Lemma 10.36.23. 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 10.36.2 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 ith 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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