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University ofCambridge

Mathematics Tripos

Part II

Galois Theory

Michaelmas, 2017

Lectures byC. Brookes

Notes byQiangru Kuang

mailto:[email protected]

Contents

Contents

0 History 2

1 Field Extensions 31.1 Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Digression on (non-)constructability . . . . . . . . . . . . . . . . 7

2 Separable, Normal and Galois Extensions 152.1 Separable Extension . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Trace & Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Fundamental Theorem of Galois Theory 253.1 Galois Group of Polynomials . . . . . . . . . . . . . . . . . . . . 283.2 Galois Theory of Finite Fields . . . . . . . . . . . . . . . . . . . . 31

4 Cyclotomic and Kummer Extensions, Cubics and Quartics, So-lution by Radicals 334.1 Cyclotomic Extensions . . . . . . . . . . . . . . . . . . . . . . . . 334.2 Kummer Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.3 Cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.4 Quartics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.5 Solubility by Radicals . . . . . . . . . . . . . . . . . . . . . . . . 44

5 Final Thoughts 505.1 Algebraic Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.2 Symmetric Polynomials & Invariant Theory . . . . . . . . . . . . 53

1

0 History

0 HistoryThe primary motivation of this course is to study polynomial equations in onevariable and to consider whether there is a formula involving roots, i.e. solutionby radicals.

Quadratics have been well understood since long long time ago and we havealready studied them at school. For cubics and quartics, it took long time beforepeople discovered how to solve them by radicals. In 1770 Lagrange studied whyit worked. However, In 1799 Ruffini claimed that there were some quintics thatcan’t be solved by radicals, i.e. there is no general formula, although his proofhad gaps.

In 1824, Abel (1802 – 1829) first accepted proof of insolubility using existingideas about permutations of roots. In 1831, Galois (1811 – 1832) first explainedwhy some polynomials are soluble by radicals and others are not. He madeuse of a group of permutations of the roots and he realised in particular theimportance of normal subgroups.

Galois’ work was not known in his lifetime — it was only published byLiouville in 1846 who realised that it fit in well with the work of Cauchy onpermutations.

Galois had submitted his work for various competitions and for entry intothe École Polytechnique. He died in a duel, leaving a 6 1

2 page letter indicatinghis thoughts about future development.

Most of this course is Galois Theory but it is presented in a slightly moremodern way in terms of field extensions.

Recall from IB Groups, Rings and Modules that if f(t) is an irreduciblepolynomial in K[t] for some field K then K[t]/(f(t)) is a field where f(t) is theideal generated by f(t). This is the starting point of this course.

This course requires quite a lot of IB Groups, Rings and Modules but nocontent about module is required except in one place where it is useful to knowthe Structural Theorem of Finitely Generated Abelian Groups.

2

1 Field Extensions

1 Field Extensions

1.1 Field Extensions

Definition (field extension). A field extension K ≤ L is the inclusion of afield K into another field L, with the same 0 and 1 and the restriction of +and · in L to K gives + and · in K.

Example. Q ≤ R,R ≤ C,Q ≤ Q(√2) = {λ + µ

√2 : λ, µ ∈ Q} and Q(i) =

{λ+ µi : λ, µ ∈ Q} ≤ C are all field extensions.

Suppose K ≤ L is a field extension. Then L is a K-vector space withaddition given by the field and scalar multiplication given by the multiplicationin the field L.

Definition (degree of extension). The degree of L over K is dimk L, thedimension of the K-vector space L. It is denoted by |L : K|. It may or maynot be finite.

Definition (finite extension). If |L : K| < ∞ the extension is finite. Oth-erwise it is infinite.

Example.

1. |C : R| = 2 since {1, i} is a basis.

2. Similarly |Q(i) : Q| = 2.

3. Q ≤ R is an infinite extension.

Theorem 1.1 (Tower Law). Suppose K ≤ L ≤ M are field extensions.Then

|M : K| = |M : L||L : K|.

Proof. Assume |M : L| < ∞, |L : K| < ∞. Then we take an L-basis {fi}bi=1

and a K-basis {ej}aj=1.Now take m ∈ M . m =

∑bi=1 µifi for some µi ∈ L. For each µi, we can

write µi =∑∞

j=1 λijej for some λij ∈ K. Thus

m =

b∑i=1

a∑j=1

λijfiej

so {fiej} span M .To show linear independence, it suffices to show that if m = 0 then each of

the λij is zero. When m = 0, the linear independence of fi forces each µi to bezero. Then the linear independence of ei forces λij to be zero as required.

The proof for infinite extensions is omitted. Observe (not very rigorously)that if M is an infinite extension of L then it is an infinite extension of K, andif L is an infinite extension of K then the larger field M must also be an infiniteextension of K.

3

1 Field Extensions

Example. ConsiderQ ≤ Q(

√2) ≤ Q(

√2, i).

Q(√2) has {1,

√2} as a Q-basis. Q(

√2, i) has {1, i} as a Q(

√2)-basis. Now

Q(√2, i) has basis {1,

√2, i, i

√2} over Q. Thus

|Q(√2, i) : Q| = 4 = 2 · 2 = |Q(

√2, i) : Q(

√2)||Q(

√2) : Q|.

What are the intermediate fields between Q and Q(i,√2)? Obviously we

have Q(√2),Q(i) and Q(i

√2). Are there any more?

To answer, this, the Galois correspondence arising in the Fundamental The-orem of Galois Theory gives an order-reversing bijection between the lattice ofintermediate subfields and the subgroups of a group of ring automorphisms ofthe extension field (Q(i,

√2) here) that fix the smaller field element-wise.

Ring automorphisms of Q(i,√2) that fixes Q include the identity e :

√2 7→√

2, i 7→ i and complex conjugation g :√2 7→

√2, i 7→ −i. Notice that i and −i

play the same role in the field Q(i,√2) — they are both roots of t2 + 1 = 0.

There is another automorphism

h :√2 7→ −

√2

i 7→ i

which switches the roots of t2 − 2 = 0. Notice that the composition

gh :√2 7→ −

√2

i 7→ −i

These form a group of order 4, which equals to |Q(i,√2) : Q|.

Q(i,√2)

Q(√2) Q(i) Q(i

√2)

Q

22

2

2 22

{e}

{e, g} {e, h} {e, gh}

{e, g, h, gh}

2 22

22

2

The recipe for producing an intermediate subfield from a subgroup is totake the elements of Q(i,

√2) which are fixed by all elements of the subgroup.

For example, Q(i√2) is the field of elements fixed by both e and gh. This

correspondence doesn’t always work for all finite field extensions: it only worksfor Galois extensions.

In the correspondence, normal extensions correspond to normal subgroups.In this example all subgroups are normal and hence the extensions are normal.

We will also prove Primitive Element Theorem, which in the context of finiteextensions of Q tells us that they are necessarily of the form Q(α) for some α.For example, Q(i,

√2) = Q(i+

√2).

We now do a review of IB Groups, Rings and Modules. Suppose K ≤ L isa field extension. Take α ∈ L. Define

Iα = {f ∈ K[t] : f(α) = 0}.

4

1 Field Extensions

Definition (algebraic, transcendental). α is algebraic over K if Iα 6= {0}.Otherwise α is transcendental over K.

Definition (algebraic extension). L is algebraic over K if for all α ∈ L, αis algebraic over K.

Remark. Iα is the kernel of the ring homomorphism

K[t]→ L

f 7→ f(α)

evaluation at α. It follows that Iα is an ideal of K[t].

Example.

1.√2 is algebraic over Q since it is a root of t2 − 2.

2. π is transendental over Q.

Lemma 1.2. Let K ≤ L be a finite field extension. Then L is algebraicover K.

Proof. Let |L : K| = n. Take α ∈ L. Consider 1, α, α2, . . . , αn. These must belinearly dependent in the n-dimensional K-vector space L so

n∑i=0

λiαi = 0

for some λi ∈ K not all zero. Then α is a root of f(t) =∑n

i=0 λiti. Thus α is

algebraic over K.

The non-zero ideal Iα (where α is algebraic over K) is principal since K[t]is a PID. Then

Definition (minimal polynomial). Iα = (fα(t)) and fα(t) can be assumedto be monic. Such a monic fα(t) is the minimal polynomial of α over K.

Remark. Multiplication by α within the field L gives a K-linear map L→ L,an automorphism of L if α 6= 0.

In IB Groups, Rings and Modules, we proved that the minimal polynomialof a linear map is unique.

Example.

1. The minimal polynomial of√2 over Q is t2 − 2.

2. The minimal polynomial of√2 over R is t−

√2.

Lemma 1.3. Suppose K ≤ L is a field extension, α ∈ L and α is algebraicover K. Then the minimal polynomial fα(t) of α over K is irreducible inK[t] and so Iα is a prime ideal.

5

1 Field Extensions

Proof. Suppose not and fα(t) = p(t)q(t). We must show that either p(t) orq(t) is a unit in K[t]. Note that 0 = f(α) = p(α)q(α) and so p(α) = 0 orq(α) = 0. Assume wlog p(α) = 0. Thus p(t) ∈ Iα and so p(t) = fα(t)r(t) sinceIα = (fα(t)). Thus fα(t) = fα(t)r(t)q(t) and so r(t)q(t) = 1 in K[t]. Thus q(t)is a unit as required.

Recall that irreducible elements in an integral domain are prime and generateprime ideals of K[t]. Thus Iα is a prime ideal.

Definition (generated field). Suppose K ≤ L is a field extension and α ∈ L.K(α) is defined to be the smallest subfield of L containing K and α. It iscalled the field generated by K and α.

Definition (simple extension). L is a simple extension if L = K(β) forsome β ∈ L.

Given α1, . . . , αn ∈ L, K ≤ L, K(α1, . . . , αn) is the smallest intermediatefield containing α1, . . . , αn. It is the field generated by K and α1, . . . , αn. Onthe other hand, K[α] = im(K[t] → L, f 7→ f(α)) is the ring generated by Kand α.

Theorem 1.4. Suppose K ≤ L is a field extension and α ∈ L is algebraic.Then

1. K(α) = K[α] = im(f 7→ f(α)).

2. |K(α) : K| = deg fα(f) where fα(t) is the minimal polynomial of αover K.

Proof.

1. Clearly K[α] ≤ K(α). We need to show that if 0 6= β ∈ K[α] then it isa unit in K[α] and so K[α] is a field. Let β = g(α) for some g(t) ∈ K[t].Since β = g(α) 6= 0, g(t) /∈ Iα = (fα(t)). Therefore fα(t) - g(t). Sincefα(t) is irreducible and K[t] is a PID, there exists r(t), s(t) ∈ K[t] with

r(t)fα(t) + s(t)g(t) = 1 ∈ K[t].

Thus s(α)g(α) = 1 ∈ K[α] and thus β = g(α) is a unit as required.

2. Let n = deg fα(t). We will show that {1, α, . . . , αn−1} is a K-vector spacebasis of K[α].

• spanning: suppose

fα(t) = tn + an−1tn−1 + · · ·+ a0

with ai ∈ K. Then

αn = −an−1αn−1 − · · · − a0.

This implies that αn is a linear combination of {αi}n−10 . An easy

induction shows that am for m ≥ n is likewise a linear combinationof {αi}n−1

0 . Spanning thus follows.

6

1 Field Extensions

• linear independence: suppose there is a relation∑n−1

i=0 λiαi = 0. Let

g(t) =∑n−1

i=0 λiti. Since g(α) = 0, we have g(t) ∈ Iα = (fα(t)).

Thus g(t) = 0 or fα(t) | g(t). The latter is not possible by degreeconsideration. Thus g(t) = 0 ∈ K[α] and all the λi’s are zero.

Corollary 1.5. If K ≤ L is a field extension and α ∈ L then α is algebraicover K if and only if K ≤ K(α) is finite.

Proof. Straightforward by combining previous results.

Corollary 1.6. Let K ≤ L be a field extension with |L : K| = n and α ∈ L.Then deg fα(t) | n.

Proof. By Tower Law on K ≤ K(α) ≤ L, we deduce that |K(α) : K| divides|L : K|. The result follows from the corollary above.

1.2 Digression on (non-)constructabilityIn schedules it mentions “other classical problems” and we are in a position totackle some of these using Corollary 1.6.

A classical problem from Greek geometry concerns the existence or otherwiseof constructions using ruler and compasses. Here “ruler” means single unmarkedstraightedge. If you’re an expert you can divide a line between two points intoarbitrarily many equal segments, bisect an angle and produce parallel lines. Inaddition, given a polygon, you can produce a square of the same area or doublethe area. However, you cannot

1. duplicate the cube (i.e. given a cube you cannot produce a cube of doublethe volume);

2. trisect the angle π/3;

3. square a circle (i.e. given a circle you cannot construct a square of thesame area).

Assume we are given a set P0 of points in R2,

• ruler operation draws a straight line through any points in P0,

• compass operation draws a circle with centre being a point in P0 and radiusbeing distance between a pair of points in P0.

Definition (constructible number). The points of intersection of any twodistinct lines or circles drawn using these operations are constructible in onestep from P0.

A point r ∈ R2 is constructible from P0 if there is a finite sequencer1, r2, . . . , rn = r such that ri is constructible in one step from P0∪{r1, . . . , ri−1}.

Exercise. Construct the midpoint of a line between two points.

7

1 Field Extensions

Let K0 be the subfield of R generated by Q and the coordinates of points inP0. Let r0 = (xi, yi). Set Ki = Ki−1(xi, yi). Then

K0 ≤ K1 ≤ · · · ≤ Kn ≤ R.

Lemma 1.7. xi, yi are both roots in Ki of quadratic polynomials in Ki−1[t].

Proof. There are three cases:

• line intersects line.

• circle intersects circle,

• circle intersects circle.

For the line intersecting circle case, the line passes through A = (p, q) andB = (r, s) and the circle centres at C = (t, u) and has radius w. Then theequation of line is

x− pr − p

=y − qs− q

and the equation of circle is

(x− t)2 + (y − u)2 = w2.

Solving gives coordinates of the intersection satisfying quadratic polynomialsover Ki−1.

The other cases are similar.

Theorem 1.8. If r = (x, y) is constuctible from a set P0 of points in R2

and if K0 is the subfield of R generated by Q and the coordinates of thepoints in P0 then the degrees |K0(x) : K0| and |K0(y) : K0| are powers of 2.

