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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?

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I do not know about these questions in particular, but Greenlees and May have worked quite hard to prove as much as possible about local (co)homology without Noetherian assumptions. –  Neil Strickland Nov 23 '10 at 7:40
A ring with ITI with respect to every ideal is not necessarily Noetherian. Indeed, a ring with the property that every proper ideal is nilpotent has ITI with respect to every ideal, hence it suffices to exhibit a non-Noetherian local ring with nilpotent maximal ideal. This we do by taking the polynomial algebra in countably many indeterminates over a field modulo the ideal generated by all products of two indeterminates. –  Fred Rohrer Nov 24 '10 at 7:39

2 Answers 2

up vote 2 down vote accepted

@ Fred: I think the following may be help you to give an example for (ITR) question:

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

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

  3. 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.

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Dear Quy, thank you for your now extended and even more interesting answer. To say it explicitly: you construct a ring without the ITR-property. One might note that this local ring has bounded torsion with respect to its maximal ideal but nevertheless fails to have ITR with respect to this ideal. –  Fred Rohrer Sep 20 '11 at 2:10
@ Fred, I think the question should as follows: Let $\frak a$ be an ideal such that Krull intersection holds, $E$ an injective $R$-module. Is the $\frak{a}$-torsion of $E$ injective. Your question may has a relation with mathoverflow.net/questions/75004/… –  Pham Hung Quy Sep 20 '11 at 7:25

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 will be found in a joint work with P.H.Quý (who gave the accepted answer), available in due time. See also this question.

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