Let $D$ be a Dedekind domain, $F$ its field of fractions, $E$ a finite dimensional extension field of $F$ and $D'$ the subring of $E$ of $D$ integral elements. Assume that $E/F$ is a finite separable field extension.
Let $P'$ be a prime ideal in $D'$ and let $r$ be a non-zero element in $P'$. Then $f(r)=0$ for the polynomial field $f(x)$ of $r$. Why does it follow that $f(x)$ is in $P'$?
Would you help, me, please? Thank you in advance.