d =p1p2 . . . , pn. Reducing mod p1 gives

0 = df(x) = a(x) · b(x).

Hence a(x) or b(x) is 0; say the first. Thus replace d by d′/p1 and a(x) by a′(x) =a(x)/p. Repeat. �

Proposition (cf. DF Prp. 9.12). Suppose f(x) ∈ Z[x] with degree n ≥ 2 and leading

coefficient an. If p is prime, p6 | an, and the image f(x) of f(x) under the “reductionmod p” map is irreducible in Fp[x], then f(x) is irreducible Q[x].

Proof. Write f(x) = anxn + · · · . Suppose f(x) were reducible in Q[x]. Then f(x)

has a nontrivial factorization f(x) = g(x)h(x) in Z[x] with deg(g) = k ≥ 1 anddeg(h) = ` ≥ 1.

Applying the “mod p” reduction homomorphism, we get

anxn + · · · = f(x) = g(x)h(x) = g(x) · h(x)

in Fp[x]. deg(f(x)) = n, so deg(g(x)) = k and deg(h(x) = `, contradicting the

irreduciblity of f(x). �

Eisenstein’s Criterion (cf. DF Cor. 9.14). Let p ∈ Z be a prime. Let f(x) =anx

n + an−1xn−1 + · · · + a1x + a0 be a polynomial in Z[x] of degree n ≥ 2. Suppose

p divides ai for all i < n, p6 | an, and p2 6 | a0. Then f(x) is irreducible in Q[x].

Proof. Suppose f(x) were reducible in Q[x]. Then f(x) has a nontrivial factorizationf(x) = g(x)h(x) in Z[x]. Write

g(x) = bkxk + bk−1x

k−1 + · · ·+ b1x+ b0

h(x) = c`x` + c`−1x

`−1 + · · ·+ c1x+ c0.

Applying the “mod p” reduction homomorphism, we get

anxn = f(x) = g(x)h(x) = g(x) · h(x), an 6= 0

in Fp[x]. This forces g(x) = bkxk and h(x) = c`x

` (by unique factorization). In

particular, b0 = c0 = 0. But then p2|a0, contradiction. �


5. January 15 – Fields of characteristic p

Proposition (DF Prp. 9.17). F a field, f(x) ∈ F [x], deg(f) = n ≥ 1 ⇒ f(x) has atmost n roots in F .

Reason: if α ∈ F is a root, then x− α is an irreducible factor of f(x) in F [x]. ByUF, f(x) can’t have more than n irreducible factors of degree 1.

Today: explore two consequences of this Proposition.

Suppose R is a commutative ring, a, b ∈ R, and p is prime. Then

(a+ b)p = ap +

(p

1

)ap−1b+

(p

2

)ap−2b2 + · · ·+

(p

p− 1

)abp−1 + bp.

For 1 ≤ i ≤ p− 1, (p

i

)=

p!

i!(p− i)!=p · (p− 1) · · · (p− i+ 1)

i · (i− 1) · · · 1.

p divides the numberator but not the denominator, so p |(pi

).

Now suppose ch(R) = p. Then

(a+ b)p = ap + 0 · ap−1b+ 0 · ap−2b2 + · · ·+ bp = ap + bp.

This is called the Freshman’s Dream.

Now suppose F is a field with ch(F ) = p. The function ϕ : F → F defined byϕ(x) = xp is called the Frobenius map (for F ).

Proposition (DF Prp. 13.35). For any field F of characteristic p, the Frobenius mapis an isomorphism from F to the subfield {xp : x ∈ F}.

Proof. First show ϕ is a ring homomorphism. We have

ϕ(x+ y) = (x+ y)p = xp + yp = ϕ(x) + ϕ(y)

ϕ(xy) = (xy)p = xpyp = ϕ(x)ϕ(y).

Thus ϕ is an endomorphism with range {xp : x ∈ F}, which must then be a subringof F . The range contains 1 and is closed under inverses, since xp 6= 0 =⇒ (xp)−1 =(x−1)p ∈ ran(ϕ), so is a subfield of F . Finally, ker(ϕ) 6= F (as 1 6∈ ker(ϕ)) soker(ϕ) = {0}, so ϕ is injective. �

Corollary. If F is a field, ch(F ) = p, then every element of F has at most one p-throot in F .

Now suppose F is a finite field with ch(F ) = p. Then ϕ must also be onto. Thus:

Corollary (DF Cor. 13.36). If F is a finite field and ch(F ) = p, then every elementof F has a unique p-th root in F .

12 R. WILLARD

Note that in this case ϕ permutes the elements of F . WLOG Fp ≤ F . Recall(Fermat) that ap = a for all a ∈ Fp. Thus the elements of Fp are fixed by ϕ.Equivalently, they are roots of xp−x. Since this polynomial can have at most p rootsin F , none of the elements of F \ Fp are fixed by ϕ.

Definition. If F is a field, then F× denotes the multiplicative group of units of F ,i.e., F \ {0}.

The next theorem refers to subgroups of F×. An example is the group

S1 = {z ∈ C : |z| = 1} ≤ C×.

Proposition (DF Prp. 9.18). Let F be a field. Any finite subgroup of F× is cyclic.

Proof. Let H ≤ F× be finite and assume that F is not cyclic. H is finite abelian;hence by the fundamental theorem,

H ∼= C1 × C2 × · · · × Ckwhere each Ci = 〈ci〉 is cyclic of order ni, and nk |nk−1 | · · · |n2 |n1. Also k ≥ 2. Letn = nk.

Each Ci contains a subgroup Di isomorphic to Ck. Every element of D1 × D2 ×· · · × Dk satisfies xn = 1. There are nk such elements. But F cannot contain morethan n roots of xn − 1, contradiction. �


6. January 17 – Fundamental construction

Section 13.1: Field extensions

Definition. A field extension is a pair of fields (F,K) with F ≤ K. Notation:K/F . (Warning: NOT a quotient.)

K is called an extension of F .

Main example: C/Q.

Given K/F , our main activity will be to consider f(x) ∈ F [x] and its roots in K.We will be particularly interested in the situation where K has “enough” roots off(x) so that f(x) factors completely in K[x].

Definition. Given K/F and f(x) ∈ F [x], we say that f(x) splits in K[x] if f(x)factors as a product of linear factors in K[x].

Example: x3 − 2 is irreducible in Q[x] (e.g., by Eisenstein), but splits in C[x]:

x3 − 2 = (x− 3√

2)(x− ω 3√

2)(x− ω2 3√

2), where ω =−1 + i

√3

2.

In fact, every f(x) ∈ Q[x] splits in C[x] (by the Fundamental Theorem of Algebra)

Suppose we replace Q with Fp. Is there an analogue for C?

We start with a strange construction. Let F be any field, let p(x) ∈ F [x] beirreducible, n = deg(p(x)), and let I = (p(x)). I is a maximal ideal of F [x], so thequotient ring K = F [x]/I is a field. The elements of K are the cosets a(x) + I of I.Note that:

(1) Every coset contains a polyomial r(x) of degree < n. (Remainder)(2) Every coset contains exactly one polynomial of degree < n.(3) Hence the elements of K are in 1-1 correspondence with Rn := {r(x) ∈ F [x] :

deg(r(x)) < n}.(4) Note that F ⊆ Rn. Let F = {a+ I : a ∈ F}. Then F ∼= F ≤ K.(5) If we identify F with F , then we can write F ≤ K.(6) This means we can evaluate polynomials f(x) ∈ F [x] at elements of K.

For example, suppose f(x) = x2 + ax+ b (a, b ∈ F ). Let θ = x+ I ∈ K.

f(θ) = (x+ I)(x+ I) + a(x+ I) + b

≡ (x+ I)(x+ I) + (a+ I)(x+ I) + (b+ I) via the identification

= (x2 + I) + (ax+ I) + (b+ I)

= (x2 + ax+ b) + I

= f(x) + I

= r(x) + I where r(x) is the remainder when f(x) is divided by p(x).

14 R. WILLARD

This works for any f(x) ∈ F [x]:

Lemma. With F, p(x), K, θ defined as above: for any f(x), g(x) ∈ F [x],

(1) f(θ) = f(x) + I.(2) f(θ) = r(x) + I where r(x) is the remainder when f(x) is divided by p(x).(3) f(θ) = g(θ) iff f(x) ≡ g(x) (mod I), i.e., iff p(x)|(f(x)− g(x)).

We can now prove:

Theorem (DF Thm. 13.3). Given any field F and f(x) ∈ F [x] with deg(f) ≥ 1,there exists K/F such that f(x) has at least one root in K.

Proof. Consider the factorization of f(x) in F [x]. Choose an irreducible factor p(x) ∈F [x]. It will suffice to find K/F such that p(x) has a root in K. Let I = (p(x)) andK = F [x]/I. We can view K as an extension of F . Let θ = x + I ∈ K. Thenp(θ) = p(x) + I = I, which is the zero element of K, so θ is a root of p(x) in K. �

By iterating, we can get

Corollary. Given any field F and f(x) ∈ F [x] with deg(f) ≥ 1, there exists K/Fsuch that f(x) splits in K[x].

Proof. By induction on deg(f). If deg(f) = 1 then f(x) already splits in F [x].Suppose deg(f) > 1; let p(x) be an irreducible factor. By the previous theorem, thereexists E/F such that p(x) has at least one root c ∈ K. Thus f(x) = (x − c)g(x)where g(x) ∈ E[x] and deg(g) = n − 1. By induction, there exists K/E such thatg(x) splits in K[x]. Then K is an extension of F and f(x) also splits in K[x]. �

Indeed, we can prove the following.

Corollary. Given any field F and any countable set S ⊆ F [x], there exists K/F suchthat every f(x) splits in K[x].

Since Fp[x] is countable, we get

Corollary. For each prime p there exists K/Fp such that every f(x) ∈ Fp[x] withdeg(f) ≥ 1 splits in K[x].


7. January 20 – Multiple roots, derivative test

Today: DF §13.5

Suppose K/F is a field extension, f(x) ∈ F [x] with deg(f) = n ≥ 1, and f(x)splits over K. Thus

f(x) = an(x− c1)(x− c2) · · · (x− cn)

where c1, . . . , cn ∈ K are the roots of f (in K).Note that some roots may be repeated, i.e., |{c1, . . . , cn}| < n. If c1, . . . , ck are the

distinct roots, then we can write

f(x) = an(x− c1)m1(x− c2)m2 · · · (x− ck)mk

where m1, . . . ,mk ≥ 1 and∑k

i=1mi = n.

Definition 7.1. mi is the multiplicity of the root ci (in K).ci is a multiple root of f(x) if mi ≥ 2.

Definition 7.2. Given K/F and f(x) ∈ F [x] such that f(x) splits over K: we saythat f(x) is separable (with respect to K) if f(x) has no multiple roots in K.

If K = R, then the roots of f(x) are places where the graph of f crosses the x-axis,and a root is a multiple root iff the graph is tangent to the x-axis there. Somethinglike this works in general.

Definition 7.3. Let f(x) = anxn + an−1x

n−1 + · · · + a2x2 + a1x + a0 ∈ F [x]. The

formal derivative of f(x), written Dxf(x) or f ′(x), is the polynomial

Dxf(x) = nanxn−1 + (n− 1)an−1x

n−2 + · · ·+ 2a2x+ a1.

Note that Dxf(x) ∈ F [x].

Example 7.4. If f(x) ∈ x4+x3+x2+x+1 ∈ F3[x], thenDxf(x) = 4x3+3x2+2x+1 =x3 + 2x+ 1.

It can be shown (exercise) that, in any field, we have

Dx(f + g) = Dx(f) +Dx(g)

Dx(f · g) = f ·Dx(g) + g ·Dx(f)

Dx(fn) = nfn−1Dx(f).

Proposition (DF Prp. 13.33). Given K/F and f(x) ∈ F [x] with f splitting over K,TFAE:

(1) f(x) has a multiple root in K (i.e., f is not separable w.r.t. K).(2) f(x) and Dx(f) have a common root in K.(3) f(x) and Dxf(x) have a common nontrivial factor in F [x].

16 R. WILLARD

Proof. (1) ⇒ (2). Suppose c ∈ K is a multiple root of f(x). Thus

f(x) = (x− c)2g(x), g(x) ∈ K[x].

Then Dxf(x) = (x− c)2Dxg(x) + 2(x− c)g(x) = (x− c)[(x− c)Dxg(x) + 2g(x)], soc is a common root of f(x) and Df (x).

(2) ⇒ (1). Suppose c is a common root of f(x) and Fxf(x). Writing f(x) =(x− c)h(x) (in K[x]) we get

Dxf(x) = (x− c)Dx(h) + h(x)

and since Dxf(c) = 0 this implies h(c) = 0; thus x − c |h(x) in K[x], so h(x) =(x− c)k(x), k(x) ∈ K[x]. So f(x) = (x− c)2k(x) and c is a multiple root of f(x).

(2) ⇒ (3). Suppose c is a common root but f(x), Dxf(x) have no nontrivialcommon factor in F [x]. Thus gcd(f(x), Dxf(x)) = 1. As F [x] is a PID, there existr(x), s(x) ∈ F [x] with f(x)r(x) + Dxf(x)s(x) = 1. Evaluating at x = c gives 0 = 1,contradiction.

(3)⇒ (2). Suppose f(x) = r(x)s(x) and Dxf(x) = r(x)t(x) with r(x), s(x), t(x) ∈F [x] and deg(r) ≥ 1. By UF in K[x], r(x) must be a product (in K[x]) of some ofthe linear factors of f(x). Thus any root of r in K is a common root of f(x) andDxf(x). �

Note that, as a consequence, whether or not f(x) ∈ F [x] is separable does notdepend on the choice of K/F .


8. January 22 – Applications of derivative test

Recall: if f(x) ∈ F [x], then TFAE:

(1) f(x) is separable (w.r.t. any extension over which f(x) splits);(2) gcd(f,Dx(f)) = 1 in F [x].

Corollary (DF Cor. 13.34). Suppose ch(F ) = 0. Every irreducible f(x) ∈ F [x] isseparable.

Proof. Let f(x) ∈ F [x] be irreducible in F [x] with deg(f) = n ≥ 1. If n = 1 thenf(x) is obviously separable, so assume n ≥ 2.

Let d(x) = gcd(f(x), Dxf(x)) ∈ F [x]. d|f so d is a unit or an associate of f .d|Dx(f) and deg(Dx(f)) = n − 1 < deg(f), so we can’t have deg(d(x)) = n. Henced(x) is a unit, i.e., gcd(f(x), Dx(f)) = 1. Hence f(x) is separable. �

Question: where in the proof did we use the assumption that ch(F ) = 0?

If ch(F ) = p, then the same proof will work, as long Dx(f) 6= 0.

When is Dx(f) = 0? Any a1xp + a0 has deriv. 0. So does a2x

2p + a1xp + a0.

In general, if ch(F ) = p then Dx(f) = 0 iff f(x) can be written g(xp) for someg(x) ∈ F [x].

Corollary. Suppose F is a finite field. Every irreducible f(x) ∈ F [x] is irreducible.

Proof. Suppose f(x) ∈ F [x] is irreducible.

Case 1: Dx(f) 6= 0. Then f(x) is separable by the above argument.

Case 2: Dx(f) = 0. Then f(x) = g(xp) for some g(x) ∈ F [x]. Write

g(x) = amxm + am−1x

m−1 + · · ·+ a1x+ a0, ai ∈ F.Because F is finite, every element of F has a unique p-th root in F . Pick b0, . . . , bm ∈F with (bi)

p = ai. Let

h(x) = bmxm + bm−1x

m−1 + · · ·+ b1x+ b0 ∈ F [x].

Then

h(x)p = (bmxm + bm−1x

m−1 + · · ·+ b1x+ b0)p

= (bmxm)p + (bm−1x

m−1)p + · · ·+ (b1x)p + (b0)p

= (bm)pxmp + (bm−1)px(m−1)p + · · ·+ (b1)pxp + (b0)p

= am(xp)m + am−1(xp)m−1 + · · ·+ a1xp + a0

= g(xp)

= f(x),

contradicting irreducibility of f(x) in F [x]. �

18 R. WILLARD

Start section 13.2. Instead of focussing on polynomials and their roots, we focus onthe roots and find their polynomials.

Definition. Given K/F and c ∈ K, we say that c is algebraic over F if c is a rootof some f(x) ∈ F [x] with deg(f) ≥ 1. Otherwise, c is transcendental over F .

Example. (1) i ∈ C is algebraic over Q as it is a root of x2 + 1 ∈ Q[x].(2) You are showing that cos(20◦) and cos(36◦) are algebraic over Q.(3) π, e are known to be transcendental over Q.

Proposition (DF Prp. 13.9). Let K/F and c ∈ K, and suppose c is algebraic/F .

(1) There exists a unique monic irreducible m(x) ∈ F [x] having c as a root.(2) m(x) has the property that for all f(x) ∈ F [x], f(c) = 0 iff m(x) | f(x) in

F [x].

