Lemma 81.16.1. In Situation 81.2.1 let $X, Y/B$ be good. Assume $Y$ integral with $\dim _\delta (Y) = k$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Then for $g \in R(Y)^*$ we have

\[ f^*\text{div}_ Y(g) = \sum m_{X', X} (X' \to X)_*\text{div}_{X'}(g \circ f|_{X'}) \]

as $(k + r - 1)$-cycles on $X$ where the sum is over the irreducible components $X'$ of $X$ and $m_{X', X}$ is the multiplicity of $X'$ in $X$.

**Proof.**
Observe that any irreducible component of $X$ dominates $Y$ (Lemma 81.9.1) and hence the composition $g \circ f|_{X'}$ is defined (Morphisms of Spaces, Section 66.47). 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

\[ \xymatrix{ U \ar[r]_ a \ar[d]_ h & X \ar[d]^ f \\ V \ar[r]^ b & Y } \]

Since $a$ is surjective and étale it follows from Lemma 81.10.3 that it suffices to prove the equality of cycles after pulling back by $a$. We can use Lemma 81.13.2 to write

\[ b^*\text{div}_ Y(g) = \sum (V' \to V)_*\text{div}_{V'}(g \circ b|_{V'}) \]

where the sum is over the irreducible components $V'$ of $V$. Using Lemma 81.11.1 we find

\[ h^*b^*\text{div}_ Y(g) = \sum (V' \times _ V U \to U)_*(h')^*\text{div}_{V'}(g \circ b|_{V'}) \]

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

\[ (h')^*\text{div}_{V'}(g \circ b|_{V'}) = \sum m_{U', V' \times _ V U} (U' \to V' \times _ V U)_*\text{div}_{U'}(g \circ b|_{V'} \circ h'|_{U'}) \]

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 81.7.1); this is an irreducible component of $X$ for example by Lemma 81.9.1. 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 81.10.4), the definitions, and Lemma 81.10.3. Combining all of what we just said we obtain

\[ a^*f^*\text{div}_ Y(g) = h^*b^*\text{div}_ Y(g) = \sum m_{\overline{a(U')}, X} (U' \to U)_*\text{div}_{U'}(g \circ (f \circ a)|_{U'}) \]

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 81.11.1 to get

\[ a^*\sum m_{X', X} (X' \to X)_*\text{div}_{X'}(g \circ f|_{X'}) = \sum m_{X', X} (X' \times _ X U \to U)_*(a')^*\text{div}_{X'}(g \circ f|_{X'}) \]

where $a' : X' \times _ X U \to X'$ is the projection. By Lemma 81.13.2 we get

\[ (a')^*\text{div}_{X'}(g \circ f|_{X'}) = \sum (U' \to X' \times _ X U)_*\text{div}_{U'}(g \circ (f \circ a)|_{U'}) \]

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.
$\square$

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