Proof. Use the previous notation Ki = Ki−1(xi, yi). By Tower Law,

|Ki : Ki−1| = |Ki−1(xi, yi) : Ki−1(xi)||Ki−1(xi) : Ki−1|.

But the previous lemma tells us that |Ki−1(xi) : Ki−1| = 1 or 2, dependingon whether the quadratic polynomial satisfied by the point of intersection isirreducible or not. Similarly yi satisfies a quadratic polynomial over Ki−1,and hence over Ki−1(xi) and so Ki−1(xi, yi) : Ki−1(xi)| = 1 or 2. Therefore|Ki : Ki−1| = 1, 2 or 4 (but in fact 4 does not happen). Therefore by TowerLaw,

|Kn : K0| = |Kn : Kn−1| · · · |K1 : K0|is a power of 2.

If r = (x, y) is contrucible from P0 then x, y ∈ Pn for some n and therefore

K0 ≤ K0(x) ≤ Kn

K0 ≤ K0(y) ≤ Kn

and the Tower Law again gives that |K0(x) : K0| and |K0(y) : K0| are powersof 2.

To use this for proofs about non-constructibility we need to be reasonablyexpert at working out minimal polynomials. Recall from IB Groups, Rings andModules

8

1 Field Extensions

Theorem 1.9 (Gauss’ Lemma). Let f(t) be a primitive integral polynomial.Then f(t) is irreducible in Q[t] if and only if it is irreducible in Z[t].

Theorem 1.10 (Eisenstein’s criterion). Let f(t) = antn+an−1t

n−1+ · · ·+a0 ∈ Z[t]. Suppose there is a prime p such that

• p - an,

• p | an−1, p | an−2, . . . , p | a0,

• p2 - a0,

then f(t) is irreducible in Z[t].

Example. Suppose p is prime. Then

f(t) = tp−1 + tp−2 + · · ·+ 1

is irreducible over Q by considering f(t+ 1) and using p as the prime in Eisen-stein.

Another method to determine reducibility is to consider an integral polyno-mials f(t) (mod p). If f(t) is reducible in Z[t] then it is reducible over Z/pZ. Soif we find a prime p such that f(t) (mod p) is irreducible then f(t) is irreduciblein Z[t].

Example. t3 + t+ 1 is irreducible mod 2: if it were reducible it would have alinear factor and so the polynomial would have a root mod 2, but 0, 1 are notroots. So t3 + t+ 1 is irreducible in Z[t] and hence irreducible in Q[t].

Remark. On example sheet we will meet an irreducible polynomials in Z[t]which is reducible mod p for all primes p.

Now back to construcibility.

Theorem 1.11. The cube cannot be duplicated by rulers and compasses.

Proof. The problem amounts to whether given a unit distance one can constructpoints distance α apart where α satisfies t3 − 2 = 0. Starting with pointsP0 = {(0, 0), (1, 0)}. Can we produce (α, 0)? No. If we could, |Q(α) : Q| wouldbe a power of 2. We show |Q(α) : Q| = 3: since α satisfies t3 − 2, whichby Eisenstein is irreducible over Z and hence over Q. Thus it is the minimalpolynomials of α over Q.

Theorem 1.12. The circle cannot be squared using ruler and compasses.

Proof. Starting with (0, 0) and (1, 0), can we construct (√π, 0) so that we have

a square of side length√π, and hence the area of the square equals to the are

of the unit circle?The answer is no since π and hence

√π is transcendental over Q.

Now return to further theory develpment.

9

1 Field Extensions

Lemma 1.13. Let K ≤ L be a field extension. Then

1. α1, . . . , αn ∈ L are algebraic over K if and only if K ≤ K(α1, . . . , αn)is a finite extension.

2. If K ≤ M ≤ L is such that K ≤ M is finite, then there existsα1, . . . , αn ∈ L such that K(α1, . . . , αn) =M .

Proof.

1. We know α is algebraic over K if and only if K ≤ K(a) is a finite extension.Suppose for each i, αi is algebraic over K, hence over K(α1, . . . , αi−1) andso |K(α1, . . . , αi) : K(α1, . . . , αi−1)| <∞. By Tower Law applied to

K ≤ K(α1) ≤ · · · ≤ K(α1, . . . αn),

we get |K(α1, . . . , αn) : K| <∞.Conversely, consider

K ≤ K(αi) ≤ K(α1, . . . , αn).

Then the Tower Law says that if |K(α1, . . . , αn) : K| < ∞ then |K(αi) :K| <∞ and so αi is algebraic over K.

2. If |M : K| = n then M is an n-dimensional K-vector space so thereexists a K-basis {α1, . . . , αn} of M . Then K(α1, . . . , αn) ≤M . However,any elements of M is a K-linear combination of α1, . . . , αn and so lies inK(α1, . . . , αn) so equality.

Definition (K-homomorphism). Suppose K ≤ L,K ≤ L′ are field exten-sions. A K-homomorphism φ : L → L′ is a ring homomorphism such thatφ|K = id.

Notation. Let

HomK(L,L′) = {K-homomorphisms L→ L′}.

A K-homomorphism φ : L→ L′ is a K-isomorphism if it is a ring isomorphism.

Notation. AutK(L) = {K-isomorphism L→ L′} which has a group structure.

Lemma 1.14. Suppose K ≤ L,K ≤ L′ are field extensions. Then

1. any K-homomorphism φ : L→ L′ is injective and K ≤ φ(L) is a fieldextension;

2. if |L : K| = |L′ : K| < ∞, then a K-homomorphism φ : L → L′ is aK-isomorphism.

Proof.

10

1 Field Extensions

1. L is a field and kerφ is an ideal of L. Note 1 7→ 1 and so kerφ 6= L, sokerφ = 0. Therefore φ(L) is a field and K ≤ φ(L) is a field extension.

2. φ is an injective K-linear map and so |φ(L) : K| = |L : K|, which, byconsidering the dimensions of K-vector spaces, is smaller than |L′ : K| =|L : K| by assumption. Thus φ : L→ L′ is a K-isomorphism. In particularif L = L′ then φ is a K-automorphism of L.

Notation. If K ≤ L is a field extension and if f(t) ∈ K[t], we denote the setof roots of f in L by Rootf (L).

Definition (splitting polynomial, splitting field). Let K ≤ L be a fieldextension and f(t) ∈ K[t]. We say f splits over L if

f(t) = a(t− α1)(t− α2) · · · (t− αn)

where a ∈ K,α1, . . . , αn ∈ L.We say L is a splitting field for f over K if L = K(α1, . . . , αn).

Remark. This is equivalent to saying that L is a splitting field for f over K ifand only if

1. f splits over L,

2. if K ≤M ≤ L and f splits over M then M = L.

Example.

1. f(t) = t3−2 over Q: Q( 3√2) is not a splitting field over Q but Q( 3

√2, ω 3√2, ω2 3

√2)

is where ω is a primitive cubic root of unity, e.g. ω = e2π3 i. Note that

Q( 3√2, ω 3√2, ω2 3

√2) = Q( 3

√2, ω).

As 3√2 satisfies t3 − 2 over Q and hence over Q(ω), we have |Q( 3

√2, ω) :

Q(ω)| ≤ 3. ω satisfies t2 + t + 1 over Q and hence over Q( 3√2) and so

|Q( 3√2, ω) : Q( 3

√2)| ≤ 2. Since t3 − 2 and t2 + t + 1 are irredcuible over

Q, |Q( 3√2) : Q| = 3 and |Q(ω) : Q| = 2 so by tower law we conclude that

equality holds, i.e. |Q( 3√2, ω) : Q(ω)| = 3 and |Q( 3

√2, ω) : Q( 3

√2)| = 2.

Q( 3√2, ω)

Q( 3√2) Q(ω)

Q

≤2

≤3

3

2

2. f(t) = (t2 − 3)(t3 − 1) over Q: the splitting field is

Q(√3,√−3, ω, ω2, 1) = Q(

√3, ω) = Q(

√3, i)

as ω = −1+i√3

2 .

11

1 Field Extensions

3. t2 − 3 and t2 − 2t− 2 both have Q(√3) as the splitting field over Q.

4. f(t) = t2 + t + 1 over F2[t]: f(t) is irreducible over F2 since it has noroots and hence no linear factors. Thus F2[t]/(t

2 + t + 1) is a field. Setα = t+ (t2 + t+ 1) ∈ F2[t]/(t

2 + t+ 1), Then F2[t]/(t2 + t+ 1) = F2(α).

The elements are 0, 1, α and 1 + α. Note that α2 = α + 1 since F2 hascharacteristic 2. Now f(t) splits over F2(α) as

f(t) = (t− α)(t− 1− α)

so F2(α) is a splitting field of f over F2.

We use the following construction to produce splitting field in general.

Theorem 1.15 (existence of splitting field). Let K be a field and f(t) ∈K[t]. Then there exists a splitting field for f over K.

Proof. Induction of deg f . If deg f = 1 then K is the splitting field of f overK. Suppose deg f > 1 and pick an irreducible factor g(t) of f(t) over K. Notethat K ≤ K[t]/(g(t)) is a field extension. Let α1 = t + (g(t)) ∈ K[t]/(g(t))so K[t]/(g(t)) = K(α1) and g(α1) = 0 in K(α1). Therefore f(α1) = 0 inK(α1) and we can write f(t) = (t − α1)h(t) in K(α1)[t]. Repeat, note thatdeg h < deg f , and we get

f(t) = (t− α1)(t− α2) . . . (t− αn)a

where a is a constant which is in K (by considering the top coefficient of f).Thus we have a factorisation of f(t) in K(α1, . . . , αn)[t]. K(α1, . . . , αn) is asplitting field for f over K.

Theorem 1.16 (uniqueness of splitting field). If K is a field and f(t) ∈K[t], then the splitting field for f over K is unique up to K-isomorphism, i.e.if there are two such splitting fields L and L′ then there is a K-isomorphismφ : L→ L′.

Proof. Suppose L and L′ are splitting fields for f(t) ∈ K[t] over K. We need toshow that there is a K-isomorphism L → L′. Suppose K ≤ M ≤ L and thereexists K ≤ M ′ ≤ L′ and a K-isomorphism ψ : M → M ′. Clearly we can takeM = K so such intermediate fields exist. Pick M so that |M : K| is maximalamong such M,M ′ and ψ.

Now we want to show that M = L and M ′ = L′. Note that if M = L thenf(t) splits over M , say

f(t) = a(t− α1) . . . (t− αm) ∈M [t].

Apply ψ we get an induced map M [t]→M ′[t] which maps f(t) to

ψ(f)(t) = ψ(a)(t− ψ(α1) . . . (t− ψ(αm))

so f(t) splits over ψ(M) = M ′. But L′ is a splitting field and L′ ≤ M ′ so wehave equality.

12

1 Field Extensions

Now suppose M 6= L and we’ll arrive at a contradition of maximality ofM . Since M 6= L, there is a root α of f(t) in L which isn’t in M . Factorisef(t) = g(t)h(t) ∈ M [t] so that g(t) ∈ M [t] is irreducible and g(α) = 0 in L.Then there exists a K-homomorphism

M [t]/(g(t))→ L

t+ (g(t)) 7→ α

The image of the map is M(α). The K-isomorphism M [t]→ M ′[t] induced byψ maps g(t) ∈M [t] to some irreducible γ(t) ∈M ′[t] so

f(t) = g(t)h(t) ∈M [t]

yieldsf(t) = γ(t)δ(t) ∈M ′[t].

We have a field extensionM ′ ≤M ′[t]/(γ(t)) and there exists anM ′-homomorphism

M ′[t]/(γ(t))→ L′

t+ (γ(t)) 7→ α′

by picking a root α′ of γ(t) in L′. However γ(t) | f(t) in M ′[t] and hence inL′[t] and so α′ is also a root of f(t) in L′.

The M ′-homomorphism gives a K-isomorphism M ′[t]/(γ(t))→M ′(α′) andwe have a K-homomorphism M(α)→M ′(α′), contradicting the maximality ofM .

Definition (normal extension). An algebraic field extension K ≤ L is nor-mal if for every α ∈ L, the minimal polynomial fα(t) of α over K splits overL.

Theorem 1.17. Let K ≤ L be a finite field extension, then K ≤ L isnormal if and only if L is the splitting field for some f(t) ∈ K[t].

Proof. Later.

Example. Let F be a finite field with |F| = m. F has characteristic p for someprime p and Fp ≤ F. Therefore m = pr for some r.

The nonzero elements of F form the multiplicative group or order n = m−1and they satisfy tn − 1 so they are roots of tn − 1. Thus tn − 1 =

∏ni=1(t− αi)

where the αi’s are the nonzero elements of F. Thus F is the splitting field fortn − 1 over Fp.

By uniqueness of splitting field, any other field with m elements is Fp-isomorphic to F. (We will later show that there exists a field of m elements.)

Theorem 1.18. Let G be a finite subgroup of the multiplicative group of afield K. Then G is cyclic.

In particular the multiplicative group of a finite field is cyclic.

13

1 Field Extensions

Proof. Let |G| = n. By the structure theorem of finite abelian groups from IBGroup, Rings and Modules,

G ∼= Cqm11× Cq

m22× · · · × Cqmr

r

with qi’s (not necessarily distinct) prime.If q = qi = qj for some i 6= j then there are at least q2 distinct solutions of

tq − 1 in K (since Cq × Cq is isomorphic to a subgroup of G). However, in afield, which is an integral domain, a polynomial of degree q has at most q roots,contradiction.

Thus all the qi’s are distinct and G is cyclic and is generated by g1 . . . grwhere each gi generates Cq

mii

.

14

2 Separable, Normal and Galois Extensions


2.1 Separable Extension

Definition (separable polynomial). Let K be a field and f(t) ∈ K[t]. Sup-pose f(t) is irreducible in K[t] and L is a splitting field of f(t) over K. Thenf(t) is separable over K if f(t) has no repeated roots in L. For general f(t),we say f(t) is separable over K if every irreducible factor in K[t] is separableover K.

All constant polynomials are deemed to be separable.

Definition (formal differentiation). If K is a field, the formal differentiationis the K-linear map

D : K[t]→ K[t]

tn 7→ ntn−1

Notation. Denote D(f(t)) by f ′(t).