Proof. Let I = {f(x) ∈ F [x] : f(c) = 0}. Certainly 0 ∈ I, and it’s easy to checkthat I is an ideal of F [x]. Thus I is principal, say I = (m(x)) as F [x] is a PID. Notethat for any f(x) ∈ F [x],

f(c) = 0 ⇐⇒ f ∈ I⇐⇒ m(x) | f(x)

So (2) is satisfied by any generator of I.By hypothesis, I contains a polynomial of degree ≥ 1. It follows that m(x) is a

nonzero member of I of least degree. By scaling, we can assume that m(x) is monic.Suppose m(x) is not irreducible in F [x]; then m(x) = f(x)g(x) for some nontrivialf, g ∈ F [x]. But m(x) ∈ I implies m(c) = f(c)g(c) = 0, so f(x) ∈ I or g(x) ∈ I,contradicting the fact that m(x) must have minimum degree among members of I.

Finally, if m′(x) were another monic irreducible polynomial having c as a root,then m(x) |m′(x) by (2), so m(x) and m′(x) are associates, but as they have thesame leading coefficient we get m(x) = m′(x). �


9. January 24 – Minimal polynomial, degree

Recall from last time:

Proposition (DF Prp. 13.9). Let K/F and c ∈ K, and suppose c is algebraic/F .

(1) There exists a unique monic irreducible m(x) ∈ F [x] having c as a root.(2) m(x) has the property that for all f(x) ∈ F [x], f(c) = 0 iff m(x) | f(x) in

F [x].

Definition. Suppose K/F and c ∈ K is algebraic/F .

(1) The unique monic irreducible polynomial m(x) ∈ F [x] from the Propositionis called the minimal polynomial of c over F .

(2) The degree of c over F , written deg(c/F ), is the degree of m(x).

Example. 3√

2 is algebraic over Q, as it is a root of x3− 2. The minimal polynomialof 3√

2 over Q is x3 − 2. Thus deg( 3√

2/Q) = 3.

Definition. Suppose K/F and c ∈ K.

(1) F [c] denotes the unique smallest subring of K containing F ∪ {c}.(2) F (c) denotes the unique smallest subfield of K containing F ∪ {c}.

What are F [c] and F (c) concretely? c ∈ F [c], so c2, c3, . . . , cn, . . . ∈ F [c]. AlsoF ⊆ F [c]; thus if a0, . . . , an ∈ F then anc

n + · · · a1c+ a0 ∈ F [c]. In other words,

f(x) ∈ F [x] =⇒ f(c) ∈ F [c].

F (c) must also be closed under multiplicative inverses. This proves part of

Proposition. If K/F and c ∈ K, then

F [c] = {f(c) : f ∈ F [x]}F (c) = {f(c)g(c)−1 : f, g ∈ F [x], g(c) 6= 0}.

More generally,

Definition. Suppose K/F and c1, . . . , cn ∈ K.

(1) F [c1, . . . , cn] denotes the unique smallest subring ofK containing F∪{c1, . . . , cn}.(2) F (c1, . . . , cn) denotes the unique smallest subfield ofK containing F∪{c1, . . . , cn}.

Proposition. Given K/F and c1, . . . , cn ∈ K:

F [c1, . . . , cn] = {f(c1, . . . , cn) : f ∈ F [x1, . . . , xn]}F (c1, . . . , cn) = {αβ−1 : α, β ∈ F [c1, . . . , cn], β 6= 0}.

Usually we want a more informative description than the general one.

20 R. WILLARD

Example. Consider Q[√

2,√

3]. Since (√

2)2 = 2 and (√

3)2 = 3, any polynomialin√

2,√

3 can be reduced to one of the form a + b(√

2) + c(√

3) + d(√

2)(√

3) witha, b, c, d ∈ Q. Thus

Q[√

2,√

3] ={a+ b(

√2) + c(

√3) + d(

√6) : a, b, c, d ∈ Q

}.

Can this be further simplified? E.g., if√

3 can be written in the form a+ b√

2 forsome a, b ∈ Q, then we could write

Q[√

2,√

3] = {a+ b(√

2) : a, b ∈ Q}.Claim.

√3 cannot be so written.

Proof. Suppose√

3 = a + b√

2 for some a, b ∈ Q. We know√

3 6∈ Q, so b 6= 0. Ifa = 0 then

√6 = 2b ∈ Q, which we also know to be false. So a, b 6= 0. Then

3 = (a2 + 2b2) + 2ab√

2√

2 =3− (a2 + 2b2)

2ab

showing√

2 ∈ Q, contradiction. �

In fact, no further simplification of any kind is possible, because every α ∈ Q[√

2,√

3]can be uniquely written in the form a+ b(

√2) + c(

√3) + d(

√6) with a, b, c, d ∈ Q.

Equivalently,

Claim. 1,√

2,√

3,√

6 are linearly independent over Q.

Proof. Assume a, b, c, d ∈ Q and a + b(√

2) + c(√

3) + d(√

6) = 0. (Must showa = b = c = d = 0.) Write

(a+ b√

2) + (c+ d√

2)√

3 = 0.

Assume first that c + d√

2 = 0. Then a + b√

2 = 0, and one can show b = 0 using√2 6∈ Q, and then a = 0. Similarly, c+ d

√2 = 0 implies c = d = 0.

Thus we can assume c + d√

2 6= 0. It follows that c − d√

2 6= 0 as well (else asimilar argument would give c = d = 0). Then

√3 =

−(a+ b√

2)(c− d√

2)

(c+ d√

2)(c− d√

2)= u+ v

√2 for some u, v ∈ Q,

contradicting the previous claim. �

Definition. Let K/F be a field extension. Assume R is a subring of K containingF . The degree of R over F , written [R : F ], is the dimension of (R,+) consideredas a vector space over F .

Example. [Q[√

2,√

3] : Q] = 4, since {1,√

2,√

3,√

6} is a basis for Q[√

2,√

3] as avector space over Q.


10. January 27 – Finite simple extensions

Recall that if F ≤ K and a1, . . . , an ∈ K, then

F [c1, . . . , cn] = the smallest subring of K containing F ∪ {c1, . . . , cm}F (c1, . . . , cn) = the smallest subfield of K containing F ∪ {c1, . . . , cm}.

Definition. F (c)/F is called a simple extension.

Also recall that if K/F then the degree of K/F , written [K : F ], is the dimensionof K as a vector space over F .

Definition. An extension K/F is finite if [K : F ] <∞.

Proposition. Suppose F ≤ K and c ∈ K.

(1) (DF Prp. 13.12) c is algebraic over F iff [F (c) : F ] <∞.(2) If c is algebraic over F and deg(c/F ) = n, then

(a) (DF Prp. 13.11) [F (c) : F ] = n.(b) (DF Cor. 13.7) F (c) = F [c] = {r(c) : ∈ F [x], deg(r) < n}.

Proof. Suppose first that [F (c) : F ] = n <∞. Consider the elements 1, c, c2, c3, . . . , cn ∈F (c). They must be linearly dependent/F , so there must exist a0, . . . , an ∈ F , notall 0, such that

a0 · 1 + a1 · c+ a2 · c2 + · · ·+ an · cn = 0.

Thus if f(x) = a0 + a1x + a2x2 + · · ·+ anx

n then f is a nonzero polynomial in F [x]having c as a root, proving c is algebraic/F of degree ≤ n. This proves (1)(⇐).

Conversely, suppose c is algebraic over F and n = deg(c/F ). Let m(x) ∈ F [x] bethe minimal polynomial of c over F .

Claim 1: F [c] = {(r(c) : ∈ F [x], deg(r) < n}.Proof. ⊇ is obvious. For ⊆, suppose f(c) ∈ F [c], f ∈ F [x]. Divide f by m,

getting quotient q(x) and remainder r(x) ∈ F [x]. Then deg(r) < deg(m) = n andf = qm+ r, so

f(c) = q(c)m(c) + r(c) = q(c)0 + r(c) = r(c),

so f(c) ∈ {r(c) : r ∈ F [x], deg(r) < n}.Claim 2: [F [c] : F ] = n.

Proof. Clearly 1, c, c2, . . . , cn−1 span F [c] by Claim 1, so it suffices to show thatthese elements are linearly independent over F . Suppose there exist a0, . . . , an−1 ∈ F ,not all 0, such that a1 · 1 + a1 · c + · · · + an−1c

n−1 = 0. This would give a nonzeropolynomial in F [x] of degree < n having c as a root; but that contradicts the factthat m(x) has minimum degree of all polys in F [x] having c as a root.

We’ll be done if we can prove F (c) = F [c].

Claim 3: F (c) = F [c].

22 R. WILLARD

Proof. It suffices to prove that F [c] is a field. Define a map ϕ : F [x] → K byϕ(f) = f(c). This is clearly a ring homomorphism and its range is F [c]. ker(ϕ) ={f ∈ F [x] : f(c) = 0} = (m(x)). Thus by the First Isomorphism Theorem for rings,

F [x]/(m(x)) ∼= F [c].

But the LHS is a field; hence F [c] is also a field. �

Let’s meditate on this last proof. F [x]/(m(x)) is our fundamental construction ofan extension of F in which the irreducible polynomial m(x) has a root θ = x + I,where I = (m(x)). The First Isomorphism theorem gives

F [x]

F [x]/IF (θ) =

F [c] = F (c)

π

ϕ

ϕ∼=

Let’s explore this isomorphism. Both ϕ and π act as the identity map on elementsof F , so ϕ(a) = a for all a ∈ F . ϕ(x) = c while π(x) = θ, so ϕ(θ) = c.

This proves the following:

Theorem (DF Thm. 13.6). Suppose K/F , c ∈ K, and c is algebraic over F withminimal polynomial m(x) ∈ F [x]. Let I = (m(x)) and θ = x + I. There exists anisomorphism α : F (c)→ F [x]/I sending c 7→ θ and fixing each element of F .

Application (if time): consider c = 3√

2 and d = ω 3√

2 where ω = −1+i√

32

. CertainlyQ(c) 6= Q(d). However, both c and d are roots of x3−2, which must be their minimalpolynomial. Hence

Q(c) ∼= Q[x]/(x3 − 2) ∼= Q(d).

Moreover, an isomorphism exists which fixes each element of Q and sends c to d.


11. January 29 – Product Theorem

Definition. K/F is algebraic if every c ∈ K is algebraic over F .

Proposition (DF Cor. 13.13). If K/F is finite, then K/F is algebraic.

Proof. Assume [K : F ] = n. Let c ∈ K be arbitrary. Then F ≤ F (c) ≤ K, so F (c)is a subspace of K (considered as vector spaces over F ), so [F (c) : F ] ≤ n. Thus c isalgebraic over F by Monday’s proposition. �

Theorem (DF Thm. 13.14). Suppose F ≤ E ≤ K.

(1) K/F is finite iff K/E and E/F are finite.(2) If K/E and E/F are finite, then [K : F ] = [K : E][E : F ].

Proof. (1) (⇒): As F -spaces, E is a subspace of K, so its dimension is less than F ’s.For the second inequality, assume that Γ = {γ1, . . . , γn} is a basis for K as a vectorspace over F . Then spanF (Γ) = K, so certainly spanE(Γ) = K, so the dimension ofK as an E-space is at most n.

(2) Assume [K : E] = n and [E : F ] = m. Let Γ = {γ1, . . . , γm} be an F -basis forE and let ∆ = {δ1, . . . , δn} be an E-basis for K. Define

Σ := {γiδj : 1 ≤ i ≤ m, 1 ≤ j ≤ n}.Claim: Σ is an F -basis for K, proving [K : F ] = mn. First, span: given any u ∈ K,

u ∈ spanE(∆) =⇒ u =n∑j=1

bjδj

for some b1, . . . , bn ∈ E. E = spanF (Γ) implies we can write bj =∑m

i=1 aijγi for somea1j . . . , amj ∈ F . Thus

u =n∑j=1

bjδj

=n∑j=1

(m∑i=1

aijγi

)δj

=

m,n∑i,j=1

aij(γiδj),

showing u ∈ spanF (Σ), so spanF (Σ) = K. Proving that Σ is F -linearly independentis a good exercise and will be on the next assignment. �

Note: if F ≤ E ≤ K ≤ L, then [L : F ] = [L : K][K : F ] = [L : K][K : E][E : F ].

Here is another useful fact.

Lemma. If K/F and c1, . . . , cn ∈ K, then F (c1, . . . , cn) = F (c1, . . . , cn−1)(cn).

24 R. WILLARD

Proof. Let E = F (c1, . . . , cn−1). We must show F (c1, . . . , cn) = E(cn). Strategy:show ⊆ and ⊇. By the Useful Fact, it is enough to prove

F ⊆ E(cn)

c1, . . . , cn ∈ E(cn)

E ⊆ F (c1, . . . , cn)

cn ∈ F (c1, . . . , cn).

The 1st, 2nd and 4th are easy. For the 3rd, recall that E = F (c1, . . . , cn−1),so to prove the inclusion it suffices to show F ⊆ F (c1, . . . , cn) and c1, . . . , cn−1 ∈F (c1, . . . , cn), both of which are obvious. �

Note: F [c1, . . . , cn] = F [c1, . . . , cn−1][cn] is proved similarly.

Corollary. Given K/F , if c1, . . . , cn ∈ K are algebraic over F , then F (c1, . . . , cn) =F [c1, . . . , cn].

Proof. By induction on n. If n = 1, the claim was proved on Monday. Assume n > 1.Let E = F (c1, . . . , cn−1); then F ≤ E ≤ K. Because cn is algebraic/F , it is alsoalgebraic/E (Prob. #2 from Ass. 3). Also, E = F [c1, . . . , cn−1] by induction. Then

F (c1, . . . , cn) = F (c1, . . . , cn−1)(cn)

= E(cn)

= E[cn] (by Monday’s result, as cn is algebraic/E)

= F [c1, . . . , cn−1][cn]

= F [c1, . . . , cn].

�

Application:√

2,√

3 are both algebraic/Q. Hence Q(√

2,√

3) = Q[√

2,√

3] =spanQ(1,

√2,√

3,√

6) as shown last Friday, so [Q(√

2,√

3) : Q] = 4.


12. January 31 – Applications of Product Theorem

Definition. K/F is finitely generated if there exist c1, . . . , cn ∈ K such thatF (c1, . . . , cn) = K.

Theorem (DF Thm. 13.17). Assume K = F (c1, . . . , cn) with each ci algebraic/F ,say of degree di. Then [K : F ] <∞; in fact, [K : F ] ≤ d1d2 · · · dn.

Proof. ⇒. Consider the chain of subfields

F ≤ F (c1) ≤ F (c1, c2) ≤ · · · ≤ F (c1, . . . , cn) = K

and write Fi = F (c1, . . . , ci), so F = F0 ≤ F1 ≤ F2 ≤ · · · ≤ Fi ≤ · · · ≤ Fn = K. Forfixed i = 1, . . . , n, consider [Fi : Fi−1]. Observe that

Fi = F (c1, . . . , ci) = F (c1, . . . , ci−1)(ci) = Fi−1(ci).

As ci ∈ K is algebraic/F and F ≤ Fi−1 ≤ K, we have deg(ci/Fi−1) ≤ deg(ci/F )(Assign. 3, prob. 2a). Hence

[Fi : Fi−1] = [Fi−1(ci) : Fi−1]

= deg(c/Fi−1)

≤ deg(c/F )

= di.

Now use the product theorem:

[K : F ] = [Fn : F0] = [Fn : Fn−1][Fn−1 : Fn−2] · · · [F1 : F0] ≤ dndn−1 · · · d1.

�

Example. Find [Q(√

3, 3√

2) : Q].Solution: Let K = Q(

√3, 3√

2). obviously deg(√

3/Q) = 2 and deg( 3√

2/Q) = 3.Thus [K : Q] ≤ 2 · 3 = 6. On the other hand, we have Q ≤ Q(

√3) ≤ K, so by

the product theorem, [K : Q] = [K : Q(√

3)][Q(√

3) : Q] = [K : Q(√

3)] · 2. Hence2 divides [K : Q]. A similar argument with Q( 3

√2) shows 3 divides [K : Q]. Hence

[K : Q] = 6.

Corollary (DF Cor. 13.18). Suppose K/F , a, b ∈ K, and a, b are both algebraic overF . Then a± b, ab, and ab−1 (if b 6= 0) are algebraic over F .

Proof. By the previous theorem, [F (a, b) : F ] < ∞. Hence F (a, b)/F is algebraic(Weds. Prop.). Since a± b, ab, ab−1 ∈ F (a, b), they must be algebraic over F . �

Corollary (DF Cor. 13.19). Given L/F , the set K := {a ∈ L : a is algebraic/F} isa subfield of L.

Example. Let K = {a ∈ C : a is algebraic over Q}. Q ≤ K ≤ C. K/Q is algebraic.But K/Q is not finite.

26 R. WILLARD

Proposition (DF Thm. 13.20). Suppose F ≤ K ≤ L. If K/F and L/K are alge-braic, then L/F is algebraic.

Proof. Let c ∈ L be arbitrary. c is algebraic over K; let m(x) = xn+an−1xn−1 + · · ·+

a1x+a0 be its minimal polynomial over K; thus a0, . . . , an−1 ∈ K, so each coefficientis algebraic over F . Consider E := F (a0, a1, . . . , an−1); we have [E : F ] <∞ and

F

E

K

F (c)

E(c)

L

Note that c is algebraic over E, as m(x) ∈ E[x]. Thus [E(c) : E] < ∞. Hence bythe product theorem, [E(c) : F ] <∞, which implies [F (c) : F ] <∞, so c is algebraicover F . �

Corollary. K/F is finite iff K/F is algebraic and finitely generated.

