injectivity of torsion submodules of injectives Local cohomology with respect to an ideal $\mathfrak{a}$ is often studied over a Noetherian ring $R$. However, the proof of a lot of basic results does not rely on noetherianity of $R$, but rather on the following two properties:

(ITI) $\mathfrak{a}$-torsion submodules of injective modules are injective.
(ITR) $\mathfrak{a}$-torsion modules have injective resolutions whose components are $\mathfrak{a}$-torsion.

If $R$ is Noetherian, then it has ITI with respect to every $\mathfrak{a}$. If $R$ has ITI with respect to $\mathfrak{a}$, then it has ITR with respect to $\mathfrak{a}$.
Is anything known about the converses of these implications? Does anybody know a ring that does not satisfy ITR?
 A: @ Fred: I think the following may be help you to give an example for (ITR) question:


*

*Choose a non-Noetherian local ring $(A, \frak{m})$ such that $\frak{m}= \frak{m}^n$ for all $n$. 

*Let $E(k)$ be the injective envelope of $k = R/\frak{m}$.

*Claim: if $E(k)$ is $\frak m$-torsion, then $E(k) = k$. Indeed, let $x \in E(k)$. There is $n$ such that $\frak{m}^n x = 0$. Hence $\frak{m} x = 0$. But $k \subseteq Rx$, so $x \in k$. So $k$ is injective.
Dear Fred, I give here that a such ring.
Let $k$ be a file and $Q^+$ be the semigroup of non-negative ration.
Let $A = k[Q^+]$ be the semigroup ring, i.e
$A = \{ \sum_{\alpha}u_{\alpha}x^{\alpha}: u_{\alpha} \in k \}$.
Let $\frak{n} = (x^{\alpha}: \alpha > 0)$ be the maximal ideal of $A$. Since for all $\alpha > 0$ we have $x^{\alpha} = x^{\alpha/2}x^{\alpha/2} \in \frak{n}^2$, so $\frak{n} = \frak{n}^2$.
Let $S = A_{\frak{n}}$.
Let $I = (x^{\alpha}: \alpha > 1)$ be an ideal of $S$.
Let $(R, \frak{m})$ be the quotient ring $S/I$. We have $\frak{m} = \frak{m}^2$ by above.
Claim: $k$ is not injective.
Proof of claim: We consider the ideal $(x) \subseteq R$.
Notice that $(x) \cong k$ as $R$-modules. If $k$ is injective, then (x) is a direct summand of local ring $R$. It is a contradiction.
A: 1) A ring has ITI with respect to an ideal $\mathfrak{a}$ if and only if it has ITR with respect to $\mathfrak{a}$.
2) A ring does not necessarily have ITI with respect to an ideal of finite type. (Note that the ideal in Quý's example is not of finite type.)
3) A ring that has ITI with respect to every ideal of finite type does not necessarily have ITI with respect to every ideal.
Proofs of the above, concrete examples, and further details on the ITI-property can be found in a joint work with P.H.Quý (who gave the accepted answer), available here. See also this question.
