lecture notes in galois theory - math.wsu.edu

Lecture Notes in Galois Theory

Lectures by Dr Sheng-Chi Liu

Throughout these notes, signifies end proof, and N signifies end ofexample.

Table of Contents

Table of Contents i

Lecture 1 Review of Group Theory 11.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Structure of cyclic groups . . . . . . . . . . . . . . . . . . . . . . 21.3 Permutation groups . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Finitely generated abelian groups . . . . . . . . . . . . . . . . . . 31.5 Group actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Lecture 2 Group Actions and Sylow Theorems 52.1 p-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Sylow theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Lecture 3 Review of Ring Theory 73.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Solutions to algebraic equations . . . . . . . . . . . . . . . . . . . 10

Lecture 4 Field Extensions 124.1 Algebraic and transcendental numbers . . . . . . . . . . . . . . . 124.2 Algebraic extensions . . . . . . . . . . . . . . . . . . . . . . . . . 13

Lecture 5 Algebraic Field Extensions 145.1 Minimal polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 145.2 Composites of fields . . . . . . . . . . . . . . . . . . . . . . . . . 165.3 Algebraic closure . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Lecture 6 Algebraic Closure 176.1 Existence of algebraic closure . . . . . . . . . . . . . . . . . . . . 17

Lecture 7 Field Embeddings 197.1 Uniqueness of algebraic closure . . . . . . . . . . . . . . . . . . . 19

Lecture 8 Splitting Fields 228.1 Lifts are not unique . . . . . . . . . . . . . . . . . . . . . . . . . 22

Notes by Jakob Streipel. Last updated December 6, 2019.

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TABLE OF CONTENTS ii

Lecture 9 Normal Extensions 239.1 Splitting fields and normal extensions . . . . . . . . . . . . . . . 23

Lecture 10 Separable Extension 2610.1 Separable degree . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Lecture 11 Simple Extensions 2611.1 Separable extensions . . . . . . . . . . . . . . . . . . . . . . . . . 2611.2 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Lecture 12 Simple Extensions, continued 3012.1 Primitive element theorem, continued . . . . . . . . . . . . . . . 30

Lecture 13 Normal and Separable Closures 3013.1 Normal closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3113.2 Separable closure . . . . . . . . . . . . . . . . . . . . . . . . . . . 3113.3 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Lecture 14 Inseparable Extensions 3314.1 Number of irreducible polynomials over finite fields . . . . . . . . 3314.2 Inseparable extensions . . . . . . . . . . . . . . . . . . . . . . . . 34

Lecture 15 Purely Inseparable Extensions 3615.1 Inseparable closures and purely inseparable extensions . . . . . . 36

Lecture 16 Galois Theory 3916.1 Galois extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Lecture 17 Artin’s Theorem 4117.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Lecture 18 Infinite Version of Artin’s Theorem 4218.1 Generalising Artin’s theorem . . . . . . . . . . . . . . . . . . . . 4218.2 Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4318.3 Lifts and Galois extensions . . . . . . . . . . . . . . . . . . . . . 43

Lecture 19 Special Kinds of Galois Extensions 4419.1 Cyclic, abelian, nilpotent, and solvable extensions . . . . . . . . . 4419.2 Examples and applications . . . . . . . . . . . . . . . . . . . . . . 46

Lecture 20 Galois Group of a Polynomial 4720.1 Galois group of polynomials . . . . . . . . . . . . . . . . . . . . . 47

Lecture 21 Examples of Galois Groups 4821.1 More examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4821.2 Roots of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Lecture 22 Cyclotomic Fields 5122.1 Proof finished . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5122.2 Quadratic reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . 5222.3 Gauss sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

TABLE OF CONTENTS iii

Lecture 23 Characters 5423.1 Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . 5423.2 Classical Galois results . . . . . . . . . . . . . . . . . . . . . . . . 5523.3 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Lecture 24 Norms and Traces 5624.1 Field norms and field traces . . . . . . . . . . . . . . . . . . . . . 5624.2 Galois theory of solvability of algebraic equations . . . . . . . . . 59

Lecture 25 Radical Extensions 6025.1 Kummer extensions . . . . . . . . . . . . . . . . . . . . . . . . . 6025.2 Radical extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Lecture 26 Solvability by Radicals 6326.1 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6326.2 Galois’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Lecture 27 Topological Groups 6727.1 Galois theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6727.2 Infinite Galois . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Lecture 28 Topological Groups 7028.1 Review of topological spaces . . . . . . . . . . . . . . . . . . . . . 70

Lecture 29 Infinite Galois Correspondence 7529.1 Infinite Galois extensions . . . . . . . . . . . . . . . . . . . . . . 75

Index 77

REVIEW OF GROUP THEORY 1

Lecture 1 Review of Group Theory

1.1 Groups

Definition 1.1.1 (Group). A group (G, ∗) is a set G with a binary operation∗ satisfying

(i) associativity, or (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G;

(ii) identity, meaning there exists an element e ∈ G such that g ∗ e = e ∗ g = gfor all g ∈ G; and

(iii) inverse, meaning that for all g ∈ G there exists an element g′ ∈ G suchthat g ∗ g′ = g′ ∗ g = e.

If g1 ∗ g2 = g2 ∗ g1 for all g1, g2 ∈ G, then G is called abelian .

Generally we will suppress the symbol ∗ for the binary operation and writesimply g1g2 in place of g1 ∗ g2.

Definition 1.1.2 (Group homomorphism). Let G and G be groups.A function f : G → H is a homomorphism is f(g1g2) = f(g1)f(g2) for

all g1, g2 ∈ G. Another way of saying this is that f preserves the structure orrespects the binary operation of the group G.

Similarly f is an isomorphism if f is a homomorphism that is one-to-oneand onto.

An important object related to a homomorphism (or indeed any map) is thekernel of f , denoted

ker f :={g ∈ G

∣∣ f(g) = eH}.

Proposition 1.1.3. A homomorphism f : G → H of groups is one-to-one ifand only if ker f =

{eG}

is trivial.

Definition 1.1.4 (Cosets). Let H be a subgroup of a group G, and let g ∈ G.A left coset of H is a set of the shape gH :=

{gh∣∣ h ∈ h} and a right coset

of H is a set of the kind Hg ={hg∣∣ h ∈ H }.

Proposition 1.1.5. Let G be a group and H a subgroup of G.

(i) Let g1, g2 ∈ G. Then g1H = g2H if and only if g−11 g2 ∈ H.

(ii) Similarly, for g1, g2 ∈ G, we have Hg1 = Hg2 if and only if g2g−11 ∈ H.

(iii) If moreover H is finite, then |gH| = |H| = |Hg| for any g ∈ G.

Definition 1.1.6 (Index). Let H be a subgroup of a group G. The index ofH in G, denoted [G : H], is the number of distinct left (or right) cosets of H inG.

Remark 1.1.7. Since each coset of H in G is of the same size, in the event thatG is finite we must therefore have |G| = [G : H] · |H|.

An immediate corollary of this is

Date: August 20th, 2019.


Theorem 1.1.8 (Lagrange). Let G be a finite group, and let H < G be asubgroup. Then |H| | |G|.

The converse of this theorem is not generally true, however in the specialcase of finite abelian groups it is. This is a consequence of the fundamentaltheorem of finite abelian groups, which we will discuss shortly.

In general left and right cosets gH and Hg are not equal, and moreover{H, g1H, g2H, . . .

}is not a group.

Definition 1.1.9 (Normal subgroup). Let H be a subgroup of a group G. Wesay that H is normal if gH = Hg for all g ∈ G (as sets; we are not saying thatgh = hg for every g ∈ G and h ∈ H).

Theorem 1.1.10. If H is a normal subgroup of G, then G/H ={gH∣∣ g ∈ G}

is a group under the (natural) binary operation

(gH)(g1H) = (gg1)H.

Proposition 1.1.11 (First isomorphism theorem). Let f : G → H be a grouphomomorphism. Then

(a) ker f is a normal subgroup of G, and

(b) G/ ker f ∼= im f .

1.2 Structure of cyclic groups

Definition 1.2.1 (Cyclic group). A group G is cyclic if G = 〈a〉 ={an∣∣ n ∈

Z}

, i.e., G is generated by a.

Theorem 1.2.2. Let G = 〈a〉 be a cyclic group.

(a) If |G| =∞, then G ∼= (Z,+).

(b) If |G| = m <∞, then G ∼= (Zm,+).

Remark 1.2.3. By Zm we mean Z/mZ, the integers modulo m. We will haveoccasion to write this quite frequently, hence the shorthand.

Proposition 1.2.4. If m = pa11 pa22 · · · pakk , where p1, p2, . . . , pk are distinct

primes, thenZm ∼= Zpa11

× Zpa22× . . .× Zpakk .

This is a consequence of the fact that if gcd(m,n) = 1, then Zmn ∼= Zm×Zn.

1.3 Permutation groups

We denote by Sn the group of all bijections σ :{

1, 2, . . . , n}→{

1, 2, . . . , n}

,called the symmetric group of n elements.

Example 1.3.1. For instance, in S4, we write things like

τ =

(1 2 3 44 1 2 3

)=(1 4 3 2

)=(1 2

) (1 3

) (1 4

),


where the first form is a notation indicating that 1 goes to 4, 2 goes to 1, and soon. The second expression is the same permutation written in cycle notation,read as a cycle where 1 goes to 4 goes to 3 goes to 2 and returns to 1. Finallywe have factored the cycle into transpositions.

As an other example showing that not everything has to be in the same greatbig cycle, consider

σ =

(1 2 3 42 1 4 3

)=(1 2

) (3 4

).

This time the permutation decomposes into two disjoint cycles—this way ofwriting a permutation as disjoint cycles is always unique (up to order of com-position, but since they’re disjoint that does not affect the result).

On the other hand, writing a permutation as transpositions is not unique,but the parity of the number of transpositions is. N

Let An denote the subgroup of Sn consisting of even permutations, calledthe alternating group.

Definition 1.3.2 (Simple group). A group G is called simple if G has noproper nontrivial subgroups.

Theorem 1.3.3. The alternating group An is simple if and only if n 6= 4.

1.4 Finitely generated abelian groups

Let G1 and G2 be two groups. Then G1 × G2 is a group under the binaryoperation

(a1, a2)(b1, b2) = (a1b1, a2b2).

Theorem 1.4.1 (Fundamental theorem of finitely generated abelian groups).Every finitely generated abelian group G is isomorphic to

Zpr11 × Zpr22 × . . .× Zprkk × Z× Z× . . .× Z

where pi are primes (not necessarily distinct) and ri ∈ N.

In the event that G is finite, of course the Z terms disappear.

1.5 Group actions

Definition 1.5.1 (Group action). An action of a group G on a set S is afunction G × S → S, (g, x) 7→ gx ∈ S, such that for all x ∈ S and g1, g2 ∈ G,we have

ex = x

and(g1g2)x = g1(g2x).

Proposition 1.5.2. Let G be a group acting on a set S.

(a) The relation on S defined by x ∼ y if and only if gx = y for some g ∈ Gis an equivalence relation.


(b) For each x ∈ S, define Gx :={g ∈ G

∣∣ gx = x}

. This is a subgroup of G,called the stabiliser of x.

Definition 1.5.3 (Orbit). Let G be a group acting on a set S, and let x ∈ S.The set Gx =

{gx∣∣ g ∈ G} ⊂ S is called the orbit of x.

Theorem 1.5.4. Let G be a group acting on S. Let x ∈ S. Then |Gx| = [G :Gx].

Proof. Consider the map G/Gx → Gx defined by gGx 7→ gx. We need to checkthat this is well-defined, that it is one-to-one, and that it is onto.

That it is one-to-one is clear: if gGx = g1Gx, then g−1g1 ∈ Gx, meaningthat g−1g1x = x, and multiplying by g we get g1x = gx.

That this is one-to-one follows immediately by the above argument, justfollowing the converse direction at each step.

Finally that this map is onto is trivial since we have the map defined for allg, and so of course we’ll cover all the gx in Gx.

Example 1.5.5. Let G be a group. Then G acts on itself by conjugation,G×G→ G defined by (g, x) 7→ gxg−1.

The orbit of x is Gx ={gxg−1

∣∣ g ∈ G}, called the conjugacy class of x.

The stabiliser Gx ={g ∈ G

∣∣ gxg−1 = x}

= CG(x) is called the centraliser ofx; it is the set of all g ∈ G that commute with x, since multiplying both sidesof gxg−1 = x by g gives gx = xg. N

Corollary 1.5.6. Let G be a finite group. Let Gx1, Gx2, . . . , Gxn be all distinctconjugacy classes of G. Then

|G| =n∑k=1

[G : CG(xk)].

Proof. This follows immediately from |Gxi| = [G : CG(xi)].

Lemma 1.5.7. Let H be a group of order pn, where p is a prime. Suppose Hacts on a finite set S. If S0 :=

{x ∈ S

∣∣ hx = x for all h ∈ H}

, then

|S| ≡ |S0| (mod p).

Proof. Write S as a disjoint union of orbits

S = S0 tHx1 tHx2 t · · · tHxk

with |Hxi| > 1 (since otherwise x1 ∈ S0) for all i = 1, 2, . . . , k. Note that

|Hxi| = [H : Hxi ] | |H| = pn,

implying that p | |Hxi|, adding the cardinalates in the disjoint union, implies

|S| ≡ |S0| (mod p)

as desired.

Theorem 1.5.8 (Cauchy). If G is a finite group with p | |G|, where p is aprime, then G contains an element of order p.

GROUP ACTIONS AND SYLOW THEOREMS 5

Lecture 2 Group Actions and Sylow Theorems

2.1 p-Groups

Proof of Theorem 1.5.8. Set |G| = n and let

S ={

(a1, a2, . . . , ap)∣∣ ai ∈ G, a1a2 · · · ap = e

}.

Note that ap = (a1a2 · · · ap−1)−1 is uniquely determined. Then |S| = np−1.Since p | n, we consequently have |S| ≡ 0 (mod p).

Let Zp act on S by cyclic permutation, i.e., for k ∈ Zp we have

k(a1, a2, . . . , ap) = (ak+1, ak+2, . . . , ap, a1, a2, . . . , ak).

It is clear that this is a group action—k = 0 clearly acts as the identity map,and the action is clearly associative.

Here S0 ={

(a, a, . . . , a) ∈ S∣∣ a ∈ G} since acting for instance by k = 1 we

see that a1 = a2, a2 = a3, et cetera. Crucially we also have |S0| > 1 since atleast (e, e, . . . e) ∈ S0.

Hence by the lemma 1 < |S0| ≡ |S| ≡ 0 (mod p), so |S0| ≥ p, implying thatthere exists some a ∈ G, a 6= e, such that a · a · · · a = ap = e, implying that ahas order p since p is a prime.

Definition 2.1.1 (p-Group). A group G is a p-group if the order of everyelement in G is some power of p, where p is a prime.

If H is a subgroup of a group G and H is a p-group, then H is called ap-subgroup.

Corollary 2.1.2. A finite group G is a p-group if and only if |G| is a power ofp.

Proof. For the forwards direction, if |G| 6= pk, then there exists some prime qwith q | |G|, so by Cauchy’s theorem there exists an element of order q, whichis a contradiction.

The converse direction follows immediately from Lagrange’s theorem.

Definition 2.1.3 (Normaliser). Let H be a subgroup of a group G. The nor-maliser of H in G is defined as

NG(H) ={g ∈ G

∣∣ gHg−1 = H}.

Remark 2.1.4. Note how H is a normal subgroup of NG(H), since for all g ∈NG(H), gH = Hg by definition.

Lemma 2.1.5. If H is a p-subgroup of a finite group G, then

[NG(H) : H] ≡ [G : H] (mod p).

Proof. Let S be the set of left cosets of H in G, so |S| = [G : H]. Let H act onS by left translation.

For x ∈ G we then have xH ∈ S0 if and only if hxH = xH for all h ∈ H ifand only if x−1hxH = H for all h ∈ H if and only if x−1hx ∈ H for all h ∈ H,which finally is equivalent to x ∈ NG(H) by definition.

Hence |S0| = [NG(H) : H], which is congruent to |S| modulo p by lemma,and |S| = [G : H] as per above.

Date: August 22nd, 2019.

GROUP ACTIONS AND SYLOW THEOREMS 6

Corollary 2.1.6. If H is a p-subgroup of a finite group G such that p | [G : H],then NG(H) 6= H.

Proof. By the lemma we have [NG(H) : H] ≡ [G : H] ≡ 0 (mod p), so [NG(H) :H] > 1, meaning the two can’t be equal.

2.2 Sylow theorems

Theorem 2.2.1 (First Sylow theorem). Let G be a group of order pn ·m withn ≥ 1, p prime, and gcd(p,m) = 1. Then G contains a subgroup of order pi

for all 1 ≤ i ≤ n, and every subgroup of pi is normal in some subgroup of orderpi+1 for i < n.

Proof. We prove the theorem by induction on i. For i = 1, we want to find asubgroup of order p1 = p, but p | |G|, so by Cauchy’s theorem there exists somesubgroup 〈a〉 of G with |〈a〉| = p.

Assume H is a subgroup of G of order pi for 1 ≤ i < n. Then p | [G : H](since there is at least one factor of p left over) and by the previous lemmaNG(H) 6= H.

Hence

1 < |NG(H)/H| = [NG(H) : H] ≡ [G : H] ≡ 0 (mod p).

Hence NG(H)/H is a group and contains an element of order p by Cauchy’stheorem. Call the subgroup generated by this element H1. Then H1/H is asubgroup of NG(H)/H of order p, so H1 is a subgroup of NG(H) of order

|H1| = |H| · |H1/H| = pi · p = pi+1.

So H < H1 < NG(H), and H is normal in NG(H), so H is normal in H1 aswell.

Definition 2.2.2 (Sylow p-subgroup). A subgroup P 6= G of a group G is calleda Sylow p-subgroup if P is a maximal p-subgroup of G, i.e., if P < H < Gand H is a p-subgroup, then P = H.

Remark 2.2.3. Note that P can be the trivial subgroup. If |G| = pn11 pn2

2 · · · pakk

and P is a Sylow p-subgroup, then |P | = pnii if p = pi for some i, and if p 6= pifor all i = 1, 2, . . . , k, then |P | = 1.

Corollary 2.2.4. Let G be a group of order pn · m where p is prime andgcd(p,m) = 1. Let H be a p-subgroup of G. Then

(a) H is a Sylow p-subgroup of G if and only if |H| = pn, and

(b) every conjugate of a Sylow p-subgroup is a Sylow p-subgroup.

Theorem 2.2.5 (Second Sylow theorem). If H is a p-subgroup of a finite groupG and P is any Sylow p-subgroup of G, then there exists some x ∈ G such thatH < xPx−1. In particular, any two Sylow p-subgroups of G are conjugate.

Proof. Let S = G/P , the set of left cosets of P in G. Let H act on S by lefttranslation. Then |S0| ≡ |S| (mod p), and |S| = [G : P ] 6≡ 0 (mod p) since Phas the maximal possible power of p for its order, so S0 6= ∅.

REVIEW OF RING THEORY 7

Hence xP ∈ S0 if and only if hxP = xP for all h ∈ H if and only if(x−1hx)P = p for all h ∈ H if and only if x−1hx ∈ P for all h ∈ H if and onlyif h ∈ xPx−1, so H < xPx−1.

Hence taking in particular H to be the largest possible p-subgroup, we seethat any Sylow p-subgroup is conjugate to it.

Theorem 2.2.6 (Third Sylow theorem). If G is a finite group and p a prime,then the number of Sylow p-subgroups of G divides |G| and is of the form kp+ 1for some integer k ≥ 0.

Proof. By the Second Sylow theorem, the number of Sylow p-subgroups is thenumber of conjugates of any one of them, say P . This number if [G : NG(P )],and [G : NG(P )] | |G|, so the number of Sylow p-subgroups divides |G|.

Secondly, let S be the set of all Sylow p-subgroups of G. Let P act on Sby conjugation. Then Q ∈ S0 if and only of xQx−1 = Q for all x ∈ P , whichis equivalent to P < NG(Q). Now Q and P are both Sylow p-subgroups ofNG(Q), and Q is normal in NG(Q) by definition, and P and Q are conjugateby the Second Sylow theorem, so Q = P , meaning that S0 =

{P}

.Hence since by our oft-used lemma

|S| ≡ |S0| (mod p)

and |S0| = 1, we have |S| = kp+ 1 for some integer k ≥ 0.

Lecture 3 Review of Ring Theory

3.1 Rings

Definition 3.1.1 (Ring). A ring (R,+, ·) is a set R with two binary operations+ and · with the properties that

(a) (R,+) is an abelian group,

(b) (ab)c = a(bc) for all a, b, c ∈ R, and

(c) a(b+ c) = ab+ ac and (a+ b)c = ac+ bc for all a, b, c ∈ R.

Example 3.1.2. The integers with ordinary addition and multiplication, writ-ten (Z,+, ·), is a ring. N

Example 3.1.3. The set of real 2× 2 matrices with the usual matrix additionand multiplication, (M2(R),+, ·), form a ring. N

We will have reason to recall rather a lot of standard definitions from ringtheory.

Definition 3.1.4 (Commutative ring). If for every a and b in a ring R we haveab = ba, then R is a commutative ring .

Definition 3.1.5 (Ring with identity). If a ring R contains an element 1R suchthat 1Ra = a1R = a for all a ∈ R, then R is called a ring with identity orring with unity or unit ring .



Definition 3.1.6 (Zero divisor). An element a 6= 0 in a ring R is called a zerodivisor if there exists some b 6= 0 in R such that ab = 0.

Definition 3.1.7 (Unit). An element a 6= 0 in a ring R with identity is calleda unit if there exists some b ∈ R such that ab = ba = 1R.

Definition 3.1.8 (Integral domain). A commutative ring with identity 1R 6= 0and no zero divisors is called an integral domain .

Definition 3.1.9 (Division ring). A ring with identity in which every nonzeroelement is a unit is called a division ring .

Definition 3.1.10 (Field). A commutative division ring is called a field .

Definition 3.1.11 (Ring homomorphism). Let R and S be rings. A functionf : R → S is a homomorphism of rings if f(a + b) = f(a) + f(b) andf(ab) = f(a)f(b) for all a, b ∈ R.

The kernel of f is defined as ker f ={r ∈ R

∣∣ f(r) = 0}

.

Definition 3.1.12 (Ideal). Let R be a ring, and I a subring of R. We call Ian ideal of R if rI < I and Ir < I for all r ∈ R.

Remark 3.1.13. We care about ideals for the same reason that we care aboutnormal subgroups: for any old subgroup H of a group G, the set of left cosetsG/H need not have group structure, but it will if H is a normal subgroup.

In the same way, for any old subring S of a ring R, R/S need not have aring structure, but it does if S is an ideal.

More precisely, if I is an ideal of R, then R/I is a ring with the naturalbinary operations,

(a+ I) + (b+ I) = (a+ b) + I

and(a+ I)(b+ I) = (ab) + I.

Note how the first of these is automatic; since (R,+) is an abelian group, as anadditive subgroup I is automatically normal.

Exercise 3.1.14. Let f : R → S be a ring homomorphism. Then ker f is anideal of R, and R/ ker f ∼= Im f as rings, not just groups.

Solution. We need only check the properties of the multiplication, since thegroup structure of (R,+) guarantees the right properties of addition.

So first let a, b ∈ ker f , i.e., f(a) = f(b) = 0. Then f(ab) = f(a)f(b) = 0, soab ∈ ker f , making ker f a subring of R.

To see that it is an ideal, we play the same game, but this time we takea ∈ ker f and r ∈ R. Then f(ar) = f(a)f(r) = 0 · f(r) = 0, and similarlyf(ra) = 0, so r ker(f) < ker(f) and ker(f)r < ker(f) for all r ∈ R.