Proof. (⇐) follows from today’s Theorem.(⇒. Assume K/F is finite. We’ve already proved that K/F is algebraic. If K = F

then obviously F/F is finitely generated. Assume K > F , pick c1 ∈ K \ F and letF1 = F (c1). If K = F1 then we’re done, so assume K > F1 and pick c2 ∈ K \F1, andlet F2 = F1(c2). Observe that F2 = F (c1)(c2) = F (c1, c2) by the previous Lemma.Continuing in this way, we get

F < F (c1) < F (c1, c2) < F (c1, c2, c3) < . . . ≤ K.

All of these fields are F -vector subspaces of K, which is finite-dimensional, so thechain must eventually reach K. �


13. February 3 – Cyclotomic polynomials

Today: DF §13.6.

Definition. Given n ≥ 1, define

Λn = {z ∈ C× : z has order n}µn = {z ∈ C× : zn = 1}.

The elements of Λn are called the primitive nth-roots of unity. µn is called the groupof nth-roots of unity.

Examples

Λ1 = {1}Λ2 = {−1}

Λ3 = {ω, ω}, ω =−1 + i

√3

2Λ4 = {i,−i}.

Also, µ2 = Λ1 ∪ Λ2, µ3 = Λ1 ∪ Λ3, µ4 = Λ1 ∪ Λ2 ∪ Λ4, etc. In general:

(1) µn is a cyclic group of order n.(2) µn =

⋃d|n Λd.

(3) |Λn| = ϕ(n).

Lemma. xn − 1 =∏

ζ∈µn(x− ζ).

Proof. Clearly µn is the set of roots of xn− 1 in C. Thus∏

ζ∈µn(x− ζ) divides xn− 1

in C[x]. Since they are both monic and have the same degree, they are equal. �

Definition. Given n ≥ 1, the nth cyclotomic polynomial Φn(x) is

Φn(x) =∏ζ∈Λn

(x− ζ) ∈ C[x].

Examples

Φ1(x) = x− 1

Φ2(x) = x+ 1

Φ3(x) = (x− ω)(x− ω) = x2 − (ω + ω)x+ ωω = x2 + x+ 1

Φ4(x) = (x− i)(x+ i) = x2 + 1.

Lemma. xn − 1 =∏

d|n Φd(x).

28 R. WILLARD

Proof.

xn − 1 =∏ζ∈µn

(x− ζ)

=∏d|n

∏ζ∈Λd

(x− ζ)

=∏d|n

Φd(x).

�

Note that if we write this last as

(†) xn − 1 =

∏d|n, d<n

Φd(x)

· Φn(x),

then we get a recursive procedure to compute Φn(x) for any n.

Example:

Φ5(x) =x5 − 1

Φ1(x)=x5 − 1

x− 1= x4 + x3 + x2 + x+ 1

Φ6(x) =x6 − 1

Φ1(x)Φ2(x)Φ3(x)=

x6 − 1

(x− 1)(x+ 1)(x2 + x+ 1)= x2 − x+ 1

Lemma (DF Lm. 13.40). Φn(x) ∈ Z[x] for all n ≥ 1.

Proof. By induction on n. Clear for n = 1. For n > 1, use (†) and Gauss’ Lemma. �

Our main goal is to prove that each Φn(x) is irreducible in Q[x]. Let ζn = e2πi/n ∈Λn. Let m(x) be the minimal polynomial of ζn over Q. Equivalently, we want toshow that m(x) = Φn(x). We will do this by showing that every ζ ∈ Λn is a root ofm(x).

Since Φn(ζn) = 0, we get m(x)|Φn(x) in Q[x], say

Φn(x) = m(x)g(x), g ∈ Q[x], monic, possibly = 1.

By Gauss’ Lemma, m(x), g(x) ∈ Z[x].

Claim: if m(ζ) = 0 and p is a prime with gcd(p, n) = 1, then m(ζp) = 0.Proof: m(ζ) = 0 implies ζ ∈ Λn. gcd(p, n) = 1 implies ζp ∈ Λn, so

0 = Φn(ζp) = m(ζp)g(ζp).

If the claim is false, then g(ζp) = 0. But m(x) is the minimal polynomial of ζ overQ, so m(x)|g(xp), say

g(xp) = m(x)h(x), h ∈ Q[x], monic.


Again by Gauss’ Lemma, h ∈ Z[x]. Thus we can reduce the above equation mod p:

(g(x))p = g(xp) = m(x)h(x) in Fp[x].

In particular, m(x) and g(x) have a common nontrivial factor in Fp[x]. Also,

Φn(x) = m(x)g(x) in Fp[x]

xn − 1 = Φn(x) · F (x) in Fp[x],

we get that xn − 1 has a repeated factor in F[x], so has a multiple root in an extensionof Fp over which xn − 1 splits.

But Dx(xn − 1) = nxn−1 and since p6 |n this is relatively prime to xn − 1 in Fp[x],

contradiction.

30 R. WILLARD

14. February 5 – Splitting fields

Recall:

µn = {z ∈ C× : zn = 1}Λn = {z ∈ µn : z has order n}

Φn(x) =∏ζ∈Λn

(x− ζ)

Φn(x) ∈ Z[x], and we are working to prove that Φn(x) is irreducible in Z[x].

We let m(x) be the minimal polynomial of ζn over Q, and have

Φn(x) = m(x)g(x), m(x), g(x) ∈ Z[x].

We have proved:

Claim: if m(ζ) = 0 and p is a prime with gcd(p, n) = 1, then m(ζp) = 0.

Now we show that every ζ ∈ Λn is a root of m(x). Let ζ ∈ Λn. Then ζ = (ζn)k

for some k ≥ 1 with gcd(k, n) = 1. If k = 1 then obviously m(ζ) = 0. Else, writek = p1p2, . . . , pt where the pi are primes. From m(ζn) = 0 we get m((ζn)p1) = 0, thenm(((ζn)p1)p2) = 0, etc., until finally m(ζ) = m((ζn)k) = 0 as required.

As explained before, this proves m(x) = Φn(x). Hence

Theorem (DF Thm. 13.41). Φn(x) is irreducible in Q[x].

It also follows that Φn(x) is the minimal polynomial over Q of each ζ ∈ Λn.

Definition. Let F be a field and f(x) ∈ F [x] with deg(f) ≥ 1. An extension E/Fis a splitting extension for f(x) if

(1) f(x) splits in E[x], and(2) E = F (c1, . . . , cr) where c1, . . . , cr are the roots of f(x) in E.

We also say that E is a splitting field for f(x) over F .

Splitting extensions always exist. Just let K/F be an extension over which f(x)splits, let c1, . . . , cr be the roots of f(x) in K, and put E = F (c1, . . . , cr).

Example. Let f(x) = x4 + x2 − 6. f(x) splits/C. Its roots in C are ±√

2,±i√

3.Hence a splitting field for f(x) over Q is

E := Q(√

2,−√

2, i√

3,−i√

3) = Q(√

2, i√

3).

We see that [E : Q] = 4.

Example. Let f(x) = x3 − 2. f(x) splits/C. Its roots in C are 3√

2, ω 3√

2 and ω 3√

2

where ω = −1+i√

32

. Thus a splitting field for x3 − 2 over Q is

E := Q(3√

2, ω3√

2, ω3√

2).


What is [E : Q]? Clearly [E : Q] ≤ 3 · 3 · 3 = 27. We can do better. ω ∈ E, soQ( 3√

2, ω) ⊆ E. Conversely, ω = ω2 ∈ Q( 3√

2, ω), so E ⊆ Q( 3√

2, ω). Thus

E = Q(3√

2, ω).

The minimal polynomial of ω is Φ3(x) = x2 + x+ 1, so deg(ω) = 2. Hence

[E : Q] ≤ 3 · 2 = 6.

Using the product theorem and

Q ≤ Q(3√

2) ≤ E and Q ≤ Q(ω) ≤ E

we get that [E : Q] is divisible by both 3 and 2, so [E : Q] = 6.

Remark: it should also be clear that ω ∈ Q(i√

3) and i√

3 ∈ Q(ω), so Q(ω) =Q(i√

3), so E = Q( 3√

2, i√

3).

32 R. WILLARD

15. February 7 – Newfoundland Theorems

Recall (January 27): if c = 3√

2 and d = ω 3√

2 where ω = −1+i√

32

, then Q(c) ∼= Q(d)via a (unique) isomorphism which fixes each element of Q and sends c to d. Thereason for the isomorphism was

Q(c) ∼= Q[x]/(x3 − 2) ∼= Q(d)

c 7→ θ ←[ d

where θ = x+ (x3 − 2). Today we will jazz this up.Suppose we have K/F and L∗/F ∗ (all are fields) and an isomorphism ϕ : F ∼= F ∗.

ϕ extends naturally to a ring isomorphism ϕ : F [x] ∼= F ∗[x], namely,

ϕ(anxn + · · ·+ a1x+ a0) := ϕ(an)xn + · · ·+ ϕ(a1)x+ ϕ(a0).

Suppose p(x) is an irreducible polynomial in F [x]; then its image p∗(x) under ϕ isirreducible in F ∗[x]. Suppose c ∈ K is a root of p(x) in K, and d ∈ L∗ is a root ofp∗(x) in L∗. Consider

F

F (c)

K

F ∗

F ∗(d)

L∗

≤

≤

≤

≤

∼=ϕ

We also have F ≤ F [x]/(p(x)) (not literally, but we’re OK with that) and we havean isomorphism α : F (c) ∼= F [x]/(p(x)) sending c 7→ x + (p(x)) =: θ and fixingevery element of F . Similarly we have F ∗ ≤ F ∗[x]/(p∗(x)) and an isomorphismβ : F ∗(d) ∼= F ∗[x]/(p∗(x)) sending d 7→ x+ (p∗(x)) =: θ∗ and fixing every element ofF ∗.

F

F (c)

K

F ∗

F ∗(d)

L∗

≤

≤

≤

≤

∼=ϕ

F [x]/(p(x)) F ∗[x]/(p∗(x))

≤ ≤α β


Finally, ϕ induces (naturally) an isomorphism ϕ : F [x]/(p(x)) ∼= F ∗[x]/(p∗(x)). (Putin picture.) In particular, ϕ extends ϕ and sends θ 7→ θ∗.

Let σ = β−1 ◦ϕ◦α. [Picture.] Then σ : F (c) ∼= F ∗(d). Moreover, it can be checked(wave hands, time permitting) that:

(1) σ�F = ϕ.(2) σ(c) = d.

Let’s record this as a theorem.

Theorem (Dummit & Foote Theorem 13.8). Suppose ϕ : F ∼= F ∗, p(x) ∈ F [x] isirreducible, c is a root of p(x) in some extension of F , and d is a root of the imagep∗(x) of p(x) under ϕ in some extension of F ∗. Then ϕ extends to an isomorphismσ : F (c) ∼= F ∗(d) which sends c 7→ d.

We can now prove the following.

Theorem (Dummit & Foote Theorem 13.27). Suppose ϕ : F ∼= F ∗, f(x) ∈ F [x]with deg(f) ≥ 1 (not necessarily irreducible), f ∗(x) its image under ϕ. Assume K/Fis a splitting extension for f(x) and L∗/F ∗ is a splitting extension for f ∗(x). Thenϕ extends to an isomorphism ψ : K ∼= L∗ (i.e., ψ�F = ϕ).

Proof. By induction on n = [K : F ]. If n = 1, then K = F , so f(x) splits in F [x], sof ∗(x) splits in F ∗[x] (under the isomorphism ϕ), so L∗ = F ∗, so we can take σ = ϕ.

Inductively, assume [K : F ] > 1. Thus f(x) does not split in F [x]. Let c1, . . . , ckbe the roots of f(x) in K, so K = F (c1, . . . , ck). Similarly let d1, . . . , d` be the rootsof f ∗(x) in L∗, so L∗ = F ∗(d1, . . . , d`).

Pick an irreducible factor p(x) ∈ F [x] of degree ≥ 2. Let p∗(x) be its image underϕ. p(x) splits over K, and WLoG we can assume that p(c1) = 0. Similarly p∗(x)splits over L∗ and WLoG we can assume p∗(d1) = 0.

34 R. WILLARD

16. February 10 – Uniqueness of splitting fields

Recall that we are proving a theorem:

Theorem (Dummit & Foote Theorem 13.27). Suppose ϕ : F ∼= F ∗, f(x) ∈ F [x]with deg(f) ≥ 1 (not necessarily irreducible), f ∗(x) its image under ϕ. Assume K/Fis a splitting extension for f(x) and L∗/F ∗ is a splitting extension for f ∗(x). Thenϕ extends to an isomorphism ψ : K ∼= L∗ (i.e., ψ�F = ϕ).

We are arguing by induction on n = [K : F ], which we assume is > 1. We have:

• K = F (c1, . . . , ck) where c1, . . . , ck are the roots of f(x) in K.• L∗ = F ∗(d1, . . . , d`) where d1, . . . , d` are the roots of f ∗ in L∗.• p(x) ∈ F [x] is an irreducible factor of f(x) of degree ≥ 2, and p∗(x) is its

image in F ∗[x] under ϕ.• p(c1) = 0 and p∗(d1) = 0.

F

F (c1)

K

F ∗

F ∗(d1)

L∗

≤

≤

≤

≤

∼=ϕ

By the previous theorem, ϕ extends to an isomorphism σ : F (c1) ∼= F ∗(d1). LetE = F (c1) and E∗ = F ∗(d1).

Key observations:

• K is a splitting field for f(x) over E.• Likewise, L∗ is a splitting field for f ∗(x) over E∗.• σ : E ∼= E∗ and σ(f) = ϕ(f) = f ∗.

In otherwords, we have all the hypotheses of the theorem. Finally,

[K : E] =[K : F ]

[E : F ]< [K : F ] since [E : F ] = deg(c1/F ) = deg(p) > 1.

Hence by induction, σ extends to an isomorphism ψ : K ∼= L∗, as required. �

The most important special case of this theorem is when F = F ∗ and ϕ = idF . Inthis case the theorem says the following:

Corollary (DF Cor. 13.28). Suppose f ∈ F [x] and K,L are two splitting fields forf(x) over F . There exists an isomorphism ψ : K ∼= L satisfying ψ(a) = a for alla ∈ F .


In particular, f(x) has the same number of distinct roots in K and in L, with thesame number of multiplicities. For if f(x) splits in K[x] as

anxn + an−1x

n−1 + · · ·+ a1x+ a0 = an(x− c1)d1 · · · (x− ck)dk

then applying ψ : K[x] ∼= L[x] to this equation we get

ψ(anxn + an−1x

n−1 + · · ·+ a1x+ a0) = ψ(an(x− c1)d1 · · · (x− ck)dk).

But ψ(f) = f since the coefficients of f are fixed by ψ.

There is an application to finite fields. Suppose F is a finite field, say of charac-teristic p. Then Fp ≤ F and [F : Fp] < ∞. If [F : Fp] = n then |F | = pn. Hence|F×| = pn − 1, so every c ∈ F× satisfies cp

n−1 = 1, so every element of F is a root ofxp

n − x. Since xpn − x can have at most pn distinct roots in any extension of Fp, it

follows that xpn − x splits over F . Hence F is a splitting field for xp

n − x over Fp.Now suppose F, F ∗ are two finite fields with |F | = |F ∗|, say = pn. By the above

remarks, F and F ∗ are both splitting fields for xpn−x over Fp; hence by the corollary

F ∼= F ∗. This proves:

Corollary. Any two finite fields of the same cardinality are isomorphic.

This argument also tells us how to “construct” finite fields of any desired size.Given pn, let f(x) = xp

n − x and let F be a splitting field of f(x) over Fp. If thereis a field of size pn, then F must be such an example. (We’ll show that it is onWednesday.)

36 R. WILLARD

17. February 12 – Existence of finite fields; automorphisms

Corollary. For every prime p and n ≥ 1 there exists a field of size pn (unique up toisomorphism by the previous Corollary).

Proof. Let f(x) = xpn − x and let K be a splitting field for f over Fp. (Exists) We’ll

show that |K| = pn.

(1) First observe that Dx(f) = −1 which is co-prime to f , so f has exactly pn

distinct roots in K.(2) Let R = {c ∈ K : f(c) = 0}. We have |R| = pn.(3) Recall the Frobenius map ϕ : K → K. It is an injective homomorphism. Since

K is finite (it is generated over Fp by finitely many algebraic elements/Fp, so[K : Fp] <∞), ϕ is an isomorphism from K to K. Thus ϕn (i.e., ϕ◦ϕ◦· · ·◦ϕ)is an isomorphism from K to K. It is easily checked that ϕn(c) = cp

nfor any

c ∈ K, and hence R = {c ∈ K : ϕn(c) = c}.(4) Using this last fact, it can be easily checked that R is a subfield of K (just

check that R is closed under +,−,×, −1 and includes 0, 1.

