Lemma 33.37.5. 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 $x^{nm} y \in B$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)