The Stacks project

Localization is exact.

Proposition 10.9.12. Let $L\xrightarrow {u} M\xrightarrow {v} N$ be 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 $xt \in \mathop{\mathrm{Ker}}(v)$. 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. $\square$

Comments (2)

There are also:

  • 5 comment(s) on Section 10.9: Localization

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00CS. Beware of the difference between the letter 'O' and the digit '0'.



View Proposition 10.9.12 as pdf