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I have a question about a proof in Fulton's Intersection theory. In the first chapter, he wants to prove that Chow groups are invariant with respect to homotopy, and the usual reductions tell us that's it's tantamount to proving it for $X$ an integral affine scheme, and for the trivial bundle $X\times \mathbb{A}^1$.

One can assume that our cycle $[V]$ in $X\times \mathbb{A}^1$ is mapped dominantly to $X$, and that the dimensions of $X$ and $V$ match.

In this case, he says that the cycle $V$ is defined by an ideal $I$, that becomes principal after tensoring with $K$, the fraction field of $A$ (with $X= \operatorname{Spec} A$).

And then he states that $$\operatorname{div}(f)=[V]-\sum [V_i]$$ where $V_i$ are varieties that are not dominantly mapped to $X$.

I don't really seem to grasp that last point. I've tried to work out on a baby example with $X=\mathbb{A}^1$, but in this case everything is trivial, and $\operatorname{div}(f)=[V]$. I have an intuition that it might have to do with the fact that $\operatorname{div}(f)$ is essentially $[f=0]-[f=\infty]$ (this is not "rigourous" but i hope you see what i mean), and thus that $[V]=[f=0]$ whence $[f=\infty]=\sum[V_i]$, but i cannot see clearly how it works.

Thank you!

Minesotafan
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  • Which theorem, exactly, are you working on? – Brenin Sep 16 '13 at 16:24
  • This is proposition 1.9, in Fulton's Book. Namely that we have isomorphisms $CH(X)\to CH(X\times \mathbb{A}^n)$ in appropriate degree. He only shows the surjectivity in this particular proposition. – Minesotafan Sep 16 '13 at 16:42

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Edit A more correct and simpler approach.

Let $k$ be the ground field. You know that $[V]$ is a principal divisor $\mathrm{div}_{\mathbb A^1_K}(f)$ on $\mathbb A^1_K$ for some rational function $f$ on $\mathbb A^1_K$.The latter has the same field of functions as $X\times_k \mathbb A^1_k$. Let $p: X\times\mathbb A^1_k\to X$ be the projection. Write $$[V]-\mathrm{div}_{X\times\mathbb A^1_k}(f)=\sum_i a_i [V_i]+ \sum_j b_j [H_j], \quad a_i, b_j\in \mathbb Z$$ with pairwise distinct irreducible closed subvarieties $V_i, H_j$ such that $p(V_i)$ is not dense in $X$ ($V_i$ is "vertical"), and $p(H_j)$ is dense in $X$ ($H_j$ is "horizontal").

Now when you tensor by $K$, you get $$0=[V]_K-\mathrm{div}_{\mathbb A^1_K}(f)=\sum_j b_j [H_j]_K.$$ As the $[H_j]_K$ are pairwise distinct ($(H_j)_K$ is dense in $H_j$), we find $b_j=0$ for all $j$, and $$[V]-\mathrm{div}_{X\times\mathbb A^1_k}(f)=\sum_i a_i [V_i].$$

The coefficients $a_i$ are not necessarily positive. But they become positive if we change the initial $f$ in a suitable way. Let $d=\dim X$. Let $Z_i=\overline{p(V_i)}$. It has $\dim Z_i\le d-1$. The fibers of $V_i\to Z_i$ are contained in $\mathbb A^1_k$, so $d=\dim V_i\le \dim Z_i+1=d$ and $\dim Z_i=d-1$. As $V_i\subseteq Z_i\times \mathbb A^1_k$ and they are of the same dimension and the latter is irreducible, they are equal. Let $p_i$ be the prime ideal of $A$ corresponding to $Z_i$. Let $a\in \prod_k p_i^N$ non-zero. Then $$\mathrm{div}_{X\times\mathbb A^1_k}(a)=\sum_i N[Z_i]+ D, \quad D\ge 0.$$ This leads to $$[V]-\mathrm{div}_{X\times\mathbb A^1_k}(f/a)=\sum_i (N+a_i) [V_i]+p^*D.$$ Taking $N$ big enough we find a postive "vertical" cycle at the right.

Cantlog
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