5.10. Infinite Galois extensions (sketch).

Back

Recall that we defined a finite extension Ω of F to be Galois over F if it is normal and

separable, i.e., if every irreducible polynomial f ∈ F[X] having a root in Ω has degf distinct

roots in Ω. Similarly, we define an algebraic extension Ω of F to be Galois over F if it is

normal and separable. Equivalently, a field Ω ⊃ F is Galois over F if it is a union of subfields

E finite and Galois over F.

Let Gal(Ω/F ) = Aut(Ω/F ), and consider the map

σ → (σ|E) : Gal(Ω/F ) →

_

Gal(E/F)

(product over the finite Galois extensions E of F contained in Ω). This map is injective,

because Ω is a union of finite Galois extensions. We give each finite group Gal(E/F) the

discrete topology and

_

Gal(E/F) the product topology, and we give Gal(Ω/F ) the subspace

topology. Thus the subgroups Gal(Ω/E), [E : F] < ∞, form a fundamental system of

neighbourhoods of 1 in Gal(Ω/F ).

FIELDS AND GALOIS THEORY 53

By the Tychonoff theorem,

_

Gal(E/F) is compact, and it is easy to see that the image

of Gal(Ω/F ) is closed—hence it is compact and Hausdorff.

Theorem 5.40. Let Ω be Galois over F with Galois group G. The maps

H → ΩH, M→ Gal(Ω/M)

define a one-to-one correspondence between the closed subgroups of G and the intermediate

fields M. A field M is of finite degree over F if and only if Gal(Ω/M) is open in Gal(Ω/F ).

Proof. Omit—it is not difficult given the finite case. See for example, E. Artin, Algebraic

Numbers and Algebraic Functions, p103.

Remark 5.41. The remaining assertions in the Fundamental Theorem of Galois Theory

carry over to the infinite case provided that one requires the subgroups to be closed.

Example 5.42. Let Ω be an algebraic closure of a finite field Fp. Then G = Gal(Ω/Fp)

contains a canonical Frobenius element, σ = (a → ap), and it is generated by it as a

topological group, i.e., G is the closure of <σ>. Endow Z with the topology for which the

groups nZ, n ≥ 1, form a fundamental system of neighbourhoods of 0. Thus two integers

are close if their difference is divisible by a large integer.

As for any topological group, we can complete Z for this topology. A Cauchy seqence in

Z is a sequence (ai)i≥1, ai ∈ Z, satisfying the following condition: for all n ≥ 1, there exists

an N such that ai ≡ aj mod n for i, j > N. Call a Cauchy sequence in Z trivial if ai → 0

as i → ∞, i.e., if for all n ≥ 1, there exists an N such that ai ≡ 0 mod n. The Cauchy

sequences form a commutative group, and the trivial Cauchy sequences form a subgroup. We

can define _Z to be the quotient of the first group by the second. It has a ring structure, and

the map sending m ∈ Z to the constant sequence m,m,m, . . . identifies Z with a subgroup

of _Z.

Let α ∈ _Z be represented by the Cauchy sequence (ai). The restriction of σ to Fpn has

order n. Therefore (σ|Fpn)ai is independent of i provided it is sufficiently large, and we can

define σα ∈ Gal(Ω/Fp) to be such that, for each n, σα|Fpn = (σ|Fpn)ai for all i sufficiently

large (depending on n). The map α → σα : _Z → Gal(Ω/Fp) is an isomorphism.

The group _Z is uncountable. To most analysts, it is a little weird—its connected components

are one-point sets. To number theorists it will seem quite natural— the Chinese

remainder theorem implies that it is isomorphic to

_

p prime Zp where Zp is the ring of p-adic

integers.

Example 5.43. Let Ω be the algebraic closure of Q in C;the n Gal(Ω/Q) is one of the

most basic, and intractible, objects in mathematics. Note that, as far as we know, it could

have every finite group as a quotient, and it certainly has Sn as a quotient group for every

n (and every sporadic simple group, and every...). We do however understand Gal(Fab/F )

when F ⊂ C is a finite extension of Q and Fab is the union of all finite abelian extensions of

F contained in C. For example, Gal(Qab/Q) ≈ _Z×. (This is abelian class field theory—see

Math 776.)

