Lemma 10.161.8 (032L)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.161: Japanese rings
Lemma 10.161.8 (cite)

Lemma 10.161.8. Let $R$ be a Noetherian normal domain with fraction field $K$. Let $L/K$ 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 10.38.6) 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

\[ M = \{ y \in L \mid \langle x_ i, y\rangle \in R, \ i = 1, \ldots , n\} \]

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. $\square$

Comments (7)

Comment #5007 by Laurent Moret-Bailly on March 30, 2020 at 12:05

If $R$ is not noetherian, the integral closure is still contained in a finite $R$ -submodule of $L$ . This can be useful, and moreover it is the key point in the proof, so it could be worth stating.

Comment #5246 by Johan on June 15, 2020 at 10:58

@#5007: OK, I am going to leave it for now. Maybe this is a lack of imagination but I can't think of a way in which this could be useful (and if it becomes useful for somebody, they would almost immediately figure it out if they look at this lemma... and your comment). Thanks for the comment!

Comment #5253 by Laurent Moret-Bailly on June 15, 2020 at 11:48

You are welcome. My view on this is that (regardless of the existence of applications) you get a better statement if you stick to what is actually proved, especially if you can save an assumption.

Comment #7426 by Jianing Li on June 15, 2022 at 21:34

The proof seems also work when R is not normal.

Comment #7439 by Zhiyu Z on June 21, 2022 at 04:39

@7426 If the proposition is true for a domain $R$ , then take $L=K$ we see the integral closure of $R$ is finite over $R$ , i.e. the normalization map for R is finite. However, the normalization map for R is not always a finite morphism: this property is called N-1 in Definition 10.161.1. See Remark 28.13.2 for an one dimensional Noetherian domain counterexample.

Comment #7445 by Jianing Li on June 22, 2022 at 12:00

@7439 Thanks for pointing out this. I now realize my mistake. The normality is necessary for the assertion that, if $x$ is integral over $R$ , then $Trace_{L/K}(x)\in R$ .

Comment #10020 by Manolis C. Tsakiris on March 01, 2025 at 22:22

The writing should be improved a bit. First, as the comments above suggest, it should be made clear where normality is used. Second, "By linear algebra" should be replaced by a more transparent argument, which also highlights where separability is used (for instance, by separabiliity the bilinear form is non-degenerate and thus $x_1,\dots,x_n$ can be chosen to be orthonormal).

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