Timeline for Is it true that if $M$ is injective then $S^{-1}M$ is also injective
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2017 at 7:08 | comment | added | Dan Petersen | Even for not-so-weird localizations, you don't get an open subset. E.g. $\mathbb Q$ is a localization of $\mathbb Z$... | |
| Mar 29, 2017 at 6:58 | comment | added | Will Chen | Ah! So the issue is with weird non-finite type localizations... Thanks! | |
| Mar 29, 2017 at 5:24 | comment | added | Dan Petersen | @oxeimon $\mathrm{Spec}\, S^{-1}A$ is not necessarily an open subset of $\mathrm{Spec}\,A$! | |
| Mar 28, 2017 at 22:44 | comment | added | Will Chen | So in Hartshorne's Algebraic Geometry, in Lemma III.6.1, he says that if $X$ is a ringed space, and $U\subset X$ open, then if $I$ is an injective object of $Mod(X)$, then $I|_U$ is an injective object of $Mod(U)$. I'm confused about how this doesn't contradict your answer here. Doesn't Hartshorne's Lemma 6.1 say that localization of injectives are injective? | |
| Sep 8, 2013 at 18:52 | comment | added | Dan Petersen | Yes, that's obvious. If M is a direct summand of the free module F, then $S^{-1}M$ is a direct summand of the free module $S^{-1}F$. | |
| Sep 8, 2013 at 14:34 | comment | added | user39070 | I have no access to Rotman's book now. But thanks anyway. What about when $M$ is projective ? Does it mean $S^{-1}M$ would be projective ? | |
| Sep 8, 2013 at 14:19 | comment | added | Dan Petersen | The second edition of Rotman's book gives a correct proof in the noetherian case. | |
| Sep 8, 2013 at 14:12 | vote | accept | CommunityBot | ||
| Sep 8, 2013 at 14:12 | comment | added | user39070 | It says (Everett's paper) the problem is true when $R$ is Noetherian. Do you know a reference ? | |
| Sep 8, 2013 at 14:04 | comment | added | user39070 | Thanks. Do you know any other example ? (other that Everett's paper) | |
| Sep 8, 2013 at 13:44 | history | answered | Dan Petersen | CC BY-SA 3.0 |