Question

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?

Answer 1

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.