Lemma 2.1. Let K be a field, f(t), g(t) ∈ K[t]. Then

(f(t)g(t))′ = f ′(t)g(t) + f(t)g′(t)

and if f(t) 6= 0, f(t) has a repeated root in a splitting field if and only iff(t) and f ′(t) have a common irreducible factor in K[t].

Proof. D is a K-linear map so we only need to check for f(t) = tn. It is left asan exercise.

Let α be a repeated root in a splitting field L. Then f(t) = (t−α)2g(t) ∈ L[t]so f ′(t) = (t − α)2g′(t) + 2(t − α)g(t) and f ′(α) = 0. Therefore the minimalpolynomial fα(t) of α in K[t] divides both f(t) and f ′(t), so it is a commonirreducible factor of f(t) and f ′(t).

Conversely, let h(t) be a common irreducible factor of f(t) and f ′(t) in K[t].Pick a root α ∈ L of h(t). Then f(α) = f ′(α) = 0. Then f(t) = (t−α)g(t) ∈ L[t]and f ′(t) = (t − α)g′(t) + g(t). Since f ′(α) = 0, we have (t − α) | f ′(t) so(t− α) | g(t). Hence (t− α)2 | f(t).

Corollary 2.2. If K is a field and f(t) ∈ K[t] is irreducible, then

1. if charK = 0 then f(t) is separable over K,

2. if charK = p > 0 then f(t) is not separable if and only if f(t) ∈ K[tp].

Proof. By Lemma 2.1, f(t) is not separable if and only if f(t) and f ′(t) havea common irreducible factor. But since f(t) is irreducible, the only possiblefactor is f(t) itself, i.e. f(t) | f ′(t). f ′(t) = 0 as it has a smaller degree. But iff(t) =

∑ni=0 ait

i then f ′(t) =∑n

i=1 iaiti−1 so f ′(t) = 0 if and only if iai = 0

for all i ≥ 1. Thus

15


1. if charK = 0, f ′(t) 6= 0 for non-constant f(t) so f(t) is separable over K,

2. if charK = p > 0, then if f ′(t) = 0 we must have iai = 0 for all i ≥ 1, i.e.f(t) is not separable if and only if f(t) ∈ K[tp].

Definition (separable element). If K ≤ L is a field extension, we say α ∈ Lis separable over K if its minimal polynomial is separable over K.

Definition (purely inseparable). If the minimal polynomial of α is fα(t) =(t − α)n = tn − αn where n is a power of charK, α is said to be purelyinseparable over K.

Definition (separable extension). Given K ≤ L, L is separable over K ifall elements of L are separable over K.

Example.

1. Let Q ≤ L be an algebraic extension. Then L is separable over Q.

2. Let L = Fp(X), the field of rational functions in X over Fp. It hasK = Fp(X

p) as a subfield. Then K ≤ L is not separable:

Proof. Observe that if f(t) = tp − Xp ∈ K[t] then f ′(t) = 0. But tp −Xp = (t − X)p ∈ L[t]. However, f(t) ∈ K[t] is irreducible: supposef(t) = g(t)h(t) ∈ K[t] ⊆ L[t], we get g(t) = (t −X)r for some 0 < r < pif the factorisation is non-trivial. But this would mean Xr ∈ K. Howeverr and p are coprime so there exist a, b ∈ Z such that ar + bp = 1, soX = (Xr)a(Xp)b ∈ K. Thus we would have X = u(Xp)/v(Xp), absurd.Thus f(t) = tp −Xp is the minimal polynomial of X over K. α is purelyinseparable over K and it follows that K ≤ L is not separable.

3. Let F be a finite field with |F| = m, a power of charF, and f(t) = tn − 1where n = m − 1. We know this is separable over Fp since we saw thatf(t) has distinct linear factors in F[t].

Remark. It is useful to have an alternative approach to separability of fieldextensions without having to check separability of minimal polynomials forall elements of the larger field. This is where we start thinking about K-homomorphisms.

Lemma 2.3. Let M = K(α) where α is algebraic over K. Let fα(t) bethe minimal polynomial of α over K. For any field extension K ≤ L, thenumber of K-homomorphisms M → L is equal to the number of distinctroots of fα(t) in L. Thus

|HomK(M,L)| ≤ deg fα(t) = |K(α) : K| = |M : K|.

16


Proof. We know that any K-homomorphism M → L is injective. Since K(α) ∼=K[t]/(fα(t)), for any root β of fα(t) in L, we can define a K-homomorphism

K[t]/(fα(t))→ L

t+ (fα(t)) 7→ β

Conversely, for any K-homomorphism φ : M → L, the image φ(α) mustsatisfy fα(φ(α)) = 0. These two maps are inverses to each other. Thus there isa one-to-one correspondence

{K-homomorphism M → L} ↔ {root of fα(t) in L}.

Example. Let K = Q, L = Q( 3√2). Then α = 3

√2 has minimal polynomial

fα(t) = t3 − 2 over Q. There is only one K-homomorphism M = K( 3√2)→ L,

i.e. the identity map.

Corollary 2.4. In the previous lemma, the number of K-homomorphismsK(α) → L equals to deg fα(t) if and only if L is large enough (so that itcontains a splitting field for fα(t)) and α is separable over K.

Proof. Trivial from Lemma 2.3.

Lemma 2.5. Let K ≤M be a field extension and M1 =M(α), where α isalgebraic over M . Let f(t) be the minimal polynomial of α over M and letK ≤ L. Let φ :M → L be a K-homomorphism. Then there is a one-to-onecorrespondence

{extension φ1 :M1 → L of φ} ↔ {root of φ(f(t)) in L}.

Remark. Lemma 2.3 is a special case where M = K and φ = idK .

Proof. f(t) ∈M [t] is irreducible so φ(f(t)) ∈ φ(M)[t] is irreducible. Any exten-sion φ1 :M1 → L of φ produces a root φ1(α) of φ(f(t)).

Conversely, given a root γ of φ(f(t)) in L,

M1 =M(α) ∼=M [t]/(f(t)) ∼= φ(M)[t]/(φ(f(t))) ∼= φ(M)(γ) ≤ L

so we get an extension φ1 of φ as required.

Corollary 2.6. Given L contains a splitting field of f(t), the number of φ1extending φ equals to the number of distinct roots of f(t) in L.

This equals to |M1 :M | if and only if α is separable over M .

Corollary 2.7. Let K ≤M ≤ N be finite field extensions and K ≤ L. Letφ :M → L be a K-homomorphism. Then the number of extensions of φ tomaps θ : N → L is less than or equal to |N :M |.

Moreover, such θ exists if L is large enough.

17


Proof. Pick α1, . . . αn so that N =M(α1, . . . , αr) and set Mi =M(α1, . . . , αi).Thus we have got

M ≤M1 ≤ · · · ≤Mr = N.

Use Lemma 2.5 repeatedly, there are at most |M1 :M | extensions φ1 :M1 → Lof φ and |Mi+1 :M | extensions φi+1 :Mi+1 → L of φi for 1 ≤ i < r so by TowerLaw the number of extensions θ : N → L of φ is less than or equal to

|Mr :Mr−1||Mr−1 :Mr−2| · · · |M1 :M | = |N :M |.

The last bit come from the proof of Lemma 2.5 since we need L to containa root.

Remark. The proof shows that the number of extensions θ of φ is |N : M | ifand only if L is large enough and αi is separable over M(α1, . . . , αi−1) for all i.

Theorem 2.8. Let K ≤ N be a field extension with |N : K| = n andN = K(α1, . . . , αi), say. Then TFAE:

1. K ≤ N is separable.

2. Each αi is separable over K(α1, . . . , αi−1).

3. If L is large enough then there are exactly n distinct K-homomorphismsN → L.

Proof.

• 1 ⇒ 2: If K ≤ N is separable then each αi is separable over K. AsK ≤ K(α1, . . . , αi−1), the minimal polynomial of αi over K(α1, . . . , αi−1)divides that over K. So if the latter has distinct roots in a splitting fieldthen the former does as well.

• 2⇒ 3: Corollary 2.7.

• 3 ⇒ 1: Suppose 3 is true and 1 is false, we shall get a contradiction.Suppose there exists β ∈ N that is not separable over K. Then there arestrictly less than |K(β) : K| K-homomorphisms φ : K(β) → L. Each φextends to at most |N : K(β)| extensions θ : N → L. Thus there arestricly less than |N : K(β)||K(β) : K| = |N : K| = n K-homomorphismsN → L. Absurd.

Corollary 2.9. A finite extension is separable if and only if it is separablygenerated.

Proof. 1⇔ 2 above.

18


Lemma 2.10. If K ≤M ≤ L are finite extensions then M ≤ L and K ≤Mare both separable if and only if K ≤ L is separable.

Proof. Example sheet.

Example. Let F be a finite field with |F| = m. Then the multiplicative groupof order n = m− 1 is cyclic. Take a generator α, then F = F(α). Since αn = 1,the minimal polynomial of α divides tn − 1.

Since tn − 1 has distinct roots (all the non-zero elements of F), the minimalpolynomial of α is separable so F = F(α) is separable over F.

Theorem 2.11 (Primitive Element Theorem). Any finite separable exten-sion K ≤ M is simple, i.e. M = K(α) for some α, which is called theprimitive element.

Proof. If K is a finite field then M is also finite. So we can take α to be generatorof the multiplicative group of M , which is cyclic.

Now assume K is an infinite field. Since K ≤ M is a finite extension,M = K(α1, . . . , αn) for some αi. It suffices to show that any field M = K(α, β)with β separable over K is of the form K(γ).

Take f(t) and g(t) to be the minimal polynomials of α and β over K respec-tively. Let L be the splitting field for f(t)g(t) over K(α, β). The distinct zerosof f(t) in L are α1 = α, α2, . . . , αa and those of g(t) are β1 = β, β2, . . . , βb. Byseparability we know b = deg g(t). Choose λ ∈ K such that all αi + λβj aredistinct (this is possible since K is infinite). Now we set γ = α+λβ (rememberα is α1 and β is β1). Let F (t) = f(γ − λt) ∈ K(γ)[t]. We have g(β) = 0 andF (β) = f(α) = 0. Thus F (t) and g(t) have a common zero. Any other commonzero would have to be βj for some j > 1. But then F (βj) = f(α + λ(β − βj)).By assumption α+ γ(β − βj) is never an αi so f(βj) 6= 0.

Now separability of g(t) says that the linear factors are all disinct. So t− βis a highest common factor of F (t) and g(t) in L[t]. However, the minimalpolynomial h(t) of β over K(γ) divides F (t) and g(t) in K(γ)[t], and hence inL[t]. This imples h(t) = t− β, and so β ∈ K(γ). Therefore α = γ−λβ ∈ K(γ).So K(α, β) ≤ K(γ). The other direction is obvious.

Exercise. In our example on page 4 we have Q ≤ Q(√2, i). We had interme-

diate subfields Q(√2),Q(i) and Q(i

√2). If we follow the procedure of the proof

of Theorem 2.11, α =√2, β = i, f(t) = t2 − 2 and g(t) = t2 + 1. Consider some√

2 + λi where ±√2 ± λi are all distinct, for example λ = 1. The proof show

that Q(√2, i) = Q(

√2 + i).

2.2 Trace & NormThese will also be used in IID Number Fields.

Definition (trace, norm). LetK ≤M be a finite field extension and α ∈M .Multiplication by a is a K-linear map θα :M →M . The trace of α over Kis

TrM/K(α) = Tr θα ∈ K

19


and the norm of α over K is

NM/K(α) = det θα ∈ K.

Note. Trace and norm depend on the ground field.

Theorem 2.12. Suppose fα(t) = ts + as−1ts−1 + · · · + a0 is the minimal

polynomial of α over K. Let r = |M : K(α)|. Then the characteristicpolynomial of θα is (fα(t))

r and

TrM/K(α) = −ras−1

NM/K(α) = ((−1)sa0)r

Proof. Regard M as a K(α)-vector space with basis β1 = 1, β2, . . . , βr. Nowtake the K-vecotr space basis 1, α, α2, . . . , αs−1 of K(α). Then {αiβj}i<s,j≤r isa K-vector space basis for M . Multiplication by α in K(α) is represented bythe matrix

A =

0 · · · −a01 0 · · · −a10 1 0 · · · −a2...

. . ....

0 0 · · · 1 −as−1

so multiplication by α in M is represented by the rs× rs matrix

A 0 · · · 00 A · · · 0...

. . ....

0 0 · · · A

whose characteristic polynomial is (fα(t))r. By inspecting the coefficients of thecharacteristic polynomial we get the trace and norm.

Theorem 2.13. Let K ≤ M be a finite separable field extension and |M :K| = n. Let α ∈ M and K ≤ L large enough so that there are n distinctK-homomorphisms σ1, . . . , σn :M → L. Then the characteristic polynomialof θα :M →M is

n∏i=1

(t− σi(α)).

Thus

TrM/K(α) =

n∑i=1

σi(α)

NM/K(α) =

n∏i=1

σi(α)

20


Proof. Let

fα(t) =

s∏i=1

(t− αi) = ts + as−1ts−1 + · · ·+ a0

be the minimal polynomial of α over K in L[t] where L is large enough thatfα(t) splits in L. There are s K-homomorphisms K(α) → L, correspondingto maps sending α to αi. Each of these extends in r = |M : K(α)| ways togive K(α)-homomorphisms M → L. However, each of such extension mappingα 7→ αi still does so. So there are r maps sending α → αi for each i. Thus ifthe n = rs distinct K-homomorphism M → L are σ1, . . . , σn then

n∑i=1

σi(α) = r(α1 + · · ·+ αs) = −ras−1 = TrM/K(α)

since the sum of roots of fα(t) is −as−1, andn∏

i=1

σi(α) = ((−1)sa0)r = NM/K(α).

Recall from Theorem 2.12 that the characteristic polynomial of θα is (fα(t))rwhere fα(t) is the minimal polynomial of α over K and r = |M : K(α)|.The characteristic polynomial is

∏sn=1(t− αi)

r where αi are the roots of fα(t)in a splitting field. We also saw that those root are σ1(α), . . . , σn(α) so thecharacteristic polynomial is

n∏i=1

(t− σi(α)).

Theorem 2.14. Let K ≤ M be a finite separable extension. We define aK-bilinear form

T :M ×M → K

(x, y) 7→ TrM/K(xy)

where xy is the product in M . Then this is non-degenerate. In particular,the K-linear map TrM/K :M → K is non-zero. Thus it is surjective.