To show the first isomorphism theorem, consider the diagram

R R/ ker f

Im f,

g

f h


where g is the projection and h is inclusion, and then f = h ◦ g. Clearly h isan injection, being the inclusion map, and it remains to show that it is alsosurjective. But the diagram commutes, and f is of course surjective, so sincef = h ◦ g must be too. �

Definition 3.1.15 (Principal ideal). An ideal I = 〈a〉 generated by a singleelement a is called a principal ideal .

A principal ideal domain , or PID is an integral domain in which everyideal is principal.

Example 3.1.16. The integers (Z,+, ·) form a principal ideal domain, specifi-cally because it possesses a division algorithm.

For the same reason, the set of polynomials k[x] over a field k is a principalideal domain. N

In the sequel, unless otherwise stated, rings under consideration will becommutative rings with identity 1 6= 0.

Definition 3.1.17 (Prime ideal). A prime ideal P 6= R in a ring R is an idealsuch that for all a, b ∈ R, if ab ∈ P , then a ∈ P or b ∈ P .

Example 3.1.18. The motivation for this definition comes from mimicking theprime numbers in Z. In Z, the primes of course are 2, 3, 5, 7, . . ., numbers withthe property that their only positive divisors are 1 and themselves. Anotherequivalent description is to say that if p | ab, with p prime, then either p | a orp | b. This is the definition we generalise.

Consider the prime ideals 〈2〉, 〈3〉, 〈5〉, 〈7〉, . . .. The division property directlytranslated to ab ∈ 〈p〉 implying that a ∈ 〈p〉 or b ∈ 〈p〉.

Note also how we require, in the definition, that a prime ideal P 6= R. One(of many) reasons for this is that otherwise R = 〈1〉 would be a prime ideal,corresponding in Z to 1 being a prime number, which we generally don’t want(for reasons like unique factorisation).

That said, note how (maybe sneakily), the trivial ideal 0 is a prime ideal inZ. N

Theorem 3.1.19. Let R be a commutative ring with identity 1 6= 0. Let P bean ideal of R. Then P is prime if and only if R/P is an integral domain.

Proof. In the forward direction, we need to show that R/P has no zero divisors.Suppose, to that end, that (a+ P )(b+ P ) = P (since P is the additive identityin R/P ). In other words, (ab) +P = P , implying that ab ∈ P . But since P is aprime ideal, this implies a ∈ P or b ∈ P , equivalent, respectively, to a+ P = Por b+ P = P .

For the converse direction, suppose ab ∈ P . This implies (ab) + P = P , so(a + P )(b + P ) = P . But R/P is an integral domain, so has no zero divisors,meaning that either a+ P = P or b+ P = P , ergo a ∈ P or b ∈ P .

Definition 3.1.20 (Maximal ideal). An ideal M in a ring R is called maximalif M 6= R and for every ideal N such that M ⊂ N ⊂ R, we have N = M orN = R.

Example 3.1.21. In Z, the ideal 〈2〉 is maximal. A good way to think of thisis to consider some element m 6∈ 〈2〉, meaning that gcd(m, 2) = 1. By Bezout’s


lemma, there must then exist k, h ∈ Z such that km + 2h = 1. So if we wereto have 〈2〉 ⊂ 〈m〉 ⊂ Z, then 1 ∈ 〈m〉, since 1 would be a linear combination ofthings in 〈m〉, so it is the whole ring.

For precisely the same reason 〈p〉 is maximal in Z for all prime numbers p.On the other hand, 〈mn〉 ⊂ 〈m〉, so indeed 〈p〉 are the only maximal ideals

in Z. N

Theorem 3.1.22. Let R be a commutative ring with identity 1 6= 0. Let M bean ideal of R. Then M is maximal if and only if R/M is a field.

Hence all maximal ideals are prime ideals, since all fields are integral do-mains.

Proof. For the forward direction, let a 6∈M , so a+M 6= M . Then 〈a〉+M = Rsince M is maximal. Since 1 ∈ R, we get ar + m = 1 for some r ∈ R andm ∈ M , meaning that (ar) + M = 1 + M , or (a + M)(r + M) = 1 + M , so(a+M) has a multiplicative inverse in R/M .

For the converse direction, suppose M ( N ⊂ R, with N an ideal. Weneed to show that N = R. Take a ∈ N \ M . Then a + M 6= M has amultiplicative inverse in R/M , being a field, i.e., there exists some r ∈ R suchthat (a+M)(r +M) = 1 +M .

But this means (ar) +M = 1 +M , implying that 1− ar ∈M ⊂ N . Hence

1 = (1− ar) + ar ∈ N

since 1− ar ∈ N and ar ∈ N (since a ∈ N and N is an ideal), so N = R.

Remark 3.1.23. This means that R is a field if and only if the zero ideal 0 = 〈0〉is maximal. The forwards direction is trivial: a field has only two ideals, 0 anditself, and maximal ideals are by definition not the whole ring, so 0 is maximal.

The converse direction is our above theorem: if 0 is maximal, then R/0 = Ris a field.

3.2 Solutions to algebraic equations

The main question we are looking to answer in this class is the following. Let kbe a field, and consider some polynomial equation

a0xn + a1x

n−1 + a2xn−2 + . . .+ an−1x+ an = 0

where ai ∈ k and a0 6= 0. Since a0 6= 0 and we are in a field, we can divide bya0 and acquire a monic polynomial.

The question is this: is it possible to solve this equation with radicals, in thesense of writing down the solutions to the equation in terms of the coefficients,multiplication, division, addition, and subtraction, along with various radicals?

Take, for instance, k = Q, k = R, or k = C.

Example 3.2.1. Let n = 1, so that we are solving x+ a = 0. Clearly x = −a,so the answer is yes. N

Example 3.2.2. If n = 2 we are solving x2 + ax + b = 0, and this time (bycompleting the square),

x =−a±

√a2 − 4b

2,

so again the answer is yes. N


Example 3.2.3. When n = 3, the equation becomes

x3 + ax2 + bx+ c = 0.

The solution then looks like

x = −a3

+ ωi3

√−q

2+

√p3

27+q2

4+ ω3−i 3

√−q

2−√p3

27+q2

4

where

p = b− a2

3, q = c− ab

3+

2a3

27, and ω =

−1 +√−3

2

for i = 0, 1, 2, so again the answer is yes. N

Example 3.2.4. If n = 4, we are solving x4 + ax3 + bx2 + cx + d = 0. Thistime

x = −a4±

√−p

2±√p2

4− r

if q = 0 and

x = −a4± A

2±√− t0

2− p

4− q

4a

if q 6= 0, where

p = b− 3a2

8, q = c− ab

2+a3

8, r = d− ac

4+a2b

16− 3a4

256,

andA =

√2t0 − p,

where finally t0 is any root of 8t3 − 4pt2 − 8rt+ 4pr − q2 = 0. N

Fascinatingly, and one of the goals of this class, there is no such formula ingeneral if n ≥ 5.

That said, there are some special cases.

Example 3.2.5. Consider x2 − 1 = 0, for which clearly x = ±1. Similarly,x3 − 1 = (x − 1)(x2 + x + 1) = 0, so we can solve for x with the quadraticformula.

The case x4−1 = 0 is uninteresting, since it factors directly as (x2−1)(x2 +1) = 0, which we can handle as above.

For x5− 1 = 0 we again factor as (x− 1)(x4 +x3 +x3 +x2 +x+ 1) = 0, andso we are left with finding roots of the second factor. The standard trick here isto notice that this looks nice if we divide through by x2, and taking y = x+ 1

xwe get

y2 + y − 1 = 0,

which we can solve with the quadratic equation again.Likewise for x7 − 1 = 0 we factor out x − 1 and we’re left with x6 + x5 +

x4 +x3 +x2 +x+ 1 = 0, which we divide through by x3 and use the same trickto get y3 + y2 − 2y − 1 = 0. N

FIELD EXTENSIONS 12

It turns out, but is highly nontrivial, that xn− 1 = 0 in general can have itsroots written in terms of radicals. This is due to Gauss around 1800.

Later on, Abel around 1820 showed that if the Galois group of the algebraicextension is abelian (all words we will make sense of shortly), then the rootscan be expressed by radicals.

More generally, Galois, around 1830, finally solved the problem in some kindof generality by showing that the roots of an algebraic equation can be expressedby radicals if and only if the Galois group of the extension is solvable.

Lecture 4 Field Extensions

4.1 Algebraic and transcendental numbers

Definition 4.1.1 (Field extension). Let k be a field and let E be a field con-taining k. Then we say that E/k is a field extension .

Of some import, E is a vector space over k, and hence it has a dimensionover k, denoted

dimk(E) = [E : k],

which we call the degree of the extension E/k.

Note that the degree of a field extension might be infinite.

Example 4.1.2. The complex numbers C is an extension of degree 2 of R. N

Example 4.1.3. The real numbers R is an extension of infinite degree overQ. N

Definition 4.1.4 (Algebraic and transcendental numbers). Let E/k be a fieldextension. An element α ∈ E is called algebraic over k if

a0αn + a1α

n−1 + . . .+ an−1α+ an = 0

for some ai ∈ k not all equal to zero. In other words, α is a root of some nonzeropolynomial

f(x) = a0xn + a1x

n−1 + . . .+ an−1x+ an ∈ k[x].

On the other hand, an element α ∈ E is said to be transcendental over kif it is not algebraic over k.

Example 4.1.5. For instance,√

2 is algebraic over Q since it is a root of x2−2,but π is transcendental over Q. This second fact is very much not obvious, andwas proved in 1882 by Lindemann. N

Let E/k be a field extension and fix an α ∈ E. Consider the evaluationhomomorphism ϕα : k[x]→ E defined by f(x) 7→ f(α). Then kerϕα is an idealof k[x], which is an integral domain, meaning that kerϕα = 〈p(x)〉 for somep(x) ∈ k[x].

There are two cases to consider. First, if kerϕα = 〈0〉. This is equivalentto α being transcendental, since it means the only polynomial in k[x] with α as


FIELD EXTENSIONS 13

a root is 0, and it is also equivalent with ϕα being injective, since its kernel istrivial.

Second, if kerϕα = 〈p(x)〉 6= 〈0〉. Then p(α) = 0, so α is algebraic over k,and by the First isomorphism theorem,

k[x]/〈p(x)〉 ∼= Imϕα ⊂ E.

Since E is a field, and hence in particular an integral domain, Imϕα mustalso be an integral domain. But we showed last time that this is equivalentwith 〈p(x)〉 being a prime ideal, which further implies p(x) is irreducible (sinceof a(x)b(x) ∈ 〈p(x)〉, either a(x) ∈ 〈p(x)〉 or b(x) ∈ 〈p(x)〉). But the idealgenerated by an irreducible polynomial must be maximal, so k[x]/〈p(x)〉 is afield.

Hence

Imϕα ={f(α)

∣∣ f ∈ k[x]}

= k[α]

={anα

n + . . .+ a1α+ a0

∣∣ ai ∈ k and n ∈ N \{

0}}

is a field.In summary, then, if α is algebraic over k, then k[α] is a field.

4.2 Algebraic extensions

Definition 4.2.1 (Algebraic extension). A field extension E/k is algebraic ifevery element of E is algebraic over k.

Proposition 4.2.2. Every finite extension is algebraic.

Proof. We have dimk(E) = [E : k] = n < ∞. Let α ∈ E and consider the set{1, α, α2, . . . , αn

}⊂ E. This set contains n+1 elements, but E is n-dimensional

as a vector space over k, so it must be k-linearly dependent. In other wordsthere exist ai ∈ k not all zero such that

a0 · 1 + a1α+ . . .+ anαn = 0,

meaning that α is algebraic over k.

The converse is not true, i.e., there are algebraic extensions that aren’t finite.

Example 4.2.3. Consider C ⊃ Q, the set of all elements algebraic over Q.This is a field. To see this, take any two elements a, b ∈ Q. Since a is algebraicover Q, Q[α] is a field by the above discussion, and so moreover is b over Q[α],so Q[α, β] contains both their inverses and products and so forth, and theseextensions are contained in Q.

Moreover, [Q : Q] =∞. The easiest way to see this is to note that pn(x) =xn − 2 is an irreducible polynomial over Q for all n = 1, 2, 3, . . . (since it isEisenstein). But pn(α) = 0 for some α ∈ C by the Fundamental theorem ofalgebra, so α ∈ Q. But that means the extension we get by just adjoining α toQ is of degree n over Q, and this is contained in Q for all n, so the degree of Qover Q is larger than all positive integers n. N

Proposition 4.2.4. Let E ⊃ F ⊃ k be a chain of field extensions. Suppose{xi∣∣ i ∈ I

}is a basis of F/k and

{yj∣∣ j ∈ J

}is a basis of E/F . Then{

xiyj∣∣ i ∈ I and j ∈ J

}is a basis of E/k.

ALGEBRAIC FIELD EXTENSIONS 14

Proof. Since{yj}

make a basis of E/F , we have for any α ∈ E and someaj ∈ F , not all zero, ∑

j∈Jajyj = α.

Then since{xi}

is a basis of F/k, we have for some bij ∈ k, not all zero,∑i∈I

bijxi = aj

for any aj ∈ F . Combining these we get∑i∈I

∑j∈J

bijxiyj = α,

so some k-linear combination of xiyj is α.

From this proposition we get immediately that

[E : k] = [E : F ] · [F : k].

Corollary 4.2.5. Let E ⊃ F ⊃ k be a chain of field extensions. The extensionE/k is finite if and only if E/F and F/k are finite.

Definition 4.2.6 (Simple extension). Let E/k be a field extension, and letα ∈ E. Define k(α) to be the smallest subfield of E containing k and α. Wecall k(α) a simple extension of k.

Note that k[α] ={f(α)

∣∣f(x) ∈ k[x]}

is the smallest subring of E containingk and α. Since k is a field, k[α] is a domain, meaning that it has a quotientfield, so

k(α) =

{f(α)

g(α)

∣∣∣∣ f(x), g(x) ∈ k[x]

}.

Hence k(α) is the quotient field of k[α], and if α is algebraic over k, k[α] = k(α).

Lecture 5 Algebraic Field Extensions

5.1 Minimal polynomials

Definition 5.1.1 (Minimal polynomial). Let E/k be a field extension and letα ∈ E. Suppose α is algebraic over k, meaning that α is the root of somepolynomial in k[x]. Then monic polynomial in k[x] of smallest degree with α asa root, denoted Irr(α; k, x), is called the minimal or irreducible polynomialof α over k.

We discussed last time how the evaluation homomorphism ϕα : k[x] → Edefined by f(x) 7→ f(α) has as kernel kerϕα = 〈p(x)〉 for some p(x) ∈ k[x], sincethere being a division algorithm on k[x] means it is a principal ideal domain.Then we must have kerϕα = 〈Irr(α; k, x)〉 by definition.

Hence also k[x]/〈Irr(α; k, x)〉 ∼= k[α], where we can view both (being isomor-phic) as field extensions of, and hence vector spaces over, k.

Date: September 3rd, 2019.


Supposing for sake of argument that Irr(α; k, x) = xn + an−1xn−1 + . . . +

a1x + a0 = 0 in k[x]/〈Irr(α; k, x)〉, we have then that{

1, x, x2, . . . , xn−1}

is abasis for k[x]/〈Irr(α; k, x)〉 over k, so

Proposition 5.1.2. Suppose α is algebraic over k. Then

[k(α) : k] = deg Irr(α; k, x).

Remark 5.1.3. Let E ⊃ F ⊃ k be a chain of fields. If α ∈ E is algebraic over k,then α is also algebraic over F . This is fairly obvious: α being algebraic over kmeans there is some polynomial with coefficients in k for which α is a root, butsince k is contained in F , that polynomial also has coefficients in F .

Proposition 5.1.4. Let E/k be a field extension. Let α1, α2, . . . , αn ∈ E bealgebraic over k. Then k(α1, α2, . . . , αn)/k is algebraic.

Proof. Working one element at a time, if α1 is algebraic over k, then [k(α1) :k] <∞. Next α2 being algebraic over k means it is also algebraic over k(α1) bythe above remark, and so [k(α1, α2) : k(α1)] <∞, and

[k(α1, α2) : k] = [k(α1, α2) : k(α1)] · [k(α1) : k] <∞,

so k(α1, α2)/k is an algebraic extension. Now repeat this argument.

Maybe counterintuitively, this is true even if we aren’t adjoining a finitenumber of α:

Proposition 5.1.5. Let E/k be a field extension, and let αλ ∈ E be algebraicover k for all λ ∈ Λ, some arbitrary index set. Then k(αλ |λ ∈ Λ)/k is algebraic.

Proof. If β ∈ k(αλ | λ ∈ Λ), then β = a1αλ1+ a2αλ2

+ . . .+ anαλn , with ai ∈ kand αλi ∈ Λ. The crucial point is that β is a k-linear combination of only finitelymany αλ.

Hence β ∈ k(αλ1 , αλ2 , . . . , αλn), which is an algebraic extension of k by theabove proposition, and so β is algebraic over k.

Though an element being algebraic by definition means there exists somepolynomial of which it is a root, it is generally speaking hard to find the poly-nomial, and comparatively easy to show that the element is algebraic.

Proposition 5.1.6. Let E ⊃ F ⊃ k be a chain of field extensions. Then E/kis algebraic if and only if E/F and F/k are algebraic.

Proof. The forward direction is trivial, and more to the point we showed it inan earlier remark.

For the converse direction we need a little bit more care. Let β ∈ E bealgebraic over F , so that

p(x) = xn + an−1xn−1 + . . .+ a1x+ a0 ∈ F [x]

with p(α) = 0. Then β is algebraic over k(an−1, an−2, . . . , a0), meaning that[k(an−1, . . . , a0, β) : k(an−1, . . . , a0)] < ∞. Each of these ai ∈ F , since F/k isalgebraic by hypothesis, are algebraic over k, so k(an−1, . . . , a0)/k is algebraic,meaning that [k(an−1, . . . , a0) : k] <∞.

Therefore finally [k(an−1, an−2, . . . , a0, β) : k] < ∞, and so β is algebraicalso over k.


Let E/k be a field extension and let α ∈ E be algebraic over k. Pick anyβ 6= 0 in k[α] = k(α) (equal since α is algebraic over k). A natural question toask is this: how would we compute the inverse β−1 in terms of α?

Given the minimal polynomial Irr(α; k, x) = p(x) ∈ k[x] of α over k, as wellas how to write β in terms of powers of α (which form a basis of k[α] over k,of course), say β = g(α) for some g(x) ∈ k[x], we can do this by essentially thesame algorithm one uses to compute inverses in Z/nZ.

Note that p(x) is irreducible and p(x) - g(x) (since otherwise g(α) = β = 0),meaning that gcd(p(x), g(x)) = 1. Hence by the Euclidean algorithm repeatedlythere exist A(x), B(x) ∈ k[x] such that

A(x)p(x) +B(x)g(x) = 1.

Evaluating this at x = α, we get B(α)g(α) = 1, so and g(α) = β, so b−1 = B(α).

5.2 Composites of fields

Definition 5.2.1 (Composite field). Let E,F ⊂ L be subfields of L. Thesmallest subfield of L that contains both E and F is called the composite ofE and F , denoted EF .

Recall how in a ring R, if A and B are ideals, then we define

AB ={ ∑

finite

aibi

∣∣∣ ai ∈ A, bi ∈ B }is the smallest ideal containing A and B. In general, however,

EF 6={ ∑

finite

eifi

∣∣∣ ei ∈ E, fi ∈ F }since the right-hand side does not include inverses, unless:

Proposition 5.2.2. Suppose we have a chain of field extensions k ⊂ E ⊂ Land k ⊂ F ⊂ L. Suppose E/k or F/k is algebraic. Then

EF ={ ∑

finite

eifi

∣∣∣ ei ∈ E, fi ∈ F }.Proof. We have, in general, EF = E(F ) = F (E), since by definition this in-cludes the inverses. Suppose, without loss of generality, that F/k is algebraic.Then

EF = E(f | f ∈ F ) = E[f | f ∈ F ] ={ ∑

finite

eifi

∣∣∣ ei ∈ E, fi ∈ F }.5.3 Algebraic closure

Let k be a field. We want to construct a field K ⊃ k such that K/k is algebraicand every irreducible polynomial in K[x] has a root in K.

Definition 5.3.1 (Algebraically closed). A field L is called algebraicallyclosed if every polynomial in L[x] has a root in L.

ALGEBRAIC CLOSURE 17

Example 5.3.2. The complex numbers C are algebraically closed. The realsR are not, however: for instance f(x) = x2 + 1 ∈ R[x] has no roots in R. N

Remark 5.3.3. (i) If L is algebraically closed, then all the roots of every poly-nomial in L[x] are in L (one is by definition; divide that factor away andrepeat). Hence

(ii) if L is algebraically closed, then every polynomial can be written as aproduct of linear factors in L.

Note that we are after an extension that is both algebraic over k and alge-braically closed. First we give a partial result:

Theorem 5.3.4. Let k be an arbitrary field. Then there exists an extensionE/k such that E is algebraically closed.

To prove it we use the following lemma:

Lemma 5.3.5. Let p(x) ∈ k[x] be irreducible. Then there exists some fieldextension E/k such that E contains a root of p(x).

Proof. Let E = k[x]/〈p(x)〉. This is a field since p(x) is irreducible, whence〈p(x)〉 is maximal.

Now k embeds naturally in E by a 7→ a+ 〈p(x)〉, which is one-to-one, so we

can identify k as a subfield of E, say k. So if p(x) = anxn + . . .+ a0 ∈ k[x], this

corresponds to (an+〈p(x)〉)xn+. . .+(a0+〈p(x)〉) ∈ k[x], so letting x = x+p(x),p(x) = (anx

n + . . .+ a0) + 〈p(x)〉 = p(x) + 〈p(x)〉 = 0 in E. Hence E has a zeroof p(x) inside of it.

Remark 5.3.6. (i) Let f(x) ∈ k[x]. Then there exists some extension E/ksuch that E has a root of f(x) in it. This follows immediately: if f(x) isirreducible, then it’s the lemma, and if it isn’t, factor it into irreducibleparts and use the lemma.

(ii) Let f1(x), f2(x), . . . , fn(x) ∈ k[x]. Then there exists a field extension E/ksuch that E has a root of fi(x) for every i.

Lecture 6 Algebraic Closure

6.1 Existence of algebraic closure

We start by proving the theorem toward the end of last lecture.

Proof. Let{fλ(x)

∣∣ λ ∈ Λ}

be the set of all irreducible polynomials of degree

> 1 in k[x]. Let{xλ∣∣ λinΛ

}be a set of variables, one for each irreducible

polynomial. Define R := k[xλ | λ ∈ Λ], and let A be the ideal of R generated byfλ(xλ) for every λ ∈ Λ.

Then A 6= R, i.e., A is a proper ideal of R. By way of contradiction, supposenot. Then

1 = gλ1fλ1

(xλ1) + gλ2

fλ2(xλ2

) + . . .+ gλrfλr (xλr )

Date: September 5th, 2019.

ALGEBRAIC CLOSURE 18

for some finite set{λ1, λ2, . . . , λr

}⊂ Λ, and gλ1

, gλ2, . . . , gλr ∈ R. By the

second remark just above, there exists an extension F of k such that F has aroot, say αλ, of fλi(xλi) for each i = 1, 2, . . . , r. But then if we evaluate theequation above at xλi = αλi , for i = 1, 2, . . . , r, we get 1 = 0, a contradiction.

Hence A is a proper ideal of R. Now R/A might not be a field, since A neednot be maximal, but there must exist a maximal ideal containing A, say M .Then E1 := R/M is a field. Letting αλ = xλ +M , then we get

fλ(αλ) = fλ(xλ) +M = 0

since fλ ∈ A ⊂ M . In other words, every irreducible polynomial in k[x] has aroot in E1.

This is not quite what we need—we want every every polynomial in E1[x]to have a root in E1. This needn’t be the case, but we can repeat this process,taking E1 in place of k, obtaining a new field E2, and so on:

k = E0 ⊂ E1 ⊂ E2 ⊂ . . . ⊂ En ⊂ . . . ,

and every irreducible polynomial in En[x] has a root in En+1 for all n ≥ 0. Nowset

E :=

∞⋃n=0

En.

