## Algebraic and integral closures

Algebraic closure. Let K/F be any field extension. The algebraic closure of F in K is the subset of K consisting of elements algebraic over F. The subfield F of K is said to be algebraically closed in K if it is its own algebraic closure in K.

Theorem. Let K/F be any field extension. The algebraic closure of F in K is a field, and is algebraically closed in K.

This is easy to prove once we can use Theorems 4.1 and 4.2 to prove that the sum and product of any elements of K algebraic over F are themselves algebraic, and that an element of K algebraic over the algebraic closure of F in K is itself algebraic over F.
Actually finding minimal polynomials for such algebraic field elements can be a nontrivial computational task. This is part of the theory of elimination and generalized resultants. That theory flourished in the latter part of the 19th century, but was approaching its practical limits just as Hilbert, Noether, and others began developing the abstract approach that still dominates modern algebra. Still, elimination theory has been making a comeback thanks to modern computer hardware and sophisticated software that greatly extend the theory's range of feasibility.

Integral closure. Now let A be a subring of B. An element u of B is said to be integral over A if there exists a monic polynomial f in A[X] such that f(u)=0. Theorem 4.1 has an analogue in this setting:

Lemma. Let A be a subring of some ring B, and u any element of B. Then u is integral over A if and only if B contains an an A[u]-module M that contains 1 and is finitely generated as an A-module.

(This is a version of "Lemma 1" of Jacobson 4.12.)

Proof: If u is integral over A then f(u)=0 for some monic polynomial f in A[X]. Let n be the degree of f. Then {1,u,u2,...,un-1} is a finite generating set for A[u]. Indeed, A[u] is generated by {um: m=0,1,2,...}; if m is greater than or equal to n, then um is an A-linear combination of um-n, um-n+1, ..., um-1; hence by induction each um is an A-linear combination of 1,u,u2,...,un-1. We may thus take M=A[u].

Suppose, conversely, that M is an A[u]-module finitely generated over A, and let x1,...,xN be generators. Then, for each i, we can write uxi as an A-linear combination of the xj. Using the coefficients of these linear combinations, construct an N-by-N matrix T over A such that ux=Tx, where x is the column vector (x1,...,xN). Let f be the characteristic polynomial of T. Then f is a monic polynomial in A[X]. Moreover, f(u)x=(f(T))(x)=0 because f(T)=0. Thus f(u)v=0 for all v in M. If moreover M contains 1 then we may take v=1 and find that f(u)=0. We have then shown that u is a root of a monic polynomial in A[X], and conclude that u is integral over A as claimed. QED

The integral closure of A in B is the subset of B consisting of elements integral over A; the subring A of B is said to be integrally closed in B if it is its own integral closure in B.

Theorem (4.23). Let A be a subring of B. The integral closure of A in B is a ring, and is integrally closed in B.

The proof is much the same as for algebraic closures. For instance, if u and v are integral over A and satisfy equations of degrees m and n, then {uivj: i=0,...,m-1, j=0,...,n-1} is a finite generating set for the A-module M=A[u,v]. Since M contains 1 and is closed under multiplication by u+v and uv, we conclude that u+v and uv are integral over A. Likewise we show that if u satisfies a monic polynomial equation f(u)=0 all of whose coefficients are integral over A, then the ring generated by u and the coefficients is finitely generated over A, whence u is algebraic over A.

Warning: It is certainly true that if K/F is a field extension, and A is a subring of K, then any element u of the integral closure of A is algebraic over F. However, it need not satisfy a monic polynomial equation over A of the same degree as the degree [F(u):F] of u over F. In general a larger degree may be necessary. This may happen even if u is in F! For instance let F=K=C, A=Z[2i]. Then i is A-integral (i2+1=0), but satisfies no monic degree-1 equation over A. Thus A is a domain (commutative ring without zero divisors) that is not integrally closed in its own fraction field (here Q(i)). It turns out that if A is integrally closed in its own fraction field F, then any A-integral element u of an extension field of F has a minimal polynomial in A[X], and thus satisfies a monic equation over A of degree [F(u):F]. We shall be able to prove this after developing Galois theory.