Remark. If K ≤M is a finite extension which is not separable then TrM/K isalways zero. It follows that T is degnerate. See example sheet.

Proof. By Primitive Element Theorem, separability implies that M = K(α)for some α. We have a K-basis {αi}n−1

i=0 of K(α) where n = |M : K|. TheK-bilinear form T is represented by the matrix

A =

TrM/K(1) TrM/K(α) · · ·TrM/K(α) TrM/K(α2) · · ·· · · · · · · · ·

Let L be a splitting field of the minimal polynomial f(t) of α over K. Thenfα(t) =

∏ni=1(t − αi) with α1, . . . , αn ∈ L. The entries in A are of the form

TrM/K(α`) which is α`1 + · · ·+ α`

n by Theorem 2.13.

21


Now consider ∆ =∏

i<j(αi − αj), the Vandermonde determinant, is thedeterminant of the Vandermonde matrix

V =

1 1 · · · 1α1 α2 · · · αn

α21 α2

2 · · · α2n

.... . .

...αn−11 αn−1

2 . . . αn−1n

Observe that Aij =

∑n α

i+j−2n , Vij = αi−1

j so

(V V T )ik = VijVkj

= αi−1j αk−1

j

=∑n

αi+k−2n

= Aik

so V V T = A. Thus 0 6= D = ∆2 = |V V T | = |A|. Thus A is non-singular andtherfore the bilinear form T is non-degenerate.

Remark. We will meetD again shortly. It is the discriminant of the polynomialfα(t).

2.3 Normal ExtensionsWe met this definition before:

Definition (normal extension). An extension K ≤M is normal if for everyα ∈M , the minimal polynomial fα(t) of α over K splits over M .

Theorem 2.15. Let K ≤ M be a finite field extension. Then K ≤ M isnormal if and only if M is the splitting field for some f(t) ∈ K[t].

Proof.

• ⇒: suppose K ≤ M is normal. Pick α1, . . . , αr ∈ M such that M =K(α1, . . . , αk). Let fαi

(t) be the minimal polynomial of αi over K. Letf(t) =

∏kn=1 fαi(t).

By normality, each fαi(t) splits over M and so is f(t). M is the splitting

field of f(t) over K since if β1, . . . , βm are the roots of f(t) then M =K(β1, . . . , βm).

• ⇐: supposeM is a splitting field for f(t) overK. ThenM = K(β1, . . . , βm)where βj are the roots of f(t) over M . Take α ∈ M . Let fα(t) be theminimal polynomial of α over K. Let M ≤ L be large enough so thatfα(t) splits over L.Now consider a K-homomorphism φ :M → L: φ(βj) is also a root of f(t)and is therefore one of the βj . Injectivity of K-homomorphisms implies

22


that φ permutes βj . However, M = K(β1, . . . , βm) and so φ is determinedby the images of βj , thus φ(M) =M .However, if αi is a root of fα(t) in L, then there is a K-homomorphism

K(α)→ K(αi)

α 7→ αi

This extends by 2.7 to a K-homomorphism φ : M → L. But φ(M) = Mso αi ∈M . Thus M is normal over K.

Remark. As the same for separability, being normal is equivalent to beingnormally generated. We will show in example sheet that K ≤ L is a normal andfinite extension if and only if L = K(α1, . . . , αr) with the minimal polynomialof each adjoined element splitting over L.

Definition (K-automorphism group). Let K ≤ M be a finite extension.Its K-automorphism group is

AutK(M) = HomK(M,M).

From before we know that such K-automorphisms are isomorphisms andthus have inverses (well, it name says so).

Lemma 2.16.|AutK(M)| ≤ |M : K|.

Proof. Corollary 2.7.

Theorem 2.17. Let K ≤M be a finite field extension, then

|AutK(M)| = |M : K|

if and only if the extension is both normal and separable.

Definition (Galois extension). A finite field extension that is normal andseparable is a Galois extension.

Definition (Galois group). Let K ≤ M be a Galois extension. Then theK-automorphism group of M is the Galois group of M over K, denoted by

Gal(M/K).

Remark. Some authors use the term “Galois group” as a synonym for auto-morphism group even when the extension is not Galois.

23


Proof of Theorem 2.17. Suppose |AutK(M)| = |M : K| = n. Let M ≤ L belarge enough. The n distinct K-automorphisms φ : M → M extend to n K-homomorphisms φ : M → L and Theorem 2.8 says that M is separable overK.

For normality, pick α ∈ M with minimal polynomial fα(t) over K. TakeM = K(α1, . . . , αm) as in the proof of Corollary 2.7 with α = α1 and L = M .We only get |M : K| extensions of the inclusion K ↪→M if each inequality in theproof is an equality. In particular, we need the number of K-homomorphismsK(α1) → M to be |K(α1) : K|. But then root correspondence says we have|K(α) : K| distinct roots of fα(t) in M . Thus fα(t) splits over M .

Conversely, suppose K ≤M is separable and normal. Then for K ≤M ≤ Lwith L large enough, separability implies there are |M : K| K-homomorphismsφ : M → L by Theorem 2.8. However, K ≤ M is normal implies that it is thesplitting field for some polynomial f(t) ∈ K[t] and thus M = K(α1, . . . , αn)where f(t) =

∏ni=1(t−αi). Note that φ(αj) is also a root of φ(f(t)) = f(t), and

is therefore one of the αj . Thus φ(M) =M . Thus |AutK(M)| = |M : K|.

Remark. In the previous proof we have shown that if K ≤ M ≤ L, φ ∈HomK(M,L) and K ≤M is normal then φ(M) =M .

Example.

1. Consider Q ≤ Q(√2, i), which is Galois:

Gal(Q(√2, i)/Q) = 〈σ :

√2 7→ −

√2, τ : i 7→ −i〉 ∼= C2 × C2.

All non-identity elements have order 2.

2. Let f(t) = t3 − 2. The splitting field of f(t) over Q is Q( 3√2, ω) where

ω is a primitive cubic root of unity. Thus Q ≤ Q( 3√2, ω) is Galois with

|Gal(Q( 3√2, ω)/Q|) = |Q( 3

√2, ω) : Q| = 6. The Galois group contains

σ :3√2 7→ ω

3√2, ω 7→ ω

τ :3√2 7→ 3

√2, ω 7→ ω2 (complex conjugation)

and is generated by these two elements. It is an exercise to show thatGal(Q( 3

√2, ω)/Q) ∼= D6

∼= S3.

24

3 Fundamental Theorem of Galois Theory


Definition (fixed field). Let K ≤ L be a field extension and H ≤ AutK(L).The fixed field of H is

LH = {α ∈ L : σ(α) = α for all σ ∈ H}.

The fixed field is a field and K ≤ LH ≤ L.

Theorem 3.1 (Fundamental Theorem of Galois Theory). Let K ≤ L be afinite Galois extension. Then

1. There is a one-to-one correspondence{intermediate subfield

K ≤M ≤ L

}←→

{subgroup

H ≤ Gal(L/K)

}M 7→ AutM (L)

LH ←[ H

2. H is a normal subgroup of Gal(L/K) if and only if K ≤ LH is normalif and only if K ≤ LH is Galois.

3. If H E Gal(L/K) then the map

θ : Gal(L/K)→ Gal(LH/K)

given by restriction to LH is a surjective group homomorphism withkernel H.

Remark.

1. Observe that M ≤ L is Galois and so we could have written Gal(L/M)instead of AutM (L). To see this,

• separability: follows from Lemma 2.10.• normality: if α ∈ L then the minimal polynomial of α over M divides

the minimal polynomial of α over K. But the latter splits over L (seeexample sheet).

2. If K ≤ M is normal then the remark after proof of Theorem 2.17 says ifσ : L→ L then σ(M) = M and so we can talk about the restriction of αto M giving an automorphism of M .

Example.

1. Q ≤ Q(√2, i). We saw in example on page 4 the lattices of intermediate

subfields and subgroups Gal(Q(√2, i)/Q) ∼= C2 × C2 which is abelian.

Thus all subgroups are normal and all intermediate subfields are normalextensions of Q.

25


2. Q ≤ Q( 3√2, ω).

Q( 3√2, ω)

Q(ω) Q( 3√2) Q(ω 3

√2) Q(ω2 3

√2)

Q

32

2 2

2 33 3

1

〈σ〉〈τ〉〈σ2τ〉〈στ〉

〈σ, τ〉 ∼= D6

32

2 2

23

3 3

The subgroup H of order 3 is normal but those of order 2 are not sothe map Gal(Q( 3

√2, ω)/Q) → Gal(Q(ω)/Q), i.e. D6 → C2, generated by

conjugation has kernel 〈σ〉.

Theorem 3.2 (Artin). Let K ≤ L be a field extension and H ≤ AutK(L)be a finite subgroup. Let M = LH . Then M ≤ L is a finite Galois extensionand H = Gal(L/M).

Remark. This implies one way of Galois correspondence:

H 7→ LH 7→ Gal(L/LH) = H.

Proof. Take α ∈ L. We first show that |M(α) : M | ≤ |H|. Let {α1, . . . , αn}be the distinct images of α under H, i.e. {φ(α) : φ ∈ H}. Define g(t) =∏n

i=1(t − αi). Each φ induces an endomorphism on L[t] under which g(t) isinvariant since φ permutes the αi. Thus the coefficients lie in LH = M and sog(t) ∈M [t]. By definition g(α) = 0 since α is one of the αi. Hence the minimalpolynomial fα(t) of α over M divides g(t). Thus

|M(α) :M | = deg fα(t) ≤ deg g(t) ≤ |H|.

This step shows that α is algebraic over M and fα(t) is separable since g(t) is.Thus M ≤ L is a separable extension.

Next we show that M ≤ L is a simple extension. Pick α ∈ L with |M(α) :M |maximal. We will show that L = M(α) for this α. Suppose β ∈ L. ThenM ≤ M(α, β) is finite and separably generated and hence is a finite separableextension. By Primitive Element Theorem M(α, β) = M(γ) for some γ. Butthen

M ≤M(α) ≤M(γ)

so by the maximality of |M(α) : M |, M(α) = M(γ). Thus β ∈ M(γ) = M(α)so L =M(α).

It follows that |L :M | ≤ |H|.

26


Finally, H ≤ AutM (L) so

|L :M | = |M(α) :M | ≤ |H| ≤ |AutM (L)| ≤ |L :M |

so we must have equality throughtout, i.e. |L : M | = |AutM (L)| = |H| soM ≤ L is a finite Galois extension and H = Gal(L/M).

Theorem 3.3. Let K ≤ L be a finite field extension. TFAE:

1. K ≤ L is Galois,

2. LH = K where H = AutK(L).

Remark. The theorem allows some authors to give yet another definition of aGalois extension.

Proof.

1. 1 ⇒ 2: Let M = LH where H = AutK(L). By Artin M ≤ L is a Galoisextension. Now we have two Galois extensions, giving the equalities

|H| = |Gal(L/K)| = |L : K|= |Gal(L/M)| = |L :M |

As K ≤ LH =M , equality.

2. 2⇒ 1: Artin.

Proof of Fundamental Theorem of Galois Theory.

1. Composing the maps

H 7→ LH

M 7→ Gal(L/M)

gives H → H by Artin. Composition the other way M 7→ Gal(L/M) 7→LH where H = Gal(L/M) gives M by Theorem 3.3.

2. Take H ≤ Gal(L/K). Then for φ ∈ Gal(L/K), LφHφ−1

= φ(LH) so by 1H E Gal(L/K) if and only if φ(LH) = LH . Let M = LH . We will showthat K ≤M is normal if and only if φ(M) =M for every φ ∈ Gal(L/K):

• ⇒: remark 2 after Theorem 3.1.• ⇐: suppose φ(M) = M for all φ ∈ Gal(L/K). Pick α ∈ M and

let fα(t) be its minimal polynomial over K. We take β as a rootfor fα(t) in L (which is possible by normality). Then there is aK-homomorphism

K(α)→ K(β)

α 7→ β

This extends to a K-homomorphism φ : L → L. Since we assumeφ(M) =M , φ(α) = β ∈M . Thus K ≤M is normal.

27


Note that K ≤M is separable since K ≤M ≤ L and K ≤ L is separable.Thus K ≤M is Galois.

3. By remark 2 after Theorem 3.1, the restriction map θ : Gal(L/K) →Gal(LH/K) is well-defined. Surjectivity follows from being able to extenda K-homomorphism LH → LH ≤ L to a K-homomorphism L → L.Clearly H ≤ ker θ. However

|L : K|| ker θ|

=|Gal(L/K)|| ker θ|

= |Gal(LH/K)| by surjectivity of θ= |LH : K| since K ≤ LH is Galois

=|L : K||L : LH |

by Tower Law

Then| ker θ| = |L : LH | = |Gal(L/LH)| = |H|

by Artin. Thus ker θ = H.

3.1 Galois Group of Polynomials

Definition (Galois group). Let f(t) ∈ K[t] be a separable polynomial andK ≤ L with L a splitting field for f(t). Then the Galois group of f(t) overK is

Gal(f) = Gal(L/K).

Since L is a splitting field for f(t), L = K(α1, . . . , αn) where α1, . . . , αn arethe roots of f(t) in L. Observe that if φ ∈ Gal(L/K) it maps bijectively rootsof f(t) to itself. Thus φ permutes the αi.

Moreover if φ fixes each αi it also fixes all elements of L and so it is theidentity map. Thus Gal(f) may be regarded as a permutation group of the nroots, so in particular admitting a permutation representation in Sn.

Lemma 3.4. Suppose separable f(t) = g1(t) · · · gs(t) with gi(t) irreduciblein K[t] is a factorisation in K[t]. Then the orbits of Gal(f) acting on theroots of f(t) correspond to the factors gj(t): two roots are in the same orbitif and only if they are roots of the same gi(t).

In particular if f(t) ∈ K[t] is irreducible, there is only one orbit, i.e.Gal(f) acts transitively on the roots of f(t).

Proof. Let αk and α` be in the same orbit under Gal(f). Thus there is φ ∈Gal(f) with α` = φ(αk). But if αk is a root of gj(t) then φ(αk) is also a root ofgj(t).

Conversely if αk and α` are roots of gj(t), then

K(αk) ∼= K[t]/(gj(t)) ∼= K(α`) ≤ L

Denote by φ0 the isomorphsim K(αk)→ K(α`). φ0 extends to φ ∈ Gal(L/K).Thus αk and α` are in the same orbit.

