Global dimension and localization - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:21:07Z http://mathoverflow.net/feeds/question/59981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59981/global-dimension-and-localization Global dimension and localization Fernando Muro 2011-03-29T15:52:35Z 2011-06-21T14:54:27Z <p>Is there any condition on a commutative ring \$R\$ so that the global dimension of \$R\$ coincides with the supremum of the global dimensions of the localizations \$R_{\mathfrak{m}}\$ at all maximal ideals \$\mathfrak{m}\subset R\$? I'm looking (if possible) for conditions which are easy to verify.</p> http://mathoverflow.net/questions/59981/global-dimension-and-localization/59983#59983 Answer by Steven Sam for Global dimension and localization Steven Sam 2011-03-29T16:29:35Z 2011-03-29T16:29:35Z <p>If one of the localizations isn't regular, then both \$R\$ and that localization have infinite global dimension, so it's trivially true in that case.</p> <p>So we can reduce to the case that \$R\$ is regular. Then Spec(\$R\$) can't have irreducible components intersecting. Since the global dimension of a regular local ring is just its dimension, we need to require that all connected components of Spec(\$R\$) have the same dimension. </p> <p>So let's focus on the case when \$R\$ is a regular domain. We need to know that all maximal ideals have the same height. If we assume that \$R\$ is a quotient of a polynomial ring over a field, then this true. There are probably counterexamples otherwise.</p> http://mathoverflow.net/questions/59981/global-dimension-and-localization/68391#68391 Answer by David White for Global dimension and localization David White 2011-06-21T14:48:13Z 2011-06-21T14:54:27Z <p>This problem is discussed at length in T.Y. Lam's <em><a href="http://books.google.com/books?id=r9VoYbk-8c4C&amp;q=5.92#v=snippet&amp;q=%2522ranges%2520over%2520a%2520complete%2520set%2520of%2520simple%2522&amp;f=false" rel="nofollow">Lectures on Modules and Rings</a></em>. The hyperlink should take you to the Theorem in question (5.92 in section 5G). The point is that for a commutative noetherian ring \$R\$ you get the result you wanted and also more:</p> <blockquote> <p>For a commutative noetherian ring \$R\$ gl.dim\$(R_m)=\$pd\$_R(R/m)\$ for all maximal ideals \$m\$. This implies gl.dim\$(R)=\sup(\$gl.dim\$(R_m)) = \sup(\$pd\$_R(S))\$ where the last supremum runs over all simple \$R\$-modules.</p> </blockquote> <p>The proof Lam gives avoids the machinery of Ext, using instead the fact that the global dimension of a commutative noetherian local ring is the injective dimension (also the projective dimension) of its residue field.</p> <p>Note that the noetherian assumption really is necessary. On page 197, Lam points out that B. Osofsky has constructed some interesting examples (he gives details) which I suspect would show this theorem fails without the noetherian hypothesis.</p>