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Tag 00CS

Chapter 10: Commutative Algebra > Section 10.9: Localization

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}

    Comments (2)

    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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