# The Stacks Project

## Tag 00HN

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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