Lemma 15.25.7. Let $A$ be a valuation ring. Let $A \to B$ be a local homomorphism which is essentially of finite type. Let $M$ be a finite $B$-module.
If $B$ is flat over $A$, then $B$ is essentially of finite presentation over $A$.
If $M$ is flat as an $A$-module, then $M$ is finitely presented as a $B$-module.
Proof. By assumption we can write $B$ as a quotient of the localization of a polynomial algebra $P = A[x_1, \ldots , x_ n]$ at a prime ideal $\mathfrak q$. In case (1) we consider $M = B$ as a finite module over $P_\mathfrak q$ and in case (2) we consider $M$ as a finite module over $P_\mathfrak q$. In both cases, we have to show that this is a finitely presented $P_\mathfrak q$-module, see Algebra, Lemma 10.6.4 for case (2).
Choose a presentation $0 \to K \to P_\mathfrak q^{\oplus r} \to M \to 0$ which is possible because $M$ is finite over $P_\mathfrak q$. Let $L = P^{\oplus r} \cap K$. Then $K = L_\mathfrak q$, see Algebra, Lemma 10.9.15. Then $N = P^{\oplus r}/L$ is a submodule of $M$ and hence flat by Lemma 15.22.10. Since also $N$ is a finite $P$-module, we see that $N$ is finitely presented as a $P$-module by Lemma 15.25.6. Since localization is exact (Algebra, Proposition 10.9.12) we see that $N_\mathfrak q = M$ and we conclude. $\square$
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