Global dimension and localization - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T11:21:07Zhttp://mathoverflow.net/feeds/question/59981http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/59981/global-dimension-and-localizationGlobal dimension and localizationFernando Muro2011-03-29T15:52:35Z2011-06-21T14:54:27Z
<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#59983Answer by Steven Sam for Global dimension and localizationSteven Sam2011-03-29T16:29:35Z2011-03-29T16:29:35Z<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#68391Answer by David White for Global dimension and localizationDavid White2011-06-21T14:48:13Z2011-06-21T14:54:27Z<p>This problem is discussed at length in T.Y. Lam's <em><a href="http://books.google.com/books?id=r9VoYbk-8c4C&q=5.92#v=snippet&q=%2522ranges%2520over%2520a%2520complete%2520set%2520of%2520simple%2522&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>