integral closures in separable extensions are finitely generated

The theorem below generalizes to arbitrary integral ring extensions (under certain conditions) the fact that the ring of integers of a number field is finitely generated over $\mathbb{Z}$. The proof parallels the proof of the number field result.

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 (http://planetmath.org/Trace2) $Tr^{L}_{K}$ 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 $Irr(\omega_{i},K)\in K[x]=x^{d}+k_{1}x^{d-1}+\ldots+k_{d}$. Choose $b\in B$ such that $bk_{i}\in B\ \forall i$. Then $(b\omega)^{d}+bk_{1}(b\omega)^{d-1}+\ldots+b^{d}k_{d}=0$ 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. ∎

References

Title integral closures in separable extensions are finitely generated IntegralClosuresInSeparableExtensionsAreFinitelyGenerated 2013-03-22 17:02:12 2013-03-22 17:02:12 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 13B21 msc 12F05 IntegralClosureIsRing