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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 $Q(A)$.

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  • $\begingroup$ The theory of minimal polynomials may not work here if $B$ doesn't have a field of fractions. $\endgroup$ – Slade Mar 2 '18 at 10:21
  • $\begingroup$ Sorry, i still don't see why. May you elaborate? We have $A \hookrightarrow K \hookrightarrow L$. as $x$ is algebraic over $A$, there exists $p \in K[t]$, $p(x) =0$. Then we are only working in $K[t]$ (?), which is an ED. $\endgroup$ – CL. Mar 2 '18 at 10:36
  • $\begingroup$ There is no such thing as $Q(B)$ if $B$ is not a domain...but there is nothing wrong with talking about algebraic elements if $B$ is not a domain; that just means they satisfy some nonzero polynomial with coefficients in $K$. $\endgroup$ – Eric Wofsey Mar 2 '18 at 17:33
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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.

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