The Stacks project

Proof. Observe that any irreducible component of $X$ dominates $Y$ (Lemma 82.9.1) and hence the composition $g \circ f|_{X'}$ is defined (Morphisms of Spaces, Section 67.47). We will reduce this to the case of schemes. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Picture

Since $a$ is surjective and étale it follows from Lemma 82.10.3 that it suffices to prove the equality of cycles after pulling back by $a$. We can use Lemma 82.13.2 to write

where the sum is over the irreducible components $V'$ of $V$. Using Lemma 82.11.1 we find

where $h' : V' \times _ V U \to V'$ is the projection. We may apply the lemma in the case of schemes (Chow Homology, Lemma 42.20.1) to the morphism $h' : V' \times _ V U \to V'$ to see that we have

where the sum is over the irreducible components $U'$ of $V' \times _ V U$. Each $U'$ occurring in this sum is an irreducible component of $U$ and conversely every irreducible component $U'$ of $U$ is an irreducible component of $V' \times _ V U$ for a unique irreducible component $V' \subset V$. Given an irreducible component $U' \subset U$, denote $\overline{a(U')} \subset X$ the “image” in $X$ (Lemma 82.7.1); this is an irreducible component of $X$ for example by Lemma 82.9.1. The muplticity $m_{U', V' \times _ V U}$ is equal to the multiplicity $m_{\overline{a(U')}, X}$. This follows from the equality $h^*a^*[Y] = b^*f^*[Y]$ (Lemma 82.10.4), the definitions, and Lemma 82.10.3. Combining all of what we just said we obtain

Next, we analyze what happens with the right hand side of the formula in the statement of the lemma if we pullback by $a$. First, we use Lemma 82.11.1 to get

where $a' : X' \times _ X U \to X'$ is the projection. By Lemma 82.13.2 we get

where the sum is over the irreducible components $U'$ of $X' \times _ X U$. These $U'$ are irreducible components of $U$ and in fact are exactly the irreducible components of $U$ such that $\overline{a(U')} = X'$. Comparing with what we obtained above we conclude. $\square$

Proof. What do we have to show? Well, suppose we are given a collection

of closed immersions, with each $W_ j$ integral of $\delta $-dimension $k + 1$ and rational functions $g_ j \in R(W_ j)^*$. Moreover, assume that the collection $\{ |i_ j|(|W_ j|)\} _{j \in J}$ is locally finite in $|Y|$. Then we have to show that

is rationally equivalent to zero on $X$. The sum on the right makes sense by Lemma 82.9.2.

Consider the fibre products

and denote $f_ j : W'_ j \to W_ j$ the first projection. By Lemma 82.11.1 we can write the sum above as

By Lemma 82.16.1 we see that each $f_ j^*\text{div}(g_ j)$ is rationally equivalent to zero on $W'_ j$. Hence each $i'_{j, *}(f_ j^*\text{div}(g_ j))$ is rationally equivalent to zero. Then the same is true for the displayed sum by the discussion in Remark 82.15.3. $\square$

Proof. What do we have to show? Well, suppose we are given a collection

of closed immersions, with each $W_ j$ integral of $\delta $-dimension $k + 1$ and rational functions $f_ j \in R(W_ j)^*$. Moreover, assume that the collection $\{ i_ j(W_ j)\} _{j \in J}$ is locally finite on $X$. Then we have to show that

is rationally equivalent to zero on $X$.

Note that the sum is equal to

Let $W'_ j \subset Y$ be the integral closed subspace which is the image of $p \circ i_ j$, see Lemma 82.7.1. The collection $\{ W'_ j\} $ is locally finite in $Y$ by Lemma 82.7.5. Hence it suffices to show, for a given $j$, that either $p_*i_{j, *}\text{div}(f_ j) = 0$ or that it is equal to $i'_{j, *}\text{div}(g_ j)$ for some $g_ j \in R(W'_ j)^*$.

The arguments above therefore reduce us to the case of a since integral closed subspace $W \subset X$ of $\delta $-dimension $k + 1$. Let $f \in R(W)^*$. Let $W' = p(W)$ as above. We get a commutative diagram of morphisms

Note that $p_*i_*\text{div}(f) = i'_*(p')_*\text{div}(f)$ by Lemma 82.8.2. As explained above we have to show that $(p')_*\text{div}(f)$ is the divisor of a rational function on $W'$ or zero. There are three cases to distinguish.

The case $\dim _\delta (W') < k$. In this case automatically $(p')_*\text{div}(f) = 0$ and there is nothing to prove.

The case $\dim _\delta (W') = k$. Let us show that $(p')_*\text{div}(f) = 0$ in this case. Since $(p')_*\text{div}(f)$ is a $k$-cycle, we see that $(p')_*\text{div}(f) = n[W']$ for some $n \in \mathbf{Z}$. In order to prove that $n = 0$ we may replace $W'$ by a nonempty open subspace. In particular, we may and do assume that $W'$ is a scheme. Let $\eta \in W'$ be the generic point. Let $K = \kappa (\eta ) = R(W')$ be the function field. Consider the base change diagram

Observe that $c$ is proper. Also $|W_\eta |$ has dimension $1$: use Decent Spaces, Lemma 68.18.6 to identify $|W_\eta |$ as the subspace of $|W|$ of points mapping to $\eta $ and note that since $\dim _\delta (W) = k + 1$ and $\delta (\eta ) = k$ points of $W_\eta $ must have $\delta $-value either $k$ or $k + 1$. Thus the local rings have dimension $\leq 1$ by Decent Spaces, Lemma 68.20.2. By Spaces over Fields, Lemma 72.9.3 we find that $W_\eta $ is a scheme. Since $\mathop{\mathrm{Spec}}(K)$ is the limit of the nonempty affine open subschemes of $W'$ we conclude that we may assume that $W$ is a scheme by Limits of Spaces, Lemma 70.5.11. Then finally by the case of schemes (Chow Homology, Lemma 42.20.3) we find that $n = 0$.

The case $\dim _\delta (W') = k + 1$. In this case Lemma 82.14.1 applies, and we see that indeed $p'_*\text{div}(f) = \text{div}(g)$ for some $g \in R(W')^*$ as desired. $\square$