Hence Fp ≤ R ≤ K. But K = Fp(R) by definition. Hence K = R. �

Exercise: Show that every finite extension [K : Fp] is simple. Conclude that forevery p and n there exists an irreducible f(x) ∈ Fp[x] of degree n.

Start chapter 14: Galois theory.

Definition. Let F ≤ K.

(1) An automorphism of K is an isomorphism ϕ : K ∼= K.(2) Aut(K) = the set of all automorphisms of K.(3) An automorphism σ ∈ Aut(K) fixes F if σ(a) = a for all a ∈ F .(4) Aut(K/F ) = {σ ∈ Aut(K) : σ fixes F}.

Note that idK ∈ Aut(K/F ) always, so Aut(K/F ) 6= ∅. In fact, Aut(K) is a groupunder composition, and Aut(K/F ) is a subgroup.

Example. Aut(C/R) = {idC, ν} where ν(z) = z.

Example. Let K = Q( 3√

2, ω) where ω = −1+i√

32

. (Recall that K is a splitting fieldfor x3 − 2 over Q.) Let’s find some elements in Aut(K/Q).

(1) We have idK ∈ Aut(K/Q).(2) To make more progress, we need a tight description of K. Let E = Q( 3

√2). By

the usual arguments, [E : Q] = 3 and [K : E] = 2. Hence by our descriptionof simple algebraic extensions, {1, 21/3, 22/3} is a basis for E over Q, and {1, ω}is a basis for K over E. Hence by the proof of the Product Theorem,

{1, ω, 21/3, 21/3ω, 22/3, 22/3ω}


is a basis for K over Q. Any ϕ ∈ Aut(K/Q) is also a Q-linear map K → K,so is determined by its action on this basis, and can be represented by a6 × 6 matrix over Q. So the question is: which 6 × 6 matrices over Q giveautomorphisms of K fixing Q?

(3) Consider ν ∈ Aut(C/R) given by ν(z) = z. ν�K is an isomorphism from Kto ν(K). Observe that if a, b, c, d, e, f ∈ Q,

ν(a+ bω + c21/3 + d21/3ω + e22/3 + f22/3ω)

= a+ bω + c21/3 + d21/3ω + e22/3 + f22/3ω

= a+ b(−1− ω) + c21/3 + d21/3(−1− ω) + e22/3 + f22/3(−1− ω)

(using ω = ω2 and ω2 + ω + 1 = 0)

= (a− b) + (−b)ω + (c− d)21/3 + (−d)21/3ω + (e− f)22/3ω + (−f)22/3ω.

From this we see that ν(K) ⊆ K, and hence K = ν(ν(K)) ⊆ ν(K), soν(K) = K, proving ν�K ∈ Aut(K/Q). Moreover, this calculation shows thatthe matrix for ν�K is

1 −1 0 0 0 00 −1 0 0 0 00 0 1 −1 0 00 0 0 −1 0 00 0 0 0 1 −10 0 0 0 0 −1

.(4) Claim: there exists σ ∈ Aut(K/Q) satisfying σ(21/3) = 21/3ω and σ(ω) = ω.

Proof. 21/3 and 21/3ω are two roots of x3 − 2. By Assign. 5 prob 5(b), thereexists σ0 ∈ Aut(K/Q) satisfying σ0(21/3) = 21/3ω. By Assign. 5 prob. 5(a), σ0

permutes the roots of x2 +x+ 1, so σ0(ω) ∈ {ω, ω}. If σ0(ω) = ω then we canset σ = σ0. If σ0(ω) = ω, then σ0(ω) = σ0(ω2) = σ0(ω)2 = (ω)2 = (ω2)2 = ω.Hence if we define σ = σ0 ◦ ν�K we get• σ ∈ Aut(K/Q).• σ(ω) = σ0(ν(ω)) = σ0(ω) = ω.• σ(21/3) = σ0(ν(21/3)) = σ0(21/3) = 21/3ω.

�

38 R. WILLARD

18. Feburary 14 – Continuation of example

Example (continued). K = Q( 3√

2, ω) where ω = −1+i√

32

. Recall:

(1) ν�K ∈ Aut(K/Q) (conjugation).(2) There exists σ ∈ Aut(K/Q) satisfying σ(21/3) = 21/3ω and σ(ω) = ω.

We will see that, in general, if σ, τ ∈ Aut(E/F ) where E = F (c1, . . . , cn), andif σ(ci) = τ(ci) for each i = 1, . . . , n, then σ = τ . Thus in our case the aboveinformation about the values of σ at 21/3 and ω should completely determine σ. Forexample, the values of σ at the basis vectors {1, ω, 21/3, 21/3ω, 22/3, 22/3ω} can becalculated as follows:

σ(1) = 1 (as σ fixes Q)

σ(ω) = ω (given)

σ(21/3) = 21/3ω (given)

σ(21/3ω) = σ(21/3)σ(ω) = (21/3ω)ω = 21/3ω2 = 21/3(−1− ω) = (−1)21/3 + (−1)21/3ω

σ(22/3) = σ(21/3)σ(21/3) = (21/3ω)(21/3ω) = 22/3ω2 = 22/3(−1− ω) = (−1)22/3 + (−1)22/3ω

σ(22/3ω) = σ(22/3)σ(ω) = 22/3(−1− ω)ω = 22/3.

Hence the matrix for σ is

M :=

1 0 0 0 0 00 1 0 0 0 00 0 0 −1 0 00 0 1 −1 0 00 0 0 0 −1 10 0 0 0 −1 0

.Since M3 = I, we see that σ has order 3 in Aut(K/Q). Since we also have an elementν�K of order 2 in Aut(K/Q), this shows that |Aut(K/Q)| ≥ 6. We will see after thebreak that |Aut(E/F )| ≤ [E : F ] for any finite extension E/F , so in this examplewe must have |Aut(K/Q)| = 6 and Aut(K/Q) is generated by {σ, ν�K}. We can alsodetermine Aut(K/Q) up to isomorphism:

(σ ◦ ν�K)(21/3) = σ(21/3) = 21/3ω

(ν�K ◦ σ)(21/3) = ν(21/3ω) = 21/3ω

so σ, ν�K do not commute. Hence Aut(K/Q) ∼= S3.

Finally, recall that K = Q(r0, r1, r2) where rk = 21/3ωk are the three roots of x3−2.Each ϕ ∈ Aut(K/Q):

• must permute r0, r1, r2 (by Assign. 5 problem 5(a));• is determined by its values at r0, r1, r2 (because K = Q(r0, r1, r2)).


This shows that the elements of Aut(K/Q) give all six possible permutations of theroots of x3 − 2.

Example. Let K be a field with |K| = pn with p > 1. Do we know any elements ofAut(K/Fp)?

Recall the Frobenius map ϕ : a 7→ ap; it is in Aut(K/Fp). Its fixed points are thesolutions in K to xp = x; i.e., the roots of xp − x. As there are at most p roots,ϕ 6= idK .ϕ2, ϕ3, . . . ∈ Aut(K/Fp). For each d ≥ 2, the fixed points of ϕd are the roots of

xpd − x. Thus we see that ϕd 6= idK if d < n. However, we know that every element

of K is a root of xpn − x; hence ϕn = idK .

This shows |Aut(K/Fp)| ≥ n. But |Aut(K/Fp)| ≤ [K : Fp] = n (we’ll prove thisafter the break), so Aut(K/Fp) is cyclic of order n, generated by the Frobenius map.

40 R. WILLARD

19. February 24 – Counting automorphisms

Given K/F finite, we want to determine |Aut(K/F )|. To do that, we need togeneralize.

Definition. Suppose F ≤ E ≤ K and F ∗ ≤ K with [K : F ] = [K : F ∗] <∞. Givenϕ : F ∼= F ∗, let

Inj(E,K/ϕ) = {σ : σ is an injective ring hom. E ↪→ K and σ�F = ϕ}.

Observe that Aut(K/F ) = Inj(K,K/idF ). In general, we will determine |Inj(K,K/ϕ)|.

First we prove the following.

Lemma. Suppose F ≤ F (c) ≤ K and ϕ : F ∼= F ∗ ≤ K with [K : F ] = [K : F ∗] <∞.Let p(x) be the minimal polynomial for c/F and let p∗(x) = ϕ(p(x)). Then

|Inj(F (c), K/ϕ)| = the number of distinct roots of p∗(x) in K.

Proof. Let the roots of p∗(x) in K be {d1, . . . , d`}. Each τ ∈ Inj(F (c), K/ϕ) isdetermined by its values on F ∪ {c}. Since τ�F is pre-determined (it must be ϕ), τis determined by its value at c.

Let p(x) = xk = ak−1xk−1+· · ·+a1x+a0. Applying τ to the equation p(c) = 0 gives

p∗(τ(c)) = 0. Thus τ(c) ∈ {d1, . . . , d`}. This already proves |Inj(F (c), K/ϕ)| ≤ `.On the other hand, given any di, we can apply the first Newfoundland Theorem fromFeb. 7 to get τi : F (c) ∼= F ∗(di) extending ϕ. Then τi ∈ Inj(F (c), K/ϕ), and sinceτ1, . . . , τ` are necessarily distinct (as they differ at c) we get |Inj(F (c), K/ϕ)| ≥ `. �

Here is a second ingredient we need. Suppose we have F ≤ E ≤ K and F ∗ ≤ Kand ϕ : F ∼= F ∗ with [K : F ] = [K : F ∗] < ∞. Each σ ∈ Inj(K,K/ϕ) restricts toan embedding σ�E : E ↪→ K. Let τ := σ�E. Note that τ ∈ Inj(E,K/ϕ). Moreover,τ : E ∼= τ(E) ≤ K, and σ ∈ Inj(K,K/τ).

Thus

Inj(K,K/ϕ) =⋃

τ∈Inj(E,K/ϕ)

Inj(K,K/τ)

so

|Inj(K,K/ϕ)| =∑

τ∈Inj(E,K/ϕ)

|Inj(K,K/τ)|.

Theorem. Suppose F, F ∗ ≤ K and ϕ : F ∼= F ∗ with [K : F ] = [K : F ∗] <∞. Then|Inj(K,K/ϕ)| ≤ [K : F ].

Proof. By induction on [K : F ]. If [K : F ] = 1 then K = F and Inj(K,K/ϕ) = {ϕ}.Assume [K : F ] = n > 1. Pick any c ∈ K \ F and let E = F (c). Let k = deg(c/F )and let m = [K : F (c)], so n = mk. Let p(x) be the minimal polynomial of c/F ,define p∗(x) as before, and let ` = the number of distinct roots of p∗(x) in K.

42 R. WILLARD

20. February 26 – Counting continued

On Monday I proved

Theorem. Suppose F, F ∗ ≤ K and ϕ : F ∼= F ∗ with [K : F ] = [K : F ∗] <∞. Then|Inj(K,K/ϕ)| ≤ [K : F ].

F

K

F ∗

K

n n

ϕ

∼=

Recall: the proof was by induction, and the induction step proceeded as follows: Pickany c ∈ K \ F and look at F (c).

F

F (c)

K

F ∗

K

k

m

n n

ϕ

∼=

I noted that if σ ∈ Inj(K,K/ϕ) and τ := σ�F (c), then

(1) τ ∈ Inj(F (c), K/ϕ).(2) τ : F (c) ∼= τ(F (c)) ≤ K.(3) σ ∈ Inj(K,K/τ).

Thus to count the number of σ ∈ Inj(K,K/ϕ), it suffices to

(1) count the number of τ ∈ Inj(F (c), K/ϕ).(2) For each such τ , count the number of σ ∈ Inj(K,K/τ).

We use the induction hypothesis to handle the second item. (Note that for each τ ,[τ(F (c)) : F ∗] = k so [K : τ(F (c))] = m = [K : F (c)].)

Hence all the “real work” is in counting the τ ’s. The key lemma was:

Lemma. In the above situation, |Inj(F (c), K/ϕ)| = the number of distinct roots ofp∗(x) in K, where p(x) is the minimal polynomial of c/F and p∗(x) is its image underϕ.


Now let’s specialize the argument to the following example: F = F ∗ = Q, ϕ = idQ,

and K = Q( 3√

2, i√

3). Hence Inj(K,K/ϕ) = Aut(K/Q).

Q

Q( 3√

2)

K

Q

K

3

2

6 6

idQ

The first step is to count the number of τ ∈ Inj(Q( 3√

2), K/idQ). By Monday’sLemma, the number of τ ’s equals the number of roots of x3 − 2 in K, which is 3. So|Inj(Q( 3

√2), K/idQ)| = 3 and the maps are

τ0 : Q(3√

2) ∼= Q(3√

2) determined by τ0(3√

2) =3√

2 (i.e., τ0 = idQ( 3√2))

τ1 : Q(3√

2) ∼= Q(3√

2ω) determined by τ1(3√

2) =3√

2ω

τ2 : Q(3√

2) ∼= Q(3√

2ω2) determined by τ2(3√

2) =3√

2ω2.

Next, we take each τi and count how many σ ∈ Inj(K,K/τi) there are. For example,when i = 1, we need to count the number of σ completing the following diagram:

Q( 3√

2)

K

Q( 3√

2ω)

Kσ

2 2

τ1

∼=F := =: F ∗

Observe that K = F (i√

3). Clearly i√

3 6∈ F , so 1 6= deg(i√

3/F ) ≤ 2, sodeg(i

√3/F ) = 2 and the minimal polynomial of i

√3/F is p(x) := x2 + 3 (same

as over Q). p∗(x) is also x2 + 3 because τ1 fixes each rational number. So by Mon-day’s Lemma, the number of extensions of τ1 to automorphisms of K is the numberof roots of x2 + 3 in K, which is 2. Similarly, the number of extensions of τ0 is 2, andthe number of extensions of τ2 is 2. So |Aut(K/Q)| = |Inj(K,K/idQ)| = 2+2+2 = 6.

Now suppose |Aut(K/F )| = [K : F ] = n. Imagine counting |Aut(K/F )| be themethod of Monday’s theorem. We can start with any c ∈ K and get the picture

44 R. WILLARD

F

F (c)

K

F

K

k

m

n n

idF

∼=

As on Monday, we know that

|Aut(K/F )| = |Inj(K,K/idF )| =∑

τ∈Inj(F (c),K/idF )

|Inj(K,K/τ)|.

By Monday’s theorem, |Inj(K,K/τ)| ≤ [K : F (c)] = m for each τ in the summation.Let

(1) p(x) be the minimal polynomial for c/F .(2) ` be the number of distinct roots of p(x) in K.

Note that p(x) has degree k and that the number of distinct τ ’s in the above sum-mation is exactly ` (by Monday’s Lemma). Thus from the above information, weget

mk = n = |Aut(K/F )| ≤∑τ

m = m`.

Hence k ≤ `. But obviously ` ≤ k, so we have ` = k. This proves:

If |Aut(K/F )| = [K : F ] < ∞, then for every c ∈ K, the minimalpolynomial of c/F is separable over F and splits in K[x].


21. March 3 – Characterizing Aut(K/F ) = [K : F ]

Last Wednesday I proved:

Proposition. Suppose K/F is finite and |Aut(K/F )| = [K : F ]. Then for everyc ∈ K, if p(x) is the minimal polynomial of c/F and k = deg(p(x)), then p(x) has kdistinct roots in K.

This is equivalent to saying p(x) splits in K[x] and is separable.

Definition. Let K/F be an algebraic extension.

(1) K/F is normal if for every c ∈ K, the minimal polynomial of c/F splits inK[x].

(2) K/F is separable if for every c ∈ K, the minimal polynomial of c/F isseparable.

Corollary. If |Aut(K/F )| = [K : F ] <∞, then K/F is normal and separable.

Today I want to prove a sufficient condition for |Aut(K/F )| = [K : F ].

Proposition (DF Prop. 14.5). Suppose K is a splitting field over F of a separablepolynomial f(x) ∈ F [x]. Then |Aut(K/F )| = [K : F ].

Proof. Not surprisingly, we will actually prove something stronger, namely:

Suppose F, F ∗ ≤ K and ϕ : F ∼= F ∗ with [K : F ] = [K : F ∗] < ∞.If there exists F0 ≤ F, F ∗ such that (i) ϕ�F0

= idF0 , and (ii) K is thesplitting field over F0 of some separable polynomial in f(x) ∈ F0[x],then Inj(K,F/ϕ) = [K : F ].

F

K

F ∗

K

n n

ϕ

∼=

F0 F0

t t

idF0

We prove this by induction on n = [K : F ]. Let c1, . . . , cN be the roots of f(x)in K. Then K = F (c1, . . . , cN). Hence in the induction step we can pick somec = ci ∈ K \ F . Consider F (c), where k = [F (c) : F ] > 1.

46 R. WILLARD

F

F (c)

K

F ∗

K

k

m

n n

ϕ

∼=

F0 F0

t t

idF0

Let p(x) be the minimal polynomial for c/F , so deg(p(x)) = k. Let p∗(x) be theimage of p(x) under ϕ.

