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The Stacks project

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

S^{\oplus m} \longrightarrow S^{\oplus n} \longrightarrow M \longrightarrow 0

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

S' = R[x_1, \ldots , x_ n]/(P_1(x_1), \ldots , P_ n(x_ n))

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

R \to R[x_1]/(P_1(x_1)) \to R[x_1, x_2]/(P_1(x_1), P_2(x_2)) \to \ldots \to S'

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


Comments (5)

Comment #747 by Fan on

Can the first paragraph of argument be replaced by lemma 6.2?

Comment #768 by Fan on

I'm just referring to the reduction to showing that S is finitely presented as an R module. It's a little more direct to use Lemma 6.2.

Comment #5425 by minsom on

In middle, "Since S′→S is surjective we see that S is of finite presentation as an S′-module (use Lemma 10.6.3)." How to use lemma 10.6.3? It looks like quite differenet arguement.

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  • 2 comment(s) on Section 10.36: Finite and integral ring extensions

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