3.2. Groups of automorphisms of fields.

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Consider fields E ⊃ F. We write Aut(E/F) for the group of F-automorphisms of E, i.e.,

automorphisms σ : E → E such that σ(a) = a for all a ∈ F.

20 J.S. MILNE

Example 3.7. (a) There are two obvious automorphisms of C, namely, the identity map

and complex conjugation. We’ll see later (last section) that by using the Axiom of Choice,

one can construct uncountably many more. They are all noncontinuous and (I’ve been told)

nonmeasurable—hence they require the Axiom of Choice for their construction.

(b) Let E = C(X). Then Aut(E/C) consists of the maps X → aX+b

cX+d , ad − bc _= 0

(Jacobson, Lectures III, p158), and so Aut(E/C) = PGL2(C). Analysts will note that this

is the same as the automorphism group of the Riemann sphere. This is not a coincidence:

the field of meromorphic functions on the Riemann sphere P1

C is C(z) ≈ C(X), and so there

is a map Aut(P1

C) → Aut(C(z)/C), which one can show is an isomorphism.

(c) The group Aut(C(X1,X2)/C) is quite complicated—there is a map

PGL3(C) = Aut(P2

C) _→ Aut(C(X1,X2)/C),

but this is very far from being surjective. When there are more X’s, the group is unknown.

(The group Aut(C(X1, . . . ,Xn)/C) is the group of birational automorphisms of Pn

C. It is

called the Cremona group. Its study is part of algebraic geometry.)

In this section, we shall be concerned with the groups Aut(E/F) when E is a finite

extension of F.

Proposition 3.8. If E is a splitting field of a monic separable polynomial f ∈ F[X],

then Aut(E/F) has order [E : F].

Proof. Let f =

_

fmi

i , with the fi monic irreducible and distinct. The splitting field

of f is the same as the splitting field of

_

fi. Hence we may assume f is a product of

distinct monic separable irreducible polynomials, and hence has deg f distinct roots in E.

Now Proposition 2.7b shows that there are [E : F] distinct F-homomorphisms E → E;the y

are automatically isomorphisms.

Example 3.9. (a) Let E = F[α], f(α) = 0; if f has no other root in E than α, then

Aut(E/F) = 1. For example, if 3

√

2 denotes the real cube root of 2, then Aut(Q[ 3

√

2]/Q) = 1.

Thus, in the proposition, it is essential that E be a splitting field.

(b) Let F be a field of characteristic p _= 0, and let a be an element of F that is not a

pth power. The splitting field of f = Xp − a is F[α] where α is the unique root of f. Then

Aut(E/F) = 1. Thus, in the proposition, it is essential that E be the splitting field of a

separable polynomial.

When G is a group of automorphisms of a field E, we write

EG = Inv(G) = {α ∈ E | σα = α, all σ ∈ G}.

It is a subfield of E, called the subfield of G-invariants of E or the subfield of E fixed by G.

We have maps

G → Inv(G) F → Aut(E/F).

Goal: Show that when E is the splitting field of a separable polynomial in F[X] and G =

Aut(E/F), then

H → Inv(H), M→ Aut(E/M)

give a one-to-one correspondence between the set of intermediate fields M, F ⊂ M ⊂ E,

and the set of subgroups H of G.

FIELDS AND GALOIS THEORY 21

Lemma 3.10 (E. Artin). Let G be a finite group of automorphisms of a field E, and let

F = EG; then [E : F] ≤ (G : 1).

Proof. Let G = {σ1 = 1, . . . , σm}, and let α1, . . . , αn be n >m elements of E. We shall

show that the αi are linearly dependent over F. In the system

σ1(α1)x1 + ・ ・ ・ + σ1(αn)xn = 0

・ ・ ・ ・ ・ ・

σm(α1)x1 + ・ ・ ・ + σm(αn)xn = 0

there are m equations and n > m unknowns, and hence there are nontrivial solutions (in E).

Choose a nontrivial solution (c1, . . . , cn) with the fewest nonzero elements. After renumbering

the αi’s, we may suppose that c1 _= 0, and then (after multiplying by a scalar) that

c1 = 1. With these normalizations, we’ll see that all ci ∈ F. Hence the first equation (recall

σ1 = 1)

α1c1 + ・ ・ ・ + αncn = 0

shows that the αi are linearly dependent over F.

If not all ci are in F, then σk(ci) _= ci for some i, k. On apply σk to the equations

σ1(α1)c1 + ・ ・ ・ + σ1(αn)cn = 0

・ ・ ・ ・ ・ ・

σm(α1)c1 + ・ ・ ・ + σm(αn)cn = 0

and using that {σkσ1, . . . , σkσm} is a permutation of {σ1, . . . , σm}, we find that

(1, . . . , σk(ci), . . . ) is also a solution to the system of equations. On subtracting it from

the first, we obtain a solution (0, . . . , ci − σk(ci), . . . ), which is nonzero (look at the ith

coordinate), but has more zeros than the first solution (look at the first coordinate)—

contradiction.

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