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Let $X$ be a smooth projective variety over an algebraically closed field and let $A,B$ be closed irreducible subvarieties of $X$. Chow's moving lemma which is proved in the book by Eisenbud and Harris ("Intersection theory and all that") shows that in this situation there is a cycle $\sum A_j$ rationally equivalent to $A$ on $X$ such that each $A_j$ meets $B$ generically transversely. This means that every irreducible component of the intersection has the expected codimension and is generically reduced, or equivalently contains a point $p$ where $A$ and $B$ are smooth and $T_p A+ T_p B = T_p X$.

My question is if it possible to find a cycle such that the intersection is transversal at every point.

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  • $\begingroup$ This would be simpler if you allowed to replace $A$ by $N\cdot A$ for $N$ a sufficiently positive and divisible integer. Use the isomorphism of $K$-theory and Chow theory, i.e., write $A$ as a polynomial in Chern classes $c_r$ of locally free sheaves $\mathcal{E}$. Using finite differences, $c_r(\mathcal{E})$ can be expressed in terms of Chern classes of $\mathcal{E}(d)$ for $d$ large. Thus, you need only find global sections of $\mathcal{E}(d)|_B$ whose determinantal locus is transversal (i.e., a Bertini-type theorem). Finally, extend those to global sections of $\mathcal{E}(d)$. $\endgroup$
    – Jason Starr
    Commented Jul 1, 2015 at 14:17
  • $\begingroup$ Hmm, actually my last suggestion won't work. The determinantal locus can be made smooth away from the locus where the $\mathcal{O}_A$-module homomorphism drops rank (further). However, it will typically be singular on that locus. $\endgroup$
    – Jason Starr
    Commented Jul 1, 2015 at 14:32
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    $\begingroup$ If you only move $A$, this is impossible. For example, take $X$ to be a projective space, $A$ a hypersurface and $B$ to be a subvariety with positive dimensional singular locus. Whatever $A$ is moved to, it is still a hypersurface and will meet $B$ in some singular points. $\endgroup$
    – Mohan
    Commented Jul 1, 2015 at 15:50
  • $\begingroup$ Thanks Mohan. What if $A$ and $B$ are both assumed to be smooth? This is actually the case. $\endgroup$
    – user115940
    Commented Jul 1, 2015 at 17:44

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