28


Lemma 3.5. The transitive subgroups of Sn for n ≤ 5 are

n2 S2

∼= C2

3 A3∼= C3, S3

4 C4, V4, D8, A4, S4

5 C5, D10,H20, A5, S5

where H20 is generated by a 5-cycle and a 4-cycle.

Theorem 3.6. Let p be a prime and f(t) ∈ Q[t] irreducible of degree p.Suppose f(t) has exactly 2 non-real roots in C. Then

Gal(f) ∼= Sp.

Proof. Gal(f) acts on the p distinct roots of f(t) in a splitting field L of f(t) (inC). By Lemma 3.4 the irreducibility of f(t) implies that Gal(f) acts transitivelyon p roots so by orbit-stabiliser p | |Gal(f)| but

|Gal(f)| ≤ |Sp| = p!

and so Gal(f) has a Sylow p-subgroup of order p, necessarily cyclic. ThusGal(f) contains a p-cycle. Supposition that there are exactly 2 non-real rootsgives that complex conjugation yields a transposition in Gal(f). The p-cycleand the transposition generate Sp.

Example. Let f(t) = t5 − 6t+ 3 ∈ Q[t], then claim Gal(f) ∼= S5:

Proof. f(t) is irreducible by Eisenstein with p = 3. We want to show that f(t)has 3 real roots (so 2 non-real roots) and apply the previous theorem.

Now we apply knowledge from analysis:

f(−2) = −17, f(−1) = 8, f(1) = −2, f(2) = 23

and f ′(t) = 5t4 − 6 which has two real roots. Thus by Intermediate ValueTheorem there are 3 real roots while by Rolle’s Theorem there are at most 3real roots.

Definition (discriminant). Let f(t) ∈ K[t] have distinct roots α1, . . . , αn

(in a splitting field) (f(t) need not be irreducible). Set ∆ =∏

i<j(αi − αj).Then the discriminant of f is

D(f) = ∆2 =∏i<j

(αi − αj)2 = (−1)

n(n−1)2

∏i 6=j

(αi − αj).

Remark. We have already met this in the proof of Theorem 2.14.

Lemma 3.7. Let f(t) ∈ K[t] be separable of degree n with charK 6= 2.Then

Gal(f) ≤ An ⇔ D(f) is a square in K.

29


Proof. Let L be a splitting field of f(t) over K. Then D(f) 6= 0 and is fixed byall elements of G = Gal(L/K) as the latter permutes the roots. Thus D ∈ Ksince LG = K by Galois correspondence.

If σ ∈ G then σ(∆) = (sgnσ)∆ where we are regarding G as a subgroup ofSn and sgn is the signature (this is where we need charK 6= 2). Thus if G ≤ An

we got that ∆ is fixed by all σ ∈ G. Thus ∆ ∈ K = LG.On the other hand if G � An we get σ(∆) = −∆ if σ is odd and so ∆ /∈

K = LG. Finally note that if D has square roots they must be ±∆.

Example.

1. n = 2: f(t) = t2 + bt+ c = (t− α1)(t− α2) has discriminant

D(f) = (α1 − α2)2 − (α1 + α2)

2 − 4α1α2 = b2 − 4c

2. n = 3: f(t) = t3 + ct+ d has discriminant

D(t) = −4c3 − 27d2

Remark. Any general monic cubic g(t) can be put into this form by asuitable substitution f(t) = g(t1 + λ) for suitable λ. Note D(f) = D(g).

Example.

• f(t) = t3− t−1 ∈ Q[t] irreducible in Z[t] since irreducible mod 2. D(f) =−23 which is not a square in Q. Thus Gal(f) ∼= S3.

• f(t) = t3 − 3t − 1 ∈ Q[t] irredicible since irreducible mod 2. D(f) = 81which is a square so Gal(f) ∼= A3.

Now we move on to irreducible quartics. We saw that the possible Galoisgroups are

V4, A4, C4, D8, S4

with the first two being subgroups of A4. From looking at the discriminant onegets information as whether the group is one of V4, A4 or one of C4, D8, S4. Weneed further methods to pin down which group we are dealing with.

Theorem 3.8 (mod p reduction). Let f(t) ∈ Z[t] be monic of degree nwith n distinct roots in a splitting field. Let p be a prime such that f(t),the reduction of f(t) mod p, also has n distinct roots in a splitting field.Let f(t) = g1(t) . . . gs(t) be the factorisation into irreducible in Fp[t] withnj = deg gj(t). Then

Gal(f) ↪→ Gal(f)

and has an element of cycle type (n1, n2, . . . , ns).

Proof. See Remark after discussion about Galois groups over finite fields. Thefact that Gal(f̄) ↪→ Gal(f) is from IID Number Fields. Look at Tony Scholl’steaching page on Galois Theory.

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Example. Given a quartic of the form f(t) = t4 + dt + e, its discriminant isD(t) = −27d4 + 256e3. Consider f(t) = t4 − t− 1, irreducible since irreduciblemod 2 and D(f) = −283 which is not a square.

Consider mod 7,

f(t) = t4 − t− 1 = (t+ 4)(t3 + 3t2 + 2t+ 5)

the second factor is irreducible over F7 since it has no roots in F7. By Theo-rem 3.8 Gal(f) contains an element of cycle type (1, 3), i.e. a 3-cycle. We deducethat Gal(f) ∼= S4 as it is the only transitive subgroup of S4 that contains anodd permutation and a 3-cycle.

3.2 Galois Theory of Finite FieldsRecall what we already know from ??: a finite field F is of characteristic p > 0where p is a prime and |F| = pr for some r. The multiplicative group of Fis cyclic. It is a splitting field for tn − 1 over Fp where n = pr − 1. By theuniqueness of splitting fields this is unique. Observe we could also describe Fas the splitting field of tpr − t over Fp.

What we haven’t shown yet is that for any pr there is a field with |F| = pr.

Definition (Frobenius automorphism). Let F be a finite field of character-istic p. The Frobenius automorphism of F is

φp : F→ Fα 7→ αp

Remark. (α+ β)p = αp + βp since all terms in binomial expansion is divisibleby p. Also Fp is fixed under this so this is an Fp-automorphism.

Since tpr − t splits as a product of linear factors (t − α) in F, we have thatFp ≤ F is a Galois extension and so we consider G = Gal(F/Fp). It has order rsince |F : Fp| = r.

Theorem 3.9 (Galois group of finite fields). Let F be a finite field with|F| = pr. Then Fp ≤ F is a Galois extension with

Gal(F/Fp) = 〈φp〉 ∼= Cr.

Proof. It remains to show that the order of Frobenius automorphism is r. Sup-pose φsp = id. Then αps

= α for all α ∈ F. But tps − t has at most ps roots inF. So we conclude s ≥ r. Observe that φrp = id since αpr

= α for all α ∈ F.

Now apply Fundamental Theorem of Galois Theory,

{intermediate fields Fp ≤M ≤ F} ↔ {subgroups H ≤ G}

where G = Gal(F/Fp) is cyclic.But we know all about subgroups of a cyclic group with generator φp with

order r. There is exactly one group of order s for each s | r generated byφr/sp . The corresponding intermediate subfields are the fixed fields F〈φr/s

p 〉 and|F : F〈φr/s

p 〉| = s. By Tower Law |F〈φr/sp 〉 : Fp| = r/s.

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Observe that all subgroups of cyclic groups are normal and therefore allour intermediate fields are normal extensions of Fp. Thus Gal(F〈φr/s

p 〉/Fp) ∼=Gal(F/Fp)/H where H = 〈φr/sp 〉.

Corollary 3.10. Let Fp ≤M ≤ F be finite fields. Then Gal(F/M) is cyclic,generated by φnp where φp is the Frobenius map and |M | = pn and M is thefixed field of 〈φnp 〉.

Proof. Set n = r/s.

Theorem 3.11 (existence of finite fields). Let p be a prime and n ≥ 1.Then there is a field of order pn unique up to isomorphism.

Proof. Consider the splitting field L of f(t) = tpn − t over Fp. Fp ≤ L is a finite

Galois extension. However the roots of f(t) form a field, the fixed field of φnp .Set L = F and |F : Fp| = n.

Remark (mod p reduction). We will discuss in IID Number Fields that Gal(f) ↪→Gal(f) if f(t) ∈ Z[t]. We factorised f(t) = g1(t) · · · gs(t) as a product of irre-ducibles (actually gs(t) lives in p-adic integers). We knew from Lemma 3.4 thatthe orbits of Gal(f) correspond to the factorisation. We know Gal(f) is cyclicgenerated by the Frobenius map, which must have cyclic type (n1, . . . , ns) wherenj = deg gj(t).

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4 Cyclotomic and Kummer Extensions, Cubics and Quartics, Solution byRadicals

4 Cyclotomic and Kummer Extensions, Cubicsand Quartics, Solution by Radicals

4.1 Cyclotomic Extensions

Definition (cyclotomic extension). Suppose charK = 0 or p is prime wherep - m. The mth cyclotomic extension of K is the splitting field L of tm − 1.

Remark. f(t) = tm − 1 and f ′(t) = mtm−1 have no common roots and so theroots of f(t) are distinct, which are the mth roots of unity. They form a finitesubgroup µm of L× and hence by (1.28) ?? a cyclic group 〈ζ〉. Thus L = K(ζ)is a simple extension.

Definition (primitive root of unity). An element ζ ∈ µm is a primitive mthroot of unity if µm = 〈ζ〉.

Choosing a primitive mth root of unity determines an isomorphism

µm → Z/mZ.

Note that ζi is a generator of µm if and only if (i,m) = 1 and so the primitiveroots of unity correspond to elements of (Z/mZ)×, the unit group of Z/mZ.

Now consider Galois groups of cyclotomic extensions. We will see that theymust be abelian. Observe f(t) = tm−1 is separable and so the extension K ≤ Lis Galois. Let G = Gal(L/K). An element σ ∈ G sends a primitive mth rootof unity ζ to another mth root of unity ζi where (i,m) = 1. Then ζ 7→ ζi

determines a K-homomorphism

K(ζ)→ K(ζ)

which is an injective map.

Definition. Define

θ : G→ (Z/mZ)×

σ 7→ i where σ(ζ) = ζi

It is a group homomorphism since if σ(ζ) = ζi, φ(ζ) = ζj then (σφ)(ζ) =σ(ζj) = ζij .

Thus G is abelian and we regard G as a subgroup of (Z/mZ)×.

Definition (cyclotomic polynomial). The mth cyclotomic polynomial is

Φm(t) =∏

i∈(Z/mZ)×(t− ζi),

the product of the linear factors of tm − 1 corresponding to the primitivemth roots of unity.

33


Remark.

f(t) = tm − 1

=∏

i∈Z/mZ

(t− ζi)

=∏d|m

Φd(t)

Example. Take K = Q, then

Φ1(t) = t− 1

Φ2(t) = t+ 1

Φ3(t) = t2 + t+ 1

Φ4(t) = t2 + 1

and from which we can deduce that

Φ8(t) = t4 + 1

since t8 − 1 = (t− 1)(t+ 1)(t2 + 1)(t4 + 1).

Lemma 4.1.

• Φm(t) ∈ Z[t] if charK = 0 (i.e. Q ↪→ K).

• Φm(t) ∈ Fp[t] if charK = p (i.e. Fp ↪→ K).

Proof. By induction on m. If m = 1 then done.For m > 1, consider

f(t) = tm − 1 = Φm(t)∏d|md 6=m

Φd(t)

Note that∏

d|m,d 6=m Φd(t) is monic and is in Z[t] or Fp[t] by induction hypoth-esis. Now

• if charK = 0 we deduce Φm(t) ∈ Q[t] by division of polynomials. ByGauss’ Lemma it is in Z[t].

• if charK = p > 0 we deduce by division that Φm(t) ∈ Fp[t].

Lemma 4.2. The homomorphism θ : G → (Z/mZ)× defined above is anisomorphism if and only if Φm(t) is irreducible.

Proof. We know from Lemma 3.4 that the orbits of G = Gal(L/K) correspondto the factorisation of f(t) in K[t]. In particular the primitive mth roots ofunity form one orbit if and only if Φm(t) is irreducible. Then θ is surjective ifand only if Φm(t) is irreducible.

34


Theorem 4.3. Let L be the mth cyclotomic extension of finite fields F = Fq

where q = pn. Then the Galois group G = Gal(L/F) is isomorphic to thecyclic subgroup of (Z/mZ)× generated by q.

Proof. We know from Corollary 3.10 that G is generated by α 7→ αpn so

θ(G) = θ(〈α 7→ αpn

〉) = 〈pn〉 = 〈q〉 ≤ (Z/mZ)×.

Remark. If (Z/mZ)× is not cyclic then θ is not surjective for any finite fieldand Φm(t) is reducible over F.

Example. Let F = F3. Then

Φ8(t) = t4 + 1 = (t2 + t− 1)(t2 − t− 1)

so t8 − 1 factorises as a product of linear and quadratic polynomials mod 3. SoL = F9, the unique field of order 9, whose multiplicative group is isomorphic toC8. As |Gal(L/F3)| = 2, it is isomorphic to C2. But

(Z/8Z)× = {1, 3, 5, 7} ∼= C2 × C2

so θ is not surjective and Φ8(t) is reducible.

So far we have worked exclusively with finite fields with non-zero character-istic. The following theorem is a result regarding Q:

Theorem 4.4. For all m > 0, Φm(t) is irreducible in Z[t] and hence inQ[t]. Thus θ is an isomorphism and

Gal(Q(ζ)/Q) ∼= (Z/mZ)×

where ζ is a primitive mth root of unity.

Remark. We already knew this when m = p for p prime by substitution andEisenstein.

Proof. Gauss’ Lemma implies that irreducibility in Z[t] gives irreducibility inQ[t]. Lemma 4.2 says that irreducibility corresponds to surjectivity of θ. Thusit is left to show that Φm(t) is irreducible in Z[t].

