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Hello,

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 $gr R = F^0 / F^1 \oplus F^1/F^2 \oplus \cdots$.

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 $q$).

Are there natural/well-known/simple conditions on $R$ ensuring that $gr R$ is graded-commutative?

Thank you very much!

Pierre

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Why is graded-commutativity natural in this context? Naturality depends on what you want to do... maybe if you explained what you expect to get out of such conditions we would be more able to help. –  Mariano Suárez-Alvarez Sep 20 '11 at 18:05
    
I mean nothing more than the fact that graded-commutativity is a condition that shows up everywhere when studying graded algebras (I'm doing algebraic topology and thinking of cohomology algebras, of course), and filtrations are a common way to produce graded algebras, so... is there a common way to produce graded-commutative algebras out of a nicely filtrated ring? –  Pierre Sep 20 '11 at 18:37
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2 Answers 2

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.

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.

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.

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$[F^i,F^j]\subset F^{i+j+1}$ ?

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That would make it commutative, rather than graded-commutative. Unless the meaning you give to the square brackets depends on i and j -- that would be an option, but it's hardly better than saying "$gr R$ is graded-commutative". I was hoping for something natural, somehow. –  Pierre Sep 20 '11 at 17:24
    
Of course, $[,]$ is the graded commutator. You were asking for well-kown and simple... –  euklid345 Sep 20 '11 at 17:59
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euklid345: if the algebra is not graded to begin with, there is no graded-commutator to compute! :) –  Mariano Suárez-Alvarez Sep 20 '11 at 18:06
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I'm sorry, you are right, in the graded commutative case this question is more subtle than I thought. My dumb answer only works if you already have a $\mathbb{Z}/2\mathbb{Z}$ grading. –  euklid345 Sep 20 '11 at 19:09
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If you know $F^0$ to $F^{i-1}$, you can use this to find out the minimum necessary for $F^i$. The minimum is most likely to give you the most interesting ring, so go with that. Then calculate the minimum for $F^{i+1}$.... If $F^0$ is well-behaved enough, you get a graded ring with the appropriate commutator property. –  Will Sawin Nov 1 '11 at 17:48
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