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 $i$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. $\square$

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.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0564. Beware of the difference between the letter 'O' and the digit '0'.