2.3. Algebraic closures.

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Recall that Ω is said to be algebraically closed if every nonconstant polynomial f(X) ∈ Ω[X]

has a root in Ω (and hence splits in Ω[X]);equ ivalently, if the only irreducible polynomials in

Ω[X] are those of degree 1. Recall also that a field Ω containing F is said to be an algebraic

closure of F if it is algebraic over F and it is algebraically closed. We want to show that

(assuming the axiom of choice) every field has an algebraic closure. The following criterion

suggests how this might be done.

Lemma 2.10. Suppose that Ω is algebraic over F and every polynomial f ∈ F[X] splits

in Ω[X]; then Ω is an algebraic closure of F.

FIELDS AND GALOIS THEORY 15

Proof. Let f ∈ Ω[X]. We know (see §1.6) how to construct a finite extension E of Ω

containing a root α of f. We want to show that α in fact lies in Ω. Write f = anXn+・ ・ ・+a0,

ai ∈ Ω, and consider the sequence of fields F ⊂ F[a1, . . . , an] ⊂ F[a1, . . . , an, α]. Because

each ai is algebraic over F, F[a1, . . . , an] is a finite field extension of F, and because f ∈

F[a1, . . . , an][X], α is algebraic over F[a1, . . . , an]. Therefore α lies in a finite extension of

F, and is therefore algebraic over F, i.e., it is the root of a polynomial with coefficients in

F. But, by assumption, this polynomial splits in Ω[X], and so all its roots lie in Ω. In

particular, α ∈ Ω.

Lemma 2.11. Let Ω ⊃ F; then

E = {α ∈ Ω | α algebraic over F}

is a field.

Proof. If α and β are algebraic over F, then F[α, β] is of finite degree over F, and so

is a field (see 1.14). Every element of F[α, β] is algebraic over F, including α Ѓ} β, α/β,

αβ, . . . .

The field E constructed in the lemma is called the algebraic closure of F in Ω. The

preceding lemma shows that if every polynomial in F[X] splits in Ω[X], then E is an algebraic

closure of F. Thus to construct an algebraic closure of F, it suffices to construct an extension

in which every polynomial in F[X] splits. We know how to do this for a single polynomial,

but passing from there to all polynomials causes set-theoretic problems.

Theorem 2.12 (*). 2Every field has an algebraic closure.

Once we have proved the fundamental theorem of algebra, that C is algebraically closed,

then we will know that the algebraic closure in C of any subfield F of C is an algebraic

closure of F. This proves the theorem for such fields. We sketch three proofs of the general

result. The first doesn’t assume the axiom of choice, but does assume that F is countable.

Proof. (First proof of 2.12) Because F is countable, it follows that F[X] is countable,

i.e., we can list its elements f1(X), f2(X), . . . . Define the fields Ei inductively as follows:

E0 = F; Ei is the splitting field of fi over Ei−1. Note that E0 ⊂ E1 ⊂ E2 ⊂ ・ ・ ・ . Define

Ω = ∪Ei;it is obviously an algebraic closure of F.

Remark 2.13. Since the Ei are not subsets of a fixed set, forming the union requires

explanation: define Ω∗ to be the disjoint union of the Ei;let a, b ∈ Ω∗, say a ∈ Ei and

b ∈ Ej;wr ite a ∼ b if a = b when regarded as elements of the larger of Ei or Ej;v erify that

∼ is an equivalence relation, and let Ω = Ω∗/ ∼.

Proof. (Second proof of 2.12) If A and B are rings containing a field F, then A⊗F B is

a ring containing F, and there are F-homomorphisms A, B → A ⊗F B. More generally, if

(Ai)i∈I is some family of rings each of which contains F, then ⊗FAi is a ring containing F,

and there are F-homomorphisms Aj →⊗FAi for each j ∈ I. It is defined to be the quotient

of the F-vector space with basis ΠAi by the subspace generated by elements of the form:

• (xi) + (yi) − (zi) with xj + yj = zj for one j ∈ I and xi = yi = zi for all i _= j.

• (xi) − a(yi) with xj = ayj for one j ∈ I and xi = yi for all i _= j.

2Results marked with an asterisk require the axiom of choice for their proof.

16 J.S. MILNE

It can be made into a ring in an obvious fashion (see Bourbaki, Alg`ebre, Chapt 3, Appendix).

For each polynomial f ∈ F[X], choose a splitting field Ef, and let Ω = (⊗fEf )/M where

M is a maximal ideal in ⊗fEf—Zorn’s lemma implies that M exists (see below). Then Ω

is a field (see 1.1), and there are F-homomorphisms Ef → Ω (which must be injective) for

each f ∈ F[X]. Since f splits in Ef , it must also split in the larger field Ω. The algebraic

closure of F in Ω is therefore an algebraic closure of F. (Actually, Ω itself is an algebraic

closure of F.)

