injectivity of torsion submodules of injectives - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:20:16Z http://mathoverflow.net/feeds/question/47043 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47043/injectivity-of-torsion-submodules-of-injectives injectivity of torsion submodules of injectives Fred Rohrer 2010-11-23T02:23:24Z 2011-12-23T08:57:40Z <p>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:</p> <blockquote> <p>(ITI) $\mathfrak{a}$-torsion submodules of injective modules are injective.</p> <p>(ITR) $\mathfrak{a}$-torsion modules have injective resolutions whose components are $\mathfrak{a}$-torsion.</p> </blockquote> <p>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 $\mathfrak{a}$.</p> <p>Is anything known about the converses of these implications? Does anybody know a ring that does not satisfy ITR?</p> http://mathoverflow.net/questions/47043/injectivity-of-torsion-submodules-of-injectives/75780#75780 Answer by Pham Hung Quy for injectivity of torsion submodules of injectives Pham Hung Quy 2011-09-18T19:39:20Z 2011-09-19T07:33:47Z <p>@ Fred: I think the following may be help you to give an example for (ITR) question:</p> <ol> <li><p>Choose a non-Noetherian local ring $(A, \frak{m})$ such that $\frak{m}= \frak{m}^n$ for all $n$. </p></li> <li><p>Let $E(k)$ be the injective envelope of $k = R/\frak{m}$.</p></li> <li><p>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.</p></li> </ol> <p>Dear Fred, I give here that a such ring.</p> <p>Let $k$ be a file and $Q^+$ be the semigroup of non-negative ration. Let $A = k[Q^+]$ be the semigroup ring, i.e</p> <p>$A = { \sum_{\alpha}u_{\alpha}x^{\alpha}: u_{\alpha} \in k }$.</p> <p>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 $\frak{n} = \frak{n}^2$.</p> <p>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.</p> <p><strong>Claim:</strong> $k$ is not injective.</p> <p><strong>Proof of claim:</strong> 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.</p> http://mathoverflow.net/questions/47043/injectivity-of-torsion-submodules-of-injectives/84146#84146 Answer by Fred Rohrer for injectivity of torsion submodules of injectives Fred Rohrer 2011-12-23T08:57:40Z 2011-12-23T08:57:40Z <p>1) A ring has ITI with respect to an ideal $\mathfrak{a}$ if and only if it has ITR with respect to $\mathfrak{a}$.</p> <p>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.)</p> <p>3) A ring that has ITI with respect to every ideal of finite type does not necessarily have ITI with respect to every ideal.</p> <p>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 <a href="http://mathoverflow.net/questions/70725/injective-modules-and-torsion-functors" rel="nofollow">this question</a>.</p>