Lemma 10.161.5. Let $R$ be a domain. Let $R \subset S$ be a quasi-finite extension of domains (for example finite). Assume $R$ is N-2 and Noetherian. Then $S$ is N-2.

Proof. Let $L/K$ be the induced extension of fraction fields. Note that this is a finite field extension (for example by Lemma 10.122.2 (2) applied to the fibre $S \otimes _ R K$, and the definition of a quasi-finite ring map). Let $S'$ be the integral closure of $R$ in $S$. Then $S'$ is contained in the integral closure of $R$ in $L$ which is finite over $R$ by assumption. As $R$ is Noetherian this implies $S'$ is finite over $R$. By Lemma 10.123.14 there exist elements $g_1, \ldots , g_ n \in S'$ such that $S'_{g_ i} \cong S_{g_ i}$ and such that $g_1, \ldots , g_ n$ generate the unit ideal in $S$. Hence it suffices to show that $S'$ is N-2 by Lemmas 10.161.3 and 10.161.4. Thus we have reduced to the case where $S$ is finite over $R$.

Assume $R \subset S$ with hypotheses as in the lemma and moreover that $S$ is finite over $R$. Let $M$ be a finite field extension of the fraction field of $S$. Then $M$ is also a finite field extension of $K$ and we conclude that the integral closure $T$ of $R$ in $M$ is finite over $R$. By Lemma 10.36.16 we see that $T$ is also the integral closure of $S$ in $M$ and we win by Lemma 10.36.15. $\square$

There are also:

• 5 comment(s) on Section 10.161: Japanese rings

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).