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Hello all,

The lemma 3.3 page 214 in Hartshorne 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 that is the lemma still true if $A$ is not Noetherian, are there any counter-example?

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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.

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I've not Hartshorne's book with me now so I cannot check the exact context of your question, anyway this is a very general fact about localization. Take a Grothendieck category $\frak C$ and a hereditary torsion subclass $\frak T$, then you have a localization of categories $Q:\frak C\to \frak C/\frak T$ (in the sense of Gabriel). There is always a fully faithful functor $S:\frak C/\frak T\to\frak C$ such that $(Q,S)$ is an adjoint pair. By the closure properties of $\frak T$, the quotient functor $Q$ is automatically exact.

Now, identifying $\frak C/\frak T$ with a subcategory of $\frak C$ via the section functor $S$, you have a very explicit construction of the localized objects using the torsion-theoretic machinery. Indeed, denote by $T:\frak C\to \frak T$ the torsion functor (for an object $C\in \frak C$, $T(C)$ is the direct union of all the sub-objects belonging to $\frak T$).

Given $X\in \frak C$, let $X'=X/T(X)$, then $Q(X)$ is isomorphic to $\pi^{-1}(T(E(X')/X'))$, where $\pi:E(X')\to E(X')/X'$ is the natural projection.

Now, in some cases you can prove that the class $\frak T$ is closed under taking injective envelopes. This is the case for example the case of localization at prime ideals in commutative Noetherian rings. This is not always the case in non-commutative (even Noetherian) rings, something can be said in the case of FBN rings. If $X$ is injective from the beginning and the torsion class is closed under taking injective envelopes, then it is an exercise to prove that $T(X)=X_t$ is again injective and so you have $X=X_t\oplus X_f$ for some torsion-free injective object $X_f$. Thus $Q(X)\cong Q(X_f)$. Furthermore, $E(X_f)/X_f=0$ and so $Q(X_f)\cong X_f$ is injective.

Anyway, not every "localization" is the localization with respect to a hereditary torsion theory and, even in that case, the torsion class is not always closed under taking injective envelopes so that the above argument does not always apply... In this moment it is not clear to me what happens for a general universal localization or for Cohn's localization.

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Without understanding everything you wrote, I do not think it is correct. It seems to imply that - without any condition on a ring $R$ - torsion submodules of injective $R$-modules are injective. This is false, cf. mathoverflow.net/questions/47043. –  Fred Rohrer May 31 '13 at 12:48
    
Ah, thank you for the explanation. So your answer does not answer the question. –  Fred Rohrer May 31 '13 at 13:02
    
Thanks for the observation, you are right, there was a gap at some point, I've edited. –  Simone Virili May 31 '13 at 13:06

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