Filtered ring giving rise to a graded-commutative ring 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
 A: There are differences in conventions and philosophy in different subjects.
Mathematically, there are two natural symmetric monoidal 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.
A: $[F^i,F^j]\subset F^{i+j+1}$ ?