Claim: p∗(x) has k distinct roots in K.Proof. p(x)|f(x) in F [x]. Thus ϕ(p(x))|ϕ(f(x)) in F ∗[x]. But ϕ(f(x)) = f(x)

because f ∈ F0[x] and ϕ�F0= idF0 , and ϕ(p(x)) = p∗(x). Hence p∗(x)|f(x) in F ∗[x],

and hence also in K[x].f(x) is separable and splits in K[x], so f(x) = aN(x − c1(x − c2) · · · (x − cN). As

p∗(x) divides f(x) and is monic and has degree k, it follows that p(x) = (x− ct1)(x−ct2) · · · (x− ctk) for some 1 ≤ t1 < t2 · · · < tk ≤ N . As ct1 , . . . , ctk ∈ K, we have thatp∗(x) has k distinct roots in K as claimed.

It follows from the Lemma from Feb 24 that |Inj(F (c), K/ϕ)| = k. Let τ1, . . . , τkbe the elements of Inj(F (c), K/ϕ). Fix and consider one of them, say τi. We have

F

F (c)

K

F ∗

τi(F (c))

K

k

m

k

m

ϕ

τi

∼=

F0 F0

t t

idF0


We can apply the inductive hypothesis to τi to get |Inj(K,K/τi)| = [K : F (c)] = m.Hence by the usual counting argument,

|Inj(K,K/ϕ)| =k∑i=1

|Inj(K,K/τi)|

=k∑i=1

m (induction)

= k ·m= n = [K : F ].

This completes the proof of the stronger claim. The claim in the statement of theProposition is the special case when F = F ∗ = F0. �

Theorem (DF Cor. Thm. 14.13). Let K/F be finite. TFAE:

(1) |Aut(K/F )| = [K : F ].(2) K/F is normal and separable.(3) K is the splitting field over F of some separable f(x) ∈ F [x].

Proof. (1) ⇒ (2). Proved Wednesday.(3) ⇒ (1). Proved above.(2) ⇒ (3). We can write K = F (c1, . . . , cn) with each ci algebraic/F . Possibly

some ci, cj have the same minimal polynomial; group them together according totheir minimal polynomial as

c1,1, . . . , c1,j1︸︷︷︸p1(x)

, c2,1, . . . , c2,j2︸︷︷︸p2(x)

, . . . , ct,1, . . . , ct,j1︸︷︷︸pt(x)

.

Let f(x) = p1(x)p2(x) · · · pm(x). Each pi(x) is separable, and no two pi(x), pj(x) canshare a root, so f(x) is separable. It remains to show that K is the splitting field off(x) over F .

Each pi(x) splits in K[x]; thus f(x) splits in K[x]. Extend each list ci,1, . . . , ci,ji tothe full list ci,1, . . . ci,mi

of roots of pi(x) in K. We have

K = F (c1, . . . , cn)

= F (c1,1, . . . , c1,j1 , c2,1, . . . , c2,j2 , . . . , ct,1, . . . , ct,jt) (rearrangement)

⊆ F (c1,1, . . . , c1,m1 , c2,1, . . . , c2,m2 , . . . , ct,1, . . . , ct,mt︸︷︷︸roots of f(x)

) ⊆ K.

Hence K is the splitting field for f(x) over F . �

48 R. WILLARD

22. March 5 – The Galois correspondence: example

Definition. (1) K/F is Galois if K/F is finite and |Aut(K/F )| = [K : F ].(2) If K/F is Galois, then Aut(K/F ) is called the Galois group of K/F .

Definition. Suppose K is a field and H ≤ Aut(K). The fixed field of H is

Fix(H) = {c ∈ K : σ(c) = c for all σ ∈ H}=

⋂σ∈H

Fix(σ).

In Assign. 6 you show that Fix(σ) ≤ K. As the intersection of subfields is asubfield, we see that Fix(H) ≤ K.

Suppose F ≤ K and H ≤ Aut(K/F ). Then a ∈ F =⇒ σ(a) = a for all σ ∈ H=⇒ a ∈ Fix(H). Hence F ≤ Fix(H) ≤ K.

This motivates the next definition.

Definition. Suppose K/F is an extension and G = Aut(K/F ). Define

Int(K/F ) = {E : F ≤ E ≤ K}Sub(G) = {H : H ≤ G}.

The rule H 7→ Fix(H) defines a map Sub(G)→ Int(K/F ).

Example. Let F = Q and K = Q( 3√

2, i√

3). Let G = Aut(K/Q). K is the splittingfield/Q of the separable polynomial x3− 2, so K/Q is Galois, i.e., |G| = [K : Q] = 6.We already knew this on Feb 14, as well as the fact that G ∼= S3.

Each σ ∈ G is determined by its values at 21/3 and i√

3. The possibilities are

σ σ(21/3) σ(i√

3) σ(ω) σ(r0) σ(r1) σ(r2)

σ0 21/3 i√

3 ω r0 r1 r2

σ1 21/3 −i√

3 ω r0 r2 r1

σ2 ω21/3 i√

3 ω r1 r2 r0

σ3 ω21/3 −i√

3 ω r1 r0 r2

σ4 ω21/3 i√

3 ω r2 r0 r1

σ5 ω21/3 −i√

3 ω r2 r1 r0

Let rk = ωk21/3 for k = 0, 1, 2. Thus r0, r1, r2 are the roots of x3 − 2. In the finalfour columns in the preceding table, we have worked out the values of each σi firstat ω, and then at r0, r1, r2.


What are the subgroups of G? They are

G

{σ0}〈σ2〉 = {σ0, σ2, σ4}〈σ1〉 = {σ0, σ1}〈σ3〉 = {σ0, σ3}〈σ5〉 = {σ0, σ5}.

What is Fix(H) for each of these subgroups?

(1) Fix({idK}) = K obviously.(2) Fix(〈σ1〉) = Fix(σ1) (easy argument). We see that 21/3 ∈ Fix(σ1) from the

table. Also Fix(σ1) 6= K since for example ω 6∈ Fix(σ1). Thus

Q(21/3) ≤ Fix(σ1) < K.

Thus Fix(σ1) = Q(21/3) by degree considerations.(3) Fix(〈σ3〉) = Q(r2) by similar arguments.(4) Fix(〈σ5〉) = Q(r1) by similar arguments.(5) Fix(〈σ2〉) = Q(ω) = Q(i

√3) by similar arguments.

(6) Fix(G) = Fix(〈σ1, σ2〉) = Fix(σ1) ∩ Fix(σ2) = Q(21/3) ∩ Q(i√

3) = Q (bydegree considerations).

Here are the subgroups of G and their fixed fields in a diagram:

{idK}

〈σ2〉〈σ1〉〈σ3〉〈σ5〉

G

2 3 3 3

3 2 2 2

Q

Q(i√

3) Q(21/3) Q(ω21/3) Q(ω21/3)

K

3 2 2 2

2 3 3 3

The diagram on the right is the image of the diagram on the left after “flipping”top-to-bottom. Note how the relative indices (of subgroups) on the left matches withthe relative degrees (of field extensions) on the right. Also note that the subgroup〈σ2〉 is normal in G, matching the fact that Fix(σ2) = Q(i

√3) is Galois over Q.

We will later see that the diagram on the right gives all the intermediate fieldsin Int(K/Q), and that the phenomena we are seeing here happens generally in anyGalois extension.

50 R. WILLARD

23. March 7 – Artin’s Theorem

Lemma. Suppose K is a field and H ≤ Aut(K), Let b1, . . . , bm ∈ K. Consider thehomogeneous system of linear equations

σ(b1)x1 + σ(b2)x2 + · · ·+ σ(bm)xm = 0 ∀σ ∈ H.If this system has a nontrivial solution in Km, then it has a nontrivial solution inFix(H)m.

Proof. LetV = {(u1, . . . , um) ∈ Km : (u1, . . . , um) is a solution}.

Km is a vector space over K and V is a vector subspace of Km of dimension ≥ 1. Vis also closed under H. Indeed, suppose (u1, . . . , um) ∈ V and τ ∈ H. Then

σ(b1)u1 + σ(b2)u2 + · · ·+ σ(bm)um = 0 ∀σ ∈ H.So applying τ to each equation we get

τ(σ(b1)u1 + σ(b2)u2 + · · ·+ σ(bm)um) = τ(0) ∀σ ∈ H,hence

τσ(b1)τ(u1) + τσ(b2)τ(u2) + · · ·+ τσ(bm)τ(um) = 0 ∀σ ∈ H.Note that as σ ranges over H, τσ also ranges over H. Hence we can rewrite the aboveas

σ′(b1)τ(u1) + σ′(b2)τ(u2) + · · ·+ σ′(bm)τ(um) = 0 ∀σ′ ∈ H,which implies (τ(u1), . . . , τ(um)) ∈ V .

Assuming there exists a nontrivial solution in Km, choose a nontrivial u ∈ V withthe fewest number of nonzero coordinates. For simplicity, assume u1, . . . , ur 6= 0 andur+1 = · · · = um = 0. Let v = u−1

r u; thus

v = (v1, . . . , vr−1, 1︸︷︷︸nonzero

, 0, 0, . . . , 0) ∈ V.

We claim that v ∈ Fix(H)m. Assume this is false. E.g., suppose v1 6∈ Fix(H).This means there exists τ ∈ H with τ(v1) 6= v1. We have

(τ(v1), . . . , τ(vr−1), 1, 0, 0, . . . , 0) ∈ V.Subtracting, we get

(v1 − τ(v1)︸︷︷︸6=0

, v1 − τ(v1), . . . , vr−1 − τ(vr−1), 0, 0, 0, . . . , 0) ∈ V.

This is a nontrivial solution with fewer nonzero coordinates than u, contradiction.Hence v ∈ Fix(H)m. �

Theorem (Artin; DF Thm. 14.9, Cor. 14.11). Suppose K is a field, H is a finitesubgroup of Aut(K), and F = Fix(H). Then K/F is Galois and H = Aut(K/F ).


Proof. Let n = |H|. We first show that [K : F ] ≤ n. Pick any b1, . . . , bn+1 ∈ K andconsider the homogeneous system

σ(b1)x1 + · · ·+ σ(bn+1)xn+1 = 0 ∀σ ∈ H.This is a system of n homogeneous linear equations in n + 1 unknowns. Hence ithas a nontrivial solution in Kn+1. By the Lemma, it follows that it has a nontrivialsolution (a1, . . . , an+1) in F n+1. One of the equations (the one with σ = idK) is

b1x1 + · · ·+ bn+1xn+1 = 0,

so∑n+1

i=1 aibi = 0. As ai ∈ F for all i and ai 6= 0 for some i, this proves b1, . . . , bn+1 arelinearly dependent over F . Since b1, . . . , bn+1 were arbitrary, this proves [K : F ] ≤ n.

In particular, [K : F ] <∞. Hence |Aut(K/F )| ≤ [K : F ] by Feb 24 theorem. Onthe other hand, note that if σ ∈ H, then

(1) σ ∈ Aut(K) (since H ≤ Aut(K));(2) For all a ∈ F , σ(a) = a (because F = Fix(H)).

Hence σ ∈ Aut(K/F ). This proves H ≤ Aut(K/F ).Putting this all together, we get

n = |H| ≤ |Aut(K/F )| ≤ [K : F ] ≤ n.

This proves H = Aut(K/F ) and |Aut(K/F )| = [K : F ]. �

52 R. WILLARD

24. March 10 – Fundamental Theorem, part 1

Recall from Friday:

Theorem (DF Thm. 14.9, Cor. 14.11). Suppose K is a field, H is a finite subgroupof Aut(K), and F = Fix(H). Then K/F is Galois and H = Aut(K/F ).

Corollary (DF Cor. 14.10). Suppose K/F is finite and G = Aut(K/F ). Then K/Fis Galois if and only if Fix(G) = F .

Proof. Let F1 = Fix(G). Then F ≤ F1 ≤ K, so

[K : F ] = [K : F1] · [F1 : F ].

We know that |G| ≤ [K : F ] (Feb 24), so G is finite. Hence K/F1 is Galois andG = Aut(K/F1) by Friday’s theorem. In particular, |G| = |K : F1]. Hence

K/F is Galois ⇐⇒ |G| = [K : F ]

⇐⇒ [K : F1] = [K : F ]

⇐⇒ F1 = F

⇐⇒ Fix(G) = F.

�

Theorem (DF Thm. 14.14, “Fund. Thm. of Galois Theory,” part 1). Suppose K/Fis Galois and G = Aut(K/F ). Recall that

Int(K/F ) = {E : F ≤ E ≤ K}Sub(G) = {H : H ≤ G}.

(0) There is a one-to-one correspondence between Sub(G) and Int(K/F ) given byH 7→ Fix(H) (H ∈ Sub(G)). The inverse map is given by E 7→ Aut(K/E)(E ∈ Int(K/F )).

(1) Both maps are inclusion reversing. That is, if E1, E2 correspond to H1, H2

respectively, then E1 ≤ E2 iff H2 ≤ H1.(2) If E1, E2 correspond to H1, H2 and E1 ≤ E2, then [E2 : E1] = [H1 : H2].

Proof. (0) Let λ : Int(K/F )→ Sub(G) and ρ : Sub(G)→ Int(K/F ) be given by

λ(E) = Aut(K/E)

ρ(H) = Fix(H).

It suffices to show ρ ◦ λ = idInt(K/F ) and λ ◦ ρ = idSub(G).To prove the second, assume H ≤ G. Then H is finite so the result follows from

Friday’s theorem (with F replaced by E).To prove the first, assume F ≤ E ≤ K.

Claim: K/E is Galois.


Proof: K is a splitting field over F of a separable polynomial; hence K is also asplitting field over E of the same polynomial, proving the claim.

Now the desired result follows from today’s Corollary (with F replaced by E andG replaced by H).

(1) This is easy. Suppose F ≤ E1 ≤ E2 ≤ K and let Hi = Aut(K/Ei). Then

σ ∈ H2 ⇐⇒ σ ∈ Aut(K/E2)

⇐⇒ σ ∈ Aut(K) and σ(c) = c ∀c ∈ E2

=⇒ σ ∈ Aut(K) and σ(c) = c ∀c ∈ E1 (as E1 ⊆ E2)

⇐⇒ σ ∈ Aut(K/E1)

⇐⇒ σ ∈ H1.

Thus H2 ⊆ H1. Similarly, if H2 ≤ H1 ≤ G then E1 = Fix(H1) ≤ Fix(H2) = E2.(2) K is Galois over both E1 and E2; hence [K : E1] = |Aut(K/E1)| = |H1| and

[K : E2] = |Aut(K/E2)| = |H2|. Hence

[E2 : E1] =[K : E1]

[K : E2]=|H1||H2|

= [H1 : H2].

�

Corollary. Supppose K/F is Galois. Then K/F has only finitely many intermediatefields.

Proof. Aut(K/F ) has only finitely many subgroups. �

Example. Let F = Q, K = Q(ζ7), and G = Aut(Q(ζ7)/Q). Q(ζ7)/Q is Galois ofdegree 6. It turns out that G is abelian of order 6, so is cyclic and has two nontrivialsubgroups of order 2 and 3.

By FTGT (the Fundamental Theorem of Galois Theory), it follows that there areonly two proper intermediate fields between Q and Q(ζ7), of degrees 3 and 2 over Qrespectively. (You will find them on Assignment 8.)

54 R. WILLARD

25. March 12 – Fundamental Theorem, part 2

Lemma. Suppose K/F is finite and normal. If c, d ∈ K have the same minimalpolynomial/F , then there exists σ ∈ Aut(K/F ) with σ(c) = d.

Proof. Since K/F is finite and normal, K is a splitting field over F of some g(x) ∈F [x] (Assign. 7 prob. 4). c, d ∈ K are roots of the same irreducible polynomial inF [x]. Hence by Assign. 5 prob. 5(b), there exists σ ∈ Aut(K/F ) with σ(c) = d. �

Theorem (DF Thm. 14.14, “Fund. Thm. of Galois Theory,” part 2). Suppose K/Fis Galois and G = Aut(K/F ). Let F ≤ E ≤ E∗ ≤ K and H := Aut(K/E) andH∗ := Aut(K/E∗). The following are equivalent:

(1) E∗/E is Galois.(2) E∗/E is normal.(3) H∗ CH.

Furthermore, if E∗/E is Galois, then Aut(E∗/E) ∼= H/H∗.

Proof. (1) ⇔ (2). Obviously E∗/E is finite since K/F is. We know that K/E isGalois (Monday), hence is separable, i.e., every c ∈ K is separable/E. In particular,every c ∈ E∗ is separable/E, so E∗/E is also separable. Thus E∗/E is Galois iffE∗/E is normal (Feb 26, Mar 3).

(2) ⇒ (3). Assume E∗/E is normal. Then by Assignment 7 prob. 4, E∗/E isstable; in particular, every σ ∈ Aut(K/E) = H satisfies σ(E∗) = E∗. Define Φ :Aut(K/E) → Aut(E∗/E) by Φ(σ) = σ�E∗ . By Assign. 7 prob. 3, Φ is a grouphomomorphism. Claim: Φ is surjective. Why? Well, given any τ ∈ Aut(E∗/E),consider

E∗

K

E∗

K

τ

K is Galois over E, so is a splitting field over E of some g ∈ E[x]. Hence K isalso a splitting field over E∗ of g(x). Let g∗(x) be the image of g(x) under τ . Theng∗(x) = g(x) because τ fixes every element of E, so certainly K is a splitting fieldover E∗ of g∗(x). Hence all the hypotheses of the 2nd Newfoundland Theorem aremet, and we can conclude that τ extends to some σ ∈ Aut(K). Clearly σ�E = idE,so σ ∈ Aut(K/E), and Φ(σ) = σ�E∗ = τ . This proves Φ is surjective.