54 J.S. MILNE

6. Transcendental Extensions

In this section we consider fields Ω ⊃ F with Ω much bigger than F. For example, we could

have C ⊃ Q.

Elements α1, ..., αn of Ω are said to be algebraically dependent over F if there is a nonzero

polynomial f(X1, ...,Xn) ∈ F[X1, ...,Xn] such that f(α1, ..., αn) = 0. Otherwise, the elements

are said to be algebraically independent over F. Thus they are algebraically independent

if

ai1,...,in

∈ F,

_

ai1,...,inαi1

1 ...αin

n = 0 =⇒ ai1,...,in = 0 all i1, ..., in.

Note the similarity with linear independence. In fact, if f is required to be homogeneous

of degree 1, then the definition becomes that of linear independence. The theory in this

section is logically very similar to a part of linear algebra. It is useful to keep the following

correspondences in mind:

Linear algebra Transcendence

linearly independent algebraically independent

A ⊂ span(B) A algebraically dependent on B

basis transcendence basis

dimension transcendence degree

Example 6.1. (a) A single element α is algebraically independent over F if and only if

it is transcendental over F.

(b) The complex numbers π and e are almost certainly algebraically independent over Q,

but this has not been proved.

An infinite set A is algebraically independent if every finite subset of A is algebraically

independent.

Remark 6.2. To say that α1, ..., αn are algebraically independent over F, is the same as

to say that the map

f(X1, ...,Xn) → f(α1, ..., αn) : F[X1, ...,Xn] → F[α1, ..., αn]

is an injection, and hence an isomorphism. This isomorphism then extends to the fields of

fractions,

Xi → αi : F(X1, ...,Xn) → F(α1, ..., αn)

In this case, F(α1, ..., αn) is called a pure transcendental extension of F. Then (see 5.28)

the polynomial

f(X) = Xn − α1Xn−1 + . . . (−1)nαn

has Galois group Sn over F(α1, ..., αn).

Let β ∈ Ω and let A ⊂ Ω. The following conditions are equivalent:

(a) β is algebraic over F(A);

(b) there exist α1, . . . , αn ∈ F(A) such that βn + α1βn−1 + ・ ・ ・ + αn = 0;

(c) there exist α0, . . . , αn ∈ F[A] such that α0βn + ・ ・ ・ + αn = 0;

(d) there exists an f(X1, . . . ,Xm, Y ) ∈ F[X1 . . . ,Xm, Y ] and a1, . . . , am ∈ F such that

f(a1, . . . , am, Y ) _= 0 but f(a1, . . . , am, β) = 0.

When these conditions hold, we say that β is algebraically dependent on A (over F). A set

B is algebraically dependent on A if each element of B is algebraically dependent on A.

FIELDS AND GALOIS THEORY 55

Theorem 6.3 (Fundamental result). Let A = {α1, ..., αm} and B = {β1, ..., βn} be two

subsets of Ω. Assume

(a) A is algebraically independent (over F);

(b) A is algebraically dependent on B (over F).

Then m ≤ n.

Proof. We first prove a lemma.

Lemma 6.4 (The exchange property). Let {α1, ..., αn} be a subset of Ω; if β is algebraically

dependent on {α1, ..., αm} but not on {α1, ..., αm−1}, then αm is algebraically dependent

on {α1, ..., αm−1, β}.

Proof. Because β is algebraically dependent on {α1, . . . , αm}, there exists a polynomial

f(X1, ...,Xm, Y ) with coefficients in F such that

f(α1, ..., αm, Y ) _= 0, f(α1, ..., αm, β) = 0.

