There are (at least) two kinds of algebraic extension or closure. In one, we start from a field F contained in some other field K, and consider all the elements of K algebraic over F. We'll show that these constitute a subfield F' of K, to be called the algebraic closure of K in F. In particular, if K is already algebraically closed, so is F'. This can be used, for instance, to construct the field of algebraic numbers as the algebraic closure of Q in C. The other kind constructs a field F' containing F starting from F and some data internal to F, with no ambient field K in which to find the new elements of F'.
Integral and algebraic closures
The first approach works almost as easily for commutative rings
with unit, so we state it in that setting. Assume that K is
a commutative ring with unit, and F is a subring with the same unit.
[We specify ``the same'' to avoid examples such as
An element x of K is said to be integral over F
if there exists a monic polynomial P in F[z] such that P(x)=0.
Note that if F is a field then ``integral over F''
is the same as ``algebraic over F''. Then TFAE:
i) x is algebraic over F.
ii) The subring F[x] of K is finitely generated as an F-module.
iii) There exists a finitely generated F-submodule M of K such that M contains 1 and xM.
Here an F-module M is ``finitely generated'' if there exist m1,...,mn in M such that every element of M is of the form a1m1+...+anmn for some ai in F. In the TFAE, the implications (i)=>(ii)=>(iii) are easy, and the key is getting from (iii) back to (i). Let v be a vector of generators, and A a matrix with entries in F such that xv=Av. By Cayley-Hamilton, we have qA(A)=0, where qA is the characteristic polynomial of A. Now observe that qA is monic, and that qA(A)=0 implies qA(x)=0, and we're done!
It quickly follows that if x and y are integral over F then so are x+y, x-y, and xy (use the F-module F[x,y]). Hence the F-integral elements of K constitute a subring F', called the integral closure of F. If F=Z then F' is the ring of algebraic integers in K. If F is a field then F' is more commonly called the algebraic closure of F. In this case, a nonzero element x of K is algebraic over F iff x-1 is, so the algebraic closure of F is also a field. Another consequence is that if x is integral over F, and y is integral over F[x], then y is integral over F (again using F[x,y]). [Actually computing polynomials satisfied by x+y, xy, etc. from the minimal polynomials of x and y is part of the classical and resurgent theory of elimination and resultants.] Moreover, if K is contained in a further commutative ring L with the same unit, then the closure of F' in L is the same as the closure of F in L. In particular, F' is integrally closed in K, and algebraically closed in K if F is a field.
The alternative approach starts with F and a polynomial P in F[z] and constructs a field F' generated by F and a root or roots of P. We've seen this already for F=R and P=z2+1 (when F'=C), and more recently when P is irreducible and F' is generated by a single root. But sometimes we'll want to allow reducible P, and/or to have F' contain all the roots of P rather than just one. We can do this recursively, but F' might then depend on which roots we adjoin, or conceivably even on the order in which we adjoin them. Fortunately, there is one important case in which P determines F' uniquely up to F-isomorphisms (isomorphisms that act trivially on F). This is the case in which F' is generated as a field by (F and) all the roots of P. Such F' is said to be the splitting field of P (over F), because P splits completely into linear factors over F'.
More precisely, one can show:
Let i be an isomorphism from F to a field F; let E/F be a splitting field of some monic polynomial P of positive degree; and let E/F be a splitting field of i(P). Then i extends to an isomorphism from E to E; moreover, the number of such extensions is at most [E:F] (that is, the dimension of E as an F-vector space), with equality if P has distinct roots in E (equivalently, if i(P) has distinct roots in E).
The proof uses the following lemma as an induction step:
Let i be an isomorphism from F to a field F, and let E,E be extensions of F,F. Suppose r is an element of E algebraic over F with minimal polynomial g(x). Then i extends to an isomorphism from F[r] to a subfield of E iff i(g) has a root in E, in which case the number of such extensions equals the number of distinct roots of i(g) in E.
Finite (a.k.a. Galois) Fields
A nice application of splitting fields is Galois' characterization of finite fields: a field Fq of q elements exists if and only if q is a power of a prime p, in which case the field is unique up to isomorphism. We have already seen ``only if''; for ``if'', we can construct Fq as the splitting field of Pq=zq-z over the ``prime field'' Fp=Z/pZ.
