most recent 30 from http://mathoverflow.net 2013-05-25T20:24:25Z http://mathoverflow.net/feeds/question/21765 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21765/projective-dimension ashpool 2010-04-18T18:09:44Z 2010-04-18T19:50:08Z

<p>Is it true that if a module has a free resolution of length $d$ then any of its submodule has a free resolution of length $\leq d$?</p>

http://mathoverflow.net/questions/21765/projective-dimension/21766#21766 Robin Chapman 2010-04-18T18:23:18Z 2010-04-18T18:23:18Z

<p>No. A submodule of a free module need not have finite projective dimension. As a simple example let $R=\mathbb{Z}/p^2\mathbb{Z}$. The free module $R$ has a submodule $p\mathbb{Z}/p^2\mathbb{Z}\cong\mathbb{Z}/p\mathbb{Z}$ which has no finite projective resolution.</p>

http://mathoverflow.net/questions/21765/projective-dimension/21767#21767 Mike Skirvin 2010-04-18T18:26:06Z 2010-04-18T18:26:06Z

<p>No, a ring will always be free viewed as a module over itself, but its ideals certainly don't have to be free.</p> <p>For example, consider the ring $R = k[t]/t^2$ and consider the submodule $I = (t),$ the ideal generated by $t$. Then $R \to I$ by multiplication by $t$ and has kernel $I$. It's then easy to see that $\ldots \to R \to R \to I \to 0$ is an infinite free resolution of $I$ where each map is multiplication by $t$.</p>

http://mathoverflow.net/questions/21765/projective-dimension/21769#21769 Jay Pottharst 2010-04-18T18:38:14Z 2010-04-18T18:54:27Z

<p>Counterexamples can even be found in a domain, by taking rings of higher dimension or singular rings&mdash;once you're no longer over a PID, ideals will suffice. Take $R = k[x,y]$, the module $M = R$ itself, and the submodule $M' = (x,y)$. Or, take the ring $S = k[x,y]/(x^3-y^2)$, the module $N = S$ itself, and the submodule $N' = (x,y)$.</p>