All the rings considered in this post are commutative with 1. Suppose that $A\subseteq B$ are two rings ($A$ is a subring of $B$). The usual definition of integrality is as follows:
Definition. We say that $B$ is integral over $A$ if for every $b\in B$, there exists a monic polynomial $f\in A[x]$ such that $f(b)=0$.
Now, let's define another notion of "integrality".
Definition. We say that $B$ is (locally) integral over $A$ if for every $b\in B$ and for every prime ideal $\mathfrak{q}\subseteq B$, the element $\overline{b}\in B/\mathfrak{q}$ is algebraic over the field $\operatorname{Frac}(A/\mathfrak{p})$ where $\mathfrak{p}:=A\cap\mathfrak{q}$ is the prime ideal of $A$.
It is clear that integral $\Rightarrow$ (locally) integral. My question is whether or not the converse holds:
If $B$ is (locally) integral over $A$, then is $B$ integral over $A$?