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The geometric counterpart of integrally closed rings (in their fraction fields) are normal varieties, as described in this MathOverflow post.

Is their a similar notion in algebraic geometry for being integrally closed in some ring $S$, when $S$ is not necessarily the fraction field?

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    $\begingroup$ If $f: X\to Y$ is a dominant morphism of algebraic varieties, one can construct the normalization $Y'$ of $Y$ in $X$: this is an integral variety such that $f$ factors through $X\to Y'\to Y$, where $Y'\to Y$ is finite and anything between $X$ and $Y'$, finite over $Y'$, is necessarily equal to $Y'$. The $X$ correspond to your $S$ and $Y$ the base ring. $\endgroup$
    – user143488
    Commented Apr 22, 2014 at 13:31
  • $\begingroup$ I should have said that $X$ and $Y$ are integral algebraic varieties. $\endgroup$
    – user143488
    Commented Apr 22, 2014 at 15:11
  • $\begingroup$ And do you mean rational maps or total maps? $\endgroup$ Commented Apr 22, 2014 at 15:13
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    $\begingroup$ Total (morphism). $\endgroup$
    – user143488
    Commented Apr 22, 2014 at 15:30

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This is the answer by cant_log with a reference. One can consider the normalization of $Y$ in $X$ if $f : X \to Y$ is a quasi-compact and quasi-separated morphism of schemes, see Section Tag 035E. The normalization of $Y$ in $X$ is a factorization $X \to Y' \to Y$ of $f$ such that for every affine open $V \subset Y$ the inverse image $V'$ of $V$ in $Y'$ is also affine and such that $$ \mathcal{O}_{Y'}(V') = \{g \in \mathcal{O}_X(f^{-1}(V)) \mid g\text{ is integral over }\mathcal{O}_Y(V)\} $$ This will at least tell you how to construct $Y'$ if $Y$ is affine and in general you just glue the affine pieces together. In particular, if $X \to Y$ is the morphism associated to a ring map $B \to A$, then $Y'$ is the spectrum of the integral closure of $B$ in $A$.

So the analogue of "$B$ being integrally closed in $A$" would be "the normalization of $Y$ in $X$ is $Y$", in other words, $Y = Y'$.

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