Let $R$ be a commutative noetherian ring, and let $M$ be an $R$-module. How I can show that if any localization $M_p$ at a prime ideal $p$ of the ring $R$ is injective over $R_p$, then $M$ is injective?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ @sasha: This is an interesting question, but quite elementary and not a research level question. See also the FAQ of mathoverflow. Therefore this question will be closed. Note that you can ask these questions on math.stackexchange.com. $\endgroup$– Martin BrandenburgCommented Mar 4, 2013 at 21:10
-
5$\begingroup$ It's not research, but I disagree it's quite elementary. There are a number of components to Paul's proof below that are not completely obvious, IMO. $\endgroup$– Todd TrimbleCommented Mar 4, 2013 at 21:18
-
$\begingroup$ Please do not only ready 'faq' but perhaps even more importantly 'how to ask'. Questions should provide (personal) motivation ("why do you want to know this?") for the question. The (repeated) lack of this in your questions contributed considerably to me casting the final vote to close on this one. $\endgroup$– user9072Commented Mar 6, 2013 at 13:01
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Let $A \to B$ be an injection of $R$-modules. We want to show that $\mathrm{Hom}(B,M) \to \mathrm{Hom}(A,M)$ is surjective. It suffices to check this locally (since localization is an exact functor and the cokernel of the map is zero iff it's zero locally). So let $p$ be a prime, and consider the localized map $\mathrm{Hom}_R(B,M) \otimes R_p \to \mathrm{Hom}_R(A,M) \otimes R_p$. Now, for $B$ and $A$ finitely presented over $R$, this is $\mathrm{Hom}_{R_p}(B_p,M_p) \to \mathrm{Hom}_{R_p}(A_p,M_p)$. Again, localization is exact so $A_p \to B_p$ is injective, and $M_p$ is injective by hypothesis, so this map is surjective.
What's left is to show that we only need to check injectivity against maps of finitely presented $R$-modules; since $R$ is noetherian, 'finitely presented' is equivalent to 'finitely generated.' In fact, there's a thing called Baer's injectivity criterion that says you only have to check injectivity against inclusions $I \to R$ for $I$ an ideal! So we're done.
