Let $R$ be a commutative ring with identity and let $S$ be a multiplicative subset of $R$. Is it true that for any injective $R$-module like $M$, $S^{-1}M$ (as the $S^{-1}R$-module) is also injective ?
2 Answers
No, this is false in general. I quote the Mathematical Review of Dade, Everett C. Localization of injective modules. J. Algebra 69 (1981), no. 2, 416–425.
Localization of modules over a commutative ring R with respect to a multiplicatively closed subset S of R is an exact functor with a large number of properties, some of which are listed in Theorem 3.76 of J. J. Rotman's book [An introduction to homological algebra, Academic Press, New York, 1979; MR0538169 (80k:18001)]. The fifth property, namely: (LI) the localization $S^{−1}E$ of any injective R-module E is an injective $S^{−1}R$-module, is false. Two examples are given here showing that arbitrary R and S need not have the property (LI). Also a positive result is given, showing that (LI) holds for certain non-Noetherian R and certain S. In particular, if R is the polynomial ring $k[x_1,x_2,\ldots]$ in a countable number of $x_n$ over a nonzero Noetherian ring k, then (LI) holds for all choices of S.
-
$\begingroup$ Thanks. Do you know any other example ? (other that Everett's paper) $\endgroup$– user39070Commented Sep 8, 2013 at 14:04
-
$\begingroup$ It says (Everett's paper) the problem is true when $R$ is Noetherian. Do you know a reference ? $\endgroup$– user39070Commented Sep 8, 2013 at 14:12
-
2$\begingroup$ The second edition of Rotman's book gives a correct proof in the noetherian case. $\endgroup$ Commented Sep 8, 2013 at 14:19
-
1$\begingroup$ 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 $S^{-1}F$. $\endgroup$ Commented Sep 8, 2013 at 18:52
-
1$\begingroup$ @oxeimon $\mathrm{Spec}\, S^{-1}A$ is not necessarily an open subset of $\mathrm{Spec}\,A$! $\endgroup$ Commented Mar 29, 2017 at 5:24
When $R$ is noetherian, yes: By Baer's criterion it suffices to prove that the map
$\hom_{S^{-1} R}(S^{-1} R,S^{-1} M) \to \hom_{S^{-1} R}(J,S^{-1} M)$
is surjective for every ideal $J \subseteq S^{-1} R$. Write $J = S^{-1} I$ for some ideal $I \subseteq R$. Since $I$ (and of course $R$) are of finite presentation, this map is isomorphic to
$S^{-1} \hom_R(R,M) \to S^{-1} \hom_R(I,M),$
which is surjective since $\hom_R(R,M) \to \hom_R(I,M)$ is surjective.
The same proof also shows the converse: If $M$ is injective locally (either in the sense that all $M_{\mathfrak{p}}$ are injective over $R_{\mathfrak{p}}$, where $\mathfrak{p}$ runs through all prime ideals, or if there is a basic open cover $\mathrm{Spec}(R) = \cup_i D(f_i)$ such that each $M_{f_i}$ is an injective $R_{f_i}$-module), then $M$ is injective. See also here.
You can also see this as a consequence of the fact that the Ext functor, when restricted to f.g. modules over a noetherian ring, commutes with localization.