Lemma 43.9.1. Let $X$ be a variety. Let $W \subset X$ be a subvariety of dimension $k + 1$. Let $f \in \mathbf{C}(W)^*$ be a nonzero rational function on $W$. Then $\text{div}_ W(f)$ is rationally equivalent to zero on $X$. Conversely, these principal divisors generate the abelian group of cycles rationally equivalent to zero on $X$.

Proof. The first assertion follows from Chow Homology, Lemma 42.18.2. More precisely, let $W' \subset X \times \mathbf{P}^1$ be the closure of the graph of $f$. Then $\text{div}_ W(f) = [W'_0]_ k - [W'_\infty ]$ in $Z_ k(W) \subset Z_ k(X)$, see part (6) of Chow Homology, Lemma 42.18.2.

For the second, let $W' \subset X \times \mathbf{P}^1$ be a closed subvariety of dimension $k + 1$ which dominates $\mathbf{P}^1$. We will show that $[W'_0]_ k - [W'_\infty ]_ k$ is a principal divisor which will finish the proof. Let $W \subset X$ be the image of $W'$ under the projection to $X$. Then $W \subset X$ is a closed subvariety and $W' \to W$ is proper and dominant with fibres of dimension $0$ or $1$. If $\dim (W) = k$, then $W' = W \times \mathbf{P}^1$ and we see that $[W'_0]_ k - [W'_\infty ]_ k = [W] - [W] = 0$. If $\dim (W) = k + 1$, then $W' \to W$ is generically finite1. Let $f$ denote the projection $W' \to \mathbf{P}^1$ viewed as an element of $\mathbf{C}(W')^*$. Let $g = \text{Nm}(f) \in \mathbf{C}(W)^*$ be the norm. By Chow Homology, Lemma 42.18.1 we have

$\text{div}_ W(g) = \text{pr}_{X, *}\text{div}_{W'}(f)$

Since $\text{div}_{W'}(f) = [W'_0]_ k - [W'_\infty ]_ k$ by Chow Homology, Lemma 42.18.2 the proof is complete. $\square$

[1] If $W' \to W$ is birational, then the result follows from Chow Homology, Lemma 42.18.2. Our task is to show that even if $W' \to W$ has degree $> 1$ the basic rational equivalence $[W'_0]_ k \sim _{rat} [W'_\infty ]_ k$ comes from a principal divisor on a subvariety of $X$.

Comment #6776 by Morten on

Why is the map $W'\rightarrow W$ generically finite? Could $W'$ not be a trivial family like $X\times \mathbb{P}^1$?

Comment #6777 by on

Ha! Yes, this could happen. Of course, in that case $[W'_0]_k - [W'_\infty]_k$ is the zero cycle. I will fix this the next time I go through all the comments.

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