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?

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, $Y = Y'$.