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 10.53.4. Let $R \to S$ be a ring map. Assume $S$ is an Artinian local ring with maximal ideal $\mathfrak m$. Then

  1. $R \to S$ is finite if and only if $R \to S/\mathfrak m$ is finite,

  2. $R \to S$ is of finite type if and only if $R \to S/\mathfrak m$ is of finite type.

  3. $R \to S$ is essentially of finite type if and only if the composition $R \to S/\mathfrak m$ is essentially of finite type.

Proof. If $R \to S$ is finite, then $R \to S/\mathfrak m$ is finite by Lemma 10.7.3. Conversely, assume $R \to S/\mathfrak m$ is finite. As $S$ has finite length over itself (Lemma 10.52.6) we can choose a filtration

\[ 0 \subset I_1 \subset \ldots \subset I_ n = S \]

by ideals such that $I_ i/I_{i - 1} \cong S/\mathfrak m$ as $S$-modules. Thus $S$ has a filtration by $R$-submodules $I_ i$ such that each successive quotient is a finite $R$-module. Thus $S$ is a finite $R$-module by Lemma 10.5.3.

If $R \to S$ is of finite type, then $R \to S/\mathfrak m$ is of finite type by Lemma 10.6.2. Conversely, assume that $R \to S/\mathfrak m$ is of finite type. Choose $f_1, \ldots , f_ n \in S$ which map to generators of $S/\mathfrak m$. Then $A = R[x_1, \ldots , x_ n] \to S$, $x_ i \mapsto f_ i$ is a ring map such that $A \to S/\mathfrak m$ is surjective (in particular finite). Hence $A \to S$ is finite by part (1) and we see that $R \to S$ is of finite type by Lemma 10.6.2.

If $R \to S$ is essentially of finite type, then $R \to S/\mathfrak m$ is essentially of finite type by Lemma 10.53.2. Conversely, assume that $R \to S/\mathfrak m$ is essentially of finite type. Suppose $S/\mathfrak m$ is the localization of $R[x_1, \ldots , x_ n]/I$. Choose $f_1, \ldots , f_ n \in S$ whose congruence classes modulo $\mathfrak m$ correspond to the congruence classes of $x_1, \ldots , x_ n$ modulo $I$. Consider the map $R[x_1, \ldots , x_ n] \to S$, $x_ i \mapsto f_ i$ with kernel $J$. Set $A = R[x_1, \ldots , x_ n]/J \subset S$ and $\mathfrak p = A \cap \mathfrak m$. Note that $A/\mathfrak p \subset S/\mathfrak m$ is equal to the image of $R[x_1, \ldots , x_ n]/I$ in $S/\mathfrak m$. Hence $\kappa (\mathfrak p) = S/\mathfrak m$. Thus $A_\mathfrak p \to S$ is finite by part (1). We conclude that $S$ is essentially of finite type by Lemma 10.53.2. $\square$


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