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*}

10.53 Homomorphisms essentially of finite type

Some simple remarks on localizations of finite type ring maps.

Definition 10.53.1. Let $R \to S$ be a ring map.

  1. We say that $R \to S$ is essentially of finite type if $S$ is the localization of an $R$-algebra of finite type.

  2. We say that $R \to S$ is essentially of finite presentation if $S$ is the localization of an $R$-algebra of finite presentation.

Lemma 10.53.2. The class of ring maps which are essentially of finite type is preserved under composition. Similarly for essentially of finite presentation.

Proof. Omitted. $\square$

Lemma 10.53.3. The class of ring maps which are essentially of finite type is preserved by base change. Similarly for essentially of finite presentation.

Proof. Omitted. $\square$

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. $\square$

The following lemma can be proven using properness of projective space instead of the algebraic argument we give here.

Lemma 10.53.5. Let $\varphi : R \to S$ be essentially of finite type with $R$ and $S$ local (but not necessarily $\varphi $ local). Then there exists an $n$ and a maximal ideal $\mathfrak m \subset R[x_1, \ldots , x_ n]$ lying over $\mathfrak m_ R$ such that $S$ is a localization of a quotient of $R[x_1, \ldots , x_ n]_\mathfrak m$.

Proof. We can write $S$ as a localization of a quotient of $R[x_1, \ldots , x_ n]$. Hence it suffices to prove the lemma in case $S = R[x_1, \ldots , x_ n]_\mathfrak q$ for some prime $\mathfrak q \subset R[x_1, \ldots , x_ n]$. If $\mathfrak q + \mathfrak m_ R R[x_1, \ldots , x_ n] \not= R[x_1, \ldots , x_ n]$ then we can find a maximal ideal $\mathfrak m$ as in the statement of the lemma with $\mathfrak q \subset \mathfrak m$ and the result is clear.

Choose a valuation ring $A \subset \kappa (\mathfrak q)$ which dominates the image of $R \to \kappa (\mathfrak q)$ (Lemma 10.49.2). If the image $\lambda _ i \in \kappa (\mathfrak q)$ of $x_ i$ is contained in $A$, then $\mathfrak q$ is contained in the inverse image of $\mathfrak m_ A$ via $R[x_1, \ldots , x_ n] \to A$ which means we are back in the preceding case. Hence there exists an $i$ such that $\lambda _ i^{-1} \in A$ and such that $\lambda _ j/\lambda _ i \in A$ for all $j = 1, \ldots , n$ (because the value group of $A$ is totally ordered, see Lemma 10.49.12). Then we consider the map

\[ R[y_0, y_1, \ldots , \hat{y_ i}, \ldots , y_ n] \to R[x_1, \ldots , x_ n]_\mathfrak q,\quad y_0 \mapsto 1/x_ i,\quad y_ j \mapsto x_ j/x_ i \]

Let $\mathfrak q' \subset R[y_0, \ldots , \hat{y_ i}, \ldots , y_ n]$ be the inverse image of $\mathfrak q$. Since $y_0 \not\in \mathfrak q'$ it is easy to see that the displayed arrow defines an isomorphism on localizations. On the other hand, the result of the first paragraph applies to $R[y_0, \ldots , \hat{y_ i}, \ldots , y_ n]$ because $y_ j$ maps to an element of $A$. This finishes the proof. $\square$


Comments (5)

Comment #673 by Keenan Kidwell on

In the last sentence of the first paragraph of the proof of 07DT, "finite -module" should be "finite -module."

Comment #2257 by Dario Weißmann on

What is the convention for "localization" in this section? Is it localization at a prime or at a multiplicative closed subset? The proof of the last Lemma (10.53.5) seems to suggest the former.

Comment #2262 by on

Localization, when not otherwise specified, just means inverting some multiplicatively closed subset. But if is a local ring and if is the localization of a ring , then of course is the localization of at a prime.

Comment #2264 by Dario Weißmann on

Thank you for your answer. I think in the proof of Lemma 10.53.5 the map does not induce an isomorphism on localization as it is not injective (). If we consider the polynomial ring without the it works fine though. Oh, and there is also a missing comma in .


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