This is a field (since any two a, b ∈ E belong to some En, which is a field).Moreover for any irreducible p(x) ∈ E[x] we must have p(x) ∈ Em[x] for somem, and so Em+1 contains a root of p(x), but Em+1 ⊂ E, so E also contains aroot of p(x). Therefore E is algebraically closed.

Definition 6.1.1 (Algebraic closure). An extension K of a field k is an alge-braic closure of k if K/k is algebraic and K is algebraically closed.

Lemma 6.1.2. Let E/k be a field extension and let A be the set of all elementsin E algebraic over k. Then A is a field.

Moreover, if E is algebraically closed, then A is algebraically closed.

Proof. Take α, β ∈ A. Then α is algebraic over k, so [k(α) : k] < ∞, and β isalgebraic over k, and hence over k(α), so [k(α, β) : k(α)] <∞. This means that

[k(α, β) : k] = [k(α, β) : k(α)] · [k(α) : k] <∞,

so k(α, β)/k is algebraic, meaning that α ± β, αβ, and α/β are all in k(α, β),so they are in A, and hence A is a field.

Now assume E is algebraically closed, and let f(x) ∈ A[x]. Then sinceA[x] ⊂ E[x], and E is algebraically closed, f(x) has a root α in E. We need toshow that such a root is algebraic over k, so that it is also in A.

Now writing f(x) = xn+an−1xn−1+. . .+a1x+a0, we have α is algebraic over

k(an−1, . . . , a0). Hence since [k(an−1, . . . , a0) : k] <∞ and [k(an−1, . . . , a0, α) :k(an−1, . . . , a0)] <∞, we have

[k(an−1, . . . , a0, α) : k] <∞

meaning that k(an−1, . . . , a0, α) is algebraic over k, so α is algebraic over k.Hence α ∈ A, and so A is algebraically closed.

Corollary 6.1.3. Let k be a field. Then there exists an extension K/k suchthat K is algebraic over k and K is algebraically closed. In other words, k hasan algebraic closure.

FIELD EMBEDDINGS 19

Lecture 7 Field Embeddings

7.1 Uniqueness of algebraic closure

Definition 7.1.1. An embedding of a field E into another field L is an injectivehomomorphism from E to L. That is, τ : E → E′ ⊂ L, with E′ = τ(E) ∼= E asfields.

Example 7.1.2. Consider the field E = Q(√

2). We can embed E into C bythe identity mapping. On the other hand, noting how

Q(√

2) ={a+√

2b∣∣ a, b ∈ Q

},

τ : Q(√

2)→ C defined by τ : a+ b√

2 7→ a− b√

2 is also an embedding of Q(√

2)into C.

These are the only two, as it turns out. To see this, suppose we have anembedding τ : Q(

√2) ↪→ C. This is a homomorphism, so τ(1) = 1, and hence

also for n ∈ Z we have τ(n) = n. It must also preserve quotients, so for anyr ∈ Q we have τ(r) = r. So τ must fix Q, meaning that

τ(a+ b√

2) = τ(a) + τ(b)τ(√

2) = a+ bτ(√

2),

and therefore τ is completely determined by its value on√

2. However

τ(√

2)2 = τ(√

2)τ(√

2) = τ(√

2√

2) = τ(2) = 2,

so τ(√

2) must be some kind of square root of 2, so the only options are τ(√

2) =√2, leading to the identity mapping, or τ(

√2) = −

√2, leading to the second

embedding above. N

Example 7.1.3. Let E = Q( 3√

2). We wish to list all embeddings of E into C.Now

Q(3√

2) ={a+ b

3√

2 + c(3√

2)2∣∣ a, b, c ∈ Q

},

where the dimension is rightly 3 since the minimal polynomial of 3√

2 over Q isx3 − 2. Now go construct an embedding τ : E ↪→ C we need only fix τ( 3

√2) by

the same reasoning as above, and similarly τ( 3√

2)3 = 2.The equation x3 − 2 = 0 has roots 3

√2, 3√

2ω, and 3√

2ω2, where

ω =−1 +

√−3

2and ω2 =

−1−√−3

2

are the nontrivial cubic roots of unity.Hence there are three possible embeddings: the identity mapping, from

τ( 3√

2) = 3√

2; the embedding from τ( 3√

2) = 3√

2ω; and the one from τ( 3√

2) =3√

2ω2. N

Example 7.1.4. When E = Q( 3√

2, ω), there are six possible embeddingsτ : E ↪→ C; there are three options for τ( 3

√2), and the minimal polynomial

of ω is of degree two, since 0 = x3 − 1 = (x − 1)(x2 + x + 1), so there are twopossible options for τ(ω). N


FIELD EMBEDDINGS 20

More generally,

Example 7.1.5. Consider the field Q(α), with α algebraic over Q, and anembedding τ : Q(α) ↪→ C. Then α has a minimal polynomial over Q, living inQ[x], so τ doesn’t affect its coefficients, meaning that τ(α) must be a root ofthe same minimal polynomial. N

Definition 7.1.6. Consider a diagram of field extensions

E L

k k′ = σ(k)

τ

σ∼=

Suppose that τ |k = σ. Then we say that τ is a lift of σ.Suppose k′ = k, σ = Id. Then we call τ a k-embedding , i.e., τ |k = Id, so

the embedding fixes k.

Proposition 7.1.7. Consider the diagram

L

E

k k′

algebraic

σ∼=

of fields, where L is algebraically closed. Then there exists a field E′ ⊂ L andan embedding τ of E into L such that τ(E) = E′ and τ |k = σ.

In other words, σ can be lifted to any algebraic extension, provided L isalgebraically closed.

Proof. This is a proof by Zorn’s lemma. To that end, let

S ={

(F, ρ)∣∣ E ⊃ F ⊃ k, ρ : F ↪→ L, ρ|k = σ

},

the family of extensions of k and lifts of σ. This family is nonempty sinceat the very least (k, σ) ∈ S. We can equip this set with a partial order by(F1, ρ1) < (F2, ρ2) if and only if F1 ⊂ F2 and ρ2|F1

= ρ1. Thus S is a partiallyordered set, and so if we can show that every chain , or totally ordered subset,of S has an upper bound, then by Zorn’s lemma S has a maximal element.

But this is not too hard: let C ⊂ S be totally ordered. Then take

(F, ρ) =⋃

(Fλ,ρλ)∈C

(Fλ, ρλ),

where by union on the fields we mean simply set unions, and by union of theembeddings we mean that ρ on an element from a field Fλ just uses the cor-responding ρλ. That this is an upper bound is clear, also E ⊂ F ⊂ k, withρ|k = σ, so (F, ρ) ∈ S.

Hence by Zorn’s lemma S has a maximal element, say M = (K, τ) ∈ S. Wenow need to show that K = E. So suppose K 6= E, so K is strictly smaller

FIELD EMBEDDINGS 21

than E. Pick an α ∈ E \K. Since E is algebraic over k, it is also algebraic overK, and so α is algebraic over K.

Let p(x) be the minimal polynomial of α over K, say p(x) = xn+an−1xn−1+

. . .+ a0 ∈ K[x]. This is irreducible and so pτ (x) = xnτ(an−1)xn−1 + . . .+ τ(a0)is also irreducible, since K ∼= τ(K). Hence

K(α) ∼=K[x]

〈p(x)〉∼=τ(K)[x]

〈pτ (x)〉∼= τ(K)[β],

where β is any root of pτ (x). Thus there exists some τα : K(α)→ τ(K)(β) ⊂ Lwith τα|k = τ , meaning that (K(α), τα) ∈ S, but then (K(α), τα) > (K, τ)strictly in our order, which contradicts the maximality of (K, τ). Hence K = E,and we are done.

Now assume E/k is algebraic and E is algebraically closed. Suppose L/k′

is also algebraic, and L is algebraically closed, and as before k ∼= k′ by σ.Then by the proposition, there exists an embedding τ of E into L such thatτ(E) = E′ ⊂ L. Since L/k′ is algebraic, L/E′ is algebraic. Moreover E isalgebraically closed, and E ∼= E′, so E′ must also be algebraically closed, butthat must mean L = E′, thus E ∼= L.

Taking k = k′, σ = Id, this discussion proves

Theorem 7.1.8 (Uniquness of algebraic closure). Suppose K and E are bothalgebraic closures of k. Then there exists an isomorphism τ : K → E such thatτ |k = Id.

In other words, the algebraic closure of a field is unique up to isomorphism.

For this reason we will routinely denote the algebraic closure (up to isomo-prhism) of a field k by k.

Example 7.1.9. For instance, Q, the set of all elements α ∈ C algebraic overQ is the algebraic closure of Q. Note that Q 6= C, since for example π is notalgebraic over Q. N

Definition 7.1.10. Let k be a field. Let Aut(k) denote the group of all auto-morphisms of k, meaning isomorphisms from k to itself, called the automor-phism group of k.

Similarly, for a field extension E/k, Autk(E) is the subgroup of Aut(E)containing all elements which restricts to the identity map on k.

Theorem 7.1.11. Let k ⊃ E ⊃ k. Suppose σ : E → k is a k-embedding intok. Then there exists τ ∈ Autk(k) such that τ |E = σ. So all embeddings of analgebraic extension can be lifted to the algebraic closure.

Theorem 7.1.12. Let k be an algebraic closure of k. Let α, β ∈ k. Then thefollowing two are equivalent:

(i) α and β are roots of the same irreducible polynomials in k[x],

(ii) there exists some ω ∈ Autk(k) such that ω(α) = β.

SPLITTING FIELDS 22

Proof. For (i) implying (ii), we have the diagram

k k

k(α) k(β)

k

ω

σ

where σ is the embedding sending α to β, since they are roots of the sameirreducible polynomial. Then the lift ω exists by the previous results.

For (ii) implying (i), note that if α is a root of p(x) ∈ k[x], then pω(x) = p(x)since ω fixes k, and β is a root of pω(x), since ω(p(α)) = pω(ω(α)) = pω(β).

Lecture 8 Splitting Fields

8.1 Lifts are not unique

We discussed previously how an embedding can always be lifted to the algebraicclosure. However, such a lift is not unique:

Example 8.1.1. Let k = Q, α = 3√

2, and β = 3√

2ω, with ω = −1+√−3

2 , asbefore. Then both α and β are roots of x3 − 2, and so

Q(α) ∼=k[x]

〈x3 − 2〉∼= Q(β).

Let σ : Q(α) → Q(β) be the Q-embedding sending α to β. We know that thiscan be lifted to τ : Q→ Q, woth τ |Q(α) = σ. Diagramatically,

Q Q

Q(α) Q(β)

Q

τ

σ∼=

This lift τ is not unique. For instance, it must send 3√

2ω to another root ofx3 − 2, i.e., 3

√2, 3√

2ω, or 3√

2ω2. It is, however, a lift of σ, and σ sends 3√

2 to3√

2ω, and being one-to-one and onto, only 3√

2 can be sent there. This leavestwo options for τ( 3

√2ω), but two is enough to make it not unique.

That is not all, however: there are many other elements in Q that τ has tosame somewhere, elements that are not in Q(α) ans so are not already fixed.For instance, τ

√2 can be either

√2 or −

√2. N

Definition 8.1.2. Let k ⊃ k and let f(x) ∈ k[x]. Then

f(x) = c(x− α1)(x− α2) · · · (x− αn)


NORMAL EXTENSIONS 23

for αi ∈ k, with αi not necessarily distinct.Let E = k(α1, α2, . . . , αn). Then E is called the splitting field of f(x).

Remark 8.1.3. The splitting field E of f(x) does not depend on the choice ofalgebraic closure k.

Lecture 9 Normal Extensions

9.1 Splitting fields and normal extensions

First, note that by the previous discussion, if E is the splitting field of somepolynomial f(x) ∈ k[x], then a k-embedding σ must have the property that forany root αi of f(x), σ(αi) is also a root of f(x). Hence σ(E) ⊂ E.

Example 9.1.1. Let k = Q and k = Q ⊂ C. Consider the polynomial f(x) =x3−2 ∈ Q[x]. The roots of f(x) are 3

√2, 3√

2ω, and 3√

2ω2, where ω = −1+

√−32.

Hence the splitting field of f(x) is E = Q( 3√

2, 3√

2ω, 3√

2ω2) = Q( 3√

2, ω). (Thatthese extensions are equal is fairly easy to see: definitely the former is containedin the latter, and since 3

√2ω2/( 3

√2ω) = ω, the latter is contained in the former.)

In similar fashion, the splitting field of the polynomial g(x) = (x3 − 2)(x2 +1) ∈ Q[x] is E = Q( 3

√2, ω,

√−1). N

Definition 9.1.2 (Normal extension). An algebraic extension E/k is normalif for all σ : E → k, k-embeddings, we have σ(E) = E.

Example 9.1.3. The extension Q( 3√

2)/Q is not normal, since σ( 3√

2) can be3√

2, 3√

2ω, or 3√

2ω2, only one of which satisfies the above condition. N

Example 9.1.4. The extension Q( 3√

2, ω)/Q is normal. This is quite clear: theimage of σ( 3

√2), as per above, is either 3

√2, 3√

2ω, or 3√

2ω2, all of which are inQ( 3√

2, ω), and similarly the σ(ω) must be ω or ω2, both of which again are inQ( 3√

2, ω). N

Remark 9.1.5. The original definition of a normal extension is this: E/k isnormal if for any L ⊃ E ⊃ k and any embedding σ : E ↪→ L, we have σ(E) =E.

It is possible to prove that any transcendental extension is not normal bythis definition, hence the requirement of E/k being algebraic in the definitionwe chose. The proof, however, is not easy.

We will make frequent use of the following lemma:

Lemma 9.1.6. Let E/k be an algebraic extension and suppose σ : E → E is ak-embedding. Then σ(E) = E.

In other words, the embedding being one-to-one must automatically be ontoif E/k is algebraic. This is not true if the extension is not algebraic:

Counterexample 9.1.7. Consider the fields C(x) ⊃ C(x2) and the map σ : C(x)→C(x2) defined by σ(x) = x2. This is a C-embedding (it fixes C), and so one-to-one. However C(x) ∼= C(x2), and the image of σ misses all terms with just x,so it is not onto. N



Proof of lemma. Let α ∈ E. We need to show that α ∈ σ(E). Let p(x) =Irr(α; k, x). Then

p(x) = (x− α1)e1(x− α2)e2 · · · (x− αr)erg(x) ∈ E[x],

where g(x) has no roots in E, and α = α1, α2, . . . , αr are distinct roots of p(x)in E.

Letting, for any f(x) = a0xn + a1x

n−1 + . . .+ an−1x+ an ∈ E[x],

fσ(x) = σ(a0)xn + σ(a1)xn−1 + . . .+ σ(an−1)x+ σ(an) ∈ E[x],

we get pσ(x) = p(x) since σ is a k-embedding, meaning it fixes k, an p hascoefficients in k. Hence

p(x) = (x− σ(α1))e1(x− σ(α2))e2 · · · (x− σ(αr))ergσ(x).

Now{α1, α2, . . . , αr

}are distinct roots of p(x) in E with total multiplicity

e1 +e2 + . . .+er, and so is{σ(α1), σ(α2), . . . , σ(αr)

}. Hence the sets are equal,

meaning α = α1 = σ(αi) for some i, and therefore σ is surjective.

Theorem 9.1.8. Let k ⊃ K ⊃ k. The following conditions are equivalent:

(i) For all σ ∈ Autk(k), σ(K) = K, i.e., K/k is normal.

(ii) K is a splitting field of a family of polynomials of k[x].

(iii) If α ∈ K and p(x) = Irr(α; k, x), then all roots of p(x) are in K.

Proof. First let us show that (i) implies (ii) and (i) implies (iii). For anyα ∈ K, let pα(x) = Irr(α; k, x). Then, say α = α1,

pα(x) = (x− α1)(x− α2) · · · (x− αn) ∈ k[x].

For any αi, there exists some σ ∈ Autk(k) such that σ(α) = αi. But (i) impliesαi ∈ K (which implies (iii)), and so K contains k(α1, α2, . . . , αn). Hence K isthe splitting field of pα(x) for α ∈ K, so this implies (ii).

Next, let us show that (ii) implies (i). AssumeK = k(αi,λ|i = 1, 2, . . . , jλ, λ ∈Λ), where

{α1,λ, α2,λ, . . . , αjλ,λ

}are the roots of fλ(x) ∈ k[x] in k. Assume

fλ(x) are all irreducible.Then for any σ ∈ Autk(k), σ(αi,λ) = α`,λ ∈ K for some `, so σ(K) ⊂ K,

and hence by the lemma σ(K) = K. Hence K/k is normal.Finally let us tackle (iii) implying (i). Let α ∈ K and take p(x) = Irr(α; k, x).

Then for any σ ∈ Autk(k), pσ(x) = p(x), so σ(α) is also a root of p(x). By (iii)this means σ(α) ∈ K, so σ(K) ⊂ K, and hence σ(K) = k by the lemma, andfinally then K/k is normal.

Example 9.1.9. Consider the chain Q ⊃ Q( 4√

2) ⊃ Q(√

2) ⊃ Q. Let us studythis one step at a time.

First, Q(√

2)/Q is normal, because Irr(√

2;Q, x) = x2−2, and both its roots√2 and −

√2 are in Q(

√2).

Second, Q( 4√

2)/Q(√

2) is also normal: Irr( 4√

2;Q(√

2), x) = x2 −√

2, theroots of which are ± 4

√2 ∈ Q( 4

√2).

What about Q( 4√

2)/Q? We have Irr( 4√

2;Q, x) = x4 − 2, the roots of whichis ± 4√

2 and ± 4√

2i, but the latter two are not in Q( 4√

2), so this extension is notnormal. N


Note that this tells us that extensions being normal is not a transitive rela-tion.

Definition 9.1.10. LetK/k be any transcendental extension. A subset{xλ∣∣λ ∈

Λ}⊂ K is called a transcendental basis of K/k if

(i){xλ∣∣ λ ∈ Λ

}is an algebraically independent set (meaning they are not

roots of any nontrivial polynomial equations, cf. linearly independentmeaning not roots of nontrivial linear equations). Another way of sayingthis, ϕ : k[yλ |λ ∈ Λ]→ k[xλ |λ ∈ Λ] defined by yλ 7→ xλ is an isomorphism(i.e., each xλ behaves as though it were a variable).

(ii) K/k(xλ | λ ∈ Λ) is an algebraic extension.

Example 9.1.11. Consider the field K = Q(√

2, e). Then{e}

is a tran-scendental basis of K/Q; Q(e)/Q is a transcendental extension, and K/Q(e) isalgebraic. N

Definition 9.1.12. If k(xλ |λ ∈ Λ) = K, then K is called a purely transcen-dental extension .

Remark 9.1.13. A purely transcendental extension K/k has no algebraic ele-ments over k not in k.

Remark 9.1.14. A transcendental basis always exists (this, like the existenceof bases in general, is proved by the standard Zorn’s lemma technique), andmoreover all such bases have the same cardinality.

This means that an transcendental extension can always be split into apurely transcendental part and an algebraic part, where the algebraic part canbe trivial if the original extension was purely transcendental.

Example 9.1.15. The extension R/Q is transcendental because, for example,π and e are transcendental over Q. However it is not purely transcendentalsince e.g.,

√2 is in R, which is algebraic over Q.

Note finally that a transcendental basis for this extension has infinite cardi-nality. N

Definition 9.1.16. A class C of field extensions is called distinguished if

(i) it is transitive, i.e.,E/k ∈ C if and only if E/F ∈ C and F/k ∈ C for everyE ⊃ F ⊃ k.

(ii) it has a lifting property, meaning if E/k ∈ C, and F/k is any extensionsuch that E,F ⊂ L for some L, then EF/F ∈ C.

(iii) it has a composite property, if E/k ∈ C and F/k ∈ C and E,F ⊂ L forsome L, then EF/k ∈ C.

Example 9.1.17. Let C be the class of all finite extensions. This is distin-guished. N

Example 9.1.18. The class C of all algebraic extensions is also distinguished.N

Example 9.1.19. The class C of all normal extensions is not distinguished—wehave already discussed how it is not transitive. However it does have both thelifting and composite properties. N

SEPARABLE EXTENSION 26

Lecture 10 Separable Extension

10.1 Separable degree

Let E/k be an algebraic extension, and let σ : k ↪→ σ(k) ⊂ σ(k) be an embeddingof k. We know from earlier that σ can be lifted to σ∗ : E → σ(k),

σ(k)

E

k σ(k)

σ∗

σ∼=

We have also learned, however, that this lift is not unique. For this reason,consider

Sσ ={σ∗ : E ↪→ σ(k) an embedding such that σ∗|k = σ

}.

Now define [E : k]s = |Sσ|, called the separable degree of E/k.Note how |Sσ| depends on σ and the choice of closure σ(k). We want to

show that [E : k]s is independent of both of these.

Lecture 11 Simple Extensions

11.1 Separable extensions

To see that [E : k]s is independent of both σ and the algebraic closure σ(k),suppose we have another embedding τ : k ↪→ τ(k), with its own algebraic closureτ(k).

Then because there exists some λ : σ(k)→ τ(k) such that λ|σ(k) = τ ◦ σ−1.In a picture,

τ(k) σ(k)

E

τ(k) k σ(k)

algebraic

λ

σ∗

στ

τ◦σ−1

This by way of saying that we want to find a lift of τ , say τ∗, correspondingto σ∗. This is a matter of following the diagram around: define τ∗ = λ ◦ σ∗;then τ∗|k = τ , since

τ∗∣∣∣∣k

= λ ◦ σ∗∣∣∣∣k

= λ ◦ σ∣∣∣∣k

= τ ◦ σ−1 ◦ σ = τ.

Date: September 19th, 2019.Date: September 24th, 2019.

SIMPLE EXTENSIONS 27

Hence given a lift σ∗ of σ we have a τ∗, and vice versa, so there is a one-to-one correspondence from Sσ to §τ defined by σ∗ 7→ τ∗ = λ ◦ σ (and inverseτ∗ 7→ σ∗ = λ−1 ◦ τ∗). Therefore |Sσ| = |Sτ |, and Sσ does not depend on thechoice of σ or closure σ(k).

In other words we can fix an algebraic closure k and take σ = Id, and|Sσ| = |SId|, and then [E : k]s is precisely the number of k-embeddings of E.

We can characterise the separable degree in other ways too. For instance,consider fields k ⊃ E ⊃ k, and suppose E = k(α) for some α ∈ E, and let p(x)be the minimal polynomial of α over k. Then σ is completely determined by itsimage of α, so σ(E) = k(β) for some root β of p(x).

Hence [E : k]s is the number of distinct roots of p(x) in k.

Example 11.1.1. Consider p(x) = x3 + x + 1 ∈ Q[x], which is irreducible (ithas no rational roots, and is of degree less than or equal to three). Let α ∈ Qwith p(α) = 0. Then [Q(α) : Q]s = 3 since p(x) has three distinct roots.

Note, maybe out of curiosity, how this is also the degree of p(x), and hence[Q(α) : Q] = 3. N

This last remark need not always be the case:

Example 11.1.2. Let Fq denote the finite field of q = pn elements, p a prime,and let k = Fq(x), x a variable, be the function field over it. Now considerp(t) = tp − x ∈ k[t]. In k, which must exist, take some α ∈ k such thatp(α) = 0. Then p(α) = αp − x = 0, so x = αp, and hence

p(t) = tp − αp = (t− α)p

in k[t], since k has characteristic p. This means p(t) has only one distinctroot in k, so [k(α) : k]s = 1, but because p(t) = tp − x is irreducible over k,[k(α) : k] = p. N

In other words, the degree of an extension is not necessarily equal to thedegree of the extension. That said, they do behave similarly in many ways.

Theorem 11.1.3. Let E ⊃ F ⊃ k be a chain of algebraic extensions. Then

[E : k]s = [E : F ]s[F : k]s.

Proof. Consider the diagram

E k

F

k

σ∗

τ∗

where σ∗ a k-embedding of E, i.e., σ∗|k = Idk, and σ∗|F = τ∗, with τ∗|k =σ∗|k = Idk.

