[Theorem 3, Nagata-Finitely]

Lemma 15.25.6. Let $A$ be a valuation ring. Let $A \to B$ be a ring map of finite type. Let $M$ be a finite $B$-module.

1. If $B$ is flat over $A$, then $B$ is a finitely presented $A$-algebra.

2. If $M$ is flat as an $A$-module, then $M$ is finitely presented as a $B$-module.

Proof. We are going to use that an $A$-module is flat if and only if it is torsion free, see Lemma 15.22.10. By Algebra, Lemma 10.57.10 we can find a graded $A$-algebra $S$ with $S_0 = A$ and generated by finitely many elements in degree $1$, an element $f \in S_1$ and a finite graded $S$-module $N$ such that $B \cong S_{(f)}$ and $M \cong N_{(f)}$. If $M$ is torsion free, then we can take $N$ torsion free by replacing it by $N/N_{tors}$, see Lemma 15.22.2. Similarly, if $B$ is torsion free, then we can take $S$ torsion free by replacing it by $S/S_{tors}$. Hence in case (1), we may apply Lemma 15.25.4 to see that $S$ is a finitely presented $A$-algebra, which implies that $B = S_{(f)}$ is a finitely presented $A$-algebra. To see (2) we may first replace $S$ by a graded polynomial ring, and then we may apply Lemma 15.25.3 to conclude. $\square$

Comment #2756 by Kestutis Cesnavicius on

For the sake of keeping track of references, it may be worthwhile to mention that part (1) of this lemma is Thoerem 3 in Nagata "Finitely generated rings over a valuation ring."

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