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)

$L \times L \longrightarrow K, \quad (x, y) \longmapsto \langle x, y\rangle := \text{Trace}_{L/K}(xy).$

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. $\square$

Comment #5007 by Laurent Moret-Bailly on

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 on

@#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

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.

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