Lemma 10.9.9 (00CR)—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 May 23, 2014 at 14:24

"for some $f' \in R$ " should be "for some $f'' \in S$ ".

Comment #588 by Wei Xu on May 23, 2014 at 15:28

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 Johan on May 23, 2014 at 20:42

@#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 $f'' \in R$ which is good enough to define the map. See here.

Comment #2147 by Johan on July 23, 2016 at 14:43

@#588: This is now addressed here.

Comment #6485 by Laurent Moret-Bailly on August 07, 2021 at 05:41

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

Comment #6557 by Johan on September 13, 2021 at 13:20

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

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