Localization of a pure-injective module is pure-injective? Hi,
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,
 A: 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. 

DETAILS
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.
