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 10.6.4. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume $R \to S$ is of finite type and $M$ is finitely presented as an $R$-module. Then $M$ is finitely presented as an $S$-module.

Proof. This is similar to the proof of part (4) of Lemma 10.6.2. We may assume $S = R[x_1, \ldots , x_ n]/J$. Choose $y_1, \ldots , y_ m \in M$ which generate $M$ as an $R$-module and choose relations $\sum a_{ij} y_ j = 0$, $i = 1, \ldots , t$ which generate the kernel of $R^{\oplus m} \to M$. For any $i = 1, \ldots , n$ and $j = 1, \ldots , m$ write

\[ x_ i y_ j = \sum a_{ijk} y_ k \]

for some $a_{ijk} \in R$. Consider the $S$-module $N$ generated by $y_1, \ldots , y_ m$ subject to the relations $\sum a_{ij} y_ j = 0$, $i = 1, \ldots , t$ and $x_ i y_ j = \sum a_{ijk} y_ k$, $i = 1, \ldots , n$ and $j = 1, \ldots , m$. Then $N$ has a presentation

\[ S^{\oplus nm + t} \longrightarrow S^{\oplus m} \longrightarrow N \longrightarrow 0 \]

By construction there is a surjective map $\varphi : N \to M$. To finish the proof we show $\varphi $ is injective. Suppose $z = \sum b_ j y_ j \in N$ for some $b_ j \in S$. We may think of $b_ j$ as a polynomial in $x_1, \ldots , x_ n$ with coefficients in $R$. By applying the relations of the form $x_ i y_ j = \sum a_{ijk} y_ k$ we can inductively lower the degree of the polynomials. Hence we see that $z = \sum c_ j y_ j$ for some $c_ j \in R$. Hence if $\varphi (z) = 0$ then the vector $(c_1, \ldots , c_ m)$ is an $R$-linear combination of the vectors $(a_{i1}, \ldots , a_{im})$ and we conclude that $z = 0$ as desired. $\square$


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