Lemma 10.35.23. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. Then $M$ is finitely presented as an $R$-module if and only if $M$ is finitely presented as an $S$-module.

**Proof.**
One of the implications follows from Lemma 10.6.4. To see the other assume that $M$ is finitely presented as an $S$-module. Pick a presentation

As $S$ is finite as an $R$-module, the kernel of $S^{\oplus n} \to M$ is a finite $R$-module. Thus from Lemma 10.5.3 we see that it suffices to prove that $S$ is finitely presented as an $R$-module.

Pick $y_1, \ldots , y_ n \in S$ such that $y_1, \ldots , y_ n$ generate $S$ as an $R$-module. By Lemma 10.35.2 each $y_ i$ is integral over $R$. Choose monic polynomials $P_ i(x) \in R[x]$ with $P_ i(y_ i) = 0$. Consider the ring

Then we see that $S$ is of finite presentation as an $S'$-algebra by Lemma 10.6.2. Since $S' \to S$ is surjective we see that $S$ is of finite presentation as an $S'$-module (use Lemma 10.6.3). Hence, arguing as in the first paragraph, it suffices to show that $S'$ is of finite presentation as an $R$-module. To see this we write $R \to S'$ as the composition

of ring maps of the form $R' \to R'[x]/(x^ d + a_1 x^{d - 1} + \ldots + a_ d)$. Again arguing as in the first paragraph of the proof it is enough to show that the $i$th ring in this sequence is of finite presentation as a module over the $(i - 1)$st one. This is true because $R'[x]/(x^ d + a_1 x^{d - 1} + \ldots + a_ d)$ is free as a module over $R'$ with basis $1, x, \ldots , x^{d - 1}$. $\square$

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