Commutative Ring of Finite Global Dimension - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:13:37Z http://mathoverflow.net/feeds/question/53138 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53138/commutative-ring-of-finite-global-dimension Commutative Ring of Finite Global Dimension arsmath 2011-01-24T22:19:13Z 2011-01-24T23:15:22Z <p>The only examples of commutative rings of finite <a href="http://en.wikipedia.org/wiki/Global_dimension" rel="nofollow">global dimension</a> I know are either:</p> <ul> <li>Dedekind domains (and fields as a degenerate special case)</li> <li>Regular local rings</li> <li>Rings constructed from the previous examples by taking direct sums, or forming the rings of polynomials over a ring of finite global dimension.</li> </ul> <p>Are there other examples? In particular, are there other examples that are finite-dimensional over a field \$k\$?</p> <p>(Examples of rings of finite global dimension are easier to come by in the noncommutative case, but I'm specifically curious about the commutative case.)</p> http://mathoverflow.net/questions/53138/commutative-ring-of-finite-global-dimension/53147#53147 Answer by Mariano Suárez-Alvarez for Commutative Ring of Finite Global Dimension Mariano Suárez-Alvarez 2011-01-24T22:29:08Z 2011-01-24T22:59:34Z <p>An artinian ring \$R\$ is a finite direct product of local artinian rings. If \$R\$ is of finite global dimension, so are the factors, and then they are regular local by Serre's theorem. As regular local rings are domains, the Jacobson radical of the factors has to be trivial (for its elements are nilpotent) and then \$R\$ is a product of fields.</p> <p>You will thus not get interesting examples of finite dimension.</p> http://mathoverflow.net/questions/53138/commutative-ring-of-finite-global-dimension/53158#53158 Answer by Donu Arapura for Commutative Ring of Finite Global Dimension Donu Arapura 2011-01-24T23:15:22Z 2011-01-24T23:15:22Z <p>At arsmath's request, I'm making this official. (This is pretty standard commutative algebra, but I realize not everyone has gone through it.)</p> <p>A commutative ring \$R\$ is regular if it's noetherian and its local rings are regular. Using Serre's theorem e.g. Matsumura Commutative Ring Theory p 156, and the fact that \$Ext\$ commutes with localization, we can see that any regular ring with finite Krull dimension has finite global dimension.</p> <p>To an algebraic geometer regular = nonsingular. So in particular, so there is a large supply of basic examples arising as coordinate rings of nonsingular affine varieties. This is a bit circular the way I'm saying it, but of course, you can test the condition using the Jacobian criterion...</p>