Suppose not and Φm(t) = g(t)h(t) in Z[t], with g(t) irreducible and monicwith deg g(t) < degΦm(t). Let Q ≤ L be the mth cyclotomic extension and ζbe a primitive mth root and unity and also a root of g(t). Claim that if p - mand p is prime then ζp is also a root of g(t) in L:

Proof. Suppose not, then ζp is also a primitive mth root of unity (since p - m)and a root of Φm(t). The supposition implies that ζp is a root of h(t). Define

r(t) = h(tp),

then r(ζ) = 0. But g(t) is the minimal polynomial of ζ over Q and so g(t) | r(t)in Q[t]. By Gauss’ Lemma r(t) = g(t)s(t) with s(t) ∈ Z[t]. Now reduce mod p,

r(t) = g(t)s(t).

35


But r(t)− h(tp) = (h(t))p. If a(t) is any irreducible factor of g(t) in Fp[t] thena(t) | (h(t))p and so a(t) | h(t). But then (a(t))2 | g(t)h(T ) = Φm(t). So Φm(t)has a repeated root and thus tm − 1 has repeated roots mod p. Absurd.

So the claim is true. Now consider a root γ of h(t). It is also a primitivemth root of unity and γ = ζi for some i coprime with m. Factorise it asi = p1 · · · pk with pj prime and not necessarily distinct and pj - m. Apply theclaim repeatedly, we get that γ is also a root of g(t) and so Φm(t) has a repeatedroot. Absurd. Thus Φm(t) is irreducible over Q.

Definition (cyclic extension). An extension K ≤ L is cyclic if the extensionis Galois and Gal(L/K) is cyclic.

Definition (abelian extension). An extension K ≤ L is abelian if the ex-tension is Galois and Gal(L/K) is abelian.

Example.

1. We saw in Corollary 3.10 that for finite fields F ≤ L is cyclic.

2. Cyclotomic extensions are abelian.

3. Theorem 4.4 says Gal(Q(ζ)/Q) ∼= (Z/mZ)× and so if m = 8, Q ≤ Q(ζ) isa non-cyclic abelian extension.

4.2 Kummer TheoryRather than consider the mth root of unity, in this section we consider Galoisextensions K ≤ L where L is the splitting field of a polynomial of the formtm − λ where λ ∈ K.

Theorem 4.5. Let f(t) = tm − λ ∈ K[t] and charK - m. Then thesplitting field L of f(t) over K contains a primitive mth root of unity ζ andGal(L/K(ζ)) is cyclic of order dividing m. Moreover f(t) is irreducible overK(ζ) if and only if |L : K(ζ)| = m.

Remark. We have the following Galois correspondence

cyclic

L|

K(ζ)|K

1|

Gal(L/K(ζ))|

Gal(L/K)

with Gal(L/K(ζ)) E Gal(L/K). By Fundamental Theorem of Galois Theory

Gal(L/K)/Gal(L/K(ζ)) ∼= Gal(K(ζ)/K)

which is abelian.

36


Proof. Since tm − λ and mtm−1 are coprime, we know that tm − λ has distinctroots, say α1, . . . αm in the splitting field L. Thus K ≤ L is Galois. Since(αiα

−1j )m = λλ−1 = 1, the elements

α1α−11 = 1, α2α

−11 , . . . , αmα

−11

are m distinct mth roots of unity in L and so

tm − λ = (t− β)(t− ζβ)(t− ζ2β) · · · (t− ζn−1β) ∈ L[t]

where β = α1 and ζ is a primitive mth root of unity. Thus

L = K(ζ, β).

Let σ ∈ Gal(L/K(ζ)). It is determined by its action on β. Note thatσ(β) is also a root of tm − λ so σ(β) = ζj(σ)β where 0 ≤ j(σ) < m. Also ifσ, τ ∈ Gal(L/K(ζ)) then

τσ(β) = τ(ζj(σ)β) = ζj(σ)τ(β) = ζj(σ)ζj(τ)β

where the second inequality comes from the fact that ζ is fixed by τ . Thus thereis a group homomorphism

Θ : Gal(L/K(ζ))→ Z/mZσ 7→ j(σ)

Note that j(σ) = 0 only if σ is the identity and so Θ is injective. ThusGal(L/K(ζ)) is isomorphic to a subgroup of Z/mZ, so a cyclic group of or-der dividing m.

Finally,|L : K(ζ)| = |Gal(L/K(ζ))| ≤ m

with equality precisely when the action of Gal(L/K(ζ)) is transitive on the rootsand that is when tm − λ is irreducible over K(ζ) by Lemma 3.4.

Example.

1. f(t) = t6 + 3 over Q. Let ζ = −ω be a primitive 6th root of unity whereω is a primitive cubic root of unity. Then

Q(ζ) = Q(ω) = Q(1

2(1 +

√−3)) = Q(

√−3).

f(t) is irreducible over Q by Eisenstein with 3. However over Q(ζ) =Q(√−3) f(t) factorises as

f(t) = (t3 +√−3)(t3 −

√−3)

so the Galois group of the splitting field L of f(t) over Q(ζ) is a propersubgroup of Z/6Z. We have the following Galois correspondence:

L 1| | 3

Q(ζ) Gal(L/Q(ζ))| | 2Q Gal(L/Q)

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Note that

Gal(L/Q(ζ)) ∼= Z/3Z ↪→ Z/6ZGal(Q(ζ)/Q) = 〈i 7→ −i〉 ∼= Z/2Z

Let β be a root of f(t). Then the roots are

β ζβ ζ2β ζ3β ζ4β ζ5β

β ω2β ωβ−ωβ −β −ω2β

where a 3-cycle is given by permuting β, ω2β, ωβ and the orbits are shownas the last two rows above. Denote the 3-cycle σ and complex conjugationτ , since

τστ−1 = σ−1,

we get dihedral relation so Gal(L/Q) is dihedral of order 6.

2. f(t) = t5−2 over Q. It is irreducible by Eisenstein. Let L be the splittingfield of f(t) over Q and ζ be a primitive 5th root of unity. We have thefollowing Galois correspondence:

L 1| | 5

Q(ζ) Gal(L/Q(ζ))| | 4Q Gal(L/Q)

Let β be a root of f(t). Then Q ≤ Q(β) ≤ L so 5 | |L : Q| and thus5 | |Gal(L/Q(ζ))|. Since we also have Gal(L/Q(ζ)) ↪→ Z/5Z, it followsthat Gal(L/Q(ζ)) ∼= Z/5Z.We can thus deduce that f(t) remains irreducible over Q(ζ) and |Gal(L/Q)| =20. By irreducibility of f(t) over Q, Gal(L/Q) is a transitive subgroup ofS5. By Lemma 3.5, it is isomorphic to H20, the subgroup generated by a5-cycle and a 4-cycle. By Fundamental Theorem of Galois Theory,

Gal(L/Q)/Gal(L/Q(ζ)) ∼= Gal(Q(ζ)/Q) ∼= (Z/5Z)×

which is cyclic and contains a 4-cycle. Thus without a priori knowledgeof H20 we know our subgroup of S5 contains a 4-cycle.

Theorem 4.5 tells us the property of the splitting field of tm − λ. Thefollowing theorem gives its converse:

Theorem 4.6. Suppose K ≤ L is a cyclic extension with |L : K| = mwhere charK - m and K contains a primitive mth root of unity. Then thereexists λ ∈ K such that tm − λ is irreducible over K, and L is the splittingfield of tm − λ over K. If β is a root of tm − λ in L then L = K(β).

Since we are going to restate the hypothesis in the theorem many times, wegive it a name

38


Definition (Kummer extension). A cyclic extension K ≤ L with |L : K| =m where charK - m and K contains a primitve mth root of unity is aKummer extension.

We need the following to prove the theorem:

Lemma 4.7 (linear independence of group characters). Let φ1, . . . , φn beembeddings of a field K into a field L. Then there do not exist λ1, . . . , λnnot all zero such that

λ1φ1(x) + · · ·+ λnφn(x) = 0

for all x ∈ K.

Proof. Example sheet 2 Q10.

Proof of Theorem 4.6. Let Gal(G/K) = 〈σ〉 which has order m. Observe that{σi}m−1

i=0 are distinct maps L → L and we can apply linear independence ofgroup characters: there exists α ∈ L such that

β = α+ ζσ(α) + · · ·+ ζm−1σm−1(α) 6= 0

where ζ is a primitive mth root of unity.Observe that σ(β) = ζ−1β = β and so β ∈ K, the fixed field of Gal(L/K).

Also σ(βm) = σ(β)m = βm. Let λ = βm ∈ K. Then

tm − λ = (t− β)(t− ζβ) · · · (t− ζm−1β) ∈ L[t]

so K(β) is the splitting field of tm − λ over K (recall ζ ∈ K).Observe also that {σi}m−1

i=0 are distinct K-automorphisms and so |K(β) :K| ≥ m. Therefore

L = K(β) = K(ζβ).

tm − λ is the minimal polynomial of β over K and hence is irreducible.

Definition (radical extension). A field extension K ≤ L is an extension byracidals if there exists

K = L0 ≤ L1 ≤ · · · ≤ Ln = L

such that each Li ≤ Li+1 is either cyclotomic or Kummer extension.

Definition (solubility by radicals). A polynomial f(t) ∈ K[t] is soluble byradicals if its splitting field lies in an extension by radicals.

39


4.3 CubicsIn this section and the following one we assume charK 6= 2 for discussion aboutdiscriminant and charK 6= 3 for cubic Kummer extension.

We have already seen that if f(t) is a monic irreducible cubic in K[t] withL its splitting field over K,

G = Gal(f) = Gal(L/K)

is either A3 or S3 since irreducibility implies that the action on roots is transitive.Thus we have the following correspondence:

L 1| | 3

K(∆) G ∩A3

| | 1 or 2K G

where ∆2 = D(f) is the discriminant of f . But to see if we can solve f byradicals we want to make use of Theorem 4.6 and so we need to adjoin theappropriate roots of unity. So we get a bigger picture:

L(ω)

K(∆, ω) L

K(∆)

K

1 or 2

1 or 23

1 or 2

where ω is a primitive cubic root of unity. From the Tower Law |L(ω) :K(∆, ω)| = 3. Hence

Gal(L(ω)/K(∆, ω)) ∼= C3.

We can now apply Theorem 4.6 to see that

L(ω) = K(∆, ω)(β)

where β is a root of an irreducible polynomial t3 − λ ∈ K(∆, ω)[t].In fact from the proof of Theorem 4.6 we see that

β = α1 + ωα2 + ω2α3

where α1, α2, α3 are roots of f(t). Now all the extensions

K ≤ K(∆) ≤ K(∆, ω) ≤ L(ω)

are either cyclotomic or Kummer. Thus f(t) is soluble by radicals.Let’s give a try to our theory. Given an irreducible cubic

f(t) = t3 + at2 + bt+ c = (t− α1)(t− α2)(t− α3),

40


we have α1+α2+α3 = −a. But we don’t actually need that many parameters:let α′ = αi+

a3 so that α′

1+α′2+α

′3 = 0 and the α′’s are roots of the polynomial

g(t) = t3 + pt+ q and most importantly,

K(α1, α2, α3) = K(α′1, α

′2, α

′3)

so the splitting field for g(t) is the same as that for f(t). Thus we could workwith g(t) instead. The discriminant of g(t) is D(g) = −4p3 − 27q2.

Set

β = α′1 + ωα′

2 + ω2α′3

γ = α′1 + ω2α′

2 + ωα′3

then

βγ = α′21 + α′2

2 + α′33 + (ω + ω2)(α′

1α′2 + α′

1α′3 + α′

2α′3)

= (α′1 + α′

2 + α′3)

2 − 3(α′1α

′2 + α′

1α′3 + α′

2α′3)

= −3p

and so β3γ3 = −27p3. Also,

β3 + γ3 = (α′1 + ωα′

2 + ω2α′3)

3

+ (α′1 + ω2α′

2 + ωα′3)

3

+ (α′1 + α′

2 + α′3)

3︸︷︷︸=0

= 3(α′31 + α′3

2 + α′33 ) + 18α′

1α′2α

′3

= −27q

since α′31 = −pα′

1 − q and so α′31 + α′3

2 + α′33 = −3q.

Thus after some messy algebra, we find that β3 and γ3 are roots of aquadratic

t2 + 27qt− 27p3

and so are

−27

2q ± 3

√−32

√−27q2 − 4p3 = −27

2q ± 3

√−32

√D.

We can solve for β3 and γ3 in K(√−3,√D) = K(ω,∆). From here we can get

β by adjoining a cubic root of β3 and set γ = − 3pβ . Finally we solve in L(ω) for

α′1, α

′2, α

′3

α′1 =

1

3(β + γ)

α′2 =

1

3(ω2β + ωγ)

α′3 =

1

3(ωβ + ω2γ)

41


4.4 QuarticsAs with the cubics, by making a substitution of the form α′

i = αi +a4 we may

assume that the sum of the roots is zero and so the t3 term vanishes, leaving ageneral quartic polynomial of the form

f(t) = t4 + bt2 + ct+ d

= (t− α1)(t− α2)(t− α3)(t− α4)

which we assume to be monic and irreducible.Let L = K(α1, . . . , α4) be the splitting field of f(t) over K.

L 1| |M G ∩ V4 E G| |

K(∆) G ∩A4

| |K G

where M is the fixed field of the normal subgroup G ∩ V4 E G, which is annormal extension of K. By Fundamental Theorem of Galois Theory,

Gal(M/K) ∼= G/(G ∩ V4).

To determine this group concretely, we use a little knowledge from group theory:there is a group isomorphism

S4/V4 ∼= S3.

Let θ : S4 → S3 denote the quotient map, which has kernel precisely V4. There-fore θ|G : G→ S3 induces an isomorphism

G/ ker θ|G = G/(G ∩ V4) ∼= im θ|G ≤ S3.

We therefore seek a cubic for which M is the splitting field. We introduceresolvent cubic: set

x = α1 + α2

y = α1 + α3

z = α1 + α4

and so

α1 =1

2(x+ y + z)

α2 =1

2(x− y − z)

α3 =1

2(−x+ y − z)

α4 =1

2(−x− y + z)

42


ThusK(α1, . . . , α4) = K(x, y, z).

Remebering that α1 + α2 + α3 + α4 = 0,

x2 = (α1 + α2)2 = −(α1 + α2)(α3 + α4)

y2 = (α1 + α3)2 = −(α1 + α3)(α2 + α4)

z2 = (α1 + α4)2 = −(α1 + α4)(α3 + α3)

These are all distinct since for example if y2 = z2 then y = ±z so α3 = α4 orα1 = α2.