Then for each τ∗ there are by definition [E : F ]s lifts σ∗, and there are[F : k]s different τ∗ to start with, giving us the right-hand side. On the otherhand, there are [E : k]s ways to do this all together.


Corollary 11.1.4. If [E : k] <∞, then [E : k]s ≤ [E : k].

Proof. Since E is a finite extension of k we can write E = k(α1, α2, . . . , αn) forsome αi ∈ E. Then

k ⊃ k(α1) ⊃ k(α1, α2) ⊂ . . . ⊂ k(α1, α2, . . . , αn) = E.

Note how, at any individual step, [k(α) : k]s ≤ [k(α) : k] since the left-hand sideis the number of distinct roots of Irr(α; k, x), whereas the right-hand side is thedegree, and hence total number of roots, with multiplicity, of same.

By the previous theorem,

[k(α1, α2, . . . , αn) : k]s = [k(α1, α2, . . . , αn) : k(α1, α2, . . . , αn−1)]s · · · [k(α1) : k]s,

and each factor is bounded their respective degrees, so the product is boundedby [k(α1, α2, . . . , αn) : k].

Corollary 11.1.5. Let E ⊃ F ⊃ k be a chain of finite extensions. Then[E : k]s = [E : k] if and only if [E : F ]s = [E : F ] and [F : k]s = [F : k].

Proof. In the proof of the previous corollary, for there to be equality each of theindividual steps in the end have to be equalities.

Definition 11.1.6. A finite extension E/k is separable if [E : k]s = [E : k].

Definition 11.1.7. Let k be a field and k an algebraic closure, and take α ∈ k.We say that α is separable over k if [k(α) : k]s = [k(α) : k].

In other words, α is separable over k if and only if Irr(α; k, x) has no repeatedroots in k.

Definition 11.1.8. A polynomial f(x) ∈ k[x] is separable if f(x) has norepeated roots in k.

Example 11.1.9. In our previous example, f(t) = tp − x ∈ Fq(x)[t] is irre-ducible over Fq(x) but not separable. N

These definitions are compatible:

Theorem 11.1.10. Let E/k be a finite extension. Then E/k is separable ifand only if all elements α ∈ E are separable over k.

Proof. For any α ∈ E,

[E : k] = [E : k(α)][k(α) : k]

and[E : k]s = [E : k(α)]s[k(α) : k]s.

For the forward direction, assume E/k is separable, so [E : k]s = [E : k].Hence for all α ∈ E, [k(α) : k]s = [k(α) : k] by the corollary above, so α isseparable over k.

For the converse direction, E/k is a finite extension so E = k(α1, α2, . . . , αn)for some α1, α2, . . . , αn ∈ E.

In general, for E ⊃ F ⊃ k, E/k being separable implies E/F is separable,so since

[E : k]s = [E : k(α1, α2, . . . , αn−1)]s · · · [k(α1) : k]s

= [E : k(α1, α2, . . . , αn−1)] · · · [k(α1) : k] = [E : k]

since each extension along the way is adjoining just a single element.


From this proof we see in particular that

Corollary 11.1.11. Let E = k(α1, α2, . . . , αn) with αi algebraic over k. ThenE/k is separable if and only if α1, α2, . . . , αn are separable over k.

In other words, it is enough to study a set of generators.All of the previous discussion have required extensions be finite, since we are

comparing degrees.

Definition 11.1.12. Let E/k be an algebraic extension, possibly infinite. Wesay that E/k is separable if every finite extension of k in E is separable.

I.e., for any F such that E ⊃ F ⊃ k with [F : k] <∞, F is separable over k.Or, another way of saying the same thing: for any α ∈ E, α is separable over k.

Theorem 11.1.13. Let E ⊃ F ⊃ k be a chain of algebraic extensions. ThenE/k is separable if and only if E/F and F/k are separable.

Proof. Using the last statement in the definition above, consider for the forwarddirection any element α ∈ E. This is separable over k, so Irr(α; k, x) has norepeated roots in E. But then it also can’t have repeated roots in F , nor canthe minimal polynomial of α over F have repeated roots in E.

Similarly for the converse.

11.2 Simple extensions

Definition 11.2.1. An algebraic extension E/k is called simple if there existssome α ∈ E such that E = k(α).

In particular, [E : k] <∞, so a simple extension is always finite.

Definition 11.2.2. Let E/k be an algebraic extension. Suppose E = k(α).Then α is called a primitive element .

This makes us wonder when, in general, a finite extension is simple.

Theorem 11.2.3 (Primitive element theorem). Let E/k be a finite extension.

(i) Suppose E/k possesses only finitely many intermediate subfields. ThenE/k is simple. The converse also holds.

(ii) If E/k is separable, then E/k is simple.

Proof. For the forward direction of (i), suppose there are only finitely manyintermediate fields F , k ⊂ F ⊂ E. Pick up any α, β ∈ E. We claim thatk(α, β) = k(γ) for some γ ∈ E.

If true, this claim solves our problem: since E/k is finite, E = k(α1, α2, . . . , αn)for some αi, and we can reduce this one at a time until it is a simple extension,using the claim.

Consider k(α+ cβ) for c ∈ k. Clearly k ⊂ k(α+ cβ) ⊂ k(α, β). By assump-tion, i.e., there being only finitely many intermediate fields, there can only befinitely many k(α + cβ), so there must exist some distinct c1, c2 ∈ k such thatk(α+ c1β) = k(α+ c2β) = K ⊂ k(α, β).

Hence (c1 − c2)β ∈ K, and c1 − c2 6= 0, so β ∈ K, and then also α ∈ K, sok(α, β) ⊂ K ⊂ k(α, β), meaning that K = k(α, β).

SIMPLE EXTENSIONS, CONTINUED 30

Lecture 12 Simple Extensions, continued

12.1 Primitive element theorem, continued

Proof continued. For the converse direction of (i), assume E = k(α) is simple,and let p(x) = Irr(α; k, x). For any field F with E ⊃ F ⊃ k, let gF (x) =Irr(α;F, x). Then gF (x) | p(x), and there are only finitely many such choices ofdivisors, since p(x) has only finitely many factors in k[x]. Define ϕ : F 7→ gF (x)with

gF (x) ∈{g(x)

∣∣ g(x) | p(x) in k[x]}.

We claim that this is one-to-one. To see this, write gF (x) = xm + β1xm−1 +

. . .+βm with βi ∈ F , and let F0 = k(β1, β2, . . . , βm) ⊂ F . Now by constructiongF (x) = Irr(α;F, x) = Irr(α;F0, x), so the degrees [F (α) : F ] = [F (α) : F0],meaning F = F0. Now F0 is completely determined by GF (x), so ϕ is one-to-one.

Lecture 13 Normal and Separable Closures

First let us finish up the overdue proof.

Proof continued. It remains to show (ii), i.e., if E/k is separable, then E is asimple extension of k.

There is a trivially easy case: if |k| < ∞, then |E| < ∞ too since E/k is afinite extension, and hence E/k has only finitely many intermediate fields, soby (i), E/k is simple.

So let us assume |k| = ∞. Then E/k is still a finite extension, being sepa-rable, so E = k(α1, α2, . . . , αn) for some αi ∈ E. By the same argument as in(i), it suffices to consider E = k(α, β), and showing that this is simple.

Let [E : k] = n = [E : k]s, since E/k is separable by assumption. Thenthere exists n distinct k-embeddings of E into k, say

{σ1, σ2, . . . , σn

}. There

are determined by their values on α and β.Define

p(x) =∏

1≤i 6=j≤n

(σi(α)− σj(α) + (σi(β)− σj(β))x).

We claim that p 6= 0. Suppose not, i.e., suppose p(x) = 0 for all x ∈ k. Thenthere exists i 6= j such that

σ(α)− σj(α) + (σi(β)− σj(β))x = 0

for infinitely many x ∈ k. This implies σi(α)−σj(α) = 0 and σi(β)−σj(β) = 0,which means σi = σj , contradicting all the σi being distinct.

By this claim there must then exist some c ∈ k such that p(c) 6= 0, so that

σi(α)− σj(α) + (σi(β)− σj(β))c 6= 0

for all i 6= j. Since c ∈ k and all our σi are k embeddings, we can bring c insidegetting σi(α+ cβ) 6= σj(α+ cβ) for all i 6= j.

Date: September 26th, 2019.Date: October 1st, 2019.

NORMAL AND SEPARABLE CLOSURES 31

Now consider the fields E = k(α, β) ⊃ k(α+ cβ) ⊃ k. Then we have

[k(α+ cβ) : k] ≤ [k(α, β) : k] = n,

and by our calculation above there exist at least n distinct k-embeddings ofk(α+ cβ), so

n ≤ [k(α+ cβ) : k]s ≤ [k(α+ cβ) : k],

together giving usn = [k(α+ cβ) : k] = [k(α, β) : k].

Hence k(α+ cβ) = k(α, β) = E, so E is simple.

13.1 Normal closure

Definition 13.1.1. Let E/k be an algebraic extension (possibly infinite), andtake k ⊃ E ⊃ k. The normal closure of E/k in k is the intersection of allfields F such that k ⊃ F ⊃ k and F/k is normal.

In other words, the normal closure of E/k in k is the smallest normal exten-sion of k containing E.

Note that k/k is normal, so F exists.

Example 13.1.2. Consider Q ⊃ E = Q( 3√

2) ⊃ Q. The normal closure of E inQ is F = Q( 3

√2, ω). N

Example 13.1.3. Suppose k ⊃ E ⊃ k and let{σi∣∣ i ∈ I } be the set of all

distinct k-embeddings of E into k.If I =

{1, 2, . . . , n

}, i.e., [E : k]s = n <∞, then σi(E) ⊂ k for each i. Then

the normal closure of E/k is

F = σ1(E)σ2(E) · · ·σn(E).

Notice how if τ : F ↪→ k is a k-embeddings, then τ(F ) ⊂ F , so τ(F ) = F , beingalgebraic, whence F/k is normal, and it contains at least the things we need,and no more. N

Example 13.1.4. Playing the same game, if |I| = [E : k]s = ∞, then thenormal closure of E/k is, again, the smallest field that contains all σi(E) for alli ∈ I, but unlike the finite case this might not be the composition. N

13.2 Separable closure

Let E/k be an algebraic extension, possibly infinite. Then, as we have preciouslydiscussed, E/k is separable if and only if every finite subextension is separable,if and only if every element of E is separable over k. Hence:

Definition 13.2.1. Let k ⊃ E ⊃ k. The separable closure F of k in E is theset of all separable elements of E over k.

Note that this is a field. The easy way to see this is to note that considersee that k(α, β) is a separable extension of k, and so all the products, sums,differences, and quotients are also contained in the separable closure.

NORMAL AND SEPARABLE CLOSURES 32

13.3 Finite fields

We want to review some basic properties of finite fields. To this end, let F bea finite field.

First, char(F ) = p, a prime, and |F | = pn for some n, and we view F as anextension of degree n of Fp, the finite field of p elements. . Customarily we letq = pn and call F = Fq.

The group of units of F , i.e., (F×, ·), is a cyclic group. To see this, notefirst by the Fundamental theorem of finitely generated abelian groups,

F× ∼= Zm1 × Zm2 × . . .× Zmk

Let d = lcm(m1,m2, . . . ,mk). Then by definition d ≤ m1m2 · · ·mk = q − 1 =|F×|. Now note how for a ∈ F×, ad = 1, and so a is a root of xd−1. Now xd−1has at most d distinct roots, and every a ∈ F× is a root of it, so |F×| ≤ d ≤ |F×|.Hence d = m1m2 · · ·mk, implying that (mi,mj) = 1 for all i 6= j, and so finally

F× ∼= Zm1m2···mk

is cyclic.In other words, F× ∼= Zq−1. Moreover all elements in F× are roots of

xq−1 − 1, so all elements in F (including 0) are roots of xq − x. Hence F is thesplitting field of xq − x over Fp.

Theorem 13.3.1 (Existence). For every prime p and every n ∈ N, there existsa field F such that |F | = pn.

Proof. Let q = pn and let f(x) = xq−x ∈ Fp[x]. Notice how f ′(x) = qxq−1−1 =−1 6= 0 since charFp = p. Hence f has no repeated roots. Let F be the splittingfield of f(x) in Fp, and let F1 be the set of all roots of f(x) in Fp. Then |F1| = q.

We claim that F1 is a field. This is routine: for α, β ∈ F1, we have αq = α and

βq = β, and hence also (αβ)1 = αqβq = αβ, for β 6= 0 we have(αβ

)q= αq

βq = αβ ,

and since we are in characteristic p, (α ± β)q = αq ± βq = α ± β. Hence all ofthese are rots of f(x), so belong to F1, so F1 is a field, and indeed F1 = F .

Theorem 13.3.2 (Uniqueness). All fields of order pn are isomorphic. In par-ticular, if Fp ⊃ F1, F2 ⊃ Fp and |F1| = |F2|, then F1 = F2.

Proof. This follows directly from things we already know: if |F1| = |F2| = pn,both are splitting fields of xp

n − x, and splitting fields of the same polynomialare isomorphic. In particular if they are under the same algebraic closure, theyare equal.

We can classify the automorphisms of such fields:

Theorem 13.3.3. If |F | = pn = q, then AutFp(F ) ∼= Zn (hence cyclic), andin particular AutFp(F ) = 〈ϕ〉, where ϕ : F → F , ϕ(x) = xp is the Frobeniusautomorphism.

Proof. Note first how ϕn(x) = xpn

= xq = x, so ϕn = IdF , meaning that theorder of ϕ is at most n.

Next let us show that it is at most n, and hence equal to n. Take y ∈ F×such that F× = 〈y〉 (since we already showed this is cyclic). In other words, the

INSEPARABLE EXTENSIONS 33

order of y is q − 1 = pn − 1, so yq−1 = 1, and moreover yq = y. Importantlythis is the smallest positive power that works, i.e., yk 6= 1 for any 0 < k < q. Soϕn(y) = yp

n

= y is the smallest power of ϕ with this property, so ϕm 6= IdF forall 0 < m < n, and hence the order of ϕ is at least m.

Hence AutFp(F ) ⊃ 〈ϕ〉 ∼= Zn.Since [F : Fp] = n = [F : Fp]s (since finite extensions of finite fields are

separable), there exists at most n distinct Fp-embeddings of F into Fp, so|AutFp(F )| ≤ n. Thus AutFp(F ) = 〈ϕ〉 ∼= Zn, as desired.

Lecture 14 Inseparable Extensions

14.1 Number of irreducible polynomials over finite fields

Theorem 14.1.1. Let k = Fp, where p is a prime. Let Nn denote the numberof irreducible polynomials in k[x] of degree n. Then

Nn =1

n

∑d|n

µ(nd

)pd,

where µ : N →{

1,−1, 0}

is the Mobius function (µ(1) = 1, and µ(n) = 0 is nis not square-free, and (−1)k if n = p1p2 · · · pk, where pi are distinct primes).

Proving is straight-forward if we first have the following lemma:

Lemma 14.1.2. Let f(x) ∈ Fp[x] be an irreducible polynomial of degree d.Then d | n if and only if f(x) | xpn − x.

Proof. First, note how for `,m ∈ N, x` − 1 | xm − 1 if and only if ` | m.For the forward direction, xm − 1 = (x`)q − 1 can have a factor of x` − 1

taken out of it. For the reverse direction, write m = `q + r, with 0 ≤ r < `.Then

xm − 1

x` − 1=x`q+r − 1

x` − 1=xr(xq` − 1)

x` − 1+xr − 1

x` − 1,

where the left-hand side is a polynomial, and so is the first term in the right-hand side, so xr−1

x`−1must be a polynomial too. But r < `, so this cannot be a

polynomial unless r = 0. Hence ` | m.Similarly, let a, `,m ∈ N, a > 1. Then a`− | am − 1 if and only if ` | m.With this we can prove the lemma: For the forward direction, assume d | n.

Then pd−1 | pn−1 by the above (with a = p), and hence xpd−1−1 | xpn−1−1.

Multiplying by x, we get xpd − x | xpn − x.

Now f(x) ∈ Fp[x] is irreducible of degree d, so let α ∈ Fp be a root of f(x),whence [Fp(α) : Fp] = d, so |Fp(α)| = pd. Hence Fp(α) is the splitting field

of xpd − x over Fp, by the uniqueness of finite fields. Therefore α is a root of

xpd − x, implying

f(x) | xpd

− x | xpn

− x.For the converse direction, assume f(x) | xpn0x, with f(x) irreducible of

degree d. Let α ∈ Fp be a root of f(x). Hence f(x) | xpn−x implies Fp(α) ⊂ Fpn ,the splitting field of xp

n − x. Therefore

n = [Fpn : Fp] = [Fpn : Fp(α)] · [Fp(α) : Fp] = [Fpn : Fp(α)] · d,Date: October 3rd, 2019.


so d | n.

14.2 Inseparable extensions

First we need two basic facts, essentially from calculus. Let k be a field, andlet f(x) ∈ k[x]. Take a ∈ k. Then a is a multiple root of f(x) if and only iff(a) = 0 and f ′(a) = 0.

To see this, write f(x) = (x− a)mg(x), and then

f ′(x) = m(x− a)m−1g(x) + (x− a)mg′(x),

where m is the multiplicity of the root a. Then we can factor out x − a fromthe derivative, so it is a root of f ′(x), if and only if m > 1.

Related to this: f(x) has a multiple root in k if and only if gcd(f(x), f ′(x)) 6=1 in k[x].

Theorem 14.2.1. Let k ⊃ k and α ∈ k. Let f(x) = Irr(α; k, x).

(i) All roots of f(x) in k have the same multiplicity.

(ii) If char(k) = 0, then f(x) = 0 is multiplicity free (i.e., f(x) has no repeatedroots, so f(x) is separable).

(iii) If char(k) = p > 0, then the common multiplicity is of the form pµ forsome µ ∈ Z, µ ≥ 0.

Moreover, β = αpµ

is separable and

[k(α) : k] = pµ[k(α) : k]s = pµ[k(β) : k]s = pµ[k(β) : k].

Proof. (i) Let α1, α2, . . . , αr be the set of all distinct roots of f(x) in k. Foreach αi, there exists σi ∈ Autk(k) such that σi(αi) = α1. Note that since σifixes k and f(x) ∈ k[x], fσi = f . Hence, if

f(x) = (x− α1)m1(x− α2)m2 · · · (x− αr)mr ,

then

fσi(x) = (x− σi(α1))m1(x− σi(α2))m2 · · · (x− σi(αi))mi · · · (x− σi(αr))mr ,

and since σi(αi) = α1, we see that mi = m1, for any i.

(ii) Let f(x) be irreducible, say f(x) = anxn + an−1x

n−1 + . . .+ a0, ai ∈ k andan 6= 0. Then

f ′(x) = nanxn−1 + (n− 1)an−1x

n−2 + . . .+ a1,

where nan 6= 0. Hence gcd(f(x), f ′(x)) = 1 since f(x) is irreducible and deg f ′ <deg f and f ′ 6= 0. Thus f(x) is separable.

(iii) If char(k) = p > 0, let m be the common multiplicity of the roots of f(x).If m = 1 then µ = 0 and f(x) is separable, so we are done.

Assume m > 1. Then, as above, if gcd(f(x), f ′(x)) 6= 1, we must havef ′(x) = 0 since deg f ′ < deg f and f(x) is irreducible. Hence f(x) must be


a polynomial in pn, so that when taking the derivative we get multiples of p,which are zero in characteristic p. In other words, there exists some g ∈ k[x]such that f(x) = g(xp). Since f is irreducible over k, so if g, and we can repeatthis process for g(x) if g(x) = 0 is not multiplicity free.

Hence there exists some µ ≥ 1 and F (x) ∈ k[x] such that F (x) is separableand f(x) = F (xp

µ

).

Write F (x) = (x− β1)(x− β2) · · · (x− βr), with βi ∈ k distinct. Then

f(x) = F (xpµ

) = (xpµ

− β1)(xpµ

− β2) · · · (xpµ

− βr)= (x− δ1)p

µ

(x− δ2)pµ

· · · (x− δr)pµ

since char(k) = p, where βi = δpµ

i . Note how δi are distinct since βi are distinct,and they are all the distinct roots of f(x), with multiplicity pµ.

So α = δ1, say, and we take β = β1 = αpµ

. Then Irr(β; k, x) = F (x),separable over k, so

[k(α) : k] = pµ · r,since r is the number of distinct roots of f(x), so

[k(α) : k] = pµ[k(α) : k]s.

But r is also the degree of F , which is separable, so in turn

pµ · r = pµ[k(β) : k] = pµ[k(β) : k]s.

The situation, in a picture, is this:

k(α)

k(β)

k.

purely inseparablepµ

separabler

We have Irr(α; k(β), x) = xpµ − β, which in k[x] factors as (x − α)p

µ

, which iswhy we call the topmost extension purely inseparable ; it has only one root,with as large a multiplicity as possible.

In this setup we also call [k(α) : k]i = pµ the inseparable degree of k(α)/k.More generally:

Definition 14.2.2. Let k be a field with char(k) = p > 0. Let E/k be a finitealgebraic extension. Then we define the inseparable degree of E/k by

[E : k]i =[E : k]

[E : k]s.

Note that if p = 0, then all finite extensions are separable, so the inseparabledegree would be uninteresting. Note also how [E : k]i is always a power of p.

Proposition 14.2.3. Let E ⊃ F ⊃ k be a chain of field extensions, with E/ka finite extension. Then

[E : k]i = [E : F ]i[F : k]i.

PURELY INSEPARABLE EXTENSIONS 36

Proof. Both the degree and the separable degree have this multiplicative prop-erty, so the inseparable degree must too.

Remark 14.2.4. Let E/k be an infinite algebraic extension. Then we define[E : k] as the cardinality of the basis of E/k as a vector space, and we defined[E : k]s as the cardinality of the number of distinct embeddings of E into k.

We don’t define inseparable degrees for infinite extensions, and the reasonis that they don’t necessarily make a great deal of sense:

Example 14.2.5. Consider k = Q and k = Q. Then [Q : Q] = |N| is countable.However [Q : Q]s = |R| is uncountable, since, for instance, every

√p, p prime,

can be sent to two different images, so the cardinality is the power set of N. N

Lecture 15 Purely Inseparable Extensions

15.1 Inseparable closures and purely inseparable exten-sions

Definition 15.1.1. Let E/k be a field extension, and let char(k) = p > 0. Letα ∈ E. Then α is called purely inseparable over k if αp

n ∈ k for some n.In this case, f(x) = xp

n −αpn ∈ k[x], with f(α) = 0. But f(x) = (x−α)pn

,meaning that Irr(α; k, x) = (x− α)p

m

for some m, whence [k(α) : k]s = 1.

Definition 15.1.2. Let char(k) = p. An algebraic extension E/k is calledpurely inseparable if every element of E if purely inseparable over k.

In other words, Eo = k where Eo is the separable closure of k in E (becausem = 0, so α ∈ k, in the above setup).

Proposition 15.1.3. Let E/k be an algebraic extension with char(k) = p.Let Eo be the separable closure of k in E. Then E/Eo is a purely inseparableextension.

Proof. For α ∈ E, there exists some µ ≥ 0 such that αpµ

= β is separable overk, so β ∈ Eo. Hence α is purely inseparable over Eo.

Proposition 15.1.4. Let K/k be a normal extension (hence also algebraic).Let Ko be the separable closure of k in K. Then Ko/k is also normal.

Proof. Let σ be a k-embedding of Ko into k. This means that σ can be liftedto an embedding τ of K, since K/k is algebraic.

Because K/k is normal, τ(K) = K, meaning that σ(Ko) ⊂ K. But everyelement α in Ko is separable over k, so σ(α) is also separable over k, meaningthat σ(α) ∈ Ko. Thus σ(Ko) ⊂ Ko, which implies σ(Ko) = Ko since Ko/k is afinite algebraic extension. Therefore Ko/k is normal.

