3.4. The fundamental theorem of Galois theory.

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Theorem 3.17 (Fundamental theorem of Galois theory). Let E be a Galois extension of

F, and let G = Gal(E/F). The maps H → EH and M → Gal(E/M) are inverse bijections

between the set of subgroups of G and the set of intermediate fields between E and F:

{subgroups of G} ↔ {intermediate fields F ⊂ M ⊂ E}.

Moreover:

(a) The correspondence is inclusion-reversing, i.e., H1 ⊃ H2 ⇐⇒ EH1 ⊂ EH2.

(b) Indexes equal degrees, i.e., (H1 : H2) = [EH2 : EH1 ].

(c) The group σHσ−1 ↔ σM, i.e., EσHσ−1 = σ(EH); Gal(E/σM) = σ Gal(E/M)σ−1.

(d) The group H is normal in G ⇐⇒ EH is normal (hence Galois) over F, in which case

Gal(EH/F) = G/H.

Proof. Let H be a subgroup of G. We first have to show that Gal(E/EH) = H. Butwe

have already observed that E is Galois over EH, and Theorem 3.12 shows that Gal(E/EH) =

H.

Next let M be an intermediate field, and let H = Gal(E/M). We have to show that

EH = M, but this is again proved in Theorem 3.12.

Thus we have proved that Inv(・) and Gal(E/・) are inverse bijections.

24 J.S. MILNE

(a) We have the obvious implications:

H1 ⊃ H2 =⇒ EH1 ⊂ EH2 =⇒ Gal(E/EH1 ) ⊃ Gal(E/EH2 ).

But Gal(E/EHi) = Hi.

(b) In the case H2 = 1, the first equality follows from (3.8) and (3.12). The general case

follows, using that

(H1 : 1) = (H1 : H2)(H2 : 1) and [E : EH1] = [E : EH2][EH2 : EH1].

(c) If H = {τ ∈ G | τα = α, all α ∈ M}, i.e., H = Gal(E/M), then σHσ−1 = {τ ∈ G |

τσα = σα, all α ∈ M}, i.e., σHσ−1 = Gal(E/σM).

(d) Assume H to be normal in G, and let M = EH. Because σHσ−1 = H for all σ ∈ G,

we must have σM = M for all σ ∈ G, i.e., the action of G on E stabilizes M. We therefore

have a homomorphism

σ → σ|M : G → Aut(M/F)

with kernel H. Let G_ be the image. Then F = MG_

, and so M is Galois over F with Galois

group G_ (by Theorem 3.12).

Conversely, assume that M is normal over F, and write M = F[α1, ..., αm]. For σ ∈ G,

σαi is a root of the minimum polynomial of αi over F, and so lies in M. Hence σM = M,

and this implies that σHσ−1 = H (by (c)).

Remark 3.18. The theorem shows that there is an order reversing bijection between

the intermediate fields of E/F and the subgroups of G. Using this we can read off more

results. For example let M1,M2, . . . ,Mr be intermediate fields, and let Hi be the subgroup

corresponding to Mi (i.e., Hi = Gal(E/Mi)). Then (by definition) M1M2 ・ ・ ・Mr is the

smallest field containing all Mi;he nce it must correspond to the largest subgroup contained

in all Hi, which is

_

Hi. Therefore

Gal(E/M1 ・ ・ ・Mr) = H1 ∩ ... ∩ Hr .

We mention two further results (they are not difficult to prove):

1. Let E/F be Galois, and let L be any field containing F. Assume L and E are contained

in some large field Ω. Then EL is Galois over L, E is Galois over E ∩ L, and the map

σ → σ|E : Gal(EL/L) → Gal(E/E ∩ L) is an isomorphism.

2. Let E1/F and E2/F be Galois, with E1 and E2 subfields of some field Ω. Then E1E2

is Galois over F, and

σ → (σ|E1, σ|E2) : Gal(E1E2/F ) → Gal(E1/F ) ×Gal(E2/F )

is injective with image {(σ1, σ2) | σ1|E1 ∩ E2 = σ2|E1 ∩ E2}.

