The Stacks project

Lemma 10.9.9. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset. Let $M$ be an $R$-module. Then

\[ S^{-1}M = \mathop{\mathrm{colim}}\nolimits _{f \in S} M_ f \]

where the preorder on $S$ is given by $f \geq f' \Leftrightarrow f = f'f''$ for some $f'' \in R$ in which case the map $M_{f'} \to M_ f$ is given by $m/(f')^ e \mapsto m(f'')^ e/f^ e$.

Proof. Omitted. Hint: Use the universal property of Lemma 10.9.7. $\square$

Comments (6)

Comment #587 by Wei Xu on

"for some " should be "for some ".

Comment #588 by Wei Xu on

And here the definition of partial ordered set is not required to have "antisymetry" (it is the definition tagged 00D3), while the definition of partial ordered set in wiki page is required to have the antisymetry property.

Comment #600 by on

@#588: Hmm... I see what you mean. We chose Definition 4.21.1 because it works well for directed limits, etc. But perhaps we should change it... not sure yet.

@#587: OK, I fixed this by requiring which is good enough to define the map. See here.

Comment #6485 by Laurent Moret-Bailly on

About the hint: one should first check that the map is well defined, i.e. independent of . Of course this can also be done via the universal property (even avoiding the use of ), so this property is applied "twice".

Comment #6557 by on

@#6485. OK, I am going to leave it as is and hope that confused readers will find your comment and benefit from it.

There are also:

  • 6 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Your email address will not be published.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 00CR. Beware of the difference between the letter 'O' and the digit '0'.