# The Stacks Project

## Tag 00CS

Localization is exact.

Proposition 10.9.12. Let $L\xrightarrow{u} M\xrightarrow{v} N$ is an exact sequence of $R$-modules. Then $S^{-1}L \to S^{-1}M \to S^{-1}N$ is also exact.

Proof. First it is clear that $S^{-1}L \to S^{-1}M \to S^{-1}N$ is a complex since localization is a functor. Next suppose that $x/s$ maps to zero in $S^{-1}N$ for some $x/s \in S^{-1}M$. Then by definition there is a $t\in S$ such that $v(xt) = v(x)t = 0$ in $M$, which means $xt \in \mathop{\rm Ker}(v)$. By the exactness of $L \to M \to N$ we have $xt = u(y)$ for some $y$ in $L$. Then $x/s$ is the image of $y/st$. This proves the exactness. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 1366–1374 (see updates for more information).

\begin{proposition}
\label{proposition-localization-exact}
\begin{slogan}
Localization is exact.
\end{slogan}
Let $L\xrightarrow{u} M\xrightarrow{v} N$ is an exact sequence
of $R$-modules. Then
$S^{-1}L \to S^{-1}M \to S^{-1}N$ is also exact.
\end{proposition}

\begin{proof}
First it is clear that $S^{-1}L \to S^{-1}M \to S^{-1}N$ is a complex
since localization is a functor. Next suppose that $x/s$ maps to zero
in $S^{-1}N$ for some $x/s \in S^{-1}M$. Then by definition there is a
$t\in S$ such that $v(xt) = v(x)t = 0$ in $M$, which means
$xt \in \Ker(v)$. By the exactness of $L \to M \to N$ we have
$xt = u(y)$ for some $y$ in $L$. Then $x/s$ is the image of $y/st$.
This proves the exactness.
\end{proof}

Comment #842 by Johan Commelin (site) on July 24, 2014 a 6:37 pm UTC

Suggested slogan: Localization of modules is exact

Comment #844 by Pieter Belmans (site) on July 24, 2014 a 7:13 pm UTC

There is a dash missing in the first sentence.

There are also 2 comments on Section 10.9: Commutative Algebra.

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