Example 3.19. We analyse the extension Q[ζ]/Q, where ζ is the primitive 7th root of

1, say ζ = e2πi/7. Then Q[ζ] is the splitting field of the irreducible polynomial

X6 + X5 + X4 + X3 + X2 + X + 1

(see 1.31), and so is Galois of degree 6 over Q. For any σ ∈ G, σζ = ζi, some i, 1 ≤ i ≤ 6,

and the map σ → i defines an isomorphism Gal(Q[ζ]/Q)→ (Z/7Z)×. Let σ be the element

of Gal(Q[ζ]/Q) such that σζ = ζ3. Then σ generates Gal(Q[ζ]/Q) because the class of 3 in

(Z/7Z)× generates it (the powers of 3 mod 7 are 3, 2, 6, 4, 5, 1). We investigate the subfields

of Q[ζ] corresponding to the subgroups < σ3 > and < σ2 >.

FIELDS AND GALOIS THEORY 25

Note that σ3ζ = ζ6 = ￣ζ (complex conjugate of ζ). The subfield of Q[ζ] corresponding to

< σ3 > is Q[ζ + ￣ζ], and ζ + ￣ζ = 2cos 2π

7 . Since < σ3 > is a normal subgroup of < σ >,

Q[ζ + ￣ζ] is Galois over Q, with Galois group < σ > / < σ3 > . The conjugates of α1 = ζ + ￣ζ

are α3 = ζ3 + ζ−3, α2 = ζ2 + ζ−2. Direct calculation shows that

_

αi =

_6

i=1

ζi = −1,

α1α2α3 = (ζ+ζ6)(ζ2+ζ5)(ζ3+ζ4) = (ζ+ζ3+ζ4+ζ6)(ζ3+ζ4) = (ζ4+ζ6+1+ζ2+ζ5+1+ζ+ζ3) = 1.

α1α2 + α1α3 + α2α3 = −2.

Hence the minimum polynomial of ζ + ￣ζ is

g(X) = X3 + X2 − 2X − 1.

The minimum polynomial of cos 2π

7 = α1

2 is therefore

g(2X)

8

= X3 + X2/2 − X/2 − 1/8.

The subfield of Q[ζ] corresponding to < σ2 > is generated by β = ζ +ζ2+ζ4. Let β_ = σβ.

Then (β − β_)2 = −7. Hence the field fixed by < σ2 > is Q[

√

−7].

Example 3.20. We compute the Galois group of the splitting field E of X5 −2 ∈ Q[X].

Recall (from the Homework) that E = Q[ζ,α] where ζ is a primitive 5th root of 1, and α is

a root of X5 −2. For example, we could take E to be the splitting field of X5 −2 in C, with

ζ = e2πi/5 and α equal to the real 5th root of 2. We have the picture:

Q[ζ,α]

N _ _ H

Q[ζ] Q[α]

G/N _ _

Q

The degrees

[Q[ζ] : Q] = 4, [Q[α] : Q] = 5.

Because 4 and 5 are relatively prime,

[Q[ζ,α] : Q] = 20.

Hence G = Gal(Q[ζ,α]/Q) has order 20, and the subgroups N and H corresponding to Q[ζ]

and Q[α] have orders 5 and 4 respectively (because N = Gal(Q[ζ,α]/Q[ζ] . . . ). Because

Q[ζ] is normal over Q (it is the splitting field of X5 − 1), N is normal in G. Because

Q[ζ] ・ Q[α] = Q[ζ,α], we have H ∩ N = 1 (see 3.18), and so G = N           θ H. We have

H ≈ G/N ≈ (Z/5Z)×, which is cyclic, being generated by the class of 2. Let τ be the

generator of H corresponding to 2 under this isomorphism, and let σ be a generator of N.

Thus σ(α) is another root of X5 −2, which we can take to be ζα (after possibly replacing σ

by a power). Hence: _

τζ = ζ2

τα = α

_

σζ = ζ

σα = ζα.

Note that τστ−1(α) = τσα = τ (ζα) = ζ2α and it fixes ζ;t herefore τστ−1 = σ2. Thus G has

generators σ and τ and defining relations

σ5 = 1, τ4 = 1, τστ − 1 = σ2.

26 J.S. MILNE

The subgroup H has five conjugates, which correspond to the five fields Q[ζiα],

σiHσ−i ↔ σiQ[α] = Q[ζiα], 1 ≤ i ≤ 5.

Definition 3.21. An extension E ⊃ F is called a cyclic, abelian, ..., solvable extension

if it is Galois with cyclic, abelian, ..., solvable Galois group.

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