Filtered ring giving rise to a graded-commutative ring - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:07:40Z http://mathoverflow.net/feeds/question/75967 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75967/filtered-ring-giving-rise-to-a-graded-commutative-ring Filtered ring giving rise to a graded-commutative ring Pierre 2011-09-20T16:02:58Z 2011-12-28T02:34:30Z <p>Hello,</p> <p>Given a ring \$R\$ with a filtration by two-sided ideals \$F^0 \supset F^1 \supset F^2 \supset\cdots\$, one can form the associated graded ring \$gr R = F^0 / F^1 \oplus F^1/F^2 \oplus \cdots\$. </p> <p>When \$R\$ is commutative, so is \$gr R\$. Now, the natural condition in the graded world is to be graded-commutative (that is \$yx = (-1)^{pq} xy\$ when \$x\$ resp \$y\$ has degree \$p\$ resp \$q\$). </p> <blockquote> <p>Are there natural/well-known/simple conditions on \$R\$ ensuring that \$gr R\$ is graded-commutative?</p> </blockquote> <p>Thank you very much!</p> <p>Pierre</p> http://mathoverflow.net/questions/75967/filtered-ring-giving-rise-to-a-graded-commutative-ring/75970#75970 Answer by euklid345 for Filtered ring giving rise to a graded-commutative ring euklid345 2011-09-20T16:14:44Z 2011-09-20T16:14:44Z <p>\$[F^i,F^j]\subset F^{i+j+1}\$ ?</p> http://mathoverflow.net/questions/75967/filtered-ring-giving-rise-to-a-graded-commutative-ring/84426#84426 Answer by Peter May for Filtered ring giving rise to a graded-commutative ring Peter May 2011-12-28T02:34:30Z 2011-12-28T02:34:30Z <p>There are differences in conventions and philosophy in different subjects. Mathematically, there are two natural symmetric monodal structures on the monoidal category of graded modules over a commutative ring under the tensor product. In algebraic topology, the natural one is the one with signs. In algebraic geometry it is (usually) the one without signs. There is another related difference. Algebraic geometers allow sums of elements of different degrees and talk of homogeneous elements for contrast. Algebraic topologists generally think of graded modules as sequences of modules and do not allow the addition of elements of different degrees. The symmetry with signs makes little sense when elements are not restricted to be homogeneous. </p> <p>To test whether you are an algebraic geometer or an algebraic topologist, ask yourself whether or not the Laurent series ring \$F[x,x^{-1}]\$ is a field, where \$F\$ is a field and \$x\$ has degree \$2\$, say (so this has nothing to do with signs). I once taught a joint course with a very fine algebraic geometer (Spencer Bloch no less) and we disagreed about the answer. </p> <p>As to your actual question, signs are unlikely to appear out of the air when passing to associated graded rings. There is no reason why they should.</p>