Let $X$ be a smooth projective variety over an algebraically closed field. The Chow group $\mathbb Q\mathrm{CH}^d(X)$ is $\mathbb Q$--linearly generated by irreducible subvarieties $Z \subseteq X$ of codimension $d$, modulo rational equivalence.

I am interested in the linear subspace of $\mathbb Q\mathrm{CH}^d(X)$ which is generated by the subvarieties $Z\subseteq X$ of codimension $d$ which are locally complete intersections, so those which are locally the zero set of exactly $d$ regular functions. Let us denote this subspace by $\mathbb Q\mathrm{CH}^d_{\mathrm{lci}}(X)$. Then the question is:

Are Chow groups generated by local complete intersections? I.e. does equality $$\mathbb Q\mathrm{CH}^d_{\mathrm{lci}}(X) = \mathbb Q\mathrm{CH}^d(X)$$ hold?

If for instance $d=1$, equality holds indeed, as $X$ is smooth. I suspect this not so in general for $d\geq 2$... but where to look for a counter example?

  • $\begingroup$ Do you allow only linear combinations of lci, or products of lci as well? In the latter case I think the answer is yes. $\endgroup$
    – Sasha
    Apr 3, 2011 at 17:02
  • $\begingroup$ @Sasha: Just linear combinations. One could of course also look at the subring of the Chowring generated by lci's... why would the answer be yes then? $\endgroup$
    – Xandi Tuni
    Apr 3, 2011 at 17:50
  • 2
    $\begingroup$ I asked Madhav Nori about this question. He said (although I may not be remembering this exactly right) that the answer is "yes" for $d \geq \mathrm{dim} \, X / 2$, and so far as he is aware, unknown in general. He also said that getting generators that are local complete intersections is no easier or harder than getting generators that are smooth. $\endgroup$ Apr 5, 2011 at 2:46

1 Answer 1


This is not an answer to the original question. Instead I will argue that the SUBRING of the Chow ring generated by lci subschemes is the whole ring.

Indeed, let us first check that Chern classes of vector bundles generate the Chow ring. Indeed, the structure sheaf of any subvariety $Z$ has a locally free resolution, hence its Chern classes (in particular the class of $Z$ itself) can be expressed as a liner combination of Chern classes of vector bundles.

So, it remains to check that Chern classes of vector bundles can be expressed as linear combinations of products of lci subschemes. Let us take a vector bundle $E$ of rank $r$. Let $O(h)$ be a very ample line bundle. Then for $n \gg 0$ the bundle $E(nh)$ is globally generated. Hence its top Chern class is represented by the zero locus of a generic section of $E(nh)$ which is lci. Considering different twists one can deduce that for any $i$ the class $c_i(E)h^{r-i}$ is represented by a linear combination of lci.

Now instead of zero loci of sections, consider degeneration schemes of morphisms $O^k \to E(nh)$. Again, the class of the degeneration scheme is $c_{r-k+1}(E(nh))$ and for generic morphism it is a smooth (and hence lci) subvariety. Taking different twists we conclude that $c_i(E)h^{r-k+1-i}$ is represented by a linear combination of lci.

  • $\begingroup$ very nice answer! $\endgroup$
    – SGP
    Apr 3, 2011 at 20:59
  • 4
    $\begingroup$ @Sasha : By Chow's moving lemma if $[Z]$ and $[Z']$ are classes in $CH(X)$, then you can find a "rational" deformation of $Z$ such that $Z$ and $Z'$ intersect properly. So the "rational" generic deformation of $Z$ will intersect $Z'$ properly. But the general "rational" deformation of $Z$ will also be lci, so that you can find a cycle $Y$ representing $[Z].[Z']$ which is the proper intersection of two lci subschemes. Hence $Y$ is also lci. As a consequence, it seems to me that you proved that $CH(X)$ is linearly generated by classes of lci subschemes. Did I make a mistake? $\endgroup$
    – Libli
    Feb 14, 2016 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.