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 15.123.5 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.

Comment #8204 by on

OK, I used Lemma 15.24.2 to show $I$ is finitely generated. (This of course just means that content ideals in valuation rings are always principal. So maybe this shouldn't be mentioned in this lemma. Oh well.) Thanks for the comment! See this change.

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