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Tag 00EO

Chapter 10: Commutative Algebra > Section 10.23: More glueing results

Zariski-local properties of modules and algebras

Lemma 10.23.2. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S$ be an $R$-algebra. Suppose that $f_1, \ldots, f_n$ is a finite list of elements of $R$ such that $\bigcup D(f_i) = \mathop{\rm Spec}(R)$ in other words $(f_1, \ldots, f_n) = R$.

  1. If each $M_{f_i} = 0$ then $M = 0$.
  2. If each $M_{f_i}$ is a finite $R_{f_i}$-module, then $M$ is a finite $R$-module.
  3. If each $M_{f_i}$ is a finitely presented $R_{f_i}$-module, then $M$ is a finitely presented $R$-module.
  4. Let $M \to N$ be a map of $R$-modules. If $M_{f_i} \to N_{f_i}$ is an isomorphism for each $i$ then $M \to N$ is an isomorphism.
  5. Let $0 \to M'' \to M \to M' \to 0$ be a complex of $R$-modules. If $0 \to M''_{f_i} \to M_{f_i} \to M'_{f_i} \to 0$ is exact for each $i$, then $0 \to M'' \to M \to M' \to 0$ is exact.
  6. If each $R_{f_i}$ is Noetherian, then $R$ is Noetherian.
  7. If each $S_{f_i}$ is a finite type $R$-algebra, so is $S$.
  8. If each $S_{f_i}$ is of finite presentation over $R$, so is $S$.

Proof. We prove each of the parts in turn.

  1. By Proposition 10.9.10 this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \mathop{\rm Spec}(R)$, so we conclude by Lemma 10.23.1.
  2. For each $i$ take a finite generating set $X_i$ of $M_{f_i}$. Without loss of generality, we may assume that the elements of $X_i$ are in the image of the localization map $M \rightarrow M_{f_i}$, so we take a finite set $Y_i$ of preimages of the elements of $X_i$ in $M$. Let $Y$ be the union of these sets. This is still a finite set. Consider the obvious $R$-linear map $R^Y \rightarrow M$ sending the basis element $e_y$ to $y$. By assumption this map is surjective after localizing at an arbitrary prime ideal $\mathfrak p$ of $R$, so it surjective by Lemma 10.23.1 and $M$ is finitely generated.
  3. By (2) we have a short exact sequence $$ 0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0 $$ Since localization is an exact functor and $M_{f_i}$ is finitely presented we see that $K_{f_i}$ is finitely generated for all $1 \leq i \leq n$ by Lemma 10.5.3. By (2) this implies that $K$ is a finite $R$-module and therefore $M$ is finitely presented.
  4. By Proposition 10.9.10 the assumption implies that the induced morphism on localizations at all prime ideals is an isomorphism, so we conclude by Lemma 10.23.1.
  5. By Proposition 10.9.10 the assumption implies that the induced sequence of localizations at all prime ideals is short exact, so we conclude by Lemma 10.23.1.
  6. We will show that every ideal of $R$ has a finite generating set: For this, let $I \subset R$ be an arbitrary ideal. By Proposition 10.9.12 each $I_{f_i} \subset R_{f_i}$ is an ideal. These are all finitely generated by assumption, so we conclude by (2).
  7. For each $i$ take a finite generating set $X_i$ of $S_{f_i}$. Without loss of generality, we may assume that the elements of $X_i$ are in the image of the localization map $S \rightarrow S_{f_i}$, so we take a finite set $Y_i$ of preimages of the elements of $X_i$ in $S$. Let $Y$ be the union of these sets. This is still a finite set. Consider the algebra homomorphism $R[X_y]_{y \in Y} \rightarrow S$ induced by $Y$. Since it is an algebra homomorphism, the image $T$ is an $R$-submodule of the $R$-module $S$, so we can consider the quotient module $S/T$. By assumption, this is zero if we localize at the $f_i$, so it is zero by (1) and therefore $S$ is an $R$-algebra of finite type.
  8. By the previous item, there exists a surjective $R$-algebra homomorphism $R[X_1,...,X_n] \rightarrow S$. Let $K$ be the kernel of this map. This is an ideal in $R[X_1,..X_n]$, finitely generated in each localization at $f_i$. Since the $f_i$ generate the unit ideal in $R$, they also generate the unit ideal in $R[X_1,...,X_n]$, so an application of (2) finishes the proof.

