The Stacks project

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 means implicitly that consists of nonzerodivisors. Perhaps it would be better to say it.

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