Lemma: Let $C$ be the integral closure of $A$ in $B$ and let ${\mathfrak a}^e$ denote the extension of $\mathfrak a$ in $C$. Then the integral closure of $\mathfrak a$ in $B$ is the radical of $\mathfrak a ^e $ .
Proposition: Let $A \subseteq B$ be integral domains, $A$ integrally closed, and $x \in B$ be integral over an ideal $\mathfrak a$ of $A$. Then $x$ is algebraic over the field of fractions $K$of $\mathfrak{a}$ and if its minimal polynomial over $K$ is $t^n+ \cdots + a_n$, then $a_i \in r(\mathfrak{a})$.
The proof is here. I did not see where $B$ being an integral domain was necessary. Am I missing something?
EDIT: here is what I think:
To make sense of $x$ being algebraic over $K=Q(A)$. We need a larger field which contains $x$. We can use the universal proeprty, to obtian a map, $ Q(A) \hookrightarrow Q(B)$. Now $x \in Q(B)$, and is algebraic over $Q(A)$.
As far as I can tell the proof works fine even if $B$ is not a domain. Beware though that the minimal polynomial of $x$ may not be irreducible in that case, and so the term "conjugates" is not really appropriate; instead you just want the roots of the minimal polynomial in some field $L$ extending $K$ where it splits.