Next we calculate ker(Φ). For any σ ∈ Aut(K/E),

σ ∈ ker(Φ) ⇐⇒ σ�E∗ = idE∗ ⇐⇒ σ ∈ Aut(K/E∗).

I.e., ker(Φ) = Aut(K/E∗) = H∗. Because kernels are always normal, we get H∗ CAut(K/E) = H, which proves (3). By the First Isomorphism Theorem we getH/ ker(Φ) = H/H∗ ∼= ran(Φ) = Aut(E∗/E), which proves the “Furthermore” claim.


(3) ⇒ (2). Assume H∗ C H. Let c ∈ E∗ be arbitrary and let p(x) ∈ E[x] be itsminimal polynomial/E. To prove E∗/E is normal, it suffices to prove that p(x) splitsin E∗[x].

Since K/E is Galois, it is normal, so p(x) splits in K[x]. Thus to prove that p(x)splits in E∗[x], it suffices to prove that every root of p(x) in K actually lies in E∗.

Let d be a root of p(x) in K. By today’s Lemma, there exists σ ∈ Aut(K/E) = Hwith σ(c) = d. Let τ ∈ H∗ be arbitrary. Observe that σ−1τσ ∈ σ−1H∗σ = H∗ (usingH∗CH), so σ−1τσ ∈ H∗ = Aut(K/E∗). Since c ∈ E∗ and σ−1τσ fixes every elementof E∗, we get σ−1τσ(c) = c; hence

τ(d) = τσ(c) = σσ−1τσ(c) = σ(σ−1τσ(c)) = σ(c) = d.

As τ was an arbitrary member of H∗, this proves d ∈ Fix(H∗) = E∗ as required. �

56 R. WILLARD

26. March 14 – Finite fields revisited

Definition. For a prime p and integer n ≥ 1, Fpn denotes the (unique up to isomor-phism) field of size pn.

Lemma. Fpn/Fp is Galois.

Proof. Fpn is a splitting field over Fp of xpn − x, so Fpn/Fp is normal. We know that

every irreducible polynomial in Fp[x] is separable (Jan 22), so Fpn/Fp is separable.Hence Fpn/Fp is Galois by the main theorem from Mar 3.

Alternatively, on Feb 14 we saw that the Frobenius map ϕ on Fpn is in Aut(Fpn/Fp)and has order n, so n ≤ |Aut(Fpn/Fp)| ≤ [Fpn : Fp] = n, so we have equality. �

Let G = Aut(Fpn/Fp). The second proof shows that G = 〈ϕ〉 is cyclic of order n.Thus for every d|n, 〈ϕd〉 is a subgroup of G n/d, and these are the only subgroups.For example, suppose n = 6. Then |G| = 6 and the subgroups look like

{id}

〈ϕ2〉〈ϕ3〉

G

2

32

3

The FTGT implies that the subfields of Fp6 are in one-one correspondence withthe subgroups of G; thus they are as shown in the diagram (below) on the left:

Fp6

Fix(ϕ2)Fix(ϕ3)

Fp

2

32

3

Fp6

Fp2Fp3

Fp

2

32

3

Clearly |Fix(ϕd)| = pd for d = 2, 3 as [Fix(ϕd) : Fp] = d; hence Fix(ϕd) = Fpd .Thus the picture of subfields of Fp6 looks like the diagram on the right (above).

In general we have:

Lemma. Fix p and n. For each d|n, Fpn has Fpd as a subfield, and has no othersubfields.

We can use this last fact to learn something about irreducible polynomials in Fp[x].

Definition. If K/F is finite, then Min(K/F ) = the set of minimal polynomials/Fof c as c ranges over K.

Lemma. If K/F is Galois, then∏

c∈K(x− c) =∏

f∈Min(K/F ) f .


(Remark: if K is infinite, both products in the statement of the above Lemma areinfinite. Let’s not get bent out of shape over this.)

Proof. Each f ∈ Min(K/F ) splits in K[x] (by normality) and has no repeated roots(by separability). Thus if we partition the elements of K according to their minimalpolynomial/F , then the product on the left can be grouped as the product on theright. �

Definition. For p prime and d ≥ 1, let Irrp(d) = the set of all monic irreduciblepolynomials in Fp[x] of degree d.

For example (Assign. 2), Irr2(4) = {x4 + x+ 1, x4 + x3 + 1, x4 + x3 + x2 + x+ 1}.

Lemma. Min(Fpn/Fp) =⋃d|n Irrp(d).

Proof. (⊆). If f(x) is the minimal polynomial/Fp of some c ∈ Fpn and d = deg(f),then f ∈ Irrp(d). Thus it remains to show d|n. Consider Fp ≤ Fp(c) ≤ Fpn . Fp(c) isa subfield of Fpn of degree d over Fp, so d|n by a previous Lemma.

(⊇). Suppose d|n and f(x) ∈ Irrp(d). Let K be a splitting field of f(x) over Fp, letc ∈ K be a root of f(x), and consider Fp(c) ≤ K. |Fp(c)| = pd, so Fp(c) ∼= Fpd ≤ Fpn .Hence f(x) also has a root c′ ∈ Fpn , and is the minimal polynomial over Fp of thisc′. �

Corollary (DF Prop. 14.18). xpn − x factors in Fp[x] as

∏d|n Irrp(d).

Example: in F2[x],

x16 − x = x(x+ 1)(x2 + x+ 1)(x4 + x+ 1)(x4 + x3 + 1)(x4 +3 +x2 + x+ 1).

58 R. WILLARD

27. March 17 – Galois groups; discriminant test

(DF §14.6)

Suppose f ∈ F [x] is separable, K is a splitting field of f(x) over F . Thus K/F isGalois. Let G = Aut(K/F ). Let deg(f) = n and let r1, . . . , rn be the roots of f(x)in K. By Assign. 5 #5(a), each σ ∈ G permutes r1, . . . , rn. Let X = {r1, . . . , rn}and let σ := σ�X ∈ SX .

It is easy to check that the map : G→ SX is a group homomorphism (similar toAssign. 7 #3(b)). Since K = F (r1, . . . , rn), it follows from Assign. 6 #1(b) that is

injective. Hence is an isomorphism from G to G := {σ�X : σ ∈ G} ≤ SX .

Definition. Assume f(x) ∈ F [x] is separable. The Galois group of f(x) (over F )

is the permutation group G ≤ SX , where X is the set of roots of f(x) in a splitting

extension K/F and G = Aut(K/F ). Gal(f(x)/F ) denotes G.

Note: When the numbering of the roots is understood, we get a natural isomorphismSX ∼= Sn and we can treat Gal(f(x)/F ) as a subgroup of Sn.

Examples

(1) If f(x) = x3 − 2 ∈ Q[x], then X = {r0, r1, r2} where ri = ωi 3√

2. We know

that if K = Q(r0, r1, r2) and G = Aut(K/Q), then |G| = 6 and so G = SX .Thus

Gal(f(x)/Q) = {id, (r0 r1 r2), (r0 r2 r1), (r0 r1), (r0 r2), (r1 r2)} = SX .

Equivalently, numbering the roots ri i + 1, we get Gal(x3 − 2/Q) = S3.In this example, (r1 r2) = τ�X where τ ∈ Aut(K/Q) is given by τ(z) = z.Under ri i+ 1, this corresponds to the 2-cycle (2 3) ∈ S3.

(2) In Assign. 6 you explored f(x) = x4−2x2+3, whose roots are α, α,−α,−α. Ifwe number them r1, r2, r3, r4 and let X = {r1, r2, r3, r4}, K = Q(r1, r2, r3, r4),and G = Aut(K/Q), then

Gal(f(x)/Q) = {idX ,(r1 r2 r3 r4) (= σ2)

(r1 r3)(r2 r4) (= σ22)

(r1 r4 r3 r2) (= σ−12 )

(r1 r2)(r3 r4) (= σ1)

(r1 r4)(r2 r3) (= σ2σ1σ−12 )

(r1 r3) (= σ2σ1)

(r2 r4)} (= σ1σ2).

Under the correspondence ri i, this group corresponds to

{id, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 2)(3 4), (1 4)(2 3), (1 3), (2 4)} ≤ S4.


The general problem, given f(x) of degree n and an enumeration of its roots, is tofind Gal(f(x)/F ) explicitly as a subgroup of Sn.

Definition. A permutation group H ≤ SX is transitive if for all i, j ∈ X thereexists σ ∈ H with σ(i) = j.

Proposition. Assume f(x) ∈ F [x] is separable. Then f(x) is irreducible in F [x] iffGal(f(x)/F ) is transitive.

Proof. (⇒) Assume f(x) is irreducible and let X be the set of its roots in its splittingfield, K, over F . By Assign. 5 #5(b), for any b, c ∈ X there exists σ ∈ Aut(K/F )with σ(b) = c. Thus σ�X ∈ Gal(f(x)/F ) and σ�X(b) = c, proving Gal(f(x)/F ) istransitive.

(⇐) Assume f(x) is not irreducible. Write f(x) = p1(x)p2(x) · · · pk(x) where eachpi(x) is irreducible in F [x]. Then k ≥ 2 as f(X) is not irreducible, and pi(x) 6= pj(x)for i 6= j, as f(x) is separable so has no repeated roots. Each pi(x) splits in K[x].Hence by Assign. 5 #5(a), every σ ∈ Aut(K/F ) permutes the roots of each p1(x).In particular, no σ ∈ Aut(K/F ) can send a root of p1(x) to a root of p2(x). ThusGal(f(x)/F ) is not transitive. �

Corollary. If f(x) ∈ F [x] is separable, irreducible, and deg(f) = 3, then Gal(f(x)/F )is either S3 or A3.

Proof. These are the only transitive subgroups of S3. �

Given separable f(x) ∈ F [x] irreducible of degree 3, how can we determine whichalternative holds?

Definition. Suppose f(x) ∈ F [x] is separable with deg(f) = 3. Let its roots in asplitting field be r1, r2, r3. The discriminant of f(x) is

D = (r1 − r2)2(r1 − r3)2(r2 − r3)2.

Also define∆ = (r1 − r2)(r1 − r3)(r2 − r3)

and observe that ∆2 = D.

Observe that D does not depend on how we have ordered the roots. That is,if r1, r2, r3 are replaced by any permutation ri1 , ri2 , ri3 , the expression for D wouldevaluate the same. In other words, D is well-defined. On the other hand, an oddpermutation of the roots would change the sign of ∆, so ∆ is defined only up to ±.

Let K be a splitting field/F of f(x), G = Aut(G/F ), and X = {r1, r2, r3}. By theabove comments, for every σ ∈ G we have σ(D) = D. Thus D ∈ Fix(G). K/F isGalois, so Fix(G) = F , proving D ∈ F . On the other hand,

σ(∆) =

{∆ if σ is even−∆ if σ is odd

60 R. WILLARD

Provided ∆ 6= −∆ (i.e., ch(F ) 6= 2), we get that ∆ ∈ Fix(G) = F iff σ is even forevery σ ∈ G; i.e., iff Gal(f(x)/F ) ⊆ A3. Hence:

Proposition (DF Prop. 14.33). If f(x) ∈ F [x] is separable and irreducible of degree3, and ch(F ) 6= 2, then

Gal(f(x)/F ) =

{A3 if D has a square-root in FS3 otherwise.


28. March 19 – Formula for the discriminant

Definition. Suppose f(x) ∈ F [x] is separable with deg(f) = n ≥ 2, with rootsr1, . . . , rn in a splitting field. The discriminant of f(x) is

D =∏i<j

(ri − rj)2

and

∆ =∏i<j

(ri − rj).

Fact (DF Prop. 14.34). D ∈ F , and ∆ ∈ F iff Gal(f(x)/F ) ⊆ An.We will derive a formula for D in case f(x) is monic and deg(f) = 3. (The method

generalizes to higher degrees).First consider the quadratic case. Suppose f(x) = x2 + bx+ c = (x− r1)(x− r2).

Then b = −(r1 + r2) and c = r1r2. Also,

D = (r1 − r2)2

= (r1 + r2)2 − 4r1r2

= b2 − 4c.

For n > 2, we need some additional notation. Suppose f(x) = x3 + ax2 + bx+ c =(x− r1)(x− r2)(x− r3). Define

t1 = r1 + r2 + r3

t2 = r21 + r2

2 + r23

...

tk = rk1 + rk2 + rk3 .

We first derive formulas for the ti’s; then we will derive a formula for D in terms ofthe ti. Observe first that

a = −(r1 + r2 + r3)

b = r1r2 + r1r2 + r2r3

c = −r1r2r3.

Observe next that t1 = −a. To find t2,

(−a)2 = (r1 + r2 + r3)2

= r21 + r2

2 + r23 + 2(r1r2 + r1r + 3 + r2r3)

= t2 + 2b.

Hence t2 = a2− 2b. We can continue in this way by brute force; however the analysisis simplified by using

62 R. WILLARD

Newton’s Identities (proved in most texts except DF):

t1 = −at2 + at1 = −2b

t3 + at2 + bt1 = −3c

t4 + at3 + bt2 + ct1 = 0

t5 + at4 + bt3 + ct2 = 0, etc.

From Newton’s identities we get

t1 = −at2 = −at1 − 2b = a2 − 2b

t3 = −at2 − bt1 − 3c = (−a)(a2 − 2b)− b(−a)− 3c

= −a3 + 3ab− 3c

t4 = −at3 − bt2 − ct1 = (−a)(−a3 + 3ab− 3c)− b(a2 − 2b)− c(−a)

= a4 − 4a2b+ 4ac+ 2b2.

How do these help calculate the discriminant of f(x)?Consider the Vandermonde matrix for r1, r2, r3:

V =

1 r1 r21

1 r2 r22

1 r3 r23

.Its determinant can be calculated by cofactor expansion along the first row:

det(V ) = 1 · det

(r2 r2

2

r3 r23

)− r1 · det

(1 r2

2

1 r23

)+ r2

1 · det

(1 r2

1 r3

)= (r2r

23 − r3r

22)− r1(r2

3 − r22) + r2

1(r3 − r2)

= (r3 − r2)[r2r3 − r2(r2 − r3) + r21]

= (r3 − r2)(r2 − r1)(r3 − r1)

= −∆.

Hencedet(V tV ) = det(V t) det(V ) = (−∆)2 = D.

What is V tV ?

V tV =

1 1 1r1 r2 r3

r21 r2

2 r23

1 r1 r21

1 r2 r22

1 r3 r23

=

3 t1 t2t1 t2 t3t2 t3 t4

.


Thus

D = det(V tV )

= 3 det

(t2 t3t3 t4

)− t1 det

(t1 t3t2 t4

)+ t2 det

(t1 t2t2 t3

)= 3(t2t4 − t23)− t1(t1t4 − t2t3) + t2(t1t3 − t22)

= (3t2 − t21)t4 − 3t33 + 2t1t2t3 − t32.Putting these together, we get

D = (3(a2 − 2b)− a2)(a4 − 4a2b+ 4ac+ 2b2)− 3(−a3 + 3ab− 3c)3

+ 2(−a)(a2 − 2b)(−a3 + 3ab− 3c)− (a2 − 2b)3

= a2(b2 − 4c)− 4b3 − 27c2 + 18abc.

Example 28.1.

(1) Consider x3 − 2 over Q. We have a = b = 0 and c = −2, so the discriminantis D = −27(−2)2 = −108. As

√−108 6∈ Q, this proves Gal(x3 − 2/Q) = S3

(which we already knew).(2) Consider f(x) = x3 − 3x + 1 over Q. It can be checked by the rational

roots theorem that f(x) is irreducible in Q[x]. By graphing, it can be seenthat f(x) has three real roots. We find D = 81. As

√81 ∈ Q, we have

Gal(f(x)/Q) = A3.(3) Let g(x) = x3 − 4x + 2. g(x) is obviously irreducible by Eisenstein, and

graphing shows it has three real roots. D = 148 = 4 · 37. As√D 6∈ Q, we

have Gal(g(x)/Q) = S3.

64 R. WILLARD

29. March 21 – Sick-Day Propositions

In Assign. 5 prob. 1(b) you showed that there exists a field E ≤ C and d ∈ Csuch that then [Q(d) : Q] = 3 but [E(d) : E] = 2. This is not possible when E/Q isGalois, as the next proposition shows.

Proposition (special case of DF Prop. 14.19). Assume F ≤ K ≤ L with K/FGalois. Also assume c ∈ L is such that [F (c) : F ] = p is prime and c 6∈ K.

F

KF (c)

K(c)

L

p

Then K(c)/F (c) is Galois and Aut(K(c)/F (c)) ∼= Aut(K/F ).

Remark: In particular, |Aut(K(c)/F (c))| = |Aut(K/F )|. Hence [K(c) : F (c)] =[K : F ] by Galoisness, and then [K(c) : K] = p by the product theorem.

