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This question is arose from the question Difference between equivalence relations on algebraic cycles
and the example 3 in lecture 1 in Mumford's book Lectures on curves on an algebraic surface.

Here is the example.

Let $E$ be an elliptic curve. A curve $C$ on $\mathbb{P}^1\times E$ is said has projection degree $(d, e)$, if the projections $C\to\mathbb{P}^1$ and $C\to E$ are of degree $d$ and $e$ respectively. In that example Mumford shows that the curves with projection degree (d,e) and $d>0$ forms a $d(e+1)$ irreducible family of curves. Clearly, these curves are algebraic equivalence. But this family is not a linear system.

I can understand why the dimension is $d(e+1)$ with the help that numerical and algebraic equivalence coincide for divisors. But why the equivalences coincide for divisors? Where is the place discussing these equivalence of algebraic cycles? Also what is the irreducible family in this example?

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up vote 5 down vote accepted

Numerical and algebraic equivalence coincide up to torsion for codimension $1$ cycles in non-singular projective varieties over $\mathbb C$. Also, the group $Alg_{\tau}^1(X)/Alg^1(X)$ can be identified with $H^2(X,\mathbb Z)_{tor}$ (here $Alg_{\tau}^1(X)$ is the group of cycles who multiple is in $Alg^1(X)$). This is discussed in 19.3.1 of Fulton's book "Intersection Theory". So you just need to show that $H^2(X,\mathbb Z)_{tor}=0$, i.e it is free abelian. But presumably this follows from the Kunneth formula.

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I meant $H^2(X,\mathbb Z)$ is free abelian. – Hailong Dao Feb 11 '10 at 17:21
Thanks for the answer. But what is the irreducible family? – Fei YE Feb 11 '10 at 19:45

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