Surjectivity of the natural map of injective module to its localization

Question

Lemma 3.3 page 214 in Hartshorne's Algebraic Geometry book states: "If $I$ is an injective module over a Noetherian ring $A$. Then for any $f\in A$, the natural map of $I$ to its localization $I_f$ is surjective." I want to ask if the lemma is still true if $A$ is not Noetherian, or are there any counter-examples?

Answer 1

A ring $A$ is said to have the ITI property with respect to an ideal $\mathfrak{a}$ if $\mathfrak{a}$-torsion submodules of injective $A$-modules are injective.

Noetherian rings have ITI with respect to every ideal, but a ring with ITI with respect to every ideal is not necessarily noetherian (e.g. absolutely flat rings have ITI). However, there are rings without ITI, even with respect to some principal ideals.

If $A$ has the ITI property with respect to (the ideal generated by) $f$ then the canonical morphism $I\rightarrow I_f$ is an epimorphism. Hence, noetheriannes is not necessary for this property to hold. But it is unknown to me (and I think also in general unknown) whether noetheriannes can be omitted.

See also this question and its answer.

ADDENDUM: Besides the ITI property, there are other hypotheses on the ring $A$ that imply that the canonical morphism $I\rightarrow I_f$ is an epimorphism for every injective $A$-module $I$. Namely, this is true if $A$ is hereditary, or if every zerodivisor of $A$ is nilpotent, hence in particular if $A$ is integral.

Answer 2

It is maybe worth pointing out that Exercise 8 of the same section on page 218 gives a counterexample: Let $A = k[x_0,x_1,x_2,\dots]$ with the relations $x_0^n x_n=0$ for $n=1,2,\dots$. Let $I$ be an injective $A$-module containing $A$. Then $I\rightarrow I_{x_0}$ is not surjective. So here, $I$ is still an injective module over $A$, but $A$ is non-Noetherian.