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, the kernel $J = \mathop{\mathrm{Ker}}(S' \to S)$ is finitely generated as an ideal by Lemma 10.6.3. Hence $J$ is a finite $S'$-module (immediate from the definitions). Thus $S = \mathop{\mathrm{Coker}}(J \to S')$ is of finite presentation as an $S'$-module by Lemma 10.5.3. Hence, arguing as in the first paragraph, it suffices to show that $S'$ is of finite presentation as an $R$-module. Actually, $S'$ is free as an $R$-module with basis the monomials $x_1^{e_1} \ldots x_ n^{e_ n}$ for $0 \leq e_ i < \deg (P_ i)$. Namely, write $R \to S'$ as the composition

This shows that the $i$th ring in this sequence is free as a module over the $(i - 1)$st one with basis $1, x_ i, \ldots , x_ i^{\deg (P_ i) - 1}$. The result follows easily from this by induction. Some details omitted. $\square$

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