Now consider the resolvent cubic

g(t) = (t− x2)(t− y2)(t− z2) ∈ K[t].

x2, y2 and z2 are permuted by G and are fixed by G ∩ V4. Thus

K(x2, y2, z2) ≤M = LG∩V4 .

We claim that we actually have equality here:

Proof. Check D(f) = D(g) — this will be addressed on example sheet — soK(∆) ≤ K(x2, y2, z2).

Now observe that

Gal(L/(K(x2, y2, z2)) = Gal(K(x, y, z)/K(x2, y2, z2))

and inK(x2, y2, z2) ≤ K(x, y2, z2) ≤ K(x, y, z2) ≤ K(x, y, z),

each extension is of degree either 1 or 2 so |K(x, y, z) : K(x2, y2, z2)| divides 8so elements of Gal(L/K(x2, y2, z2)) have order dividing 8. But since

Gal(L/K(x2, y2, z2)) ≤ G ∩A4,

by checking the order of elements in A4 we see that

Gal(L/K(x2, y2, z2)) ≤ G ∩ V4.

Therefore by Fundamental Theorem of Galois Theory

M = K(x2, y2, z2).

Consider the coefficients of g(t): its roots x2, y2, z2 are permuted by G andso it has coefficients in K. We can actually write down these coefficients:

x2 + y2 + z2 = −2bx2y2 + x2z2 + y2z2 = b2 − 4d

xyz = −cx2y2z2 = c2

sog(t) = t3 + 2bt2 + (b2 − 4d)t− c2.

We know how to solve cubics so we can solve for x2, y2, z2, from which we cansolve for x, y, z. Finally we use formula α1 = 1

2 (x + y + z) etc to recover theroots for the quartic. Whew, done!

43


Remark. In our mapθ : S4 → S3,

we have restrictionθ|A4 : A4 → A3

∼= C3

andθ|G∩A4

: G ∩A4 → A3

soG ∩A4/(G ∩ V4) ≤ A3

which is either 1 or A3, corresponding to M/K(∆). If the resolvent cubic isirreducible then it is isomorphic to A3. Otherwise it is the trivial group.

Example. Let f(t) = t4 + 4t2 + 2. As an aside even if we didn’t study Galoistheory we could solve it. Then the resolvent cubic g(t) = t3 + 8t2 + 8t. Notethat c = 0 in f(t) so g(t) has no constant term and thus reducible. It followsthat M = K(∆).

4.5 Solubility by RadicalsNow suppose we have a Galois extension K ≤ L with

K = L0 ≤ L1 ≤ · · · ≤ Lm = L

such that each Li ≤ Li+1 is either cyclotomic or Kummer extension.Let G = Gal(G/K). There is a corresponding chain of subgroups of G

G = G0 ≥ G1 ≥ · · · ≥ Gm = 1

with Gi = Gal(L/Ki) and, by Fundamental Theorem of Galois Theory, Li =LGi . However each extension Li ≤ Li+1 is Galois and we know

Gi+1 = Gal(L/Li+1) E Gal(L/Li) = Gi

andGi/Gi+1

∼= Gal(Li+1/Li)

which is abelian if Li ≤ Li+1 is cyclotomic and cyclic if Li ≤ Li+1 is Kummer.It is the perfect time to introduce some group theory:

Definition (soluble group). A group is soluble if there is a chain of sub-groups

1 E Gm E Gm−1 E · · · E G1 E G0 = G (∗)

with Gi/Gi+1 abelian.

Remark. Note that some authors use “cyclic” in the definition. While we willprove shortly that they are equivalent for a finite group G, in general they aredifferent.

Example.

1. S3 is soluble since1 E 〈(123)〉 E S3.

44


2. S4 is soluble since1 E V4 E A4 E S4.

and A4/V4 ∼= C3, S4/A4∼= C2.

3. A5 is not soluble: we have proved in IA Groups and again in IB Groups,Rings and Modules that A5 is simple, and therefore any normal subgroupis 1 or A5. Then any chain of normal subgroups would have non-abelianquotients and thus A5 is not soluble.

Lemma 4.8. A finite group G is soluble if and only if

1 = Gm E Gm−1 E · · · E G1 E G0 = G (†)

with Gi/Gi+1 cyclic.

This says that we only have to consider cyclic extensions.

Proof. The if part is easy. For the only if part, from Structural Theorem ofFinitely Generated Abelian Groups, if A is abelian then there is a chain

0 = Ar E Ar−1 E · · · E A0 = A

with Ar/Ar+1 cyclic. Thus if we have a chain (∗) with abelian factors Gi/Gi+1,we can refine it to one of the form (†) by Third Isomorphism Theorem.

Definition (derived subgroup). The derived subgroup G′ of a group G isthe subgroup generated by all the commutators

g1g2g−11 g−1

2

for g1, g2 ∈ G.

Note that it is the subgroup generated by all such elements since it is notobvious that they are closed.

Lemma 4.9. Let K E G. Then G/K is abelian if and only if G′ ≤ K.

Proof. G/K is abelian if and only if

Kg1Kg2Kg−11 Kg−1

2 = K

for all g1, g2 ∈ G, if and only if

g1g2g−11 g−1

2 ∈ K

if and only if G′ ≤ K.

Remark. Some interesting asides:

1. In IID Representation Theory we will prove Burnside’s Theorem: if |G| =paqb with p, q distinct primes then G is soluble.

45


2. Feit-Thompson Theorem: if |G| is odd then G is soluble.

3. There is an analogue of Sylow’s Theorem due to Philip Hall: given a finitegroup G, for every m and n coprime and |G| = mn, there is a subgroupof order m if and only if G is soluble.

Definition (derived series). The derived series {G(m)} of G is defined in-ductively as

G(0) = G

G(i+1) = (G(i))′

Thus· · · E G(2) E G(1) E G(0) = G

with G(i)/G(i+1) abelian.

Lemma 4.10. Given a finite group G, G is soluble if and only if G(m) = 1for some m.

Proof. If G(m) = 1 then the derived series gives a chain of the form (∗) as inthe definition of solubility.

Conversely, if there is a chain of the form (∗)

1 E Gm E · · · E G1 E G

withGi/Gi+1 abelian then induction shows thatG(j) ≤ Gj and soG(m) = 1.

Remark. The derived series is the fastest descending chain with abelian factors.

Lemma 4.11.

1. Let H ≤ G. If G is soluble then H is soluble.

2. Let H E G. Then G is soluble if and only if H and G/H are bothsoluble.

Proof.

1. G is soluble so there is some m such that G(m) = 1. But H(m) ≤ G(m)

and so H is soluble.

2. Let H E G. Suppose G is soluble. Then by the previous line H is soluble.Let G(m) = 1, observed that

(G/H)′ = G′H/H ≤ G/H

and inductively(G/H)(j) = G(j)H/H ≤ G/H.

Thus (G/H)(m) = H/H = 1 so G/H is soluble.

46


Conversely, suppose both H and G/H are soluble, i.e. H(r) = 1 and(G/H)(s) = H/H. But from above (G/H)(s) = G(s)H/H so G(s)H = Hand thus G(s) ≤ H. Therefore

G(r+s) ≤ H(r) = 1

so G is soluble.

Example. S5 is not soluble since its subgroup A5 is not soluble.

Theorem 4.12. Let K be a field with charK = 0 and f(t) ∈ K[t]. Thenf(t) is soluble by radicals over K if and only if Gal(f) over K is soluble.

Remark. We don’t need to restrict to charK = 0. What we need to do fora particular f(t) is to avoid a finite number of bad characteristics (to avoidcharacteristics smaller than deg f(t)).

Corollary 4.13. If f(t) is a monic irreducible polynomial in K[t] withcharK = 0 and Gal(f) ∼= A5 or S5 then f(t) is not soluble by radicals.

Example. In the example after Theorem 3.6 on page 29, f(t) = t5−6t+3 ∈ Q[t]has Galois group S5 over Q: recall that it has three real roots and so complexconjugation gives a transposition. f(t) is irreducible and so 5 | |Gal(f)| andso there is a 5-cycle. Together they generate S5. Thus f(t) is not soluble byradicals.

Proof of Theorem 4.12. Suppose f(t) is soluble by radicals. Then the splittingfield of f(t) over K, L, lies in an extension of K by radicals:

K = L0 ≤ L1 ≤ · · · ≤ Lm

with each Li ≤ Li+1 either cyclotomic or Kummer.

Lemma 4.14. If K ≤ N is an extension by radicals then there existsN ≤ N ′ with K ≤ N ′ an extension of radicals and a Galois extension.

Proof. Suppose we have a sequence as above and want to extend this into aGalois extension of the same form (assume charK = 0). By Primitive ElementTheorem Lm = K(α1) for some α1. Let g(t) be the minimal polynomial of α1

over K with splitting field M . Thus M = K(α1, α2, . . . , αn) where α1, . . . , αn

are roots of g(t). There are K-homomorphisms φi :M →M,α1 7→ αi extendingK-homomorphisms K(α1)→ K(αi) ≤M . The tower

K ≤ φ1(K) ≤ φ2(L1) ≤ · · · ≤ φ1(Lm) = K(α1)

has cyclotomic or Kummer extensions at each step as before. Consider

Lm = K(α1) ≤ φ2(L1)(α1) ≤ φ2(L2)(α1) ≤ · · · ≤ φ2(Lm)(α1) = K(α1, α2).

Consider the extension φ2(Lj)(α1) ≤ φ2(Lj+1)(α1) in this tower. There are twocases:

47


• if Lj ≤ Lj+1 is cyclotomic then all the roots of unity adjoined are now inLm = K(α1) and so φ2(Lj)(αi) = φ2(Lj+1)(α1).

• if Lj ≤ Lj+1 is Kummer then we obtain Lj+1 by adjoining roots of anelement of Lj and so we obtain φ2(Lj+1) by adjoining roots of an elementin φ2(Lj). Hence we get from φ2(Lj)(α1) to φ2(Lj+1)(α1) by adjoiningroots of an element of φ2(Lj). So this is a Kummer extension.

Now continue to get a suitable chain K(α1, α2) ≤ · · · ≤ K(α1, α2, α3) etc.Thus we get a suitable chain from K to K(α1, . . . , αn) = M . Observe thatK ≤M is Galois.

Assuming this lemma and so we may assume K ≤ Lm is Galois. By Funda-mental Theorem of Galois Theory there is a corresponding chain of subgroupsGal(Lm/K). Previous discussion shows that Gal(Lm/K) is soluble. So to sumup we have K ≤ L ≤ Lm with Lm Galois so by Fundamental Theorem of GaloisTheory

Gal(L/K) ∼= Gal(Lm/K)/Gal(Lm/L)

But quotients of soluble group are soluble so Gal(L/K) is soluble.Conversely, assume charK = 0 and suppose G = Gal(f) over K is soluble.

Let L be the splitting field of f(t) over K and so |G| = |L : K| = n. Set m = n!and let ζ be a primitive mth root of unity and consider L(ζ).

L(ζ)

K(ζ) L

K

≤n

n

Our proof is similar to that used for cubics. Observe that |L(ζ) : K(ζ)| ≤ n:by Primitive Element Theorem L = K(α) for some α with minimal polynomialg(t) of degree n. Then L(ζ) = K(ζ)(α) and the minimal polynomial of α overK(ζ) divides g(t) and so has degree less than or equal to n.

Note that Gal(L(ζ)/L) is abelian since the extension is cyclotomic. ThenGal(L(ζ)/K) is soluble since the subgroup Gal(L(ζ)/L is soluble and the quo-tient group Gal(L/K) ∼= Gal(L(ζ)/K)/Gal(L(ζ)/L) is also soluble. Then thesubgroup Gal(L(ζ)/K(ζ)) is soluble. Thus there is a chain of subgroups

1 = Gm E · · · E G1 E G0 E Gal(L(ζ)/K(ζ))

with Gi/Gi+1 cyclic (since for a finite group cyclic is equivalent to abelian inthe definition of a soluble group). Now use Fundamental Theorem of GaloisTheory to get a corresponding chain of fields

L(ζ) = Km ≥ · · · ≥ K1 ≥ K(ζ)

with each Ki ≤ Ki+1 Galois with cyclic Galois group.By Theorem 4.6 all those extensions are Kummer (note all the extensions

are of degree less than or equal to n and so we have the appropriate roots ofunity, this is why we choose m = n!). Thus we have embedded L in an extensionof K by radicals.

48


Example.

1. Recall previously we had f(t) = t4+4t2+2 which is irreducible by Eisen-stein over Q. The resolvent cubic g(t) = t3 + 8t2 + 8t is reducible. Itactually has roots 0,−4± 2

√2. We thus have

L = K(√−4 + 2

√2,√−4− 2

√2)

K(−4 + 2√2,−4− 2

√2) = K(

√2)

K

4

2

so |L : K| = 8. The Galois group is a transitive subgroup of S4 of order8. It is the dihedral group.

In particular the roots ±√−2±

√2 are two conjugate pairs so complex

conjugation gives double transposition.

2. f(t) = t4+2t+2 which is irreducible by Eisenstein over Q. A quick (hmm)calculation shows that the discriminant is 101 · 42 which is not a square.Thus the Galois group contains an odd permutation.The resolvent cubic g(t) = t3−8t−4 is irreducible mod 5. Therefore thereis a 3-cycle in the Galois group. We deduce that Gal(f) = S4.

3. Finally a quintic: f(t) = t5−t−1 is irreducible over Q since it is irreduciblemod 5 and so the Galois group contains a 5-cycle and is a transitive sub-group of S5. Reduction mod 2, f(t) factorises as a product of irreduciblecubic and an irreducible quadractic

f(t) = (t3 + t2 + 1)(t2 + t+ 1)

so Gal(f) is generated by an element of cycle type (3, 2). Thus Gal(f)also contains an element of cycle type (3, 2). Then g3 is a transposition.A subgroup of S5 containing a 5-cycle and a transposition must be S5.Therefore f(t) is not solvable by radicals.

49

5 Final Thoughts

5 Final Thoughts

5.1 Algebraic Closure

Definition (algebracially closed). A field L is algebraically closed if anyf(t) ∈ L[t] splits into a product of linear factors in L[t].

Remark. This is equivalent to saying that any f(t) ∈ L[t] has a root in L, orthat any algebraic extension of L is L itself.

Definition (algebraic closure). An extension K ≤ L is an algebraic closureof K if K ≤ L is algebraic and L is algebraically closed.

