The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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

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

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 10.35.2 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

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

Then we see that $S$ is of finite presentation as an $S'$-algebra by Lemma 10.6.2. Since $S' \to S$ is surjective 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

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

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 $1, x, \ldots , x^{d - 1}$. $\square$


Comments (3)

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.


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'.