Proof. Pick f(x) ∈ F [x] separable such that K is a splitting field/F for f(x). Ifr1, . . . , rn are the roots of f(x) inK, thenK = F (r1, . . . , rn), soK(c) = F (c, r1, . . . , rn) =F (c)(r1, . . . , rn), so K(c) is a splitting field/F (c) for f(x). As f(x) is separable, thisproves that K(c)/F (c) is Galois.

Suppose σ ∈ Aut(K(c)/F (c)). Then σ ∈ Aut(K(c)/F ). Focussing on F ≤ K ≤K(c), we have that K/F is Galois, so is normal, so is stable (Assign. 7 prob. 4(c)).Hence σ(K) = K. Thus σ�K ∈ Aut(K/F ). This proves that we have a map

�K : Aut(K(c)/F (c))→ Aut(K/F )

and as usual it is a group homomorphism. We’ll prove it’s one-to-one and onto.

One-to-one: check its kernel. For σ ∈ Aut(K(c)/F (c)),

σ ∈ ker(�K) ⇐⇒ σ ∈ Aut(K(c)/F (c)) and σ�K = idK

⇐⇒ σ ∈ Aut(K(c)) and Fix(σ) ⊇ F (c) ∪K⇐⇒ σ ∈ Aut(K(c)) and Fix(σ) = K(c)

⇐⇒ σ = idK(c).

Hence �K is one-one.

Onto: let H = ran(�K) ≤ Aut(K/F ) and n = |Aut(K(c)/F (c))| = |H|. Apply theFTGT to {idK} ≤ H ≤ Aut(K/F ):


{idK}

H

Aut(K/F )

n

=⇒

F

Fix(H)

K

F (c)

K(c)

n

n

p

Observe that [K(c) : F (c)] = |Aut(K(c)/F (c)| = n, so [K(c) : F ] = np. Since[K(c) : K] > 1 by hypothesis, we must have [K(c) : K] = p and [Fix(H) : F ] = 1,i.e., Fix(H) = F . Again by FTGT, this means H = Aut(K/F ). �

The next two results concern subfields of C. Given a ∈ C and n ≥ 2, n√a denotes

your favorite complex n-th root of a.

Proposition (cf. DF Prop. 14.36). Suppose F ≤ C, p is prime, ζp ∈ F , and a ∈ Cwith a ∈ F but p

√a 6∈ F . Then F ( p

√a)/F is Galois and Aut(F ( p

√a)/F ) is cyclic of

order p.

Proof. Let f(x) = xp − a ∈ F [x] and K = F ( p√a). The roots of f(x) are (ζp)

i p√a for

i = 0, 1, . . . , p− 1. If F contained any one root, then F would contain all the roots,which is false as p

√a 6∈ F . Thus F contains no roots of xp − a. It follows from the

Bonus problem to Assign. 1 that xp − a is irreducible in F [x] and hence [K : F ] = p.Obviously K is a splitting field for f(x) over F and hence K/F is Galois, meaning|Aut(K/F )| = [K : F ] = p. As p is prime, Aut(K/F ) must be cyclic. �

The final result is a converse to the previous result.

Proposition (cf. DF Prop. 14.37). Suppose F ≤ K ≤ C, p is prime, ζp ∈ F , K/F isGalois, and [K : F ] = p. Then there exists a ∈ F such that p

√a 6∈ F and K = F ( p

√a).

Proof. Let G = Aut(K/F ). G is a cyclic group of order p. Let σ ∈ G be a generator.Then σp = idK .

Recall that we can consider K as a finite-dimensional vector space over F , andfrom this point of view σ : K → K is an F -linear transformation. Recall from linearalgebra that the minimal polynomial of σ is the monic polynomial m(x) ∈ F [x]of least degree such that m(σ) = 0K . Because σp = idK , the minimal polyno-mial for σ must be a factor of xp − 1. Thus the eigenvalues for σ must be among1, ζp, (ζp)

2, . . . , (ζp)p−1. It follows that:

(1) F contains all the eigenvalues of σ (because ζp ∈ F ).

66 R. WILLARD

(2) The minimal polynomial for σ has no repeated roots (as it is a factor of xp−1,which has no repeated roots).

It follows that σ is diagonalizable. Since σ 6= idK , it cannot be that 1 is the onlyeigenvalue of σ. Hence there exists 1 ≤ i < p such that (ζp)

i is an eigenvalue of σ.Let b ∈ K be an eigenvector for (ζp)

i; thus b 6= 0 and σ(b) = (ζp)ib. In particular,

σ(b) 6= b, so b 6∈ F (as σ fixes every element of F ). Let a = bp. Then b = (ζp)j p√a for

some j. Clearly p√a 6∈ F and p

√a ∈ K (since ζp ∈ F and b ∈ K \ F ). It follows that

K = F ( p√a) (since [K : F ] = p is prime). Observe that

σ(a) = σ(bp) = σ(b)p = ((ζp)ib)p = bp = a.

Thus a ∈ Fix(σ) = Fix(〈σ〉) = Fix(G) = F . �


30. March 24 – Solving cubic equations by radicals

Highlights from last Friday (“Sick-Day Propositions”):

Proposition (special case of DF Prop. 14.19). Assume F ≤ K ≤ L with K/FGalois. Also assume c ∈ L is such that [F (c) : F ] = p is prime and c 6∈ K.

F

KF (c)

K(c)

p

Then K(c)/F (c) is Galois and Aut(K(c)/F (c)) ∼= Aut(K/F ).

Proposition (cf. DF Prop. 14.36). Suppose F ≤ C, p is prime, ζp ∈ F , and a ∈ Cwith a ∈ F but p

√a 6∈ F . Then F ( p

√a)/F is Galois and Aut(F ( p

√a)/F ) is cyclic of

order p.

Proposition (cf. DF Prop. 14.37). Suppose F ≤ K ≤ C, p is prime, ζp ∈ F , K/F isGalois, and [K : F ] = p. Then there exists a ∈ F such that p

√a 6∈ F and K = F ( p

√a).

The remainder of the course explores the “algebraic” solution of polynomial equa-tions in C[x]. Start with the monic quadratic case: suppose f(x) = x2 + bx+ c (withb, c ∈ C) and let its roots be r1, r2. We want to express r1, r2 “in terms of b, c.” Iguess we could use the quadratic formula, but let’s pretend we don’t know it.

Let F = Q(b, c), let K be the splitting field/F of f(x), and let G = Gal(f(x)/F ).The interesting case is when f(x) is irreducible in F [x]. G is a transitive subgroupof S2, so G = S2, so [K : F ] = |S2| = 2. Note that ζ2 = −1 ∈ F . Thus by the 3rd

Sick-Day Propositions, there exists A ∈ F such that K = F (√A). In fact, we can

choose A = D = b2 − 4c, since√D = ±(r1 − r2) ∈ K but

√D 6∈ F (else we would

have G ⊆ A2 = {idK}). So r1, r2 ∈ F (√D) = Q(b, c,

√D), implying r1, r2 can be

written as rational expressions in b, c,√D. (Of course, we already know this.)

Let’s try the same thing with a monic cubic. Suppose f(x) = x3 +ax2 +bx+c. Letits roots be r1, r2, r3, put F = Q(a, b, c), let K be its splitting field/F of f(x), andlet G = Gal(f(x)/F ). If f(x) were reducible/F its solution reduces to the quadraticcase, so assume f(x) is irreducible/F . There are essentially four cases.

Case 1: G = A3 and ζ3 (= ω) ∈ F .

By the 3rd Sick-Day Proposition, K = F ( 3√A) for some A ∈ F . Hence r1, r2, r3

can be written as rational expressions in a, b, c, 3√A, where A itself can be written

rationally in a, b, c.

Case 2: G = A3 and ζ3 6∈ F .

68 R. WILLARD

Note that F (ζ3) = F (√−3), so

√−3 6∈ F . Since [K : F ] = 3 we have

√−3 6∈ K.

Let F ′ = F (√−3) and K ′ = K(

√−3). Consider

F

KF (√−3)

K(√−3)

23

Now we have ζ3 ∈ F ′ and K ′/F ′ is Galois with [K ′ : F ′] = 3 (1st Sick-Day

Proposition). Thus K ′ = F ′( 3√A) for some A ∈ F ′. As r1, r2, r3 ∈ K ′, this means

they can be written as rational expressions in a, b, c,√−3, 3√A for some A rationally

expressible in terms of a, b, c,√−3.

Case 3: G = S3.G has a normal subgroup H C G of index 2; hence K/F has an intermediate

field E with E/F Galois and [E : F ] = 2. Again by the 3rd Sick-Day Proposition,

E = F (√A) for some A ∈ F ; in fact, E = F (

√D) where D = a2(b2 − 4c) − 4b3 −

27c2 + 18abc.We know that K/E is Galois with [K : E] = 3, so we can proceed as in the first

two cases, depending on whether or not ζ3 ∈ E.

Conclusion: For every cubic f(x) ∈ C[x], the roots of f(x) can be written asrational expressions involving the coefficients, square-roots of such expressions, andcube-roots of such expressions.

By applying the above argument to a “generic” cubic polynomial, one can showthat there is a single (complicated) formula that works for all cubic polynomials.This result is essentially due (independently) to Scipione del Ferro (died 1526) andNiccolo Tartaglia (died 1557), but first published by Gerolamo Cardano in 1545, soknown as Cardano’s formula.

In 1540, Lodovico Ferrari announced a reduction of degree-4 polynomial equationsto cubic equations. Combined with Cardano’s formulas for the cubic, this reductiongives in principal a formula for the roots of a degree-4 polynomial equation. Thisformula again gives the roots as rational expressions involving the coefficients, squareroots of such expressions, and cube roots of such expressions. If one doesn’t requirethe formula, but is interested only in proving the fact that the roots can be ex-pressed in this fashion, then it is possible to prove this fact by arguments similar tothose above for the cubic (though with many, many more cases to consider).

Cardano’s formula for the cubic, and Ferrari’s reduction of the quartic to thecubic, can be found in DF in section 14.7. The formulas are not nearly as useful asthe quadratic formula for degree-2 polynomials is.


31. March 26 – Radical extensions, solvable groups

Definition. Suppose F ≤ K ≤ C and u ∈ C.

(1) K/F is a simple radical extension if K = F ( n√a) for some a ∈ F and

n ≥ 2.(2) K/F is a radical extension if there exists a chain F = F0 ≤ F1 ≤ · · · ≤

Fm = K such that each Fi+1/Fi is a simple radical extension.(3) u is expressible by radicals over F if u belongs to some radical extension

of F .

Example: If deg(u/F ) = 2 or 3, then u is expressible by radicals over F . (Also trueif deg(u/F ) = 4.) We’ll see more examples on Friday.

Lemma. Suppose F ≤ K ≤ L ≤ C and u ∈ C.

(1) If K/F and L/K are radical extensions, then L/F is a radical extension.(2) If u is expressible by radicals over F , then u is expressible by radicals over K.(3) If K/F is a radical extension, then the chain F = F0 ≤ F1 ≤ · · · ≤ Fm = K

can be chosen so that each Fi+1/Fi is an extension by a p-th root for someprime p.

Proof. (1) Obvious from the definition.(2) Suppose E is a radical extension of F containing u. Thus we can write F =

F0 ≤ F1 ≤ · · · ≤ Fm = E where

• F1 = F0( n0√a0) for some a0 ∈ F0 and n0 ≥ 2.

• F2 = F1( n1√a1) for some a1 ∈ F1 and n1 ≥ 2.

• etc.

Define K0 = K and Ki+1 = Ki( ni√ai). Inductively ai ∈ Ki for all i < m, so

K = K0 ≤ K1 ≤ · · · ≤ Km is a radical extension, and Fi ≤ Ki for all i. In partic.,u ∈ Fm ⊆ Km.

(3) The point is that mk√a = m

√k√a. Thus a simple radical extension by a composite

root F0 ≤ F1 where F1 = F0( mk√a) (with a ∈ F0) can be refined to F0 ≤ E ≤ F1 with

E = F0( k√a) and F1 = E( m

√b) where b = k

√a ∈ E. �

Proposition. Suppose K/F is Galois and let G = Aut(K/F ). If G is abelian andζp ∈ F for every prime divisor of |G|, then K/F is a radical extension.

Proof. The proof is by induction on |G|. If |G| = 1 then K = F = F ( 2√

1). Assuming|G| = n > 1, let p be a prime divisor. By the Fundamental Theorem of finite abeliangroups, we know that G has a subgroup of size n/p. Thus

70 R. WILLARD

{idK}

H

G

n/p

p

=⇒

F

E

K

p

n/p

Note that H C G (as G is abelian) so E/F is Galois. Also, ζp ∈ F by hypothesis.Thus by the third Sick-Day Proposition, E/F is a simple radical extension. ObviouslyK/E is Galois (with Galois group H), and E contains ζq for every prime divisor qof |H|. Hence by induction, K/E is a radical extension. By transitivity, K/F is aradical extension. �

Remark. The hypothesis that G is abelian can be weakened. All that we needis that there exist subgroups G = H0 ≥ H1 ≥ H2 ≥ · · · ≥ Hm = {1} such thatHi+1 C Hi and [Hi : Hi+1] is prime for all i < m. Finite groups with this propertyare said to be solvable.

Corollary. Suppose K/F is Galois and let G = Aut(K/F ). If G is solvable andζp ∈ F for every prime divisor of |G|, then K/F is a radical extension.


32. March 28 – Strong radical extensions; Galois’ 1st Theorem

Recall from last time:

Corollary. Suppose K/F is Galois and let G = Aut(K/F ). If G is solvable andζp ∈ F for every prime divisor of |G|, then K/F is a radical extension.

Recall that G is solvable if there exist subgroups G = H0 ≥ H1 ≥ H2 ≥ · · · ≥Hm = {1} such that Hi+1 CHi and [Hi : Hi+1] is prime for all i < m.

Recalling the proof from Wednesday, in the inductive step we built F ≤ E ≤ Kwith E = Fix(H1); hence E/F is Galois and [E : F ] = p (for some prime p dividing|G|). E/K is a simple radical extension by the 3rd Sick-Day Proposition. Then weapplied the inductive hypothesis to the extension K/E. In effect, this builds a radicalextension from E up to K by marching down the subgroup chain from H1 to {1}.

Definition. Say that K/F is a strong radical extension if there exists a chainF = F0 ≤ F1 ≤ · · · ≤ Fm = K such that for all i < m, Fi+1/Fi is a simple radicalextension “because of the 3rd SDP.” I.e.,

• Fi+1/Fi is Galois,• [Fi+1 : Fi] = pi for some prime pi, and• ζpi ∈ Fi.

What Wednesday’s argument actually showed is the following:

Corollary. Suppose K/F is Galois and let G = Aut(K/F ). If G is solvable andζp ∈ F for every prime divisor of |G|, then K/F is a strong radical extension.

We will eventually get rid of the assumption about the ζp’s. First, let’s exploresolvable groups a bit.

(1) In general, a group G is defined to be solvable if there exist subgroups G =H0 ≥ H1 ≥ · · · ≥ Hm = {1} such that each Hi+1 CHi with Hi/Hi+1 abelian.When G is finite, this can be shown to be equivalent to the definition above.

(2) In particular, every abelian group is solvable.(3) Every subgroup of a solvable group is solvable.(4) Every homomorphic image of a solvable group is solvable.(5) If G is a group and there exists N C G such that N and G/N are solvable,

then G is solvable.(6) Sn is solvable for n ≤ 4.(7) More generally, every finite group of order pαqβ (with p, q primes) is solvable

(Burnside, 1904).(8) Every finite group of odd order is solvable (Feit-Thompson, 1963).(9) Sn is not solvable for any n ≥ 5. We can see this for S5, since the only

normal subgroups are S5 and A5, and A5 is simple (has no nontrivial normalsubgroups).

72 R. WILLARD

We will prove the next result on Monday.

Proposition. For every F ≤ C and n ≥ 2, there exists a strong radical extensionF ∗/F such that {ζk : k ≤ n} ⊆ F ∗.

Theorem (Galois). Suppose F ≤ K ≤ C with K/F Galois. If Aut(K/F ) is solvable,then there exists a strong radical extension L/F satisfying K ⊆ L. Hence every c ∈ Kis expressible by radicals/F .

Proof. Let G = Aut(K/F ) and n = |G|. Choose a strong radical extension F ∗/F bigenough to contain ζp for all primes p|n. Thus we have F = F0 ≤ F1 ≤ · · · ≤ Fm = F ∗

where each Fi+1/Fi is Galois, [Fi+1 : Fi] = pi, ζpi ∈ Fi, and hence Fi+1 = Fi( pi√ai)

for some ai ∈ Fi. Let K0 = K and Ki+1 = Ki( pi√ai), and put L := Km.

F0

F1

F2

F3

· ··

Fm

K =K0

K1

K2

K3

· ··

Km= L

= F

= F ∗

We claim that for each i = 0, . . . ,m, Ki/Fi is Galois and Aut(Ki/Fi) is isomorphicto a subgroup of G. We’ll prove this by induction on i. In the inductive step thereare two cases:

Case 1: pi√ai 6∈ Ki.

