Theorem 1   Let $B$ be an integrally closed Noetherian domain with field of fractions $K$ . Let $L$ be a finite separable extension of $K$ , and let $A$ be the integral closure of $B$ in $L$ . Then $A$ is a finitely generated $B$ -module.

Proof. We first show that the trace $Tr^L_K$ maps $A$ to $B$ . Choose $u\in A$ and let $f=Irr(u,K)\in K[x]$ be the minimal polynomial for $u$ over $K$ ; assume $f$ is of degree $d$ . Let the conjugates of $u$ in some splitting field be $u=a_1,\ldots,a_d$ . Then the $a_i$ are all integral over $B$ since they satisfy $u$ 's monic polynomial in $B[x]$ . Since the coefficients of $F$ are polynomials in the $a_i$ , they too are integral over $B$ . But the coefficients are in $K$ , and $B$ is integrally closed (in $K$ ), so the coefficients are in $B$ . But $Tr^L_K(u)$ is just the coefficient of $x^{d-1}$ in $f$ , and thus $Tr^L_K(u)\in B$ . This proves the claim.

Now, choose a basis $\omega_1,\ldots,\omega_d$ of $L/K$ . We may assume $\omega_i\in A$ by multiplying each by an appropriate element of $B$ . (To see this, let $Irr(\omega_i,K)\in K[x] = x^d+k_1x^{d-1}+\ldots+k_d$ . Choose $b\in B$ such that $bk_i\in B\ \forall i$ . Then $(b\omega)^d+bk_1(b\omega)^{d-1}+\ldots+b^dk_d=0$ and thus $b\omega\in A$ ). Define a linear map $\varphi:L\rightarrow K^d:a\mapsto(Tr^L_K (a\omega_1),\ldots,Tr^L_K (a\omega_d))$ .

$\varphi$ is 1-1, since if $u\in\ker\varphi, u\neq 0$ , then $Tr(uL)=0$ . But $uL=L$ , so $Tr^L_K$ is identically zero, which cannot be since $L$ is separable over $K$ (it is a standard result that separability is equivalent to nonvanishing of the trace map; see for example [1], Chapter 8).

But $Tr^L_K:A\rightarrow B$ by the above, so $\varphi:A\hookrightarrow B^d$ . Since $B$ is Noetherian, any submodule of a finitely generated module is also finitely generated, so $A$ is finitely generated as a $B$ -module.