is there some work on localization of pure-injective modules? Is a localization of a pure-injective module pure-injective?

By localization I mean the standard localization defined for any multiplicatively closed subest S of the ring R.

I'm interested in this question for modules over commutative (noetherian) rings.

Thank you,

  • $\begingroup$ what is your setting? What do mean by "localization"? This could be an interesting question, it would be nice to have some more details (do you refer to a left or right Ore localization? are you thinking to modules over a commutative ring? ...). $\endgroup$ – Simone Virili Dec 18 '12 at 18:33

I believe the answer in your setting is yes: localizations of pure injective modules are pure injective. I don't seem to use Noetherian below, but I am using the fact that $R$ is commutative all over the place. I don't know what happens without this hypothesis.

Recall that $M$ is pure injective if for every pure submodule $P$ of $M$, maps $P\to N$ extend to maps $M\to N$. Recall also that submodules of $S^{-1}M$ take the form $S^{-1}P$ for $P$ a submodule of $M$. Here is a sketch of the proof, with details to be filled in below. Suppose $S^{-1}P$ is a pure submodule of $S^{-1}M$ and let $f:S^{-1}P \to N$. Then we have $f\circ \phi: P \to S^{-1}P \to N$ and this will extend to some $g:M\to N$ because $P$ is a pure submodule (see below) and $M$ is pure injective. So we have the commutative diagram below, where $f \circ \phi$ takes $S$ to units, so $g$ takes $S$ to units:

\begin{array}{ccc} P & \to & M \\\ \downarrow & & \downarrow \\\ S^{-1}P & \to & N \end{array}

By the universal property of localization, there exists a unique $S^{-1}g: S^{-1}M \to N$. By construction, this map extends $f$ and will fit into the commutative diagram above on the right hand side. This proves $S^{-1}M$ is pure injective.


The formal way to phrase the universal property of localization for modules (i.e. to understand what is meant by ``taking $S$ to units") requires a shift from thinking of $S$ as a subset of $R$ into thinking of $S$ as a collection of endomorphisms of $M$. For each $s\in S$ let $\mu_s:M\to M$ be multiplication by $s$. The universal property then says that if $f:M\to N$ takes every map $\mu_s$ to an isomorphism then there exists a unique map $g:S^{-1}M \to N$. It a fun exercise to prove this is equivalent to the universal property as taught in a standard first year algebra course, using the fact that $S^{-1}M \cong M\otimes_R S^{-1}R$ and the universal properties for tensor product and for localizations of rings.

Secondly, we have to justify the statement that $P$ is pure in $M$ above. Actually, I don't know if $S^{-1}P$ pure in $S^{-1}M$ implies $P$ pure in $M$, but I know it implies it $S$-locally (i.e. for any $S$-local modules $X$, $f\otimes 1_X: P\otimes X \to M\otimes X$ is injective). The reason is simple: a module $X$ is $S$-local if $X\cong S^{-1}R\otimes Y$ for some $Y$. Since $P\otimes X \cong P\otimes (S^{-1}R \otimes Y) \cong S^{-1}P\otimes Y$, we see that the map $f\otimes 1_X$ can be written as $S^{-1}f \otimes 1_Y$, which is injective because $S^{-1}P$ is pure in $S^{-1}M$. The $S$-local version of the purity is enough because we only need to know that maps $P\to N$ taking $S$ to units extend to maps $M\to N$. So basically, I'm avoiding the issue by moving to the $S$-local category. If someone can prove $P$ is pure in $M$ without this shift I'd like to hear it.

  • $\begingroup$ By the way, there might be a different proof which doesn't use universal properties and homological algebra. The simpler proof would be to use the fact that $M$ is pure injective iff $M$ is algebraically compact. Then you just need to understand how localization affects systems of equations. See en.wikipedia.org/wiki/Pure_injective_module. I prefer homological algebra myself, and mainly wrote this answer to try to brush up on it and have some fun. For my tastes, the approach via equations wouldn't have been as much fun $\endgroup$ – David White Dec 29 '12 at 18:40
  • $\begingroup$ I think that your definition of puré injective is not correct. Your conditions are equivalent to pure injectivity for $N$, not for $M$. $\endgroup$ – Fernando Muro May 5 '13 at 7:08
  • $\begingroup$ Hi Fernando. I can't remember anything about this problem, since it was so long ago. I see my links above were to wikipedia, so maybe there was some ambiguity there. A better source for info on pure injectives appears to be eprints.ma.man.ac.uk/1148/01/covered/MIMS_ep2008_83.pdf, and it confirms that you're right. Anyway, this hasn't been a very popular post, and I don't think the OP ever came back so I'm not going to bother trying to tweak this answer to prove $S^{-1}N$ is pure injective. I think the same methods and types of considerations should work. $\endgroup$ – David White May 5 '13 at 14:14

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