Make lemma iff

THanks to Rankeya
https://stacks.math.columbia.edu/tag/0333#comment-5006

algebra.tex

@@ -43598,18 +43598,25 @@ \section{Japanese rings}
 
 \begin{lemma}
 \label{lemma-characterize-N-1}
-Let $R$ be a Noetherian domain.
-Assume
+Let $R$ be a Noetherian domain. Then $R$ is N-1 if and only if the following
+two conditions hold
 \begin{enumerate}
 \item there exists a nonzero $f \in R$ such that $R_f$ is normal, and
 \item for every maximal ideal $\mathfrak m \subset R$
 the local ring $R_{\mathfrak m}$ is N-1.
 \end{enumerate}
-Then $R$ is N-1.
 \end{lemma}
 
 \begin{proof}
-Let $K$ be the fraction field of $R$.
+First assume $R$ is N-1. Let $R'$ be the integral closure of $R$ in its
+field of fractions $K$. By assumption we can find $x_1, \ldots, x_n$ in $R'$
+which generate $R'$ as an $R$-module. Since $R' \subset K$ we can find
+$f_i \in R$ nonzero such that $f_i x_i \in R$. Then $R_f \cong R'_f$
+where $f = f_1 \ldots f_n$. Hence $R_f$ is normal and we have (1).
+Part (2) follows from Lemma \ref{lemma-localize-N}.
+
+\medskip\noindent
+Assume (1) and (2). Let $K$ be the fraction field of $R$.
 Suppose that $R \subset R' \subset K$ is a finite
 extension of $R$ contained in $K$. Note that $R_f = R'_f$ since
 $R_f$ is already normal. Hence by Lemma \ref{lemma-openness-normal-locus}