Write

f(X1, ...,Xm, Y ) =

_

i

ai(X1, ...,Xm−1, Y )Xim

and observe that, because f(α1, . . . , αm, Y ) _= 0, at least one of the polynomials

ai(α1, ..., αm−1, Y ), say ai0, is not the zero polynomial. Because β is not algebraically dependent

on {α1, ..., αm−1}, ai0(α1, ..., αm−1, β) _= 0. Therefore, f(α1, ..., αm−1,Xm, β) is not the

zero polynomial. Since f(α1, ..., αm, β) = 0, this shows that αm is algebraically dependent

on {α1, ..., αm−1, β}.

Lemma 6.5 (Transitivity of algebraic dependence). If C is algebraically dependent on B,

and B is algebraically dependent on A, then C is algebraically dependent on A.

Proof. The argument in the proof (2.10) shows that if γ is algebraic over a field E which

is algebraic over a field F, then γ is algebraic over F (if a1, . . . , an are the coefficients of the

minimum polynomial of γ over E, then the field F[a1, . . . , an, γ] has finite degree over F).

Apply this with F(A ∪ B) for E and F(A) for F.

Proof. (of the theorem). We now prove the theorem. Let k be the number of elements

that A and B have in common. If k = m, then A ⊂ B, and certainly m ≤ n. Suppose that

k < m, and write B = {α1, ..., αk, βk+1, ..., βn}. Since αk+1 is algebraically dependent on

{α1, ..., αk, βk+1, ..., βn} but not on {α1, ..., αk}, there will be a βj, k + 1 ≤ j ≤ n, such that

αk+1 is algebraically dependent on {α1, ..., αk, βk+1, ..., βj} but not {α1, ..., αk, βk+1, ..., βj−1}.

The exchange lemma then shows that βj is algebraically dependent on

B1 =df B ∪ {αk+1} − {βj}.

Therefore B is algebraically dependent on B1, and so A is algebraically dependent on B1

(by the last lemma). If k + 1 < m, repeat the argument with A and B1. Eventually we’ll

achieve k = m, and m ≤ n.

Definition 6.6. A transcendence basis for Ω over F is an algebraically independent set

A such that Ω is algebraic over F(A).

Lemma 6.7. If Ω is algebraic over F(A), and A is minimal among subsets of Ω with this

property, then it is a transcendence basis for Ω over F.

56 J.S. MILNE

Proof. If α1, . . . , αm ∈ A are not algebraically independent, then one is algebraically

dependent on the remainder, and it follows from (6.5) that Ω will still be algebraic over

F(A) after it has been dropped from A.

Theorem 6.8. If there is a finite subset A ⊂ Ω such that Ω is algebraic over F(A), then

Ω has a finite transcendence basis over F. Moreover, every transcendence basis is finite, and

they all have the same number of elements.

Proof. In fact, any minimal subset A_ of A such that Ω is algebraic over F(A_) will be a

transcendence basis. The second statement follows from Theorem 6.3.

The cardinality of a transcendence basis for Ω over F is called the transcendence degree of

Ω over F. For example, the pure transcendental extension F(X1, . . . ,Xn) has transcendence

degree n over F.

Example 6.9. Let p1, . . . , pn be the elementary symmetric polynomials in X1, . . . ,Xn.

The field F(X1, . . . ,Xn) is algebraic over F(p1, . . . , pn), and so {p1, p2, . . . , pn} contains a

transcendence basis for F(X1, . . . ,Xn). Because F(X1, . . . ,Xn) has transcendence degree

n, the pi’s must themselves be a transcendence basis.

Example 6.10. Let Ω be the field of meromorphic functions on a compact complex manifold

M.

(a) The only meromorphic functions on the Riemann sphere are the rational functions in

z. Hence, in this case, Ω is a pure transcendental extension of C of transcendence degree 1.

(b) If M is a Riemann surface, then the transcendence degree of Ω over C is 1, and Ω is

a pure transcendental extension of C ⇐⇒ M is isomorphic to the Riemann sphere

(c) If M has complex dimension n, then the transcendence degree is ≤ n, with equality

holding if M is embeddable in some projective space.

