Lemma 38.25.6. If in Situation 38.25.1 the ring $A$ is a valuation ring then the lemma holds.

**Proof.**
Recall that an $A$-module is flat if and only if it is torsion free, see More on Algebra, Lemma 15.22.10. Let $T \subset N$ be the $A$-torsion. Then $u(T) = 0$ and $N/T$ is $A$-flat. Hence $N/T$ is finitely presented over $B$, see More on Algebra, Lemma 15.25.6. Thus $T$ is a finite $B$-module, see Algebra, Lemma 10.5.3. Since $N/T$ is $A$-flat we see that $T/\mathfrak m_ A T \subset N/\mathfrak m_ A N$, see Algebra, Lemma 10.39.12. As $\overline{u}$ is injective but $u(T) = 0$, we conclude that $T/\mathfrak m_ A T = 0$. Hence $T = 0$ by Nakayama's lemma, see Algebra, Lemma 10.20.1. At this point we have proved two out of the three assertions ($N$ is $A$-flat and of finite presentation over $B$) and what is left is to show that $u$ is universally injective.

By Algebra, Theorem 10.82.3 it suffices to show that $N \otimes _ A Q \to M \otimes _ A Q$ is injective for every finitely presented $A$-module $Q$. By More on Algebra, Lemma 15.124.3 we may assume $Q = A/(a)$ with $a \in \mathfrak m_ A$ nonzero. Thus it suffices to show that $N/aN \to M/aM$ is injective. Let $x \in N$ with $u(x) \in aM$. By Lemma 38.19.6 we know that $x$ has a content ideal $I \subset A$. Since $I$ is finitely generated (More on Algebra, Lemma 15.24.2) and $A$ is a valuation ring, we have $I = (b)$ for some $b$ (by Algebra, Lemma 10.50.15). By More on Algebra, Lemma 15.24.3 the element $u(x)$ has content ideal $I$ as well. Since $u(x) \in aM$ we see that $(b) \subset (a)$ by More on Algebra, Definition 15.24.1. Since $x \in bN$ we conclude $x \in aN$ as desired. $\square$

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