Lemma 38.19.6. Let $A \to B$ be a local ring map of local rings which is essentially of finite type. Let $N$ be a finite $B$-module which is flat as an $A$-module. If $A$ is a valuation ring, then any element of $N$ has a content ideal $I \subset A$ (More on Algebra, Definition 15.24.1). Also, $I$ is a principal ideal.

**Proof.**
The final statement follows from the fact that $I$ is a finitely generated ideal by More on Algebra, Lemma 15.24.2 and Algebra, Lemma 10.50.15.

Proof of existence of $I$. Let $A \subset A^ h$ be the henselization. Let $B'$ be the localization of $B \otimes _ A A^ h$ at the maximal ideal $\mathfrak m_ B \otimes A^ h + B \otimes \mathfrak m_{A^ h}$. Then $B \to B'$ is flat, hence faithfully flat. Let $N' = N \otimes _ B B'$. Let $x \in N$ and let $x' \in N'$ be the image. We claim that for an ideal $I \subset A$ we have $x \in IN \Leftrightarrow x' \in IN'$. Namely, $N/IN \to N'/IN'$ is the tensor product of $B \to B'$ with $N/IN$ and $B \to B'$ is universally injective by Algebra, Lemma 10.82.11. By More on Algebra, Lemma 15.123.6 and Algebra, Lemma 10.50.17 the map $A \to A^ h$ defines an inclusion preserving bijection $I \mapsto IA^ h$ on sets of ideals. We conclude that $x$ has a content ideal in $A$ if and only if $x'$ has a content ideal in $A^ h$. The assertion for $x' \in N'$ follows from Lemma 38.19.5 and Algebra, Lemma 10.89.6. $\square$

## 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Your email address will not be published.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 (2)

Comment #8087 by Laurent Moret-Bailly on

Comment #8204 by Stacks Project on