Lemma 5.1. If K ≤ L is algebraic and every polynomial in K[t] splitscompletely over L then L is an algebraic closure of K.

Proof. We need to show that L is algebraically closed. Suppose L ≤ L(α) is afinite extension and fα(t) = tn + an−1t

n−1 + · · ·+ a0 is the minimal polynomialof α over L. Let M = K(a0, . . . , an−1). Then M ≤ M(α) is a finite extension.But each ai is algebraic over K and so |M : K| < ∞. Hence |M(α) : K| < ∞by Tower Law and so α is algebraic over K.

The minimal polynomial of α over K must split over L and so α ∈ L. Thusany algebraic extension of L is L itself.

Example (algebraic number). Let A = {α ∈ C : α algebraic over Q}. It isa subfield of C: if α and β are algebraic over Q then |Q(α, β) : Q| < ∞.So if γ = α + β, α − β, αβ or α/β for β 6= 0 we get Q(γ) ≤ Q(α, β) and so|Q(γ) : Q| < ∞ so γ is algebraic over Q and so γ ∈ A. Therefore A = Q is analgebraic closure of Q.

Like construction of many other objects in this course, we want to prove theexistence and uniqueness of algebraic closures. In general we need to appeal toZorn’s Lemma. This will be covered in IID Logic and Set Theory, and it fact itis equivalent to both Axiom of Choice and Well-ordering Principle.

We will do a quick ad hoc statement of Zorn’s Lemma here and relegate themain discussion to IID Logic and Set Theory.

Definition (partial order). (ζ,≤) is a partial order on a set ζ is

1. reflexive: x ≤ x for all x ∈ ζ.

2. transitive: x ≤ y and y ≤ z implies x ≤ z.

3. antisymmetric: if x ≤ y and y ≤ x then x = y.

Definition (total order). A partial order (ζ,≤) is a total order if it is total:for any x, y ∈ ζ, either x ≤ y or y ≤ x.

50

5 Final Thoughts

Definition (chain). A chain in a partially ordered set (ζ,≤) is a totallyordered subset.

Theorem 5.2 (Zorn’s Lemma). Let (ζ,≤) be a non-empty partially orderedset. Suppose that any chain has an upper bound. Then ζ has a maximalelement.

Recall a result derived from Zorn’s Lemma in IB Groups, Rings and Modules:

Lemma 5.3. Let R be a ring. Then R has a maximal ideal.

Proof. Let ζ be the set of proper ideals of R. {0} is a proper ideal so ζ isnon-empty. Let ζ be partially orderd by inclusion.

Note that an ideal I E R is proper if and only if 1 /∈ I. Any chain of properideals has an upper bound in ζ, namely the union of the chain, so by Zorn’sLemma ζ has a maximal element, i.e. a maximal ideal of R.

Theorem 5.4 (existence of algebraic closure). Any field has an algebraicclosure.

Proof. Let

ζ = {(f(t), j) : f(t) irreducible monic in K[t], 1 ≤ j ≤ deg f}.

For each pair s = (f(t), j) ∈ ζ, introduce an indeterminate Xs = Xf,j . Considerthe polynomial ring K[Xs : s ∈ ζ] and set

f̃(t) = f(t)−deg f∏j=1

(t−Xf,j) ∈ K[Xs : s ∈ ζ][t].

Let I E K[Xs : s ∈ ζ] be the ideal generated by coefficients of all the f̃(t)’s,which we denote by af,` for 0 ≤ ` < deg f . Claim that I 6= K[Xs : s ∈ ζ]:

Proof. Suppose 1 ∈ I for contradiction. We thus have a relation

b1af1,`1 + · · ·+ bNafN ,`N = 1 ∈ K[Xs : s ∈ ζ]. (∗)

Let L be a splitting field for the product f1(t) · · · fN (t).For each i, fi splits overL so write

fi(t) =

deg fi∏j=1

(t− αij).

Define a K-linear ring homomorphism which is identity on K

θ : K[Xs : s ∈ ζ]→ L

Xfi,j 7→ αij

Xs 7→ 0 otherwise

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This induces a map θ : K[Xs : s ∈ ζ][t]→ L[t]. Then

θ(f̃i(t)) = θ(fi(t))−deg fi∏j=1

θ(t−Xfi,j)

= fi(t)−deg fi∏j=1

(t− αij)

= 0

so θ(afi,j) = 0 since afi,j are the coefficients of f̃i(t). But applying θ to (∗)shows 0 = 1. Absurd.

We have thus shown that I E K[Xs : s ∈ ζ] is proper. By Zorn’s Lemma,there is a maximal ideal P E K[Xs : s ∈ ζ] containing I. Set

L1 = K[Xs : s ∈ ζ]/P,

which is a field. We thus have a field extension K ≤ L1. Claim that L1 is analgebraic closure of K:

Proof. We first show K ≤ L1 is algebraic. L1 is generated by the images xf,jof the Xf,j . However f̃(t) has coefficients in I and so its image in L1[t] is thezero polynomial. Thus in L1[t],

f(t) =

deg f∏j=1

(t− xf,j) (†)

and so f(xf,j) = 0. Thus the xf,j ’s are algebraic. Any element of L1 involvesonly finitely many of the xf,j ’s and so is algebraic over K.

Moreover, from (†) any f(t) ∈ K[t] splits completely over L1. The claimthus follows from Lemma 5.1.

Theorem 5.5. Suppose θ : K → L is a ring homomorphism and L isalgebraically closed. Suppose K ≤ M is an algebraic extension, then θ canbe extended to a homomorphism φ :M → L, i.e. φ|K = θ.

Proof. Let

ζ = {(N,φ) : K ≤ N ≤M : φ : N → L homomorphism extending θ}

and define a partial order ≤ on it by

(N1, φ1) ≤ (N2, φ2) if

{N1 ≤ N2

φ2|N1= φ1

(K, θ) ∈ ζ so it is non-empty. If there is an ascending chain

(N1, φ1) ≤ (N2, φ2) ≤ · · ·

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5 Final Thoughts

then set N =⋃

λNλ, which is a subfield of M and we can define a homo-morphism N → L as follow: if α ∈ N then α ∈ Nλ for some λ and we sendα 7→ φλ(α). This is well-defined. Thus this is an upper bound for the chain inζ. Zorn’s Lemma thus says that there is an maximal element of ζ, (N,φ).

We now show that N =M . Given α ∈M , it is algebraic over K and henceover N . Let fα(t) be its minimal polynomial over N . φ(fα(t)) ∈ L[t] and sosplits completely since L is algebraically closed. Write

φ(fα(t)) = (t− β1) · · · (t− βr).

Since φ(fα(βj)) = 0 there is a map

N(α) ∼= N [t]/(fα(t))→ L

α 7→ β1

extending φ. Maximality of (N,φ) implies that N(α) = N . Thus α ∈ N andN =M .

With this theorem we can finally show

Theorem 5.6 (uniqueness of algebraic closure). If K ≤ L1,K ≤ L2 aretwo algebraic closures of K, there exists an isomorphism φ : L1 → L2.

Proof. By the previous theorem there is a homomorphism φ : L1 → L2 extend-ing the embedding of K in L2. Since K ≤ L2 is algebraic, so is K ≤ φ(L1). ButL1 is algebraically closed and so φ(L1) is algebraically closed. Thus L2 = φ(L1)and φ is an isomorphism.

5.2 Symmetric Polynomials & Invariant TheoryIn the build-up of Fundamental Theorem of Galois Theory we met Artin, whichsays that if K ≤ L, H is a finite subgroup of AutK(L) and M = LH thenM ≤ L is a Galois extension and H = Gal(L/M).

Example. Let L = K(X1, . . . , Xn). Let Sn be the permutation group on theXi’s. These permutations induce K-automorphisms of L. By Artin, M =LSn ≤ L is Galois and Gal(L/M) = Sn. Thus Sn is a Galois group of somefield extension.

Since we can regard any finite group G as a subgroup of some Sn, any finitegroup is a Galois group of some finite field extension

Q ≤ K.

Finding the field extension is the inverse Galois problem.Using notation from the previous example, let

f(t) =

n∏i=1

(t−Xi) = tn − s1tn−1 + s2tn−2 + · · ·+ (−1)nsn ∈M [t]

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5 Final Thoughts

where

s1 = X1 + · · ·+Xn

s1 =∑i<j

XiXj

...sn = X1X2 · · ·Xn

Definition (elementary symmetric polynomial). The si’s are the elemen-tary symmetric polynomials.

Definition (algebraic independence). α1, . . . , αn are algebraically indepen-dent over K if the ring homomorphism

K[Y1, . . . , Yn]→ K[α1, . . . , αn] ≤ LYi 7→ αi

is an isomorphism where K[Y1, . . . , Yn] is the polynomial ring in Y1, . . . , Yn.

Theorem 5.7. The fixed field M = LSn equals to K(s1, . . . , sn) and thesi’s are algebraically independent over K in L.

Proof. Certainly the si’s are fixed by Sn so M1 = K(s1, . . . , sn) ≤M . Observethat L is the splitting field for f(t) =

∏ni=1(t − Xi) over M1. But f(t) has

degree n and so by a proposition we proved in example sheet, the degree of thesplitting field over M1 is at most n!. Artin implies that |L : M | = |Sn| = n! soM1 =M .

To show the algebraic independence of si’s, we make use of the idea of tran-scendence bases and transcendence degree: we may consider the algebraicallyindependent subsets of L and partially order them by inclusion. Note that theunion of a chain is algebraically independent and thus an upper bound of thechain. By Zorn’s Lemma there is a maximal algebraically independent subset,which is a transcendence basis. The transcendence degree of L over K is thecardinality of such a set, which is well-defined by Steinitz Exchange Lemma forinfinite sets, and is denoted trdegK(L).

Since L is algebraic over M , we have trdegK(L) = trdegK(M). If si’swere algebraically dependent then trdegK(M) < n as M would be algebraicover a subfield generated by fewer elements and trdegK K(α1, . . . , αn) ≤ n byinduction. Absurd.

Polynomial invariant theory studies K[X1, . . . , Xn]H for a finite subgroup

H ≤ Sn and as an aside, rather than confining ourselves to permutations ofthe variables, we may also consider H ≤ GL(V ) where V is the K-vector spacegenerated by X1, . . . , Xn.

Question.

1. Is K[X1, . . . , Xn]H finitely generated?

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5 Final Thoughts

2. Is K[X1, . . . , Xn]H isomormphic to any polynomial algebra?

We have shown above that the fixed field of K(X1, . . . , Xn) is K(s1, . . . , sn).A similar result holds for the ring:

Theorem 5.8.K[X1, . . . , Xn]

Sn = K[s1, . . . , sn].

Definition (symmetric polynomial). The elements of K[s1, . . . , sn] are thesymmetric polynomials.

Proof. Let f(X1, . . . , Xn) ∈ K[X1, . . . , Xn]Sn . The proof is by induction on the

total degree of f . If the total degree is 0 then f is a constant polynomial andthus in K, so in K[s1, . . . , sn].

Suppose the total degree of f is positive. Let

θ : K[X1, . . . , Xn]→ K[X1, . . . , Xn−1]

g(X1, . . . , Xn) 7→ g(Xn, . . . , Xn−1, 0)

i.e. the projection of the first n − 1 coordinates. Then ker θ = (Xn). Sincef(X1, . . . , Xn) is fixed by Sn, by slight abuse of notation θ(f(X1, . . . , Xn)) =f(X1, . . . , Xn−1) is fixed under the subgroup Sn−1, i.e. the subgroup that fixesn. Note that in particular

θ(sj(X1, . . . , Xn)) = sj(X1, . . . , Xn−1)

for j ≤ n − 1 where sj is the jth elementary symmetric polynomial. Applyinduction,

θ(f(X1, . . . , Xn)) = p(s1(X1, . . . , Xn−1), . . . , sn−1(X1, . . . , Xn−1))

where p is a polynomial. Rearrange,

θ(f(X1, . . . , Xn)− p(s1(X1, . . . , Xn), . . . , sn−1(X1, . . . , Xn))) = 0

so Xn divides the polynomial

f(X1, . . . , Xn)− p(s1(X1, . . . , Xn), . . . , sn−1(X1, . . . , Xn)) (∗)

which is a symmetric polynomial since it is fixed under Sn. It follows that Xi

divides (∗) for all i. However, K[X1, . . . , Xn] is a UFD and the Xi’s are coprimeso the product X1 · · ·Xn divides (∗). Write

f(X1, . . . , Xn) = g(X1, . . . , Xn)X1 · · ·Xn+p(s1(X1, . . . , Xn), . . . , sn−1(X1, . . . , Xn)).

Observe that the total degree of g(X1, . . . , Xn) is smaller that that of f(X1, . . . , Xn)and that g(X1, . . . , Xn) is fixed under Sn. Applying induction to g(X1, . . . , Xn)and so it is a polynomial in si’s. Thus f(X1, . . . , Xn) ∈ K[s1, . . . , sn].

Example. K[X1, . . . , Xn]An is generated by si’s and

∆(X1, . . . , Xn) =∏i<j

(Xi −Xj).

Emmy Noether in 1920s consider other subgroups of Sn and showed thatthe invariant rings are Noetherian.

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5 Final Thoughts

Theorem 5.9 (Chevalley-Shephard-Todd). C[X1, . . . , Xn]H where H ≤

GL(V ) finite is isomorphic to a polynomial algebra if and only if H isgenerated by pesudoreflections, whose 1-eigenspace has codimension 1.

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Index

K-homomorphism, 10

algebraic closure, 50algebraic independence, 54algebraically closed, 50Artin’s Theorem, 26

constructible number, 7cyclotomic polynomial, 33

derived series, 46differentiation, 15discriminant, 29

elementary symmetric polynomial,54

field extension, 3abelian, 36algebraic, 5cyclic, 36cyclotomic, 33finite, 3Galois, 23Kummer, 39

normal, 13, 22radical, 39separable, 16simple, 6

fixed field, 25Frobenius automorphism, 31

Galois group, 23, 28

inverse Galois problem, 53

minimal polynomial, 5

norm, 19

separable, 15, 16solubility, 39soluble group, 44splitting field, 11splitting polynomial, 11symmetric polynomial, 55

trace, 19transcendence basis, 54

Zorn’s Lemma, 51

57

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