Normality
For an integral scheme, being normal (integrally closed in ones own fraction field) satisfies this property. Indeed, suppose that $a, b \in A = \Gamma(X, O_X)$ are such that $a/b$ satisfy some polynomial $p(x) \in A[x]$. Then $a|_U, b|_U$ satisfy the same polynomial after restriction to each (affine) set $U \subseteq X$.Of course this shows up in many applications of things like Stein factorization.
Semi-normality
A reduced ring $R$ is seminormal if for any finite extension $R \subseteq S$ satisfying the following two properties is an isomorphism.
- The induced map on $\text{Spec}$'s is an isomorphism
- The induced residue field extensions $k(r) \subseteq k(s)$ are isomorphisms for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.
The typical example of a seminormal ring is a node, the cusp $k[x^2,x^3]$ is not seminormal
Equivalently, $R$ is seminormal if and only if for any $a/b$ in the total ring of fractions of $R$, one has that if $(a/b)^2, (a/b)^3 \in R$ then $(a/b) \in R$ (see a paper by Swan, he might be assuming finitely many minimal primes, I forget the details). It follows similarly that seminormality satisfies this property.
Weak normality
Weak normality is similar to semi-normality. A reduced ring is called weakly normal if for any finite *birational* extension $R \subseteq S$ satisfying the following properties is an isomorphism:- The induced map on $\text{Spec}$'s is an isomorphism
- The induced residue field extensions $k(r) \subseteq k(s)$ are purely for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.
I do NOT know if weakly normal rings satisfy the sort of property asked for. I do not think it is in the literature (but perhaps I am wrong). I remember I convinced myself that they did not several years ago, but never wrote down an example carefully.