Observe first that for any field F of characteristic p, the roots of Pq in F are automatically distinct (because the formal derivative P' equals -1, so P'(x) is nonzero for any root x), and constitute a subfield of F! In particular, if F is the splitting field of Pq over Fp then F consists entirely of the roots of Pq, whence F is a q-element field. It remains to show that Fq is unique up to isomorphism. But in any field of q elements, each element must satisfy zq=z. (This can be proved by generalizing one standard proof of Fermat's little theorem (the case q=p): if z=0 then clearly Pq(z)=0; otherwise, multiplication by z permutes F*, so the product of all the elements of F* is preserved by multiplication by zq-1, whence zq-1=1 and zq=z as claimed.) Since a q-element field automatically contains Fp, it must be the splitting field of Pq over Fp, and we're done.
Further finite field factoids (some of which require more Galois theory, see below):
Let q' be another prime power. Then Fq' contains a copy of Fq iff q' is a power of q, say q'=qf. Then the group of Fq-isomorphisms of Fq' is cyclic of order f, generated by the map taking x to xq. The trace and norm from Fq' to Fq are: tr(x)=x+xq+xq2+...+xqf-1, N(x)=x·xq·xq2·...·xqf-1=x(q'-1)/(q-1). It follows that trace:Fq'->Fq and norm:Fq'*->Fq* are surjective, because their kernels consist of roots of polynomials of degrees q'/q and (q'-1)/(q-1) respectively.
What is the structure of the finite groups (Fq,+) and (Fq*,·)? The former is easy: if q=pf then (Fq,+) is the direct sum of f cyclic groups of order p. As to (Fq*,·), this group is cyclic; this is a bit harder to prove, and can be done by generalizing the standard proof that there's a ``primitive residue'' mod p for each prime p. In fact, any finite subgroup G of the multiplicative group of any field F must be cyclic. This can be proved by induction on |G|. The key fact is that in a field the equation xn=1 has at most n distinct roots.
Separable polynomials, separable extensions, and Galois groups
A monic polynomial P over a field is said to be separable if its roots in a splitting field of P are distinct. This is equivalent to the condition that P and its formal derivative P' have no common factors. For instance, our polynomial xq-x from the previous section is separable.
If F is a subfield of a field K, The Galois group Gal(K/F) is the group of F-isomorphisms of K. If G is a group of isomorphisms of K, the set of elements of K invariant under G is a subfield, denoted KG and called the invariant field of G. Clearly F is contained in KGal(K/F); if F' is a subfield of K, then Gal(K/F) contains Gal(K/F'); and if G contains G' then KG is contained in KG'. In fact, more precise and satisfying results hold for separable extensions of finite degree.
We have already seen in effect that if P in F[z] is separable and K is a splitting field of P then |Gal(K/F)|=[K:F]. On the other hand, for any field K, and any finite group G of automorphisms of K, [K:KG] is at most |G|. This is Artin's Lemma, and can be proved using linear algebra. Assume on the contrary that [K:KG]>|G|. Let xi be n>|G| elements of K that are linearly independent over F. Let V be the K-vector subspace of Kn consisting of (bi) such that the sum of big(xi) vanishes for each g in G. Then V is a vector space of positive dimension stable under the action of G on Kn. Choose a vector b in V that minimizes the number of nonzero bi's, and scale so that one of them equals 1. We claim that all the bi are then in F, which yields a linear dependence among the xi. Indeed, if this were not true, there would be some index j and some Galois group element g such that g(bj) does not equal bj. Then b-g(b) would be a nonzero vector in V with strictly fewer nonzero entries. This completes the proof of Artin's Lemma.
One can now prove the key results of Galois theory.
Let K/F be an extension of fields. We say K/F is a
separable (algebraic) extension if each element of K
is a root of a separable monic polynomial in F[x].
(This condition is automatically satisfied for algebraic extensions
in characteristic zero.) It is a normal extension
if every polynomial in F[x] that has a root in K splits completely in K.
i) K is a splitting field over F of a separable polynomial;
ii) F=KG for some finite group G of automorphisms of K;
iii) K is finite dimensional, normal, and separable over F.
Moreover, if these conditions are satisfied then the G of (ii) is Gal(K/F) and K=FGal(K/F). Finally:
Suppose K is finite dimensional, normal, and separable over F, with Galois group G. Then there is a bijection between subgroups H of G and intermediate fields E that contain F and are contained in K. This bijection takes H to KH and E to Gal(K/E), reverses inclusions between fields and between subgroups, and takes normal subgroups of G to fields E normal over F and vice versa.