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).

Proof. 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 15.123.5 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. $\square$

Comment #8087 by Laurent Moret-Bailly on

It follows from Comment 8086 that the content ideal is principal.

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).