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

Lemma 10.23.1. Let $R$ be a ring.

  1. For an element $x$ of an $R$-module $M$ the following are equivalent
    1. $x = 0$,
    2. $x$ maps to zero in $M_\mathfrak p$ for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$,
    3. $x$ maps to zero in $M_{\mathfrak m}$ for all maximal ideals $\mathfrak m$ of $R$.
    In other words, the map $M \to \prod_{\mathfrak m} M_{\mathfrak m}$ is injective.
  2. Given an $R$-module $M$ the following are equivalent
    1. $M$ is zero,
    2. $M_{\mathfrak p}$ is zero for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$,
    3. $M_{\mathfrak m}$ is zero for all maximal ideals $\mathfrak m$ of $R$.
  3. Given a complex $M_1 \to M_2 \to M_3$ of $R$-modules the following are equivalent
    1. $M_1 \to M_2 \to M_3$ is exact,
    2. for every prime $\mathfrak p$ of $R$ the localization $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$ is exact,
    3. for every maximal ideal $\mathfrak m$ of $R$ the localization $M_{1, \mathfrak m} \to M_{2, \mathfrak m} \to M_{3, \mathfrak m}$ is exact.
  4. Given a map $f : M \to M'$ of $R$-modules the following are equivalent
    1. $f$ is injective,
    2. $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is injective for all primes $\mathfrak p$ of $R$,
    3. $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is injective for all maximal ideals $\mathfrak m$ of $R$.
  5. Given a map $f : M \to M'$ of $R$-modules the following are equivalent
    1. $f$ is surjective,
    2. $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is surjective for all primes $\mathfrak p$ of $R$,
    3. $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is surjective for all maximal ideals $\mathfrak m$ of $R$.
  6. Given a map $f : M \to M'$ of $R$-modules the following are equivalent
    1. $f$ is bijective,
    2. $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is bijective for all primes $\mathfrak p$ of $R$,
    3. $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is bijective for all maximal ideals $\mathfrak m$ of $R$.

Proof. Let $x \in M$ as in (1). Let $I = \{f \in R \mid fx = 0\}$. It is easy to see that $I$ is an ideal (it is the annihilator of $x$). Condition (1)(c) means that for all maximal ideals $\mathfrak m$ there exists an $f \in R \setminus \mathfrak m$ such that $fx =0$. In other words, $V(I)$ does not contain a closed point. By Lemma 10.16.2 we see $I$ is the unit ideal. Hence $x$ is zero, i.e., (1)(a) holds. This proves (1).

Part (2) follows by applying (1) to all elements of $M$ simultaneously.

Proof of (3). Let $H$ be the homology of the sequence, i.e., $H = \mathop{\mathrm{Ker}}(M_2 \to M_3)/\mathop{\mathrm{Im}}(M_1 \to M_2)$. By Proposition 10.9.12 we have that $H_\mathfrak p$ is the homology of the sequence $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$. Hence (3) is a consequence of (2).

Parts (4) and (5) are special cases of (3). Part (6) follows formally on combining (4) and (5). $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 4062–4117 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-zero-local}
    Let $R$ be a ring.
    \begin{enumerate}
    \item For an element $x$ of an $R$-module $M$ the following are equivalent
    \begin{enumerate}
    \item $x = 0$,
    \item $x$ maps to zero in $M_\mathfrak p$ for all $\mathfrak p \in \Spec(R)$,
    \item $x$ maps to zero in $M_{\mathfrak m}$ for all maximal ideals
    $\mathfrak m$ of $R$.
    \end{enumerate}
    In other words, the map $M \to \prod_{\mathfrak m} M_{\mathfrak m}$
    is injective.
    \item Given an $R$-module $M$ the following are equivalent
    \begin{enumerate}
    \item $M$ is zero,
    \item $M_{\mathfrak p}$ is zero for all $\mathfrak p \in \Spec(R)$,
    \item $M_{\mathfrak m}$ is zero for all maximal ideals $\mathfrak m$ of $R$.
    \end{enumerate}
    \item Given a complex $M_1 \to M_2 \to M_3$
    of $R$-modules the following are equivalent
    \begin{enumerate}
    \item $M_1 \to M_2 \to M_3$ is exact,
    \item for every prime $\mathfrak p$ of $R$ the localization
    $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$
    is exact,
    \item for every maximal ideal $\mathfrak m$ of $R$ the localization
    $M_{1, \mathfrak m} \to M_{2, \mathfrak m} \to M_{3, \mathfrak m}$
    is exact.
    \end{enumerate}
    \item Given a map $f : M \to M'$ of $R$-modules the following are equivalent
    \begin{enumerate}
    \item $f$ is injective,
    \item $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is injective
    for all primes $\mathfrak p$ of $R$,
    \item $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is injective
    for all maximal ideals $\mathfrak m$ of $R$.
    \end{enumerate}
    \item Given a map $f : M \to M'$ of $R$-modules the following are equivalent
    \begin{enumerate}
    \item $f$ is surjective,
    \item $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is surjective
    for all primes $\mathfrak p$ of $R$,
    \item $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is surjective
    for all maximal ideals $\mathfrak m$ of $R$.
    \end{enumerate}
    \item Given a map $f : M \to M'$ of $R$-modules the following are equivalent
    \begin{enumerate}
    \item $f$ is bijective,
    \item $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is bijective
    for all primes $\mathfrak p$ of $R$,
    \item $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is bijective
    for all maximal ideals $\mathfrak m$ of $R$.
    \end{enumerate}
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Let $x \in M$ as in (1). Let $I = \{f \in R \mid fx = 0\}$.
    It is easy to see that $I$ is an ideal (it is the
    annihilator of $x$). Condition (1)(c) means that for
    all maximal ideals $\mathfrak m$ there exists an
    $f \in R \setminus \mathfrak m$ such that $fx =0$.
    In other words, $V(I)$ does not contain a closed point.
    By Lemma \ref{lemma-Zariski-topology} we see $I$ is the unit ideal.
    Hence $x$ is zero, i.e., (1)(a) holds. This proves (1).
    
    \medskip\noindent
    Part (2) follows by applying (1) to all elements of $M$ simultaneously.
    
    \medskip\noindent
    Proof of (3). Let $H$ be the homology of the sequence, i.e.,
    $H = \Ker(M_2 \to M_3)/\Im(M_1 \to M_2)$. By
    Proposition \ref{proposition-localization-exact}
    we have that $H_\mathfrak p$ is the homology of the sequence
    $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$.
    Hence (3) is a consequence of (2).
    
    \medskip\noindent
    Parts (4) and (5) are special cases of (3). Part (6) follows
    formally on combining (4) and (5).
    \end{proof}

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