Proposition 15.1.5. Let E/k be an algebraic extension, char(k) = p. Thenthe following are equivalent:

(i) [E : k]s = 1;

(ii) every element of E is purely inseparable over k;

Date: October 8th, 2019.


(iii) let α ∈ E. Then Irr(α; k, x) = xpn − α for some a ∈ k, n ≥ 0; and

(iv) there exists{αi∣∣ i ∈ I } ⊂ E such that E = k(αi | i ∈ I), where αi are

purely inseparable over k.

Proof. For (i) implying (ii), take α ∈ E. The separable degree has a multiplica-tive property, so 1 = [E : k]s = [E : k(α)]s[k(α) : k]s. Hence [k(α) : k]s = 1 aswell, whence Irr(α; k, x) has only one distinct root α, so Irr(α; k, x) = (x−α)p

µ

=xp

µ − αpµ . Thus α is inseparable over k.For (ii) implying (iii), we are almost immediately done: α ∈ E being purely

inseparable over k means αpµ

= a ∈ k.For (iii) implying (iv), we have αp

n

= a ∈ k, so every α ∈ E works in (iv).Finally, let us attack (iv) implying (i). If σ is a k-embedding of E, then σ is

determined by σ(αi). Since αi is purely inseparable, σ(αi) = αi for every i ∈ I,.Hence σ = IdE , so there is only one distinct embedding of E into k, whence[E : k]s = 1.

Definition 15.1.6. Let E be a field of characteristic p. By Ep we mean theset of all p-powers of elements in E. That is, Ep =

{αp∣∣ α ∈ E }.

Note how Ep is a subfield of E. The only nontrivial part is to show thatsums are in there, but in characteristic p, αp ± βp = (α± β)p.

Proposition 15.1.7. Let char(k) = p, and let E/k be a finite extension. ThenEpk = E if and only if E/k is separable.

Proof. First note that Epk = E if and only if Epn

k = E for every n ≥ 1. Thereverse direction is trivial—take n = 1. For the forward direction, note howEp

2

k = (Ep)pk = (Epk)pk since kp ⊂ Ep. But Epk = E by assumption, so thisis (Epk)pk = Epk = E. Repeating this, we get the result.

For the forward direction of the proposition, let Eo be the separable closureof k in E. We want to show that Eo = E.

Being a finite extension of k, E = k(α1, α2, . . . , αn). For each i, there exists

some µi such that αpµi

i ∈ Eo. Let n = max{µ1, µ2, . . . , µn

}. Then αp

n

i ∈ Eofor all i. Hence αp

n ∈ Eo for all E, so Epn ⊂ Eo. By assumption, Ep

n

k = E,so E = Ep

n

k ⊂ Eok = Eo, so E = Eo since by construction Eo ⊂ E.For the converse direction, assume E/k is separable. For α ∈ E, αp ∈

Ep ⊂ Epk. This implies E/Epk is purely inseparable. But E/k being separableimplies E/Epk is separable, so E/Epk is both separable and purely inseparable,meaning that E = Epk.

Proposition 15.1.8. Let K/k be a normal extension, and let char(k) = p. LetG = Autk(K). Define KG :=

{a ∈ K

∣∣ ag = g for all g ∈ G}

be the set ofall elements of K fixed by G (by ag we mean g(a)). This is a subfield (sinceautomorphism,s being homomorphisms, preserve the field operations). Let Ko

be the separable closure of k in K.Then

(i) KG is the set of all purely inseparable elements of K over k,

(ii) KG ∩Ko = k,

(iii) K/KG is separable,


(iv) K = KoKG.

Proof. (i) Let α ∈ K be purely inseparable over k. Then for σ ∈ G, σ(α) = α,meaning that α ∈ KG. Conversely, for β ∈ KG, there exists some µ ≥ 0 suchthat βp

µ

= γ is separable over k. If we can show that γ ∈ k we are done, sincethem β is both separable and purely inseparable over k.

To see this, suppose γ 6∈ k. Then there exists a k-embedding σ such thatσ(γ) = γ′ 6= γ. But σ can be lifted to σ : K → K, where σ(β) = β, soσ(βp

µ

) = βpµ

, meaning that σ(γ) = γ, a contradiction.

(ii) This part follows from (i). An element α ∈ Ko ∩KG is both separable andpurely inseparable over k, so α ∈ k.

(iii) This proof is due to E. Artin. Let α ∈ K, [k(α) : k] <∞. Let{σ1, σ2, . . . , σn

}be a maximal set of embeddings of K/k such that

{σ1(α), σ2(α), . . . , σn(α)

}are all distinct. In other words, this is the set of all distinct roots of Irr(α; k, x).

Let f(x) =∏

(x− σi(α)), and take σ ∈ Autk(K). Then

fσ(x) =∏

(x− σσi(α)) = f(x),

so f(x) ∈ KG[x] since all σ ∈ G fix the coefficients. Hence f(x) is separable,and α is a root of f(x), so α is separable over KG.

(iv) Consider the diagram

K

KoKG

Ko KG

k

purely inseparable separable

Being purely inseparable over a smaller field means purely inseparable over abigger field, so K/KoK

G is purely inseparable. Similarly, being separable overa smaller field means separable over a bigger one, so K/KoK

G is separable.Hence K = KoK

G.

Definition 15.1.9. A field k is perfect if every algebraic extension is separable.

Example 15.1.10. The rationals Q is a perfect field because char(Q) = 0. Forthe same reason, any field with characteristic 0 is perfect.

All finite fields are perfect because the elements are roots of xpn − x, which

has no repeated roots. N

GALOIS THEORY 39

Lecture 16 Galois Theory

16.1 Galois extensions

Definition 16.1.1 (Galois extension). A field extension K/k is Galois if it isnormal (hence algebraic) and separable.

Let G = Autk(K) and let KG ={a ∈ K

∣∣ σ(a) = a for all σ ∈ G}

be thesubfield of K consisting of all elements fixed by G. Trivially, k ⊂ KG.

Let H < G be a subgroup. Then similarly define KH ={a ∈ K

∣∣ σ(a) =

a for all σ ∈ H}

.

Theorem 16.1.2 (Fundamental theorem of Galois theory). Let K/k be a finiteGalois extension and let G = Autk(K). Let F be the set of all subfields F suchthat K ⊃ F ⊃ k and let H be the set of all subgroups of G.

Then there is a one-to-one correspondence between F and H.

Remark 16.1.3. Note that |F| < ∞ by the Primitive element theorem, sinceK/k is both finite and separable.

Moreover, |G| < ∞ since there are only finitely many k-embeddings of K,and hence |H| <∞.

To demonstrate that the conditions are necessary, consider the followingcounterexamples.

Counterexample 16.1.4. Let k be a field of characteristic p, and let t and ube variables. Then k(u, t) ⊃ k(up, tp) is a normal but not separable extension.This is a finite extension—[k(u, t) : k(up, tp)] = p2, but there are infinitely manyintermediate subfields. N

Counterexample 16.1.5. The extension Q/Q is normal and separable, andhence Galois. On the one hand |Q| is countable, but |G| is uncountable, whereG = AutQ(Q).

But |F| and |H| are both uncountable, however there isn’t a one-to-onecorrespondence.

As it happens, essentially the same theorem is still true, only for it to workin the infinite case we need a topology (namely the Krull topology) and thecorrespondence is with closed subgroups (and indeed all subgroups are closedin the finite case). N

We will spend a fair amount of time working our way up to this fundamentaltheorem, starting with

Proposition 16.1.6. Let K/k be a Galois extension (possibly infinite), and letG = Autk(K). Then KG = k.

Proof. First, k ⊂ KG by definition.Suppose there exists some α ∈ KG \ k. Then k(α)/k is a finite separable

extension, so [k(α) : k]s = [k(α) : k] > 1 since α 6∈ k. This means thereexists some σ 6= Id such that σ : k(α) ↪→ k is a k-embedding (so in particularσ(α) 6= α).


GALOIS THEORY 40

Since K/k(α) is algebraic, we can lift σ to τ : K ↪→ k, also a k-embedding,and since K/k is normal, τ(K) = k, so τ ∈ G.

But τ(α) 6= α since it restricts to σ, and this contradicts α ∈ KG. HenceKG = k.

Definition 16.1.7 (Galois group). Let K/k be Galois. The Galois group ofK/k is defined as Gal(K/k) := Autk(K).

Theorem 16.1.8. Let K/k be a Galois extension. Let F , k ⊂ F ⊂ K bean intermediate subfield. Then K/F is also Galois. Moreover, letting H =Gal(K/F ), KH = F .

Proof. If σ : K ↪→ k is an F -embedding, then it is also a k-embedding sincek ⊂ F . Since K/k is normal, this means σ(K) = K, and so K/F is also normal.

Since the class of separable extensions is distinguished, K/k being separableimplies K/F is separable. Hence K/F is Galois.

For the second part, simply use the previous proposition.

The situation, so far, is this: if K/k is Galois and G = Gal(K/k), thenstarting with an intermediate subfield F , k ⊂ F ⊂ K, we have a subgroupH = GF = Gal(K/F ) of G, and by the above theorem we can pull this back toKH = F .

Hence, in the Galois correspondence, we know that F → H is one-to-one,and the opposite direction H → F is onto.

The big question, then, is if we can go the other way: if we start with asubgroup H < G, move to KH , and then pull back to H ′ = Gal(K/KH) on thegroup side, we have H ′ ⊃ H, but are they equal?

In the finite case, the answer turns out to be yes. In the infinite case theyneed not be, but H ′ is the closure of H in the Krull topology.

There are some simple results we can establish about these objects that wewill make use of shortly. Let K/k be Galois, and let k ⊂ Ki ⊂ K, i = 1, 2, andlet Hi = AutFi(K). Then

(i) AutF1F2(K) = H1 ∩H2;

(ii) letting H = 〈H1, H2〉, then KH = F1 ∩ F2; and

(iii) F1 ⊃ F2 if and only if H1 ⊂ H2.

Theorem 16.1.9 (Artin). Let K be a field. Let G be a finite subgroup ofAut(K). Then K/KG is Galois, and G = Gal(K/KG).

To prove the second part of this theorem we will make use of the followinglemma:

Lemma 16.1.10. Let E/k be a separable extension. Suppose [k(α) : k] ≤ n forall α ∈ E. Then [E : k] ≤ n.

Proof. Take α ∈ E such that [k(α) : k] = m = maxβ∈E

[k(β) : k].

We claim that E = k(α). To see this, suppose E 6= k(α), i.e., there existssome β ∈ E \ k(α). Then

[k(α, β) : k] = [k(α, β) : k(α)][k(α) : k] > m

ARTIN’S THEOREM 41

since β 6∈ k(α) means [k(α, β) : k(α)] > 1 and [k(α) : k] = m. But k(α, β)/kis a finite and separable extension, so by the Primitive element theorem it issimple, meaning that k(α, β) = k(γ) for some γ ∈ E. But this contradicts themaximality of [k(α) : k].

Proof of Theorem 16.1.9. By definition KG ={a ∈ K

∣∣g(a) = a for all g ∈ G}

.

For any α ∈ K, let{σ1(α), σ2(α), . . . , σr(α)

}be a maximal set of distinct

elements obtained by applying σ ∈ G.Note that since |G| = n is finite by assumption, r ≤ n.Take σ1(α) = α, and let

fα(x) =

r∏i=1

(x− σi(α)).

Note that for σ ∈ G,{σσ1(α), . . . , σσr(α)

}={σ1(α), . . . , σr(α)

}, so σ just

permutes the roots. Hence fσα = fα, meaning that fα ∈ KG[x]¿ Moreover fα isseparable by construction, so α is a root of a separable polynomial in KG[x], soα is separable over KG. Hence K/KG is separable.

Since K is the splitting field of a family{fα(x) ∈ KG[x]

∣∣ α ∈ K }, K/KG

is normal.Hence in all, K/KG is Galois.Secondly we wish to show that G = Gal(K/KG). For α ∈ K, Irr(α;KG, x) |

fα(x). Now deg fα = r ≤ n = |G|, so deg Irr(α;KG, x) ≤ n, meaning that[KG(α) : KG] ≤ n for all α ∈ K.

By the lemma, this means [K : KG] ≤ n = |G|, so

G ⊂ Gal(K/KG) = AutKG(G),

but|Gal(K/KG)| = [K : KG]s = [K : KG] = n,

since K/KG is separable, and n ≤ |G|, so G and Gal(K/KG) have the sameorder, so they are equal.

Lecture 17 Artin’s Theorem

17.1 Examples

Definition 17.1.1. Let K be any field, and let ko be the smallest subfieldcontaining 1. Then ko is called the prime field of K.

In particular, if char(K) = 0, then ko ∼= Q, and if char(K) = p, then ko ∼= Fp.

In this setting, Aut(K) = Autko(K) by definition. In Artin’s theorem, K/komay be transcendental.

Example 17.1.2. Let k = C and let K = C(x), where x is a variable. ThenK/k is transcendental, and

Autk(K) =

{σ : x 7→ ax+ b

cx+ d

∣∣∣∣ a, b, c, d ∈ C and ad− bc 6= 0

}∼= PGL2(C).


INFINITE VERSION OF ARTIN’S THEOREM 42

If G = Autk(K), then KG = k, and K/KG is not Galois. This does notcontradict Artin’s theorem, because G is not finite.

If G = 〈σ〉, σ : x 7→√−1x, then σ4 = Id, so |G| = 4, and indeed G ∼= Z4.

Then KG = C(x4), and K/KG = C(x)/C(x4) is Galois. To see this, noticehow

f(t) = Irr(x;KG, t) = t4 − x4 ∈ KG[x],

and in C(x) this factors at (t− x)(t+ x)(t−√−1x)(t+

√−1x). This splits in

C(x) and is separable over C(x4), so the extension is normal and separable. N

Lecture 18 Infinite Version of Artin’s Theorem

18.1 Generalising Artin’s theorem

Notice how in our proof of Artin’s theorem, the only time we used the finitenessof |G| was to establish that the maximal set

{σ1(α), . . . , σr(α)

}is finite.

This immediately tells us we can generalise Artin’s theorem to infinite groupsG, provided we know this set is again finite. An easy way to do so is to justrequire K/k be algebraic, since then σ(α), σ ∈ G are all roots of a certainirreducible polynomial, and that polynomial has only finitely many roots.

Theorem 18.1.1. Assume K/k is algebraic. Let G ≤ Autk(K) (possibly infi-nite). Then K/KG is Galois.

Proof. For any α ∈ K, the set{σ(α)

∣∣ σ ∈ G} is finite, since all σ(α) are rootsof Irr(α; k, x) because K/k is algebraic.

Take{σ1(α), σ2(α), . . . , σr(α)

}to be a maximal set of distinct elements and

consider fα(x) = (x− σ1(α)) · · · (x− σr(α)). Notice how fσα (x) = fα(x) for allσ ∈ G, meaning that fα(x) ∈ KG[x]. Now as in the proof of the finite versionof Artin’s theorem, this means K/KG is Galois.

Remark 18.1.2. Notice how supα∈K

deg fα may be infinite, soK/KG may be infinite

Galois.Similarly, we do get G ≤ Gal(K/KG). In the finite case, we compared the

order of these groups to conclude that they are equal. In the infinite case wecannot do that, and indeed the groups need not be equal.

The situation, as we have discussed at some point in the past, is this: LetK/k be Galois and let F be the set of all intermediate subfields and let H bethe set of all subgroups of G = Gal(K/k).

Define Γ: F → H by F 7→ Gal(K/F ) and define Φ: H → F by H 7→ KH .Then we have already showed Φ ◦ Γ = IdF .For the reverse direction, we have to consider two cases.First: K/k is finite Galois. Then H ≤ G is a finite subgroup. Standard Artin

then tells us K/KH is Galois and H = Gal(K/KH), so in this case Γ◦Φ = IdH.In other words, in this case Γ and Φ are one-to-one and onto correspondences.Second: K/k is infinite Galois. In this case, by infinite Artin, we know

K/KH is Galois and H ≤ Gal(K/KH) = H ′. Since H and H ′ are not neces-sarily the same, Γ ◦ Φ 6= IdH.

Date: October 22nd, 2019.

INFINITE VERSION OF ARTIN’S THEOREM 43

That said, if H ≤ G is a finite subgroup, then we do have Γ ◦ Φ(H) = H,and K/KH is a finite Galois extension (with [K : KH ] = |H|).

This gives us a modified version of Galois’ theorem:

Theorem 18.1.3. Let Fo be the set of all intermediate subfields F such that[K : F ] < ∞. Let Ho be the set of all finite subgroups of G. Then Fo ↔ Ho isa one-to-one and onto correspondence.

18.2 Conjugation

Definition 18.2.1 (Conjugatation). Let ‖ ⊃ E ⊃ k and let σ ∈ Autk(k).The field σ(E) is called a conjugation of E and for α ∈ E, σ(α) is called aconjugate of α.

Proposition 18.2.2. Let K/k be Galois and let G = Gal(K/k). Let σ ∈ G.Let k ⊂ F ⊂ K, and take H = Gal(K/F ). Then Gal(K/σ(F )) = σHσ−1.

In other words, conjugate fields have conjugate Galois groups.

Proof. Let τ ∈ Gal(K/σ(F )). For all a ∈ F , τ(σ(a)) = σ(a), so σ−1 ◦ τ ◦σ(a) =a, meaning that σ−1 ◦ τ ◦ σ ∈ H, so τ ∈ σHσ−1.

On the other hand, for h ∈ H, a ∈ F , we have σhσ−1(σ(a)) = σh(a) = σ(a),so σhσ−1 ∈ Gal(K/σ(F )), so σHσ−1 ⊂ Gal(K/σ(F )).

Proposition 18.2.3. Let K/k be Galois. Let G = Gal(K/k) and let K ⊃ F ⊃k. Then F/k is Galois if and only if H = Gal(K/F ) is a normal subgroup ofG.

Moreover, Gal(F/k) ∼= G/H.

Proof. The extension F/k is Galois if and only if F/k is normal (separability isalways true since K/k is separable) if and only if σ(F ) = F for every σ ∈ G, ifand only if σHσ−1 = H for every σ ∈ G, if and only if H is normal in G.

For the isomorphism, consider ϕ : G→ Gal(F/k) defined by ρ 7→ ρ∣∣F

. Thisis well-defined since ρ(F ) = F because F/k is normal. It is clear that ϕ is agroup homomorphism.

Suppose ρ ∈ kerϕ. This means ρ∣∣F

= IdF , i.e., ρ fixes F , meaning thatρ ∈ Gal(K/F ) = H. So kerϕ = H.

Since K/F is Galois, every σ ∈ Gal(F/k) can be lifted to Gal(K/k), so ϕ isalso onto.

Hence by the isomorphism theorem, G/ kerϕ ∼= Imϕ, or in other wordsG/H ∼= Gal(F/k).

18.3 Lifts and Galois extensions

Proposition 18.3.1. Let K/k be Galois. Let F/k be any extension. Then

(i) K/(K ∩ F ) is Galois,

(ii) KF/F is Galois, and

(iii) Gal(KF/F ) ∼= Gal(K/(K ∩ F )).

Proof. (i) This is trivial: being Galois over a small field implies Galois over abigger field.

SPECIAL KINDS OF GALOIS EXTENSIONS 44

(ii) Both normality and separability are preserved by lifting, so this, too, isclear.

(iii) Let G = Gal(KF/F ) and H = Gal(K/(K ∩ F )). Define ϕ : G → Hby ρ 7→ ρ

∣∣K

. This fixes K ∩ F since it fixes F , which is bigger. This is a

homomorphism, and if ρ ∈ kerϕ, then ρ∣∣K

= IdK , but ρ∣∣F

= IdF . Hence

ρ = IdKF , so kerϕ ={

Id}

, so ϕ is one-to-one.

We claim that ϕ is onto.

First, consider the case where K/k is a finite Galois extension. Then Imϕ ={σ∣∣K

∣∣σ ∈ G} = H ′ ≤ H. We want to show that H ′ = H. If we can show that

KH′ , then H = Gal(K/(K ∩ F )) = H ′, so the claim follows.

Note that clearlyK∩H ⊂ KH′ . Suppose there exists some α ∈ KH′\(K∩F ).Then F (α) ) F , but σ(α) = α for every σ ∈ H ′, so σ(α) = α for every σ ∈ G.Hence α ∈ (KF )G = F , which is a contradiction.

Second, consider the case where K/k is an infinite Galois extension. Let{Kλ

∣∣ λ ∈ Λ}

be the set of all intermediate fields Kλ such that K ⊃ Kλ ⊃ kand [Kλ : k] <∞ and Kλ/k is Galois.

Note thatK =⋃λ∈Λ

Kλ, and by the first case Gal(Kλ/(Kλ∩F )) ∼= Gal(KλF/F ).

We want to show that any ρ ∈ Gal(K/(K∩F )) can be lifted to Gal(KF/F ).Given ρ ∈ Gal(K/(K ∩ F )), define ρλ = ρ

∣∣Kλ∈ Gal(Kλ/(Kλ ∩ F )) for λ ∈ Λ.

By the finite case, we can lift ρλ to σλ ∈ Gal(KλF/F ). For α ∈ KF , wemust have α ∈ FKλ for some λ, and so define σ(α) = σλ(α).

We claim that σ ∈ Gal(KF/F ). First of all, it is well-defined—if we haveanother Kλ′ , then Kλ and Kλ′ , with their corresponding ρλ and ρλ′ , may havean intersection. So do the lifts σλ and σλ′ have the same image on α? Theanswer is yes—look at KλKλ′ = Kβ , and consider the lift σβ of λβ . Thenrestricted to Kλ this becomes σλ, and restricted to Kλ′ it becomes σλ′ .

Lecture 19 Special Kinds of Galois Extensions

19.1 Cyclic, abelian, nilpotent, and solvable extensions

Definition 19.1.1. Let K/k be Galois and let G = Gal(K/k). If G is cyclic(abelian, nilpotent, solvable) then we say that the extension K/k is cyclic(abelian , nilpotent , solvable).

Let us recall what the latter of these two mean:

Definition 19.1.2. Let G be a group and let

Z(G) ={z ∈ G

∣∣ gz = zg for all g ∈ G}

be the centre of G. This is a normal subgroup. Consider the map ϕ1 : G →G/Z(G), and consider the pullback Z2(G) = ϕ−1

1 (Z(G/Z(G))), which, beingthe pullback of a normal subgroup, is normal. Repeat, i.e., consider ϕ2 : G →G/Z2(G), define Z3(G) = ϕ−1

2 (Z(G/Z2(G))), and so on.



Then we have a sequence of normal subgroups{1}< Z(G) < Z2(G) < . . . < Zn(G) < . . . < G.

We say that G is nilpotent if G = Zn(G) for some n.

Definition 19.1.3. A group G is solvable if there exists a sequence

G = G0 > G1 > G2 > . . . > Gs ={

1}

of subgroups such that Gi is normal in Gi−1 and Gi−1/Gi is abelian.

Exercise 19.1.4. If G is nilpotent (solvable), then every subgroup of G isnilpotent (solvable).

(The same is true, but very trivial, for cyclic and abelian, of course.)

Proposition 19.1.5. If K/k is cyclic (abelian, nilpotent, solvable) and k ⊂ F ⊂K is an intermediate subfield, then K/F is cyclic (abelian, nilpotent, solvable).

Proof. Because Gal(K/F ) < Gal(K/k), all of this is immediate.

Proposition 19.1.6. If K/k is cyclic (abelian) and F is an intermediate sub-field. Then F/k is cyclic (abelian).

Proof. This follows from noting that Gal(K/F ) is normal in Gal(K/k) is K/kis cyclic (abelian), and then Gal(F/k) ∼= Gal(K/k)/Gal(K/F ).

Theorem 19.1.7. Let K1/k and K2/k be Galois extensions and let Gi =Gal(Ki/k), i = 1, 2. Then K1K2/k is Galois and Gal(K1K2/k) is isomorphicto a subgroup of G1 ×G2.

Moreover if K1 ∩K2 = k, then Gal(K1K2/k) ∼= G1 ×G2.

Proof. Normality and separability are preserved by composition, so K1K2/kbeing Galois is trivial.