Ki/Fi is Galois by the inductive hypothesis, so the hypotheses of the first Sick-DayProposition are met for

Fi

Ki

Fi+1

Ki+1

= Fi( pi√ai)

= Ki( pi√ai)

pi


so applying the proposition we get that Ki+1/Fi+1 is Galois and Aut(Ki+1/Fi+1) ∼=Aut(Ki/Fi). Since Aut(Ki/Fi) is isomorphic to a subgroup of G by the inductivehypothesis, so is Aut(Ki+1/Fi+1).

Case 2: pi√ai ∈ Ki.

In this case Ki+1 = Ki and Fi ≤ Fi+1 ≤ Ki, so the picture looks like

Fi

Fi+1

Ki

= Fi( pi√ai)

= Ki+1

pi

Since Ki/Fi is Galois (by the inductive hypothesis), it follows that Ki/Fi+1 is alsoGalois and Aut(Ki/Fi+1) is a subgroup of Aut(Ki/Fi). As Aut(Ki/Fi) is isomorphicto a subgroup of G (by the inductive hypothesis) and Aut(Ki+1/Fi+1) ≤ Aut(Ki/Fi),it follows that Aut(Ki+1/Fi+1) is also isomorphic to a subgroup of G. This completesthe inductive step of the proof of the claim.

When i = m, the claim gives that L/F ∗ is Galois and Aut(L/F ∗) is isomorphic toa subgroup of G. Since G is solvable and |G| = n, it follows that

(1) Aut(L/F ∗) is solvable.(2) |Aut(L/F ∗)| is a divisor of n.

By construction, F ∗ contains ζp for every prime divisor of n. Hence by today’sCorollary, L/F ∗ is a strong radical extension. By transitivity, L/F is a strong radicalextension, and L contains K. �

.

74 R. WILLARD

33. March 31 – Expressing primitive roots

Last Friday I proved the following theorem of Galois:

Theorem. Suppose F ≤ K ≤ C with K/F Galois. If Aut(K/F ) is solvable, thenthere exists a strong radical extension L/F satisfying K ⊆ L.

We still need to prove one proposition:

Proposition. For every F ≤ C and n ≥ 2, there exists a strong radical extensionF ∗/F such that {ζk : k ≤ n} ⊆ F ∗.

Lemma. For all F ≤ C and n ≥ 2, F (ζn)/F is Galois and Aut(F (ζn)/F ) is abelian.

Proof. LetK = F (ζn) and let f(x) be the minimal polynomial/F of ζn. So f(x)|Φn(x)in F [x]. The roots of f(x) are among {ζ in : 0 ≤ i < n} ⊆ K, so f(x) splits in K[x],so K/F is Galois.

Suppose σ, τ ∈ Aut(K/F ). Then σ(ζn) = (ζn)k and τ(ζn) = (ζn)` for some 0 ≤k, ` < n. It follows that σ((ζn)i) = (ζn)ki and τ((ζn)i) = (ζn)`i for all i ∈ Z. Inparticular,

σ(τ(ζn)) = σ((ζn)`) = (ζn)k`

τ(σ(ζn)) = τ((ζn)k) = (ζn)`k.

Hence σ ◦ τ and τ ◦ σ agree at ζn, so are equal, proving Aut(K/F ) is abelian. �

Proof of Proposition. We prove this by induction on n.

Base case: n = 2. We can let F ∗ = F .

Inductive step: Assume n > 2 and we have a strong radical extension E/Fcontaining {ζk : k < n}. Consider E(ζn). E(ζn)/E is Galois and its Galois group Gis abelian. Furthermore, |G| = [E(ζn) : E] ≤ ϕ(n) ≤ n−1, so E contains ζp for everyprime p dividing |G|. Hence by the improved Corollary from March 28, E(ζn)/E isa strong radical extension. Let F ∗ = E(ζn). Since both F ∗/E and E/F are strongradical extensions, so is F ∗/F . And clearly {ζk : k ≤ n} ⊆ F ∗. �

On Wednesday we will prove a strong converse to Galois’ theorem. First we needto prove that every radical extension can be extended to a strong radical extension.

Proposition. For every radical extension K/F there exists K ≤ L ≤ C such thatL/F is a strong radical extension.

Proof. Suppose K/F is a radical extension. Using the Lemma from March 26, choosea tower of fields F = F0 ≤ F1 ≤ · · · ≤ Fm = K and primes pi such that Fi+1 =Fi( pi√ai) for some ai ∈ Fi. Choose n larger than all the pi and let F ∗/F be a strong

radical extension with {ζp : p ≤ n} ⊆ F ∗. Then define F ∗0 = F ∗ and F ∗i+1 = F ∗i ( pi√ai)

and let L = F ∗m. Clearly K ≤ L, so it suffices to show that L/F is a strong radicalextension.


F0

F1

F2

F3···

Fm

F ∗0

F ∗1

F ∗2

F ∗3

···

F ∗m

F =

K =

= F ∗

= L

Let i < m and consider the simple extension F ∗i+1/F∗i .

Case 1: pi√ai ∈ F ∗i .

Then F ∗i+1 = F ∗i . We will do nothing in this case.

Case 2: pi√ai 6∈ F ∗i .

Clearly ai ∈ F ∗i and ζpi ∈ F ∗i . Hence by the second Sick-Day Proposition, F ∗i+1/F∗i

is Galois with [F ∗i+1 : F ∗i ] = pi, so F ∗i+1/F∗i has the form of a single step of a strong

radical extension.

Thus if “weed out” the adjacent intermediate fields where F ∗i = F ∗i+1, we are leftwith a chain F ∗ = F ?

0 < F ?1 < · · · < F ?

k = L witnessing the fact that L/F ∗ is a strongradical extension. Since F ∗/F is also a strong radical extension, it follows that L/Fis a strong radical extension. �

76 R. WILLARD

34. April 2 – Galois’ Main Theorem

Here is an example that addresses a question from Monday.

Example. Let f(x) = x14 + 4 and let K be the splitting field for f(x) over Q. Theroots of f(x) are the 14th-roots of −4, which are rk := 21/7eiπ(1+2k)/14 for 0 ≤ k < 14.In particular, r0 = 7

√2i and r13 = 7

√−2i, and r0/r13 = ei2π/14 = ζ14, so

K = Q(r0, r1, . . . , r13) = Q(r0, ζ14) = Q(r0, r13).

What is [K : Q]? Maple tells us that f(x) is irreducible in Q[x], so deg(rk/Q) = 14for each k, proving 14|[K : Q]. Similarly, ζ14 ∈ K and deg(ζ14/Q) = 6 tell us that6|[K : Q]. On the other hand, [K : Q] ≤ deg(r0/Q) · deg(ζ14/Q) = 14 · 6 = 84. Thus[K : Q] = 42 or 84.

Note that (r0)7 = 2i so i ∈ K. Using these facts we can show that K/Q is aradical extension, via Q < Q(i) < Q(i, r0) < Q(i, r0, r13) = K. K/Q is also Galoisand Aut(K/Q) is solvable (because the smallest nonsolvable groups have order 60,120, 168, 180, . . . ).

However K/Q is not a strong radical extension (I think). The idea behind aconjectured proof is easy: 3|[K : Q], so if K/Q were a strong radical extension thenthere would exist Q ≤ F ≤ F ′ ≤ K with F ′/F Galois, [F ′ : F ] = 3, and ζ3 ∈ F , orequivalently

√−3 ∈ F . But it is my strong suspicion that

√−3 6∈ K.

Key Proposition. Suppose K/F is a strong radical extension. Then there existsL/K such that L/F is Galois and is a strong radical extension.

The proof will be in an appendix to this lecture.

Theorem (Galois). Suppose F ≤ C and f(x) ∈ F [x] is irreducible. Let E be thesplitting field for f(x)/F . The following are equivalent:

(1) Some root of f(x) is expressible by radicals/F .(2) Every root of f(x) is expressible by radicals/F .(3) There exists K/E such that K/F is a radical extension.(4) There exists K/E such that K/F is a strong radical extension.(5) Aut(E/F ) is solvable.

Proof. Clearly (4) ⇒ (3) ⇒ (2) ⇒ (1). (5) ⇒ (4) is the Theorem from March 28.Thus it remains to prove (1) ⇒ (5).

Suppose c is a root of f(x) and c is expressible by radicals/F . This means thereexists a radical extension K/F with c ∈ K. By Monday’s Proposition, there existsan extension L/K such that L/F is a strong radical extension. Then by today’sunproved Proposition there exists M/L such that M/F is Galois and is a strongradical extension. Of course E,M ≤ C.


F

F (c)

E K

L

M

M/F is Galois, so is normal. As f(x) ∈ F [x] and f(x) has a root in M , it followsthat f(x) splits in M [x]. Hence M contains all the roots of f(x) in C. Since E isgenerated over F by these roots, it follows that E ≤M .M/F is a strong radical extension, so choose witnessing fields F = F0 ≤ F1 ≤· · · ≤ Fm = M so that each Fi+1/Fi is Galois and [Fi+1 : Fi] = pi is prime (andζpi ∈ Fi). Let G = Aut(M/F ) and for each i let Hi = Aut(M/Fi).

F0

F1

F2

Fm−1

Fm

...

p0

p1

pm−1

F =

M = H0

H1

H2

Hm−1

Hm

...

p0

p1

pm−1

= G

= {idM}

By the FTGT, each Hi+1 CHi and [Hi : Hi+1] = pi. Hence G is solvable.Again M/F is Galois, and F ≤ E ≤ M . Let H = Aut(M/E), the subgroup of G

corresponding to E. E/F is Galois, so is normal, so H CG by the FTGT. Moreover,Aut(E/F ) ∼= G/H. Since G is solvable, and quotients of solvable groups are solvable,it follows that Aut(E/F ) is solvable. �

Example (I.N. Herstein, Topics in Algebra, 2nd ed., Wiley, 1975, p. 258). Let f(x) =2x5− 10x+ 5. f(x) is irreducible in Q[x] and has 3 real roots. Let E be the splittingfield of f(x)/Q and put G = Gal(f(x)/Q). G is a transitive subgroup of S5. It can bededuced from the above information that G contains a 5-cycle (because 5|[E : Q]) anda 2-cycle (complex conjugation switches the two nonreal roots). But any 5-cycle and

78 R. WILLARD

2-cycle generate all of S5, proving G = S5. Hence G is not solvable, so Aut(E/Q) isnot solvable. This proves that the roots of f(x) cannot be expressed by radicals overQ. As the coefficients of f(x) are in Q, this proves the impossibility of expressing theroots of a degree-5 polynomial in terms of the coefficients using “algebraic” operations(+, ×, −, ÷, and n

√). This impossibility is known as the Abel-Ruffini theorem,

generally credited to Abel in 1823, about 10 years before Galois’ discoveries.

34.1. Appendix.

Lemma. Suppose F ≤ E ≤ K ≤ C with K/F Galois, and E ≤ E∗ ≤ C with E∗/EGalois, [E∗ : E] = p prime, and ζp ∈ E.

F

E

E∗K

p

Then there exists a strong radical extension L/K with E∗ ⊆ L such that L/F isGalois.

Proof. If E∗ ⊆ K then we can take L = K, so assume E∗ 6⊆ K. Pick a ∈ E such thatE∗ = E( p

√a). Let f(x) be the minimal polynomial of p

√a/F and let c1, c2, . . . , cn be

the roots of f(x) in C with c1 = p√a. Set K1 = K(c1) and L = K(c1, . . . , cn).

F

E

E∗K

K1

L

= E(c1)

= K(c1)

= K(c1, . . . , cn)

p

Obviously E∗ ⊆ L. We are given that K/F is Galois, so let g(x) ∈ F [x] be suchthat K is the splitting field for g(x)/F . Then it is easy to see that L is the splittingfield for f(x)g(x)/F . This proves L/F is Galois.

It remains to show that L/K is a strong radical extension. We first show

Claim: (ci)p ∈ K for all i = 1, . . . , n.


Proof of Claim. Since L/F is Galois and c1, ci have the same minimal polynomial/F ,we can choose σ ∈ Aut(L/F ) with σ(c1) = ci. Since K/F is Galois it is stable, soσ(K) = K. Now compute:

(ci)p = σ(c1)p = σ((c1)p) = σ(a) ∈ σ(E) ⊆ σ(K) = K.

Returning to the proof of the Lemma, for 1 < i ≤ n define Ki = Ki−1(ci). Alsodefine K0 = K. So K = K0 ≤ K1 ≤ K2 ≤ · · · ≤ Kn = L. For some i it may be thatKi−1 = Ki. We’ll ignore such trivial steps and show that the nontrivial steps havethe form of one step in a strong radical extension.

So assume Ki−1 < Ki. Let ai = (ci)p; thus ci = (ζp)

ki p√ai for some ki. As

ζp ∈ Ki−1 we have Ki−1( p√ai) = Ki−1(ci) = Ki and ai ∈ K ⊆ Ki−1 by the Claim.

Since Ki−1 < Ki we have p√ai 6∈ Ki−1. Since ζp ∈ Ki−1 we can apply the 2nd Sick-Day

Proposition to conclude that Ki/Ki−1 is Galois and [Ki : Ki−1] = p as required. �

Key Proposition. Suppose K/F is a strong radical extension. Then there existsL/K such that L/F is Galois and is a strong radical extension.

Proof. Suppose F = F0 ≤ F1 ≤ · · · ≤ Fm = K witnessing that K/F is a strongradical extension. We argue by induction on m. If m = 0 then K = F and we canchoose L = F . For the inductive step, assume that m > 0 and let E = Fm−1. Weknow that K/E is Galois, [K : E] = p for some prime p, and ζp ∈ E. We also knowthat E/F is a strong radical extension and hence by induction there exists M/E suchthat M/F is Galois and is a strong radical extension.

F

E

KM

p= Fm

Fm−1 =

This is exactly the situation to apply the previous Lemma. Applying it gives us astrong radical extension L/M with K ⊆ L such that L/F is Galois. Since L/M andM/F are strong radical extensions, so is L/F . �

80 R. WILLARD

35. April 4 – Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every polynomial in C[x]splits in C[x]. Most proofs use complex analysis. We will give an elementary proofusing algebra. The only facts about C we need are:

(A1) C = R(i) where i is a root of x2 + 1.(A2) Every complex number has a complex square-root.(A3) Every f(x) ∈ R[x] of odd degree has a root in R.

(A1) is a definition, (A2) follows from De Moivre’s theorem, and (A3) can beproved using the intermediate value theorem of calculus.

Lemma 1. [C : R] = 2.

Proof. By (A1). �

Lemma 2. There does not exist K/C with [K : C] = 2.

Proof. If [K : C] = 2, pick c ∈ K \ C; then deg(c/C) = 2. Let x2 + αx + β be its

minimal polynomial/C and let D = α2 − 4β. There exists√D ∈ C by (A2); then

(−α±√D)/2 are roots of x2 + αx+ β in C, contradiction. �

Lemma 3. There does not exist K/R with [K : R] = m odd and m > 1.

Proof. If such K existed, choose c ∈ K \ R and let k = deg(c/R). Then k > 1, andsince R ≤ R(c) ≤ K we have k|m so k is odd. The minimal polynomial of c/R thushas odd degree greater than 1. But that is impossible, because every such polynomialhas a root in R by (A3). �

Theorem (Fundamental Theorem of Algebra). Every f(x) ∈ C[x] splits in C[x].

Proof. It suffices to show that every irreducible polynomial in C[x] has degree 1.Suppose for a contradiction there exists f(x) ∈ C[x] is irreducible with degree > 1.There exists an extension F/C such that F contains a root θ of f(x). ConsiderR ≤ C ≤ C(θ). θ is algebraic/C, so [C(θ) : C] < ∞, and [C : R] = 2 by Lemma 1.Hence [C(θ) : R] < ∞, so C(θ)/R is algebraic; in particular, θ is algebraic/R. Letg(x) ∈ R[x] be the minimal polynomial of θ/R and note that f(x)|g(x) in C[x].

Let K be a splitting field of g(x) over C. Thus K = C(θ1, . . . , θk) where θ1, . . . , θkare the roots of g(x) in K. Hence K = R(i, θ1, . . . , θk), which proves that K is asplitting field/R of (x2+1)g(x). Hence K/R is Galois. Also note that C < K, becauseotherwise g(x) would split in C[x] and hence f(x) would split in C[x], contrary toour assumption.

Note that [K : R] is even; write [K : R] = 2αm with m odd and α ≥ 1. Thus


R

C

K

2

2α−1m

Let G = Aut(K/R). Thus |G| = 2αm. Let P be a Sylow 2-subgroup of G. LetE1 = Fix(P ).

{idK}

P

G

2α

m

=⇒

R

E1

K

Cm

2α2α−1m

2

By Lemma 3, we must have m = 1. Thus

R

C

K

2

2α−1

Note that we cannot have α = 1, as that would imply K = C which is false. Soα ≥ 2. Note that K/C is Galois; let G1 = Aut(K/C). Then |G1| = 2α−1. By Assign.#1 prob. #4(c), G1 has a subgroup of order 2β for every β ≤ α − 1. In particular,G1 has a subgroup H of order 2α−2. Let E2 = Fix(H).

{idK}

H

G1

2α−2

2

=⇒

C

E2

K

2

2α−2

This contradicts Lemma 2. �

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