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I have heard several times that the proof of the second part of the Chow's moving lemma (of algebraic geometry), is problematic; and that this is the reason Fulton, Intersection theory, does not use it to define intersection product.

Has anyone heard similar claims, and if so, what/where is the gap ?

For completeness, I mean the moving lemma, say as stated in J.Roberts, Appendix 2(Algebraic Geometry, 1970, ed. Oort). The problem claimed is with the 'Moreover' part.

Lemma (Chow's moving lemma). - Let $W$ and $W'$ be cycles on X. Then there exists a cycle $W''$ rationally equivalent to $W'$ such that $W \cdot W''$ is defined. Moreover, there exists an algebraic family $\{Z(t)\}_{t \in P^1(k)}$ deforming $W'$ into $W$ such that $Z(t)\cdot W$ is defined for almost all t.

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  • $\begingroup$ This is not the reason that the moving lemma is not used in Fulton's book. The way intersection theory is developed there is more general, since it requires no quasi-projectivity assumptions, and much more powerful, since he constructs a "refined" intersection product which is impossible using the moving lemma. $\endgroup$
    – naf
    Commented Dec 17, 2012 at 13:30
  • $\begingroup$ I think that the proof of Roberts is correct for the case where the ground field is algebraic closed. May be the gap, if any, happens when k is not algebraic closed. By the way, in the second part, the family Z(t) is rational ($t\in \mathbb{P}^1(k)$), which is stronger than that you stated. $\endgroup$
    – nono
    Commented Dec 17, 2012 at 15:23
  • $\begingroup$ Nono, thanks, I corrected the statement. I should have been more clear: the claim I heard was that there is no complete proof in the literature, or rather that some experts think so. $\endgroup$
    – Andrew
    Commented Dec 17, 2012 at 16:30
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    $\begingroup$ By the way, the first part says that you can find a cycle W" which is rationally equivalent to W' such that W.W" is well-defined. If you accept the first part, then I think you must accept the second part as well, at least in principle. This is because to say that W" is rationally equivalent to W', you must have a family W(t) with $t\in \mathbb{P}^1(k)$ so that $W(0)=W'$ and $W(\infty )=W"$. Now this family intersects properly with $W$ at one point $\infty$, it should do so for a dense open set in $\mathbb{P}^1$. $\endgroup$
    – nono
    Commented Dec 17, 2012 at 23:25
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    $\begingroup$ There is another problem with using the moving lemma to define intersection products, and that is well-definedness. That is to say, if $\alpha, \beta$ are given, and $\alpha'$ and $\alpha''$ are rationally equivalent to $\alpha$ and meet $\beta$ properly, then one should check that $\alpha' \cdot \beta = \alpha''\cdot \beta$. Moreover, one wants the intersection product to factor through rational equivalence, so one should vary $\beta$ as well. This method is used in the Stacks Project, and all details are carefully carried out there. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 20, 2017 at 17:20

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