Define ϕ : Gal(K1K2/k)→ G1 ×G2 by ρ 7→ (ρ∣∣K1, ρ∣∣K2

).

Suppose ρ ∈ kerϕ. Then ρ∣∣K1

= IdK1 and ρ∣∣K2

= IdK2 , meaning that

ρ∣∣K1K2

. Hence ϕ is one-to-one. This means Gal(K1K2/k) is a subgroup ofG1 ×G2.

Now assume K1 ∩K2 = k. By the lifting property,

Gal(K1/k) ∼= Gal(K1K2/K2) ⊂ Gal(K1K2/k).

Since Gal(K1/k) embeds into G1 × G2 by g 7→ (g, IdK2), meaning that G1 ×{

IdK2

}⊂ Im(ϕ), and similarly

{IdK1

}×G2 ⊂ Im(ϕ). Hence G1×G2 ⊂ Im(ϕ),

so ϕ is onto in this case, meaning that Gal(K1K2/k) ∼= G1 ×G2.

Theorem 19.1.8. Let Ki/k be Galois and Gi = Gal(Ki/k), i = 1, 2, . . . , n.Assume Ki ∩ (K1K2 · · ·Ki−1) = k, i = 1, 2, . . . , n. Then

Gal(K1K2 · · ·Kn/k) ∼= G1 ×G2 × . . .×Gn.

Proof. Let E = K1K2 · · ·Kn−1. Then E ∩Kn = k, so

Gal(EKn/k) ∼= Gal(E/k)×Gn

by the previous theorem. Now repeat on E.


Theorem 19.1.9. Let K/k and L/k be abelian. Then KL/k is abelian.

Proof. Since Gal(KL/k) ↪→ Gal(K/k) × Gal(L/k), and the two factors on theleft are abelian, the product is abelian, and so is any subgroup of it.

Definition 19.1.10. Let k ⊃ k. We define the abelian closure kab as thelargest abelian extension of k in k. (This exists because, by the theorem, anycomposition of abelian groups is abelian, and there is at least one abelian ex-tension (namely k itself), so there is a maximal one by Zorn.) In fact, kab is thecomposition of all abelian extensions E/k.

That is to say, let{Eλ∣∣ k ⊃ Eλ ⊃ k, Eλ/k abelian

}. Then

kab =∏λ

Eλ

andGal(kab/k) ↪→

∏λ

Gal(Eλ/k),

where the right-hand side is abelian.

Example 19.1.11. Consider Q ⊃ Q. There is almost nothing known aboutGal(Q/Q), but we do know something about Qab/Q. In particular,

Gal(Qab/Q) ∼=∏p

Z×p ,

where Zp is the p-adic integers. N

19.2 Examples and applications

In the proceeding discussion, we will always let k denote a base field and k analgebraic closure of k.

Let f(x) ∈ k[x] be monic and separable, deg f = n. Then

f(x) = (x− α1)(x− α2) · · · (x− αn)

for αi ∈ k.Let K = k(α1, α2, . . . , αn). Then K/k is Galois, and let G = Gal(K/k).The group G permutes

{α1, α2, . . . , αn

}, and any σ ∈ G is determined by its

image on this set. So G is a permutation of n elements, meaning that G ↪→ Sn,the symmetric group of n elements. In particular this means |G| ≤ n!.

Example 19.2.1. Let deg f = 2, the quadratic case. Assume that char(k) 6= 2,so that we can complete the square,

f(x) = x2 + αx+ β =(x− α

2

)2

+ β − α2

4.

Then we may assume f(x) = x2 + a ∈ k[x] by a change of variables.Note how f(x) is irreducible if and only if

√a 6∈ k, i.e., a isn’t the square of

any element in k, so f has no root in k. Then f(x) = (x−√a)(x+

√a) in k[x],

and G ↪→ S2∼= Z2. So either |G| = 1 or |G| = 2. If |G| = 1, then both roots of

f in k are equal, so f is not separable, which is a contradiction. Hence |G| = 2,so G ∼= S2

∼= Z2.In the char(k) = 2 case, note that in F2, the only irreducible quadratic

polynomial if x2 + x+ 1. N

GALOIS GROUP OF A POLYNOMIAL 47

Example 19.2.2. Now consider the cubic case, deg f = 3, and for the samereason assume char(k) 6= 2, 3. We may assume f(x) = x3 + ax + b ∈ k[x] by achange of variables, and f(x) is irreducible. Then

f(x) = (x− α1)(x− α2)(x− α3)

for αi ∈ k, and G ↪→ S3. Now |S3| = 3! = 6, and [K : k] = |G|, but [k(α1) : k] =3, so 3 | |G|, implying that G ∼= A3 or G ∼= S3, since either |G| = 3 or |G| = 6.

To determine which of the two is the case, let δ = (α1−α2)(α1−α3)(α2−α3),and set

∆ = δ2 =∏

1≤i<j≤3

(αi − αj)2.

Recall how G permutes{α1, α2, α3

}. Suppose σ ∈ G is a transposition, i.e., it

switches two of them and keeps the third fixed. Then σ(δ) = −δ.Hence in general, if σ is an odd permutation, σ(δ) = −δ, and if σ is an even

permutation, then σ(δ) = δ.In either case, this means σ(∆) = ∆ for all σ ∈ G. But that means G fixes

∆, so ∆ ∈ k.There are two cases to consider. If ∆ is a square in k, i.e., ∆ = ε2 for some

ε ∈ k, then ∆ = ε2 = δ2, so δ = ±ε ∈ k. Hence for σ ∈ G, σ(δ) = δ, so G doesnot contain an odd permutation, meaning that G ∼= A3.

On the other hand, if ∆ is not a square in k, then δ 6∈ k, so there existsa σ ∈ G with σ(δ) 6= δ, whence σ(δ) = −δ. Therefore G contains an oddpermutation, so G ∼= S3.

In conclusion, for f(x) = x3 + ax+ b ∈ k[x] with char(k) 6= 2, 3, let

∆ =∏

1≤i<j≤3

(αi − αj)2,

the discriminant of f .If ∆ is a square in k, then G ∼= A3, and if ∆ is not a square in k, then

G ∼= S3.This is easy to check, because as it happens ∆ = −4a3 − 27b2. This comes

from comparing x3 + ax+ b = (x− α1)(x− α2)(x− α3).Hence for example, consider f(x) = x3−x+1 ∈ Q[x]. Then ∆ = −4(−1)3−

27(1)2 = −23, which is not a square in Q, so G ∼= S3.On the other hand, if f(x) = x3 − 3x+ 1 ∈ Q[x], ∆ = −4(−3)3 − 27(1)2 =

81 = 92, so G ∼= A3. N

Lecture 20 Galois Group of a Polynomial

20.1 Galois group of polynomials

Let f(x) ∈ k[x] be separable, and let K be the split field of f . Then K/k isGalois, and we write Gal(f) = Gal(K/k).

Example 20.1.1. Consider f(x) = x4 − 2 inQ[x]. By Eisenstein’s criterionthis is irreducible. We want to figure out what Gal(f) is.


EXAMPLES OF GALOIS GROUPS 48

In the splitting field, f(x) = (x − 4√

2)(x − 4√

2i)(x + 4√

2)(x + 4√

2i), so thesituation looks like

K = Q( 4√

2, i)

Q( 4√

2)

Q(i)

Q

Galois

8

4

2

where K/Q( 4√

2) is Galois because Q(i)/Q is Galois, and it lifts. We also haveQ = Q( 4

√2) ∩Q(i), since the former is real and the latter is complex.

Hence |Gal(f)| = 8. Thus to identify what group Gal(f) is, note how thegroups of order 8 are Z8, Z4×Z2, Z2×Z2×Z2 (all abelian) and D4 (the dihedralgroup) and Q8 (the quaternion group), the latter two being nonabelian.

Now G contains an automorphism τ sending i to −i and 4√

2 to itself, whenceτ2 = Id is of order two, and an automorphism σ sending i to i and 4

√2 to i 4

√2,

so σ4 = Id is of order 4.Hence G ⊃ 〈τ, σ〉 ∼= D4, meaning that G = D4. N

Lecture 21 Examples of Galois Groups

21.1 More examples

Example 21.1.1. Let k be a field and let t1, t2, . . . , tn be variables. Let K =k(t1, t2, . . . , tn).

The symmetric group Sn acts on K/k by permuting{t1, t2, . . . , tn

}, so

G = Sn ⊂ Autk(K), and |Sn| = n!, so G = Sn is finite. Let F = KG. ByArtin’s theorem, K/F is Galois, and Gal(K/F ) = G.

So, aside from F = KG, how might we describe F?Elements in KG ⊂ K are rational functions in t1, t2, . . . , tn, and they have

to be fixed by all σ ∈ G, so they must be symmetric polynomials.The simplest symmetric polynomials are the elementary symmetric poly-

nomials of t1, t2, . . . , tn,

s1 = t1 + t2 + . . .+ tn

s2 = t1t2 + t1t3 + . . .+ t1tn + t2t3 + . . .+ t2tn + . . .+ tn−1tn,

...

sn = t1t2 · · · tn.

All of these are fixed by G. Hence, letting E := k(s1, s2, . . . , sn), since sσi = sifor every σ ∈ G, E ⊂ F .

Date: October 31st, 2019.


We claim E = F . One way to see this is to note that [K : F ] = n!, so if wecan show that [K : E] ≤ n!, then E = F . Consider

f(x) = (x−t1)(x−t2) · · · (x−tn) = xn−s1xn−1+s2x

n−2−s3xn−3+. . .+(−1)nsn.

Therefore f(x) ∈ E[x], so K = E(t1, t2, . . . , tn) is the splitting field of f(x) overE, and deg f(x) = n, so [K : E] ≤ n! (and in fact it divides n!). N

Remark 21.1.2. Every symmetric polynomial can be expressed as a polynomialof elementary symmetric polynomials. Hence if f1, f2, . . . , fm are symmetricpolynomials.

Remark 21.1.3. A large reason we care about this is this: Every finite group His a subgroup of Sn for some n. Hence there exists a Galois extension K/F suchthat Gal(K/F ) = H.

Let f(x) ∈ k[x] be separable, and write f(x) = (x−α1) · · · (x−αn) in somealgebraic closure. Define

δ = δ(f) =∏

1≤i<j≤n

(αi − αj)

and∆ = ∆(f) = δ2 =

∏1≤i<j≤n

(αi − αj)2,

the discriminant if f .The degree three discussion we had previously generalises in one case, and

not quite in the other:

Theorem 21.1.4. Let f(x) ∈ k[x] be separable. Let K be the splitting field off over k, and G = Gal(K/k). Let n = deg f .

Then G ↪→ An if and only of ∆ is a square in k.

Proof. Note that G ↪→ Sn, and ∆ = δ2 ∈ k since ∆ ∈ KG.For the forward direction, suppose G ↪→ An. This means σ(δ) = δ for all

σ ∈ G, whence δ ∈ k, so ∆ = δ2 is a square in k.For the converse direction, suppose ∆ = ε2 for ε ∈ k. But ∆ = δ2, whence

δ = ±ε ∈ k, so σ(δ) = δ for all σ ∈ G. Thus σ is even for all σ ∈ G, else itswitches sign.

Theorem 21.1.5. Let p be a prime and f(x) ∈ Q[x] be irreducible of degree p.Suppose f(x) has exactly 2 nonreal roots. Then Gal(f) ∼= Sp.

Proof. Recall how Sp is generated by one transposition and one p-cycle, andalso note how Gal(f) ↪→ Sp.

Let{α1, α2, α3, . . . , αp

}be the roots of f(x), and suppose α1, α2 are the

nonreal roots.Let τ be the complex conjugate. Then τ(α1) = α2, τ(α2) = α1, since

complex roots of real polynomials come in complex conjugate pairs, and τ(αi) =αi for all i ≥ 3. Hence τ = ( 1 2 ) in Sp.

Second, Gal(f) acts transitively on the roots{α1, α2, . . . , αp

}, so the per-

mutationσ : α1 7→ α2 7→ α3 7→ . . . 7→ αp 7→ α1,

i.e., σ = ( 1 2 ··· p ) ∈ Gal(f). Hence Sp = 〈τ, σ〉 ↪→ Gal(f), and Gal(f) ↪→ Sp,so Gal(f) ∼= Sp.


Example 21.1.6. Let f(x) = x3 − 2 ∈ Q[x]. This has two nonreal roots, soGal(f) ∼= S3. N

Example 21.1.7. Consider f(x) = x5 − 4x + 2 ∈ Q[x]. This is irreducibleby Eisenstein’s criterion, and has two nonreal roots. Its degree is prime, soGal(f) =∼= S5. N

21.2 Roots of unity

Let k be a field. The roots of xn − 1 are called the nth roots of unity .Suppose char(k) = 0. Then f(x) = xn − 1 is trivially separable over k.Suppose char(k) = p. If p | n, then n = pm and xn − 1 = (xm)p − 1 =

(xm − 1)p. So xn − 1 = 0 if and only if xm − 1 = 0, hence we should assumep - n.

In that case, if p - n, f(x) = xn − 1 and f ′(x) = nxn−1 have no commonfactors (the latter has only 0 as roots, and the former doesn’t have 0 as a root),so gcd(f(x), f ′(x)) = 1 and f(x) is separable.

From now on, we will therefore assume we are in one of those separablesituations.

We will denote by µn = µ[n] the set of all the nth roots of unity. This is anabelian group of order n, and in fact µn is a cyclic group (because it is a finite

subgroup of k×

).Let kn = k(µn) be the splitting field of xn − 1. Then kn/k is Galois.Since µn is cyclic, let ξ be a primitive nth roots of unity, so that µn = 〈ξ〉,

and ord(ξ) = n. Then kn = k(ξ).This raises a natural question: what is G = Gal(Kn/k)?Certainly σ ∈ G is determined by σ(ξ) = ξi (since all elements in µn look

like ξi for some i). But we must have 〈ξi〉 = 〈ξ〉 = µn, so we need gcd(i, n) = 1.Hence we have G ↪→ (Z/nZ)× by σ 7→ i, where σ(ξ) = ξi.

Theorem 21.2.1. We have [Q(µn) : Q] = ϕ(n) and Gal(Q(µn)/Q) ∼= (Z/nZ)×.

Proof. Write µn = 〈ξ〉. Let f(x) = Irr(ξ;Q, x). We can clear the denominatorto get f(x) ∈ Z[x] (strictly speaking a different f , but redefine it to be so). Weclaim that deg f = ϕ(n).

Note how f(x) | xn − 1, implying that xn − 1 = f(x)h(x) for some h(x) ∈Q[x]. In fact, by Gauss lemma we can guarantee h(x) ∈ Z[x].

Note that since f(ξ) = 0, for any prime p - n, f(xp) = 0. Otherwise, iff(xp) 6= 0, then

0 = (ξp)n − 1 = f(ξp)h(ξp),

where f(ξp) 6= 0, so h(ξp) = 0. Hence ξ is a root of h(xp), meaning thatf(x) | h(xp), because f = Irr(ξ;Q, x).

Now consider this modulo p, say xn − 1 = f(x)h(x) in Fp. Now modulo p,

h(xp) = h(x)p, so f(x) | h(x)

p. Thus every root of f(x) is a root of h(x), so

whilst the left-hand side of xn − 1 = f(x)h(x) is separable (since p - n), so ithas no repeated roots, the right-hand side does have repeated roots.

CYCLOTOMIC FIELDS 51

Lecture 22 Cyclotomic Fields

22.1 Proof finished

Proof continued. Similarly, for another prime q - n, (ξp)q = ξpq is a zero of f(z).Hence ξn

′is a zero of f(x) for any n′ coprime to n, meaning that deg f(x) ≥

ϕ(n).On the other hand, [Q(µn) : Q] ≤ ϕ(n), so together deg f(x) = ϕ(n) =

[Q(µn) : Q], and so Gal(Q(µn)/Q) ∼= (Z/nZ)×. (The left-hand side embeds intothe right-hand side, and they are of the same order.)

There is good reason we specified k = Q in this theorem:

Counterexample 22.1.1. Let k = R, and let ξ be a primitive 5th root ofunity, i.e., a root of x5− 1. Then Gal(R(ξ)/R) ↪→ (Z/5Z)×, and |(Z/5Z)×| = 5.

However

x5−1 = (x−1)(x4+x3+x2+x+1) = (z−1)(x2+

1 +√

5

2x+1

)(x2+

1−√

5

2x+1

)over R, so [R(ξ) : R] = |Gal(R(ξ)/R)| = 2. N

Counterexample 22.1.2. Let k = Fq, where q = pm is a prime power. Let ξbe a primitive nth root of unity. Then K = k(ξ) is the splitting field of xn − 1,and Gal(K/k) = 〈σ〉 where σ : a 7→ aq is the Frobenius automorphism.

Then Gal(K/k) = 〈σ〉 ↪→ (Z/nZ)× by σ 7→ q, so the order of σ is the orderof q in (Z/nZ)×. But q may or may not be a generator in (Z/mZ)×. N

Definition 22.1.3. A splitting field K of xn − 1 ∈ k[x] is called a cyclotomicextension of order n, or just a cyclotomic field if we don’t specify the order.

Let us recall some properties of the Euler φ function. First, ϕ(pn) = pn −pn−1 for p prime. Second, if (m,n) = 1, then ϕ(mn) = ϕ(m)ϕ(n), i.e., φ ismultiplicative.

Proposition 22.1.4. Suppose (m,n) = 1. Then Q(µm) ∩Q(µn) = Q.

Proof. We have the following situation:

Q(µm)Q(µn)

Q(µm) Q(µn)

K = Q(µm) ∩Q(µn)

Qϕ(m) ϕ(n)

First, notice how Q(µm)Q(µn) = Q(µmµn)−Q(µmn). The last step is a conse-quence of the fact that

(Z/mZ)× × (Z/nZ)× ∼= (Z/mnZ)×

Date: November 5th, 2019.


if (m,n) = 1. Second, since Q(µn)/Q is Galois, so is Q(µn)/K, and we can liftthis to Q(µmn)/Q(µm), and Gal(Q(µm)/K) ∼= Gal(Q(µmn)/Q(µn)). Now

|Gal(Q(µmn)/K)| = ϕ(mn)

ϕ(n)= ϕ(m)

since (m,n) = 1. Hence K = Q.

22.2 Quadratic reciprocity

Let p be an odd prime, and let v ∈ Z with (v, p) = 1. Define(v

p

)=

{1, if there exists x ∈ Z such that x2 ≡ v (mod p),

−1, otherwise.

If(vp

)= 1, then v is called a quadratic residue modulo p, and if

(vp

)= −1,

then v is called a quadratic nonresidue modulo p.Note how, since x2 = (−x)2, squaring the numbers

{1, 2, . . . , p − 1

}=

(Z/pZ)×, we get exactly half of them back, i.e., exactly p−12 of them are

quadratic residues.This quadratic symbol has a couple of basic properties:

(i)(pq

)(qp

)= (−1)

p−12

q−12 , called the law of quadratic reciprocity;

(ii)(−1p

)= (−1)

p−12 ;

(iii)(

2p

)= (−1)

p2−18 ;

(iv)(mnp

)=(mp

)(np

), so it is completely multiplicative; and

(v)(a2

p

)=(ap

)2= 1.

22.3 Gauss sums

Let p be an odd prime and let ξ be a primitive pth root of unity (e.g., ξ = e2πip ).

We define the Gauss sum

S :=∑

1≤v≤p−1

(v

p

)ξv.

Proposition 22.3.1. S2 =(−1p

)p, and hence S = ±√p or S = ±

√−p.

Proof. The proof is essentially a change of variables. We have

S2 =( ∑

1≤v≤p−1

(v

p

)ξv)( ∑

1≤u≤p−1

(u

p

)ξv)

=∑v,u

(uv

p

)ξv+u

=∑u

(∑v

(uv

p

)ξv+u

).


Notice how, since u 6= 0, uv just permutes v modulo p, so making the change ofvariables v 7→ uv we don’t change the sum, so

S2 =∑u

(∑v

(u2v

p

)ξvu+u

)=∑u

(∑v

(v

p

)ξ(v+1)u

).

We split this sum into two parts, one where v = −1 = p − 1, and one wherev 6= −1. So

S2 =

(−1

p

)(p− 1) +

∑u

∑v 6=−1

(v

p

)ξ(v+1)u.

Let us focus on the second sum for a moment. Changing the order of summation,∑u

∑v 6=−1

(v

p

)ξ(v+1)u =

∑v 6=−1

(v

p

) ∑1≤u≤p−1

ξ(v+1)u,

where again v + 1 6= 0 on the inside, so the inside sum is just∑u

ξu = ξ + ξ2 + . . .+ ξp−1 = −1,

whence, putting this back,

S2 =

(−1

p

)(p− 1)−

∑v 6=−1

(v

p

)=

(−1

p

)p−

∑v

(v

p

)=

(−1

p

)p.

Remark 22.3.2. Gauss proved that

S =

{√p, if p ≡ 1 (mod 4)√−p, if p ≡ 3 (mod 4).

We will not prove that here, because we don’t need it.

Example 22.3.3. We have, for instance, for p = 3

S = ξ − ξ2,

where ξ = e2πi3 , and if p = 7, take ξ = e

2πi7 and we have

S = ξ + ξ2 − ξ3 + ξ4 − ξ5 − ξ6. N

Theorem 22.3.4. Let K be a field such that [K : Q] = 2 (i.e., K is a quadratic

extension of Q). Then there exists some m such that K ⊂ Q(µm) = Q(e2πim ).

In other words, every quadratic extension of Q is contained in a cyclotomicextension.

Something more general, but much harder to prove, is true:

Theorem 22.3.5 (Kronecker–Weber). Let K/Q be an abelian extension. Thenthere exists an m such that K ⊂ Q(µm).

CHARACTERS 54

Lecture 23 Characters

23.1 Cyclotomic extensions

Let us start by proving Theorem 22.3.4 from last lecture.

Proof. We have a degree two extension K/Q, which, being finite and sincechar(Q) = 0, must be simple according to the Primitive element theorem. Ergo,K = Q(α) for some α, and α is the root of a second degree polynomial.

By completing the square we can make a change of variable and assume thepolynomial looks like x2 + 1, and by clearing denominators we can moreoverassume a ∈ Z. Therefore Q(

√a).

We can also assume a is square free, since otherwise we would just factorthem out, so a = p1p2 · · · pr or a = −p1p2 · · · pr, where pi are distinct primes.

Consequently√a =√p1√p2 · · ·

√pr or

√a =√−1√p1√p2 · · ·

√pr.

Recall how if p is an odd prime,

S =

p−1∑v=1

(v

p

)ξv,

with ξ = e2πip ∈ Q(µp), then S = ±√p ∈ Q(µp) or S = ±

√−1√p ∈ Q(µp).

Also, Q(µ4) = Q(e2πi4 ) = Q(

√−1), so

√−1 ∈ Q(µ4).

Moreover, for the prime p = 2, we have Q(µ8) = Q(e2πi8 ), whence e

2πi8 =

(1+i)√

22 ∈ Q(µ8), and by the reflexive property (1−i)

√2

2 ∈ Q(µ8). Hence their

sum is in Q(µ8), so√

2 ∈ Q(µ8).Now remember that, if (m,n) = 1, then Q(µmµn) = Q(µmn).So if all pi in a are odd, then

K ⊂ Q(µ4µp1 · · ·µpr ) = Q(µ4|a|),

and if p1 = 2, then

K ⊂ Q(µ8µp2 · · ·µpr ) = Q(µ4|a|),

so in both cases K ⊂ Q(µ4|a|).

We will not prove the more general Kronecker–Weber theorem here, for itrequires local class field theory, which we will not discuss.

Definition 23.1.1. Let

Φn(x) =

ϕ(n)∏i=1

(x− ξi),

where{ξi∣∣ i = 1, 2, . . . , ϕ(n)

}is the set of all primitive nth roots of unity.

Then by our proof of the fact that Gal(Q(µn)/Q) ∼= (Z/nZ)× it follows thatΦn(x) ∈ Q[x] is irreducible, and clearly deg Φn = ϕ(n).

