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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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