# Geometric meaning of being integrally closed in some overring

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?

• 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.
– user143488
Commented Apr 22, 2014 at 13:31
• I should have said that $X$ and $Y$ are integral algebraic varieties.
– user143488
Commented Apr 22, 2014 at 15:11
• And do you mean rational maps or total maps? Commented Apr 22, 2014 at 15:13
• Total (morphism).
– user143488
Commented Apr 22, 2014 at 15:30

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, $Y = Y'$.