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I was looking in Ravi Vakil's notes on Intersection Theory, Class 20, where he introduces the bivariant intersection theory, in particular the Chow ring $A^\ast (X)$.

On p. 2, he writes the following which I had no idea about:

Alarming fact: This ring is apparently not known to be commutative in general, because the argument requires resolution of singularities. (It is known to be commutative in characteristic 0, and for smooth things in positive characteristic, and a few more things.) I think it should be possible to show that the ring is commutative in general using technology not available when this theory was first developed, using Johan de Jong’s “alteration theorem” in positive characteristic. If you would like to patch this hole, then come talk to me.

Vakil's notes for this class are from 2004, so plenty could have happened since then. Is this still an open problem?

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up vote 6 down vote accepted

De Jong's theorem proves that the ring is commutative when tensored with $\mathbb Q$. For the torsion part, I am pretty sure that the question is still open.

[Edit:] Mikhail is absolutely right, with Gabber's theorem (http://www.math.u-psud.fr/~illusie/refined_uniformization3.pdf) one can show that the bivariant Chow ring tensored with $\mathbb Z[1/p]$ is commutative in characteristic $p > 0$.

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Probably one can apply Gabber's result instead; then it suffices to invert $p$. –  Mikhail Bondarko Jan 25 '13 at 18:36
    
@Angelo: Besides a small spelling correction in "torsion", I might add for what it's worth that torsion (but not commutativity) is a serious issue in prime characteristic when computing cohomology of line bundles on flag varieties. This is clear from old work of H.H. Andersen. The Chow ring might be better behaved in general, but one has to be cautious. –  Jim Humphreys Jan 26 '13 at 14:44
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