## Tag `032L`

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.8. 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) $$ L \times L \longrightarrow K, \quad (x, y) \longmapsto \langle x, y\rangle := \text{Trace}_{L/K}(xy). $$ 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 10.37.6) 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 $$ M = \{y \in L \mid \langle x_i, y\rangle \in R, ~i = 1, \ldots, n\} $$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 42634–42639 (see updates for more information).

```
\begin{lemma}
\label{lemma-Noetherian-normal-domain-finite-separable-extension}
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 $R$.
\end{lemma}
\begin{proof}
Consider the trace pairing
(Fields, Definition \ref{fields-definition-trace-pairing})
$$
L \times L \longrightarrow K,
\quad (x, y) \longmapsto \langle x, y\rangle := \text{Trace}_{L/K}(xy).
$$
Since $L/K$ is separable this is nondegenerate
(Fields, Lemma \ref{fields-lemma-separable-trace-pairing}).
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 \ref{lemma-minimal-polynomial-normal-domain})
and because $\text{Trace}_{L/K}(x)$ is an
integer multiple of one of these coefficients
(Fields, Lemma \ref{fields-lemma-trace-and-norm-from-minimal-polynomial}).
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
$$
M = \{y \in L \mid \langle x_i, y\rangle \in R, \ i = 1, \ldots, n\}
$$
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.
\end{proof}
```

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