Lemma 6.11. Suppose that A is algebraically independent, but that A∪{β} is algebraically

dependent. Then β is algebraic over F(A).

Proof. The hypothesis is that there exists a nonzero polynomial f(X1, ...,Xn, Y ) ∈

F[X1, ...,Xn, Y ] such that f(a1, ..., an, β) = 0, some distinct a1, ..., an ∈ A. Because A is

algebraically independent, Y does occur in f. Therefore

f = g0Y m + g1Y m−1 + ・ ・ ・ + gm, gi ∈ F[X1, ...,Xn], g0 _= 0, m≥ 1.

As g0 _= 0 and the ai are algebraically independent, g0(a1, ..., an) _= 0. Because β is a root of

f = g0(a1, ..., an)Xm + g1(a1, ..., an)Xm−1 + ・ ・ ・ + gm(a1, ..., an),

it is algebraic over F(a1, ..., an) ⊂ F(A).

Proposition 6.12. Every maximal algebraically independent subset of Ω is a transcendence

basis for Ω over F.

Proof. We have to prove that Ω is algebraic over F(A) if A is maximal among algebraically

independent subsets. But the maximality implies that, for every β ∈ Ω, A∪{β} is

algebraically dependent, and so the lemma shows that β is algebraic over F(A).

Theorem 6.13 (*). Every field Ω containing F has a transcendence basis over F.

FIELDS AND GALOIS THEORY 57

Proof. Let S be the set of algebraically independent subsets of Ω. We can partially order

it by inclusion. Let T be a totally ordered subset, and let B = ∪{A | A ∈ T }. I claim that

B ∈ S, i.e., that B is algebraically independent. If not, there exists a finite subset B_ of

B that is not algebraically independent. But such a subset will be contained in one of the

sets in T , which is a contradiction. Now we can apply Zorn’s lemma to obtain a maximal

algebraically independent subset A.

It is possible to show that any two (possibly infinite) transcendence bases for Ω over F

have the same cardinality.

Proposition 6.14. Any two algebraically closed fields with the same transcendence degree

over F are F-isomorphic.

Proof. Choose transcendence bases A and A_ for the two fields, and choose a bijection

ϕ : A → A_. Then ϕ extends uniquely to an F-isomorphism ϕ : F[A]→ F[A_], and hence to

an isomorphism of the fields of fractions F(A) → F(A_). Use this isomorphism to identify

F(A) with F(A_). Then the two fields in question are algebraic closures of the same field,

and hence are isomorphic (Theorem 2.16).

Remark 6.15. Any two algebraically closed fields with the same uncountable cardinality

and the same characteristic are isomorphic. The idea of the proof is as follows. Let F and

F_ be the prime subfields of Ω and Ω_;we can identify F with F_. Then show that when Ω

is uncountable, the cardinality of Ω is the same as the cardinality of a transcendence basis

over F. Finally, apply the proposition.

Remark 6.16. What are the automorphisms of C? If we assume the axiom of choice,

then it is easy to construct many: choose any transcendence basis A for C over Q, and

choose any permutation α of A;the n α defines an isomorphism Q(A) → Q(A) that can be

extended to an automorphism of C. On the other hand, without the axiom of choice, there

are probably only two, the identity map and complex conjugation. (I have been told that

any other is nonmeasurable, and it is known that the axiom of choice is required to construct

nonmeasurable functions.)

Theorem 6.17 (L¨uroth’s theorem). Any subfield E of F(X) containing F but not equal

to F is a pure transcendental extension of F.

Proof. See, Jacobson, Lectures in Abstract Algebra III, p157.

Remark 6.18. This fails when there is more than one variable—see the footnote on p38

and Noether’s conjecture 5.30. The best that is true is that if [F(X, Y ) : E] < ∞ and F

is algebraically closed of characteristic zero, then E is a pure transcendental extension of F

(Theorem of Zariski, 1958).__