Lemma 15.14.4. Let $A$ be a ring. Let $S \subset A$ be a multiplicative subset consisting of nonzerodivisors. If $S^{-1}A$ is absolutely integrally closed and $A \subset S^{-1}A$ is integrally closed in $S^{-1}A$, then $A$ is absolutely integrally closed.

Proof. Omitted. $\square$

## Comments (2)

Comment #6486 by Laurent Moret-Bailly on

Writing $A\subset S^{-1}A$ means implicitly that $S$ consists of nonzerodivisors. Perhaps it would be better to say it.

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