# The Stacks Project

## Tag 032H

Lemma 10.155.4. Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$. Same for N-2.

Proof. Assume $R_{f_i}$ is N-2 (or N-1). Let $L$ be a finite extension of the fraction field of $R$ (equal to the fraction field in the N-1 case). Let $S$ be the integral closure of $R$ in $L$. By Lemma 10.35.11 we see that $S_{f_i}$ is the integral closure of $R_{f_i}$ in $L$. Hence $S_{f_i}$ is finite over $R_{f_i}$ by assumption. Thus $S$ is finite over $R$ by Lemma 10.23.2. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42547–42552 (see updates for more information).

\begin{lemma}
\label{lemma-Japanese-local}
Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the
unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$.
Same for N-2.
\end{lemma}

\begin{proof}
Assume $R_{f_i}$ is N-2 (or N-1).
Let $L$ be a finite extension of the fraction field of $R$ (equal to
the fraction field in the N-1 case). Let $S$ be the integral
closure of $R$ in $L$. By Lemma \ref{lemma-integral-closure-localize}
we see that $S_{f_i}$ is the integral closure of $R_{f_i}$ in $L$.
Hence $S_{f_i}$ is finite over $R_{f_i}$ by assumption.
Thus $S$ is finite over $R$ by Lemma \ref{lemma-cover}.
\end{proof}

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