This polynomial Φn(x) is called the nth cyclotomic polynomial .


CHARACTERS 55

Since xn − 1 has as its roots all nth roots of unity, not just the primitiveones, we must have

xn − 1 =∏d|n

Φd(x),

and consequently

Φn(x) =xn − 1∏d|nd 6=n

Φd(x)

For instance, Φ1(x) = x− 1 and Φ2(x) = x+ 1. Next

Φ3(x) =x3 − 1

x− 1= x2 + x+ 1,

and

Φ4(x) =x4 − 1

(x− 1)(x+ 1)= x2 + 1.

Since p = 5 is a prime, we see simply how

Φ5(x) = x4 + x3 + x2 + x+ 1,

but more interestingly, for instance,

Φ6(x) = x2 − x+ 1

andΦ12(x) = x4 − x2 + 1.

23.2 Classical Galois results

A classical question in Galois theory is this: can we find a Galois extension ofQ with corresponding Galois group Sn?

Below are two answers.First, by Schur, let

En(x) = 1 + x+x2

2!+ · · ·+ xn

n!∈ Q[x].

Then degEn = n and

Gal(En) =

{An, if n ≡ 0 (mod 4),

Sn, otherwise.

Another answer: let

Hm(x) = (−1)mex2

2dm

dxm

(e−

x2

2

)∈ Q[x]

be the mth Hermite polynomial . Then Hm(x) is even if m is even, and oddif m is odd, and therefore we can write

H2n(x) = K(0)n (x2)

andH2n+1(x) = xK(1)

n (x2).

Then Gal(K(i)n ) ∼= Sn for i = 0, 1 and all n > 12.

NORMS AND TRACES 56

23.3 Characters

Definition 23.3.1. Let G be a monoid1 and let K be a field. A homomorphismϕ : G→ K× is called a character of G to K.

Very often we will just take G to be a group.

Theorem 23.3.2 (Artin). Let G be a monoid and K a field. Let χ1, χ2, . . . , χnbe distinct characters of G to K. Then

{χ1, χ2, . . . , χn

}is linearly independent

over K.

Proof. Suppose the set{χ1, χ2, . . . , χn

}is linearly dependent. Then

a1χ1 + a2χ2 + · · ·+ amχm = 0

for some ai ∈ K, ai 6= 0. In particular, take m to be the smallest such number,i.e., we have the smallest nontrivial set of linearly dependent characters.

Since χ1 6= χ2, there exists some z ∈ G so that χ1(z) 6= χ2(z), and conse-quently consider the system

a1χ1 + a2χ2 + · · ·+ amχm = 0

a1χ1(z)χ1 + a2χ2(z)χ2 + · · ·+ amχm(z)χm = 0.

Taking χ1(z) times the first equation and subtracting the second, we get

a2(χ1(z)− χ2(z))χ2 + · · ·+ am(χ1(z)− χm(z))χm = 0,

but χ1(z) − χ2(z) 6= 0, so we have produced a smaller linearly dependent set,which contradicts the minimality of m.

Lecture 24 Norms and Traces

First let us state a simple consequence of Artin’s theorem from last time:

Corollary 24.0.1. Let E/k be separable of degree n. Let{σ1, σ2, . . . , σn

}be

distinct k-embeddings of E into k. Then{σ1, σ2, . . . , σn

}is a linearly indepen-

dent set.

Proof. Just consider σi : E× → k

×, which is a character.

24.1 Field norms and field traces

Definition 24.1.1. Let E/k be a separable extension of degree n. Let{σ1, σ2, . . . , σn

}be the set of all k-embeddings of E into k.

For α ∈ E, we define the (field) normnorm of α as

NE/k(α) = σ1(α)σ2(α) · · ·σn(α)

and the (field) trace

TrE/k(α) = σ1(α) + σ2(α) + · · ·+ σn(α).

When there is no risk of ambiguity we will omit the subscript and just writeN and Tr.

1Meaning a semigroup with identity, i.e., a group but elements do not necessarily haveinverses.


NORMS AND TRACES 57

Remark 24.1.2. Both NE/k(α) and TrE/k(α) are in k. To see this, notice howapplying any σ to either of them just permutes the terms, so they are fixed byall σ.

Now suppose there exists some α ∈ E, fixed by all σ, but not in k. Thenσ must send α to another root of Irr(α; k, x), but σ(α) = α, so k(α) would bepurely inseparable, which is a contradiction to E/k being separable.

Remark 24.1.3. Suppose E ⊃ F ⊃ k. Then we can compute norms over thelarge extension piecewise,

NE/k(α) = NF/k(NE/F (α)),

and likewiseTrE/k(α) = TrF/k(TrE/F (α)).

Example 24.1.4. Suppose E = k(α) is a simple extension of k. Write

f(x) = Irr(α; k, x) = xn + a1xn−1 + · · ·+ an,

which in k[x] factors as

f(x) = (x− σ1(α))(x− σ2(α)) · · · (x− σn(α)).

Hence NE/k(α) = (−1)nan and TrE/k(α) = −a1. N

Example 24.1.5. Let E be a separable extension of k and let β ∈ E. Thenthere is a subfield F = k(β) between k and E, and by the above example weunderstand the norm of β on F/k. Specifically, let

f(x) = Irr(β; k, x) = xm + a1xm−1 + · · ·+ am,

and so NF/k(β) = (−1)mam. Then

NE/k(β) = NF/k(NE/F (β)),

but β ∈ F , so all F -embeddings fix is, meaning that NE/F (β) = β[E:F ]. Hence

NE/k(β) = NF/k(β[E:F ]) = NF/k(β)[E:F ] = ((−1)mam)[E:F ].

Similarly,

TrE/k(β) = TrF/k(TrE/F (β)) = TrF/k([E : F ] · β)

= [E : F ] TrF/k(β) = −[E : F ]a1. N

Remark 24.1.6. In the above we have used the following properties:

(i) N(αβ) = N(α) N(β), i.e., the norm is multiplicative, since the σ involvedare homomorphisms, and hence also multiplicative.

Similarly, Tr(α+ β) = Tr(α) + Tr(β) is additive, for the same reason.

(ii) In particular, NE/k(a · α) = an NE/k(α) for a ∈ k, if [E : k] = n.

Similarly, TrE/k(a · α) = aTrE/k(α). Hence trace is a k-linear map.

NORMS AND TRACES 58

Theorem 24.1.7. Define a k-bilinear form 〈·, ·〉 : E × E → k by 〈x, y〉 =TrE/k(xy). Then this bilinear form is non-degenerate2. Hence E and itsdual space E∗ = Homk(E, k) are identified.

Proof. Suppose 〈·, ·〉 is degenerate, i.e., there exists some x 6= 0 such that〈x, y〉 = 0 for all y ∈ E. That is, TrE/k(xy) = 0 for all y ∈ E, so TrE/k(xE) = 0,meaning that TrE/k(E) = 0.

This means that we have

σ1(x) + σ2(x) + · · ·+ σn(x) = 0

for all x ∈ E, but this implies{σ1, σ2, . . . , σn

}is a linearly dependent set,

which contradicts Artin’s theorem.Hence 〈·, ·〉 is non-degenerate, and therefore the map E → E∗ defined by

x 7→ φx, with φx(y) = 〈x, y〉 = TrE/k(xy) is one-to-one.But these are finite dimensional vector spaces (over k), so one-to-one implies

onto, so this is an isomorphism.

Definition 24.1.8. Let E/k be separable of degree n. Let{w1, w2, . . . , wn

}be a basis of E/k (as a vector space). Then

{w′1, w

′2, . . . , w

′n

}is called the dual

basis of{w1, w2, . . . , wn

}if (w′i, wj) = δij , where

δij =

{1, if i = j,

0, if i 6= j.

Here we use w′i to also correspond to an element 〈w′i, ·〉 in the dual space.

Theorem 24.1.9. Let E/k be separable of degree n. Let α ∈ E so that E = k(α)(note how E/k is finite and separable, so by the Primitive element theorem thisis possible). Let f(x) = Irr(α; k, x), and let

{α1, α2, . . . , αn

}be the distinct

roots of f(x).Then

{1, α, α2, . . . , αn−1

}is a basis of E/k and its dual basis is given by{

δ0, δ1, . . . , δn−1

}where

δi =βi

f ′(α)

for i = 0, 1, 2, . . . , n− 1, and βi are the coefficients of

f(x)

x− α= βn−1x

n−1 + βn−2xn−2 + · · ·+ β0 ∈ E[x].

To prove this we need the following lemma:

Lemma 24.1.10 (Lagrange interpolation theorem). With the same assump-tions as in the previous theorem, we have for any 0 ≤ r ≤ n− 1

n∑i=1

f(x)

x− αiαri

f ′(αi)= xr.

2That is, for x ∈ E, 〈x, y〉 = 0 for all y ∈ E if and only if x = 0.

NORMS AND TRACES 59

Proof. We have f(x) = (x−α1)(x−α2) · · · (x−αn). Then by the product rule

f ′(αi) =∏j 6=i

(αi − αj).

Consider

g(x) = xr −n∑i=1

f(x)

x− αiαri

f ′(αi).

This is a polynomial of degree at most n− 1.But on the other hand, if we plug αk into g, only the i = k term in the sum

survives, and

g(αk) = αrk −∏j 6=k

(αk − αj)αrk

f ′(αk)= 0

for k = 1, 2, . . . , n. So it has more zeros than its degree, meaning that g = 0.

With this we are ready to prove the theorem:

Proof of Theorem 24.1.9. Notice first how the terms

f(x)

x− αiαri

f ′(αi)

for i = 1, 2, . . . , n are all conjugates of one another, since our embeddings sendαi to other αj . Hence the sum in the previous lemma is

TrE/k

( f(x)

x− ααr

f ′(α)

)= xr,

where by taking the trace of a polynomial we mean apply the trace to thecoefficients.

Hence, since f(x)x−α = βn−1x

n−1 + · · ·+ β0, we have

TrE/k

((βn−1x

n−1 + · · ·+ β0)αr

f ′(α)

)= xr.

Comparing coefficients here, we see that the xith coefficient of the left-hand sideis

TrE/k

(βi

αr

f ′(α)

)= TrE/k(δiα

r) = δir,

so TrE/k(δiαr) = 〈δi, αr,=〉δir, so we have a dual basis 〈δi, ·〉.

24.2 Galois theory of solvability of algebraic equations

Theorem 24.2.1 (Hilbert’s theorem 90). Let K/k be a cyclic extension ofdegree n. Let β ∈ K. Then N(β) = 1 if and only if there exists some α ∈ Ksuch that β = α

σ(α) , where Gal(K/k) = 〈σ〉.

Proof. The converse direction is trivial. Suppose β = ασ(α) . Then

N(β) = N( α

σ(α)

)=

N(α)

N(σ(α)=

N(α)

N(α)= 1

RADICAL EXTENSIONS 60

since σ(α) just permutes the terms in the norm of α.For the forward direction we need to be much more careful. We will use

the notation σ(α) = ασ. By Artin’s theorem,{

1 = σ0, σ, σ2, . . . , σn−1}

is alinearly independent set. Therefore

β0σ0 + βσ + β1+σσ2 + β1+σ+σ2

σ3 + · · ·+ β1+σ+σ2+···+σn−2

σn−1

is a non-zero map. In the σ(β) = βσ notation, note how this means, for example,

β1+σ = β1βσ = βσ(β)

andβ1+σ+σ2

= β1βσβσ2

= βσ(β)σ2(β).

This being a non-zero map, there must exists some θ ∈ K such that, whenevaluating at θ, we get

α := θ + βθσ + β1+σθσ2

+ · · ·+ β1+σ+···+σn−2

θσn−1

6= 0.

Apply σ and multiply by β and we get

βασ = β(θσ + βσθσ

2

+ βσ+σ2

θσ3

+ · · ·+ βσ+σ2+···+σn−1

θσn).

Note two things: first, σn = Id, so θσn

= θ. Second, multiplying through by βthe last term becomes

ββσ+σ2+···+σn−1

= β1+σ+σ2+···+σn−1

= N(β) = 1.

Hence the above is nothing but

θ + βθσ + β1+σθσ2

+ · · ·+ β1+σ+···+σn−2

θσn−1

= α,

so βασ = α, and we are done.

Lecture 25 Radical Extensions

25.1 Kummer extensions

From now on out, until the end of these notes, we will assume extensions areseparable unless otherwise stated.

Theorem 25.1.1 (Kummer extensions). Assume k contains a primitive nthroot of unity.

(i) Let K/k be a cyclic extension of degree n. Then there exists some α ∈ Ksuch that K = k(α) and Irr(α; k, x) = xn − a ∈ k[x]. In particular,αn = a ∈ k.

(ii) Let α be a root of f(x) = xn−a ∈ k[x]. Then k(α)/k is a cyclic extensionof degree d with d | n, and αd = b ∈ k, with Irr(α; k, x) = xd − b.



This theorem therefore classifies cyclic extensions of degree n if the base fieldk contains a primitive nth root of unity.

Proof. (i) Let ξ be a primitive nth root of unity, which is in k by assumption.Then

NK/k(ξ−1) = (ξ−1)n =1

ξn= 1

since ξ ∈ k, and hence ξ−1 ∈ k, means its norm is just itself to the powern = [K : k]. By Hilbert 90 this means that there exists some α ∈ K so that

1

ξ=

α

σ(α)

where Gal(K/k) = 〈σ〉, which is cyclic by assumption. Therefore σ(α) = αξ.

Notice how σ2(α) = σ(αξ) = σ(α)ξ = αξ2, and in general σi(α) = αξi fori = 0, 1, . . . , n− 1, which are all distinct since ξ is a primitive nth root of unity.

Hence

f(x) = Irr(α; k, x) = (x− α)(x− σ(α)) · · · (x− σn−1(α)),

which is of degree n, and therefore K = k(α) since K ⊃ k(α) and both aredegree n extensions of k.

To show that f(x) = Irr(α; k, x) = xn − 1, we will show that αn = a ∈ k,so that f(x) | xn − a. Since they are of the same degree, this means they areequal.

So let a = αn. To show a ∈ k is suffices to show that σ(a) = a, since thismeans it is fixed by the entire Galois group. Now

σ(a) = σ(αn) = (αξ)n = αnξn = αn = a

where ξn = 1 since it is a (primitive) nth root of unity. Hence f(x) = xn − a.

(ii) Since α is a root of xn − a, we know that Irr(α; k, x) | xn − a. Moreover,we know that

xn − a =

n−1∏i=0

(x− αξi).

Since ξ ∈ k means ξi ∈ k, the field k(α) is a splitting field of both Irr(α; k, x)and xn − a. Hence k(α)/k is Galois, and so let H = Gal(k(α)/k).

For τ ∈ H, we we have τ(α) = αωτ , where

ωτ ∈ 〈ξ〉 = Set1, ξ, ξ2, . . . , ξn−1 ∼= Z/nZ.

Define φ : H → 〈ξ〉 ∼= Z/nZ by φ : τ 7→ ωτ . Then φ is one-to-one, so H isa subgroup of Z/nZ, and consequently H is cyclic, making k(α)/k a cyclicextension. Moreover if |H| = d, we must have d | n because H is a subgroup ofZ/nZ, which is of order n.

Finally, Irr(α; k, x) has degree d. Let b = αd. Then

τ(b) = τ(αd) = τ(α)d = (αωτ )d = αdωdτ = αd = b,

so b ∈ k. Hence, as above, Irr(α; k, x) = xd − b.

To study the same question—that of classifying cyclic extensions of degreen—when the base field k does not contain any primitive nth roots of unity ismuch harder, and to do this we need Galois cohomology to do Kummer theory.


25.2 Radical extensions

Definition 25.2.1. We call E/k a radical extension (of height 1) if

(i) E = k(α) for some α ∈ E and αp ∈ k with p prime; and

(ii) αp 6= βp for all β ∈ k. In particular, α 6∈ k, so E is a proper extension.

Example 25.2.2. Let k = Q, α = e2πip , a primitive pth root of unity, and

consider E = Q(α). Notice how αp = 1 = 1p ∈ Q, and 1 ∈ Q. Does this meanE is not a radical extension? We can’t tell—there is not sufficient informationhere. It is possible that E = Q(α) = Q(β) for some other generator β whichdoes satisfy the conditions.

For example, take p = 3 and ω = −1+√−3

2 , and consider Q(ω)/Q. Here

ω3 = 1 ∈ Q, so we are in the above situation. However Q(ω) = Q(√−3), so

taking α =√−3 we have ω2 = −3 6= β2 for any β ∈ Q. So Q(ω)/Q is a radical

extension. N

All by way of saying: it is difficult to determine if an extension is radical.In particular, note that the first part of the definition is generally not too hardto check—if αp ∈ k—but the second part is. That is, checking whether αp is apth power of something in the base field k.

There are special situations where we don’t have to check this:

Example 25.2.3. Let p be a prime. Then xp−a ∈ k[x] is irreducible if a 6∈ kp.Hence assume it is irreducible.

Let α ∈ k so that αp = a, i.e., α is a root of xp − a. Let E = k(α). Then[E : k] = p (which, for the record, means there are no intermediate subfields,the degree being prime).

Let θ be a primitive pth root of unity in k. Then

xp − a = (x− α)(x− αθ) · · · (x− αθp−1).

Note how αθi 6∈ k for i = 0, 1, . . . , p− 1 since xp − a is irreducible.Assume θ ∈ k. Then if αp = βp for some β ∈ k, we must have (αβ−1)p = 1,

so αβ−1 is a pth root of unity. Hence αβ−1 = θi for some i, meaning thatα = βθi ∈ k, contradicting α 6∈ k. N

So if the base field contains a primitive pth root of unity, then we don’t haveto check the harder part—we have αp 6= βp for all β ∈ k automatically.

Remark 25.2.4. If the prime p is replaced by an integer n, then xn − a, a 6∈ kn,need not be irreducible, so we cannot apply this argument.

Definition 25.2.5. We say that E/k is a radical extension (of height s) ifthere exists a chain

k = E0 ⊂ E1 ⊂ R2 ⊂ · · · ⊂ Es = E

such that each Ei/Ei−1, for i = 1, 2, . . . , s, is a radical extension fo height 1with respect to some prime pi. So [E : k] = p1p2 · · · ps.

SOLVABILITY BY RADICALS 63

The situation looks like

E = Es = Es−1(αs)

...

E2 = E1(α2)

E1 = E0(α1)

k = E0

p2

p1

where at each step there are no intermediate subfields, though it is possiblethere are other subfields between E and k not in the tower.

Example 25.2.6. Consider Q( 4√

2)/Q. This is a rational extension, but ofheight 2. Consider Q ⊂ Q(

√2) ⊂ Q( 4

√2).

The first extension is radical, since√

22

= 2 ∈ Q and Q contains a primitive2nd root of unity (namely −1), and

√2 6∈ Q. For the same reason, the second

extension is radical since 4√

22

=√

2. N

Remark 25.2.7. Let E/k be a radical extension and let F be an intermediatesubfield, i.e., E ⊃ F ⊃ k. Then E/F is radical, but in general F/k need not be.

To see that E/F is radical, consider the tower for E/k above, where at eachstage we adjoin an αi.

Now adjoining the same elements to form a tower over F , so F ⊂ F (α1) ⊂F (α1, α2) ⊂ · · · ⊂ E, we see that E/F is radical.

On the other hand, if both E/F and F/k are radical, then E/k is radical—just combine the two chains.

Remark 25.2.8. Composites of radical extensions are not radical in general.For example, consider k = Q, E1 = Q( 7

√2), and E2 = Q(

√72e

2πi7 ). Then

E1/Q and E2/Q are both radical, but E1E2/Q is not.

Lecture 26 Solvability by Radicals

26.1 Solvable groups

Definition 26.1.1. Let f(x) ∈ k[x]. We say that f is solvable by radicals ifthere exists a radical extension E/k such that f splits in E[x].

(So in other words, f splits in E[x] and everything in E, hence including theroots of f , can be written as radical expressions of things in k.)

Remark 26.1.2. A polynomial f being solvable by radicals does not mean thesplitting field of f over k is radical—its splitting field is just contained in someradical extension of k.



Remark 26.1.3. In fact, every root of f(x) is contained in a radical extensionof k if and only if f is solvable by radicals, i.e., all roots are contained in oneradical extension of k.

Theorem 26.1.4. Let E/k be an extension such that there exists u ∈ E withun ∈ k and E = k(u) for some n. Assume k contains a primitive nth root ofunity. Then E/k is a radical extension (of some height).

Remark 26.1.5. For convenience, we say k/k is radical of height 0.

Proof. We use induction on n. If n = 1, then E = k and we are done.Suppose n > 1, and write n = p1p2 · · · pr where pi are primes, not necessarily

distinct. Write n = p1 · n1 and set u1 = up1 . So un11 = up1n1 = un ∈ k, and k

contains a primitive n1th root of unity since n1 | n.By the induction hypothesis, E1 = k(u1)/k is radical. Now up1 = u1 ∈ E1,

E1 contains a primitive p1th root of unity, and p1 < n so E = E1(u)/E1 isradical by the induction hypothesis.

Putting these together, E/k is radical as well.

Recall how, in general, compositions of radical extensions are not radical. Inspecial cases, they might be:

Proposition 26.1.6. Suppose E1/k and E2/k are radical. Assume k containssufficiently many roots of unity. Then E1E2/k is radical.

Proof. By assumption E1/k is radical. If we can show that E1E2/E1 is radical,we are done. Since E2/k is radical, there exists a chain k ⊂ k(u1) ⊂ k(u1, u2) ⊂· · · ⊂ k(u1, u2, . . . , un) = E2, where each step is radical for some prime pi.

Then certainly we have a chain

E1 ⊂ E1(u1) ⊂ E1(u1, u2) ⊂ · · · ⊂ E1(u1, u2, . . . , un) = E1E2.

We need to show that each step in this new chain is radical. This is the case if,at each step, Ei contains an pith root of unity, by the last theorem. Hence if kitself contains all those pith roots of unity, the composition is indeed radical.

Recall that a group G is solvable if there exists

G = G0 > G1 > G2 > · · · > Gs ={

1}

such that Gi is normal in Gi−1 and Gi−1/Gi is abelian.Note how if G is a finite group, we can refine the sequence

{Gi}

suchthat Gi−1/Gi is cyclic (by the Fundamental theorem of finitely abelian groups,essentially).

Proposition 26.1.7. (i) Every homomorphic image and subgroup of a solv-able group is solvable (because homomorphisms send normal subgroups tonormal subgroups, and abelian subgroups to abelian subgroups).

(ii) G is solvable if and only if G/N and N are solvable for some normalsubgroup N ⊂ G.


Consider a tower of extensions E ⊃ F ⊃ k, where E/F and F/k are solvable.We don’t know if E/k is solvable—we don’t even know if E/k is Galois.

However if E/k is Galois, then ley G = Gal(E/k) and N = Gal(E/F ), sothat G/N = Gal(F/k). So E/k is Galois.

In the first lecture we mentioned how Gauss proved that xm − 1 is solvableby radicals. We are finally ready to prove this:

Theorem 26.1.8 (Gauss). Let char(k) = 0. Let F be any intermediate fieldsuch that k ⊃ F ⊃ k (i.e., F/k is algebraic). Let η be a primitive mth root ofunity. Then there exists a radical extension M/F such that M ⊃ F (η).

Proof. We use induction on m. If m = 1, then M = F = k and we are done.Assume the lemma holds for all ` < m. Let 1 6= `1 < m, and take η1

to be a primitive `1st root of unity. By the induction hypothesis, there existsF1 ⊃ F (η1) such that F1/F is radical.

Now use F1 as the new base field, letting `2 6= 1, `1, `2 < m, and taking η2

to be a primitive `2nd root of unity. Then by the induction hypothesis thereexists F2 ⊃ F1(η2) and F2/F1 is radical, whence F2/F is radical.

