## Tag `032I`

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.5. 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 $K = f.f.(R) \subset L = f.f.(S)$. Note that this is a finite field extension (for example by Lemma 10.121.2 (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 10.122.15 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 10.155.3 and 10.155.4. 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 $f.f(R)$ and we conclude that the integral closure $T$ of $R$ in $M$ is finite over $R$. By Lemma 10.35.16 we see that $T$ is also the integral closure of $S$ in $M$ and we win by Lemma 10.35.15. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 42564–42571 (see updates for more information).

```
\begin{lemma}
\label{lemma-quasi-finite-over-Noetherian-japanese}
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.
\end{lemma}
\begin{proof}
Let $K = f.f.(R) \subset L = f.f.(S)$. Note that this is
a finite field extension (for example by
Lemma \ref{lemma-isolated-point-fibre} (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 \ref{lemma-quasi-finite-open-integral-closure}
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 \ref{lemma-localize-N} and \ref{lemma-Japanese-local}.
Thus we have reduced to the case where $S$ is finite over $R$.
\medskip\noindent
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 $f.f(R)$ and we conclude that the integral closure $T$ of $R$ in
$M$ is finite over $R$. By Lemma \ref{lemma-integral-closure-transitive}
we see that $T$ is also the integral closure of $S$ in $M$ and we win by
Lemma \ref{lemma-integral-permanence}.
\end{proof}
```

## Comments (0)

## Add a comment on tag `032I`

Your email address will not be published. Required fields are marked.

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.

There are also 3 comments on Section 10.155: Commutative Algebra.