Lemma 10.161.8. Let $R$ be a Noetherian normal domain with fraction field $K$. Let $K \subset L$ be a finite separable field extension. Then the integral closure of $R$ in $L$ is finite over $R$.

**Proof.**
Consider the trace pairing (Fields, Definition 9.20.6)

Since $L/K$ is separable this is nondegenerate (Fields, Lemma 9.20.7). Moreover, if $x \in L$ is integral over $R$, then $\text{Trace}_{L/K}(x)$ is in $R$. This is true because the minimal polynomial of $x$ over $K$ has coefficients in $R$ (Lemma 10.38.6) and because $\text{Trace}_{L/K}(x)$ is an integer multiple of one of these coefficients (Fields, Lemma 9.20.3). Pick $x_1, \ldots , x_ n \in L$ which are integral over $R$ and which form a $K$-basis of $L$. Then the integral closure $S \subset L$ is contained in the $R$-module

By linear algebra we see that $M \cong R^{\oplus n}$ as an $R$-module. Hence $S \subset R^{\oplus n}$ is a finitely generated $R$-module as $R$ is Noetherian. $\square$

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