$\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 4145–4169 (see updates for more information).

    \begin{lemma}
    \label{lemma-cover}
    \begin{slogan}
    Zariski-local properties of modules and algebras
    \end{slogan}
    Let $R$ be a ring. Let $M$ be an $R$-module. Let $S$ be an $R$-algebra.
    Suppose that $f_1, \ldots, f_n$ is a finite list of
    elements of $R$ such that $\bigcup D(f_i) = \Spec(R)$
    in other words $(f_1, \ldots, f_n) = R$.
    \begin{enumerate}
    \item If each $M_{f_i} = 0$ then $M = 0$.
    \item If each $M_{f_i}$ is a finite $R_{f_i}$-module,
    then $M$ is a finite $R$-module.
    \item If each $M_{f_i}$ is a finitely presented $R_{f_i}$-module,
    then $M$ is a finitely presented $R$-module.
    \item Let $M \to N$ be a map of $R$-modules. If $M_{f_i} \to N_{f_i}$
    is an isomorphism for each $i$ then $M \to N$ is an isomorphism.
    \item Let $0 \to M'' \to M \to M' \to 0$ be a complex of $R$-modules.
    If $0 \to M''_{f_i} \to M_{f_i} \to M'_{f_i} \to 0$ is exact for each $i$,
    then $0 \to M'' \to M \to M' \to 0$ is exact.
    \item If each $R_{f_i}$ is Noetherian, then $R$ is Noetherian.
    \item If each $S_{f_i}$ is a finite type $R$-algebra, so is $S$.
    \item If each $S_{f_i}$ is of finite presentation over $R$, so is $S$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We prove each of the parts in turn.
    \begin{enumerate}
    \item By Proposition \ref{proposition-localize-twice}
    this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \Spec(R)$,
    so we conclude by Lemma \ref{lemma-characterize-zero-local}.
    \item For each $i$ take a finite generating set $X_i$ of $M_{f_i}$.
    Without loss of generality, we may assume that the elements of $X_i$ are
    in the image of the localization map $M \rightarrow M_{f_i}$, so we take
    a finite set $Y_i$ of preimages of the elements of $X_i$ in $M$. Let $Y$
    be the union of these sets. This is still a finite set.
    Consider the obvious $R$-linear map $R^Y \rightarrow M$ sending the basis
    element $e_y$ to $y$. By assumption this map is surjective after localizing
    at an arbitrary prime ideal $\mathfrak p$ of $R$, so it surjective by
    Lemma \ref{lemma-characterize-zero-local}
    and $M$ is finitely generated.
    \item By (2) we have a short exact sequence
    $$
    0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0
    $$
    Since localization is an exact functor and $M_{f_i}$ is finitely
    presented we see that $K_{f_i}$ is finitely generated for all
    $1 \leq i \leq n$ by Lemma \ref{lemma-extension}.
    By (2) this implies that $K$ is a finite $R$-module and therefore
    $M$ is finitely presented.
    \item By Proposition \ref{proposition-localize-twice}
    the assumption implies that the induced morphism
    on localizations at all prime ideals is an isomorphism, so we conclude
    by Lemma \ref{lemma-characterize-zero-local}.
    \item By Proposition \ref{proposition-localize-twice} the assumption
    implies that the induced
    sequence of localizations at all prime ideals is short exact, so we
    conclude by Lemma \ref{lemma-characterize-zero-local}.
    \item We will show that every ideal of $R$ has a finite generating set:
    For this, let $I \subset R$ be an arbitrary ideal. By
    Proposition \ref{proposition-localization-exact}
    each $I_{f_i} \subset R_{f_i}$ is an ideal. These are all
    finitely generated by assumption, so we conclude by (2).
    \item For each $i$ take a finite generating set $X_i$ of $S_{f_i}$.
    Without loss of generality, we may assume that the elements of $X_i$
    are in the image of the localization map $S \rightarrow S_{f_i}$, so
    we take a finite set $Y_i$ of preimages of the elements of $X_i$ in
    $S$. Let $Y$ be the union of these sets. This is still a finite set.
    Consider the algebra homomorphism $R[X_y]_{y \in Y} \rightarrow S$
    induced by $Y$. Since it is an algebra homomorphism, the image $T$
    is an $R$-submodule of the $R$-module $S$, so we can consider the
    quotient module $S/T$. By assumption, this is zero if we localize
    at the $f_i$, so it is zero by (1) and therefore $S$ is an
    $R$-algebra of finite type.
    \item By the previous item, there exists a surjective $R$-algebra
    homomorphism $R[X_1,...,X_n] \rightarrow S$. Let $K$ be the kernel
    of this map. This is an ideal in $R[X_1,..X_n]$, finitely generated
    in each localization at $f_i$. Since the $f_i$ generate the unit ideal
    in $R$, they also generate the unit ideal in $R[X_1,...,X_n]$, so an
    application of (2) finishes the proof.
    \end{enumerate}
    \end{proof}

    Comments (1)

    Comment #1653 by Matthieu Romagny on September 24, 2015 a 9:09 pm UTC

    Suggested slogan: Zariski-local properties of modules and algebras

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