Repeat this argument and we get some F ′/F that is radical, and F ′ containsall `th roots of unity for ` < m. Write

m = pe11 pe22 · · · perr ,

with pi distinct primes. There are two cases to consider: if r ≥ 2, then F ′

contains ηpe11 and ηpe22 ···perr

. Since those two orders are coprime, F ′ contains aprimitive mth root of unity, namely their product.

Second, if r = 1, i.e., m = pe11 , then consider G = Gal(F ′(ηm)/F ′) ↪→(Z/mZ)×, so |G| ≤ ϕ(m) < m. So η

ϕ(m)m ∈ F ′, but F ′ contains a primitive

ϕ(m)th root of unity, hence by the previous theorem F ′(ηm)/F ′ is radical.Hence F ′(ηm)/F is radical, and we are done.

26.2 Galois’ theorem

Theorem 26.2.1 (Galois). Assume char(k) = 0. Let f(x) ∈ k[x]. Let K bea splitting field of f(x), and let G = Gal(K/k). Then every root of f(x) iscontained in a radical extension of k if and only if G is solvable.

Proof. We prove the converse first. Suppose G is solvable, and let θ be a prim-itive nth root of unity with n sufficiently large (more on this in a moment).

We have the following situation:

K(θ)

K k(θ)

F = K ∩ k(θ)

k

solvable


Here, K(θ)/k(θ) is Galois and solvable, since Gal(K/F ) is a subgroup of Gal(K/k),which is solvable, and H = Gal(K(θ)/k(θ)) = Gal(K/F ) by Galois lifting.

Since K(θ)/k(θ) is solvable, we have

K(θ){

1}

= Hk

......

E2 H2

E1 H1

k(θ) H = H0

meaning there exists a sequence

H = H0 ⊃ H1 ⊃ · · · ⊃ Hk ={

1}

such that Hi−1/Hi is cyclic (since the original extension is finite). For n suf-ficiently large, we can assume k(θ) contains all primitive |Hi−1/Hi|th roots ofunity.

Hence Ei/Ei−1 is cyclic with Galois group Gal(Ei/Ei−1) ∼= Hi/Hi−1. ByKummer’s theorem and our previous theorem, Ei/Ei−1 is therefore radical.

Hence K(θ)/k(θ) is radical. So it K(θ)/k were radical we would be done,but in general that is not the case.

The trick now is to invoke Gauss lemma. By Gauss lemma there exists someM/k, which is radical, with k(θ) ⊂M . This extends our diagram to

MK(θ)

K(θ) M

E = K(θ) ∩M

K k(θ)

F = K ∩ k(θ)

k

radical by Gauss

and by Galois lifting and repeating the argument we used to show K(θ)/k(θ)is radical we can show, in the same way, that MK(θ)/M is radical. HenceMK(θ)/k is radical, and the converse direction is done.

TOPOLOGICAL GROUPS 67

Lecture 27 Topological Groups

27.1 Galois theorem

Proof, continued. For the forward direction, assume f(x) is irreducible in k[x].(If not, suppose f(x) = f1(x)f2(x) with the two factors irreducible. Then ifK1 is the splitting field of f1 over k and K2 is the splitting field of f2 over k,we have that K1/k is Galois and solvable, and so is K1K2/K1 since it is thesplitting field of f2 over K1. Now the large extension K1K2/k will be solvableif it is Galois, which is the case: it is the splitting field of f .)

Let α be a root of f(x). Then there exist

k = E0 ⊂ E1 ⊂ E2 ⊂ · · · ⊂ Es,

with each extension being radical of height 1 and α ∈ Es (by which we mean thechain depends on α). We claim there exists a solvable extension M/k such thatM ⊃ Es for all α. Hence M contains all roots of f(x), meaning that K ⊂ M ,so G is solvable.

We prove this by induction on s. In other words, suppose such M exist fors− 1.

By definition of Es/Es−1 being radical there exists some us ∈ Es such thatupss ∈ Es−1. By the induction hypothesis there exists some M/k such thatM ⊃ Es−1 and M/k is solvable.

Let η be a primitive psth root of unity, and let M ′ = M(η), and let M ′′ =M ′(us). Then upss ∈ Es−1 ⊂ M ′, so by Kummer’s theorem M ′′/M ′ is a cyclicGalois extension, so in particular it is solvable (cyclic implies abelian impliessolvable). In the same way, M ′/M is solvable.

The situation is as follows:

M ′′ = M ′(us)

M(η) = M ′

Es = Es−1(us) M

Es−1

...

E1

k = E0

solvable

solvable

ps

solvable

Date: November 21st, 2019.


Hence if M ′′/k is Galois we would be done, since then M ′′/k is solvable, butin general this is not the case.

Instead we enlarge slightly by letting M ′′ be the Galois closure of M ′′/k.

Then, by the Lemma below, Gal(M ′′/k) is solvable, and we are done.

The lemma we need is this:

Lemma 27.1.1. Let k ⊂ F ⊂ D ⊂ D, where D is the Galois closure of D/k

(i.e., the smallest Galois extension D/k such that D ⊂ D ⊂ k.) Assume D/F

and F/k are solvable Galois extensions. Then Gal(D/k) is solvable.

Proof. Let G = Gal(D/k) and N = Gal(D/F ). Since Gal(F/k) ∼= G/N , G/Nis solvable.

Similarly, since Gal(D/F ) ∼= N/H, N/G is solvable.Set

V =⋂g∈G

Hg,

where Hg = gHg−1 is the conjugate.Then V ⊂ H is a normal subgroup in G, so fixed field E of V is Galois over

k.We have the following Galois correspondence:

D{

1}

E V

D H

F N = Gal(D/F )

k G = Gal(D/k)

But D is the smallest Galois extension of k containing D, so by minimalityE = D, and V =

{1}

. Hence D/k is Galois.

Exercise 27.1.2. Suppose A and B are normal subgroups of a group G. As-sume G/A and G/B are solvable. Then G/AB is also solvable.

Now H is normal in N , so Hg ∩ N is normal in N , and N/H is solvable.Hence N/N ∩Hg is solvable. Repeat this for all g ∈ HG, and we conclude

N/⋂g∈G

Hg

is solvable, i.e., N/V = N/{

1}

= N is solvable. Hence since G/N and N aresolvable, G is solvable.

This immediately gives:


Theorem 27.1.3 (Abel). Let k be a field with char(k) = 0. For a generalf(x) ∈ k[x] with deg f = n, f(x) is solvable by radicals if and only if n ≤ 4.

(By this we mean, all polynomials of degree at most 4 are solvable by radicals,but for higher degree this is not true.)

Proof. This follows from the fact that Sn is solvable if and only if n ≤ 4.

Theorem 27.1.4. Let k ⊂ R. Let f(x) = x3 + ax2 + bx + c ∈ k[x] be irre-ducible. Suppose all three roots of f(x) are real. Then any root of f(x) cannotbe expressed using only real radicals.

Proof. Write f(x) = (x − α1)(x − α2)(x − α3), so K = k(α1, α2, α3) is thesplitting field of f(x) over k.

Supposek = E0 ⊂ E1 ⊂ E2 ⊂ · · · ⊂ Es

are radical extensions with α1 ∈ E2, and suppose that this chain is minimal inlength for α1 specifically (if not, just relabel).

So we have α1 ∈ Es, but α1 6∈ Es−1, and indeed α2, α3 6∈ Es−1 either.Since Es−1 don’t contain the roots of f(x), it is irreducible in there, so

[Es−1(α1) : Es−1] = deg f = 3. On the other hand, we can write Es = Es−1(us)for some us, where upss ∈ Es−1 and [Es : Es−1] = ps is prime. Hence ps = 3,and Es = Es−1(α1).

Now let E be the Galois closure of Es/Es−1. Since E contains Es, whichcontains α1, and E/Es−1 is Galois, E contains all roots α1, α2, α3.

Hence E contains the splitting field of f(x) over Es−1. Let u3s = a ∈ Es−1,

and let g(x) = x3 − a ∈ Es−1[x].Then us ∈ Es ⊂ E, so g(x) = (x− us)(x− usω)(x− usω2) in E, where

ω =−1 +

√−3

2

is a primitive 3rd root of unity. In particular, ω = usωus∈ E, so E 6⊂ R. Hence

the radical extension containing the splitting field of f(x) is not real.

27.2 Infinite Galois

We will finally revisit something hinted at long ago: what happens when Galoisgroups are infinite?

Let K/k be Galois (possibly infinite). Let G = Gal(K/k). Let F ={F∣∣k ⊂

F ⊂ K}

be the set of all intermediate subfields and let H ={H∣∣H < G

}be

the set of all subgroups.Let Γ: F → H be defined by Γ(F ) = Gal(K/F ), and let Φ: H → F be

defined by Φ(H) = KH , the fixed field of H.We know already that Φ(Γ(F )) = F , i.e., Φ ◦ F = IdF .However Γ ◦ Φ 6= IdH in general, though if G is finite, then Γ ◦ Φ = IdH.To fix this, we will define a topology on G, and in that topology let Ho ={

H∣∣H < G closed subgroup

}. We will show that Γ ◦ Φ

∣∣∣Ho

= IdHo .

Definition 27.2.1. A group G is said to be a topological group if

(i) G is a topological space and


(ii) G × G → G defined by (x, y) 7→ xy and G → G defined by x 7→ x−1 areboth continuous (i.e., the group actions are compatible with the topology).

Remark 27.2.2. If G is a topological group and U(1) is the system of neighbour-hoods of 1, then U(1) satisfies

(i) u ∈ U(1) implies 1 ∈ u;

(ii) u1, u2 ∈ U(1) implies u1 ∩ u2 ∈ U(1);

(iii) u ∈ U(1) and u ⊂ v, v open, implies v ∈ U(1);

(iv) for all u ∈ U(1) there exists w ∈ U(1) such that w · w ⊂ u;

(v) for all u ∈ U(1), u−1 ∈ U(1); and

(vi) for all u ∈ U(1) and all g ∈ G, gug−1 ∈ U(1).

Conversely, if a group G possesses a set U of subsets of G satisfying (i)–(vi),then G can be given the structure of a topological group with its fundamentalsystem of neighbourhoods of 1 given by U .

Lecture 28 Topological Groups

28.1 Review of topological spaces

Definition 28.1.1. A topology on a set X is a collection F of subsets of Xsatisfying

(i) ∅, X ∈ F ;

(ii) if Aα ∈ F , α ∈ Λ, then⋃α∈ΛAα ∈ F ; and

(iii) if A1, A2, . . . , An ∈ F , then A1 ∩A2 ∩ · · · ∩An ∈ F .

The tuple (X,F) is called a topological space . The elements in F arecalled open sets.

Note that arbitrary unions of open sets are open, but only finite intersectionsof open sets are open.

Definition 28.1.2. Let (X,FX) and (Y,FY ) be topological spaces. A functionf : X → Y is continuous if f−1(A) ∈ FX for every A ∈ FY .

Let K/k be Galois, and let G = Gal(K/k). Let

F ={Kλ

∣∣K ⊃ Kλ ⊃ k, Kλ/k is finite Galois}

={Kλ

∣∣ λ ∈ Λ}.

We then have the following correspondence

K{

1}

Kλ Nλ = Gal(K/Kλ)

k G

finite Galois

Date: December 3rd, 2019.


Now since Kλ/k is finite Galois, Nλ is normal in G, and so Gal(Kλ/k) = G/Nλis a finite group, since the Kλ/k is finite. Let N =

{Nλ∣∣ λ ∈ Λ

}.

Our goal is to use N to generate a topology on G.

Lemma 28.1.3. (i)⋂λ∈ΛNλ =

{1}

.

(ii) For Nλ, Nµ ∈ N , there exists some Nν ∈ N such that Nν = Nλ ∩Nµ.

(iii) For Nλ ∈ N , there exists some Nµ ∈ N such that N−1µ ⊂ Nλ.

(iv) For Nλ ∈ N , there exists Nν ∈ N such that NµNµ ⊂ Nλ.

(v) For Nλ ∈ N and g ∈ G, there exists Nµ ∈ N such that gNµg−1 ⊂ Nλ.

Proof. (i) It suffices to show for σ 6= 1 there exists some Nλ such that σ 6∈ Nλ.Since 1 6= σ : K → K, there must exist some α ∈ K such that σ(α) 6= α.Now k(α)/k might not be Galois, but if not we simply take its Galois closure

k(α). Hence k(α)/k is finite Galois, so it is equal to Kλ for some λ, meaning itcorresponds to Nλ = Gal(K/Kλ). Then σ(α) 6= α ∈ Kλ implies σ 6∈ Nλ sinceanything in Nλ must fix Kλ.

(ii) We have Nλ = Gal(K/Kλ) and Nµ = Gal(K/Kµ). Then Nλ ∩ Nµ =Gal(K/KλKµ). Considering the diagram

KλKµ

Kλ Kµ

k

finite Galois

we can lift to see that KλKµ/Kµ is also finite Galois. This gives us a new graph

K

KλKµ H = Gal(K/KλKµ) = Nλ ∩Nµ

k G

Galois

Here since H is the intersection of two normal subgroups in G, it is itself normal,so KλKµ/k is finite Galois, meaning it Kν for some ν, and we are done.

For the last three parts, using Nλ = Nµ suffices.

We use N ={Nλ∣∣ λ ∈ Λ

}as a basis of the system of neighbourhoods of

1. Let N (1) denote this system of neighbourhoods of 1.For x ∈ G, let N (x) =

{xN

∣∣N ∈ N (1)}

.In particular, if G = Gal(K/k), then this endows G with the structure of a

topological group.

Proposition 28.1.4. G is a T0-topological space.3

3Meaning for any two distinct points, we can find a neighbourhood of one not containingthe other.


Proof. For g1, g2 ∈ G, g1 6= g2, we can shift to 1, g−11 g2. Recall how⋂

λ∈Λ

Nλ ={

1}

so there exists some Nλ such that 1 ∈ Nλ but g−11 g2 6∈ Nλ. Shifting back,

g1 ∈ g1Nλ and g2 6∈ g1Nλ.

Topological groups in general, not specifically Galois groups, have remark-able structure:

Proposition 28.1.5. Let G be a general topological group.

(i) Any T0-topological group is a T2-space.4

(ii) Any T2-topological group is regular5

(iii) Any open subgroup of a topological group is closed.

(iv) Any closed subgroup of finite index is open.

(v) Any subgroup that contains an open subgroup is open.

Proof. We prove some of these.

(i) Without loss of generality, take 1, g ∈ G. There exists an open set U suchthat 1 ∈ U and g 6∈ U since by hypothesis the space is T0. Now take an openneighbourhood V of 1 such that V V −1 ⊂ U (this is always possible, since wecould take for instance V = U ∩ U−1). Then gV is an open neighbourhood ofg. We claim gV ∩ V = ∅.

If not, then there exist v, u ∈ V such that gv = u, meaning that g = uv−1 ∈V V −1 ⊂ U , contradicting g 6∈ U .

(iii) Let H < G be an open subgroup. We can decompose G into cosets, say

H =⊔

g∈G/H

gH

where by t we mean a disjoint union. Since H is open, so are gH, and we canrearrange this into

H = G \(⊔g 6=1

gH),

so H is the complement of some arbitrary union of open sets, so it is is thecomplement of an open set, so it is closed.

(iv) As above, but this time we need finite index to ensure we get a finiteintersection.

4Also known as a Hausdorff space, meaning that for any two distinct points we can findneighbourhoods of both that do not intersect.

5Meaning we can separable not only two distinct points by neighbourhoods, but any pointand any closed set.


(v) The idea is similar: if G ⊃ H ⊃ J , with J open, then we can write

H =⋃h∈H

hJ

with hJ open, so H is a union of open sets, so open.

Back to Galois groups G = Gal(K/k) in particular.

Proposition 28.1.6. The subgroup{

1}

is closed. Hence any single point inG is closed.

Proof. Recall {1}

=⋂λ∈Λ

Nλ.

where Nλ is open. Open subgroups are closed, as discussed above, so Nλ isclosed. Hence

{1}

is an arbitrary intersection of closed sets, so is closed.

In the special case where K/k is finite Galois, G = Gal(K/k) is a finitegroup. This means that for any g ∈ G,

{g}c

= G \{g}

is closed, because it is

finite, so{g}

is also open. Hence G is a discrete group (i.e., its topology isdiscrete—every set is open and closed.).

Theorem 28.1.7. Let K/k be infinite Galois, with G = Gal(K/k). Then wehave

(i) A subgroup H < G is open if and only if there exists K ⊃ F ⊃ k, [F :k] <∞, with H = Gal(K/k).

(ii) A subgroup H < G is closed if and only if H =⋂α Uα where Uα are open

subgroups.

Proof. (i) For the forward direction, there exists Nλ such that Nλ ⊂ H sinceH is open. We have the situation

K{

1}

Kλ Nλ

F H open

H ′ = Gal(K/F )

k G

finite Galois


If we focus on the lower part of this, we can reframe it as

Kλ

{1}

H/Nλ

F H ′/Nλ

k G/Nλ

finite Galois

Then H = H ′ because H/Nλ = H ′/Nλ, since both come from finite Galoisextensions, where the correspondence is one-to-one always.

For the converse direction, things fall out pretty much immediately. We have

K{

1}

F = Kλ Nλ

F H = Gal(K/F )

k G

finite

so H contains an open subgroup, meaning that H is open.

(ii) For the forward direction we use the following lemma, to be proved mo-mentarily:

Lemma 28.1.8. Let H be asubgroup of G. Then H =⋂λ∈ΛHNλ.

With this, it follows immediately that if H is closed,

H = H =⋂λ∈Λ

HNλ,

where HNλ is open.

For the converse,

H =⋂α

Uα,

where Uα are open subgroups, meaning they’re closed, so H is an arbitraryintersection of closed sets, so it is closed.

We now prove the above lemma:

Proof of lemma. We have σ ∈ H if and only if for all λ ∈ Λ, σNλ ∩H 6= ∅, ifand only if for all λ ∈ Λ, there exists nλ ∈ Nλ and hλ ∈ H such that σnλ = hλ.This is equivalent to σ ∈ HNλ (using the inverse property) for all λ ∈ Λ, whichis true if and only if σ ∈

⋂λ∈ΛHNλ.

INFINITE GALOIS CORRESPONDENCE 75

Lecture 29 Infinite Galois Correspondence

29.1 Infinite Galois extensions

In the last theorem, note how the intermediate field F/k is finite Galois, meaningthat it is finite and separable. The Primitive element theorem tells us thisis simple, so F = k(α) for some α ∈ K. In other words, if H < G is asubgroup, then H is open if and only if there exists some α ∈ K such thatH = Gal(K/k(α)).

Consequently:

Proposition 29.1.1. Let K/k be infinite Galois. Let G = Gal(K/k). LetK ⊃ F ⊃ k by any intermediate field. Then

Γ(F ) = Gal(K/F ) =⋂α∈F

Γ(k(α))

is closed.

Proof. It is clear: Γ(k(α)) is open, but open subgroups are closed, so this is anarbitrary intersection of closed sets, and so closed.

Let Ho be the set of all closed subgroups of G. Let F be the set of allintermediate fields. Then as before we have a map Γ: F → Ho taking theGalois group and a map back Φ: Ho → F taking the fixed field.

Theorem 29.1.2 (Krull). Let K/K be (infinite) Galois, and G = Gal(K/k).Let H < G be a subgroup. Then

(i) H is dense in Γ(Φ(H)) (i.e., H = Γ(Φ(H))).

(ii) H is closed if and only if H = Γ(Φ(H)).

Proof. First note how (ii) follows immediately from (i), so let us focus on thefirst one.

It suffices to show that, for any σ ∈ Γ(Φ(H)), and any neighbourhood σNλof σ, σNλ contains an element of H (so σNλ ∩H 6= ∅).

We have the following, quite complicated, diagram:

K{

1}

KλM Mλ

Kλ M = Φ(H) H Nλ

Γ(Φ(H))

k G

finite Galois

Date: December 5th, 2019.

INFINITE GALOIS CORRESPONDENCE 76

There are some comments to be made here. Any neighbourhood of 1 lookslike Nλ, so shifting it σNλ is a neighbourhood of σ. On the other hand, Nλcorresponds to an extension Kλ/k that is finite Galois.

Moreover, we can lift this to KλM/M , whence this is also finite Galois, andso this corresponds to some Mλ ⊂ Nλ.

Now consider a map ϕ : Γ(Φ(H)) → Gal(KλM/M) defined by ρ 7→ ρ∣∣KλM

(note how Γ(Φ(H)) = Gal(K/M), so this makes sense). Then ϕ is a surjectivehomomorphism, since automorphisms can be lifted to K, being an algebraicextension.

The fixed field of ϕ(H) is M , so ϕ(H) = Gal(KλM/M). To see this, considerthe Galois correspondence

K{

1}

KλM

M H

and restrict the correspondence to the lower half of the diagram.Consequently ϕ(σ) = σ

∣∣KλM

= ϕ(h) ∈ Gal(KλM/M) for some h ∈ H. So

we have h : K → K, and we can lift σ : K → K, and hence σ−1 ◦h ∈ Gal(K/k).On the other hand, σ−1 ◦ h

∣∣KλM

= IdKλM , so σ−1 ◦ h ∈ Gal(K/KλM) =Mλ ⊂ Nλ.

Hence h ∈ σNλ, but h ∈ H, so H ∩ σNλ 6= ∅.

This finishes our Galois correspondence:

Theorem 29.1.3 (Galois correspondence). Let K/k be (infinite) Galois and letG = Gal(K/k). Let F be the set of all intermediate fields and let Ho be the setof all closed subgroups of G. Then Γ: F → Ho and Φ: Ho → F are one-to-oneand onto correspondences.

In particular, if we have the diagram

K{

1}

F H

k G

then F/k is Galois if and only if H is a closed normal subgroup of G.

Index

abeliangroup, 1

abelian closure, 46algebraic, 12

extension, 13algebraic closure, 18algebraically closed, 16algebraically independent, 25alternating group, 3automorphism group, 21

centraliser, 4centre, 44chain, 20character, 56composite field, 16conjugacy class, 4conjugate, 43conjugation, 43continuous, 70coset

left, 1right, 1

cyclotomic extension, 51cyclotomic field, 51cyclotomic polynomial, 54

degreeof extension, 12

discrete group, 73discriminant, 47, 49distinguished class, 25division ring, 8dual basis, 58

embedding, 19, 20extension

abelian, 44cyclic, 44nilpotent, 44solvable, 44

field, 8field extension, 12

simple, 14Frobenius automorphism, 32fundamental system, 70

Fundamental theorem of Galois theory,39

Galois extension, 39Galois group, 40Gauss sum, 52group, 1

cyclic, 2simple, 3

group action, 3

Hausdorff space, 72Hermite polynomial, 55homomorphism

of groups, 1of rings, 8

ideal, 8maximal, 9prime, 9principal, 9

index, 1inseparable degree, 35integral domain, 8isomorphism

of groups, 1

kernel, 1, 8

lift, 20

minimal polynomial, 14monoid, 56

nilpotent, 45non-degenerate

bilinear form, 58normal, 23normal closure, 31normaliser, 5

open set, 70orbit, 4

p-group, 5p-subgroup, 5perfect field, 38permutations

even, 3

77

INDEX 78

PIDsee principal ideal domain, 9

prime field, 41primitive element, 29primitive element theorem, 29principal ideal domain, 9purely inseparable, 35

element, 36extension, 36

purely transcendental extension, 25

quadratic nonresidue, 52quadratic residue, 52

radical extensiongeneral, 62height 1, 62

regular, 72ring, 7

commutative, 7with identity, 7

root of unity, 50

semigroup, 56separable

element, 28extension, 28, 29polynomial, 28

separable closure, 31separable degree, 26simple extension, 29solvable, 45solvable by radicals, 63solvable group, 64splitting field, 23stabiliser, 4subgroup

normal, 2Sylow p-subgroup, 6Sylow theorems, 6–7symmetric group, 2

topological group, 69topological space, 70topology, 70

basis, 71trace, 56transcendental, 12transcendental basis, 25transposition, 3

unit, 8

zero divisor, 8

lecture notes in galois theory - math.wsu.edu

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