Lemma 33.36.5. Let $B$ be a semi-local Noetherian domain of dimension $1$. Let $B'$ be the integral closure of $B$ in its fraction field. Then $B'$ is a semi-local Dedekind domain. Let $x$ be a nonzero element of the Jacobson radical of $B'$. Then for every $y \in B'$ there exists an $n$ such that $x^ n y \in B$.

**Proof.**
Let $\mathfrak m_ B$ be the Jacobson radical of $B$. The structure of $B'$ results from Algebra, Lemma 10.120.18. Given $x, y \in B'$ as in the statement of the lemma consider the subring $B \subset A \subset B'$ generated by $x$ and $y$. Then $A$ is finite over $B$ (Algebra, Lemma 10.36.5). Since the fraction fields of $B$ and $A$ are the same we see that the finite module $A/B$ is supported on the set of closed points of $B$. Thus $\mathfrak m_ B^ n A \subset B$ for a suitable $n$. Moreover, $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is surjective (Algebra, Lemma 10.36.17), hence $A$ is semi-local as well. It also follows that $x$ is in the Jacobson radical $\mathfrak m_ A$ of $A$. Note that $\mathfrak m_ A = \sqrt{\mathfrak m_ B A}$. Thus $x^ m y \in \mathfrak m_ B A$ for some $m$. Then $x^{nm} y \in B$.
$\square$

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