Injective modules and torsion functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:29:12Z http://mathoverflow.net/feeds/question/70725 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70725/injective-modules-and-torsion-functors Injective modules and torsion functors Fred Rohrer 2011-07-19T10:44:07Z 2011-12-23T08:58:04Z <p>(<a href="http://mathoverflow.net/questions/47043/injectivity-of-torsion-submodules-of-injectives" rel="nofollow">This</a> is a related question.)</p> <p>Local cohomology is studied mostly over Noetherian rings. Parts of the machinery do in fact not rely on Noetherianness, but on some weaker properties, for example the following:</p> <blockquote> <p>(ITI) \$\mathfrak{a}\$-torsion submodules of injective modules are injective.</p> </blockquote> <p>Noetherian rings have the ITI-property with respect to every ideal \$\mathfrak{a}\$; this is usually proven on use of the Artin-Rees Lemma or on use of Matlis' structure theory for injective modules. Rings with the ITI-property with respect to every ideal are not necessarily Noetherian; a (somewhat silly) example can be found in my comment to <a href="http://mathoverflow.net/questions/47043/injectivity-of-torsion-submodules-of-injectives" rel="nofollow">this question</a>.</p> <p>Of course, in order for this property to be really useful one should know if certain classes of rings have the ITI-property with respect to certain classes of ideals. A particular choice of rings and ideals yields the following question:</p> <blockquote> <p>Does every ring have the ITI-property with respect to every principal ideal generated by a non-zero-divisor?</p> </blockquote> <p>(If this is the case, then it follows of course that integral rings have the ITI-property with respect to finitely generated ideals, and that polynomial algebras over arbitrary rings have the ITI-property with respect to finitely generated monomial ideals - hence local cohomology would not behave too bad in these two large and interesting settings.)</p> http://mathoverflow.net/questions/70725/injective-modules-and-torsion-functors/84147#84147 Answer by Fred Rohrer for Injective modules and torsion functors Fred Rohrer 2011-12-23T08:58:04Z 2011-12-23T08:58:04Z <p>If \$R\$ is a valuation ring whose maximal ideal \$\mathfrak{m}\$ is of finite type, then \$R\$ has ITI with respect to \$\mathfrak{m}\$ if and only if \$R\$ is Noetherian. Since there exists a non-Noetherian valuation ring whose maximal ideal is of finite type, the answer to the question is no. Moreover, it follows that integral rings do not necessarily have the ITI property with respect to ideals of finite type.</p> <p>A proof of the above, concrete examples, and further details on the ITI-property will be found in a joint work with P.H.Quý, available in due time.</p>