Lemma 2.14 (Zorn’s). Let (S,≤) be a nonempty partially ordered set (reflexive, transitive,

anti-symmetric, i.e., a ≤ b and b ≤ a =⇒ a = b). Suppose that every totally ordered

subset T of S (i.e., for all s, t ∈ T, either s ≤ t or t ≤ s) has an upper bound in S (i.e.,

there exists an s ∈ S such that t ≤ s for all t ∈ T ). Then S has a maximal element (i.e., an

element s such that s ≤ s_ =⇒ s = s_).

Zorn’s lemma is equivalent to the Axiom of Choice.

Lemma 2.15 (*). Every nonzero commutative ring A has a maximal ideal.

Proof. Let S be the set of all proper ideals in A, partially ordered by inclusion. If T is

a totally ordered set of ideals, then J =

 

I∈T I is again an ideal, and it is proper because

if 1 ∈ J then 1 ∈ I for some I in T. Thus J is an upper bound for T . Now Zorn’s lemma

implies that S has a maximal element, which is a maximal ideal in A.

Proof. (Third proof of 2.12) First show that the cardinality of any field algebraic over F

is the same as that of F. Next choose an uncountable set Ξ of cardinality greater than that

of F, and identify F with a subset of Ξ. Let S be the set triples (E,+, ・) with E ⊂ S and

(+, ・) a field structure on E such that (E,+, ・) contains F as a subfield and is algebraic over

it. Write (E,+, ・) ≤ (E_,+_, ・_) if the first is a subfield of the second. Apply Zorn’s lemma to

show that S has maximal elements, and then show that a maximal element is algebraically

closed. (See Jacobson, Lectures in Algebra, III, p144 for the details.)

There do exist naturally occurring fields, not contained in C, that are uncountable. For

example, for any field F there is a ring F[[T ]] of formal power series

_

i≥0 aiT i, ai ∈ F, and

its field of fractions is uncountable even if F is finite.

Theorem 2.16 (*). Let Ω be an algebraic closure of F, and let E be an algebraic extension

of F; then there is an F-homomorphism E → Ω. If E is also an algebraic closure of

F, then any such map is an isomorphism.

Proof. Suppose first that E is countably generated over F, i.e., E = F[α1, ..., αn, . . . ].

Then we can extend the inclusion map F → Ω to F[α1] (map α1 to any root of its minimal

polynomial in Ω), then to F[α1, α2], and so on.

The uncountable case is a straightforward application of Zorn’s lemma.

Let S be the set of pairs (M, ϕM) with M a field F ⊂ M ⊂ E and ϕM an F-homomorphim

M → Ω. Write (M, ϕM) ≤ (N,ϕN) if M ⊂ N and ϕN|M = ϕM. This makes S into

a partially ordered subset. Let T be a totally ordered subset of S. Then M_ = ∪M∈TM

is a subfield of E, and we can define a homomorphism ϕ_ : M_ → Ω by requiring that

ϕ_(x) = ϕM(x) if x ∈ M. The pair (M_, ϕ_) is an upper bound for T in S. Hence Zorn’s

lemma gives us a maximal element (M, ϕ) in S. Suppose that M _= E. Then there exists an

element α ∈ E, α /∈ M. Since α is algebraic over M, we can apply (2.2) to extend ϕ to M[α],

FIELDS AND GALOIS THEORY 17

contradicting the maximality of M. Hence M = E, and the proof of the first statement is

complete.

If E is algebraically closed, then every polynomial f ∈ F[X] splits in E and hence in ϕ(E),

i.e., f(X) =

_

(X −αi), αi ∈ ϕ(E). Let α ∈ Ω, and let f(X) be the minimum polynomial of

α. Then X−α is a factor of f(X) in Ω[X], but, as we just observed, f(X) splits in ϕ(E)[X].

Because of unique factorization, this implies that α ∈ ϕ(E).

The above proof is a typical application of Zorn’s lemma: once we know how to do

something in a finite (or countable) situation, Zorn’s lemma allows us to do it in general.

Remark 2.17. Even for a finite field F, there will exist uncountably many isomorphisms

from one algebraic closure to a second, none of which is to be preferred over any other. Thus

it is (uncountably) sloppy to say that the algebraic closure of F is unique. All one can say

is that, given two algebraic closures Ω, Ω_ of F, then, thanks to the axiom of choice, there

exists an F-isomorphism Ω → Ω_.

18 J.S. MILNE

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