## 42.25 Intersecting with an invertible sheaf and push and pull

In this section we prove that the operation $c_1(\mathcal{L}) \cap -$ commutes with flat pullback and proper pushforward.

Lemma 42.25.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Assume $Y$ is integral and $n = \dim _\delta (Y)$. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$. Then we have

$f^*\text{div}_\mathcal {L}(s) = \sum n_ i\text{div}_{f^*\mathcal{L}|_{X_ i}}(s_ i)$

in $Z_{n + r - 1}(X)$. Here the sum is over the irreducible components $X_ i \subset X$ of $\delta$-dimension $n + r$, the section $s_ i = f|_{X_ i}^*(s)$ is the pullback of $s$, and $n_ i = m_{X_ i, X}$ is the multiplicity of $X_ i$ in $X$.

Proof. To prove this equality of cycles, we may work locally on $Y$. Hence we may assume $Y$ is affine and $s = p/q$ for some nonzero sections $p \in \Gamma (Y, \mathcal{L})$ and $q \in \Gamma (Y, \mathcal{O})$. If we can show both

$f^*\text{div}_\mathcal {L}(p) = \sum n_ i\text{div}_{f^*\mathcal{L}|_{X_ i}}(p_ i) \quad \text{and}\quad f^*\text{div}_\mathcal {O}(q) = \sum n_ i\text{div}_{\mathcal{O}_{X_ i}}(q_ i)$

(with obvious notations) then we win by the additivity, see Divisors, Lemma 31.27.5. Thus we may assume that $s \in \Gamma (Y, \mathcal{L})$. In this case we may apply the equality (42.24.3.1) to see that

$[Z(f^*(s))]_{k + r - 1} = \sum n_ i\text{div}_{f^*\mathcal{L}|_{X_ i}}(s_ i)$

where $f^*(s) \in f^*\mathcal{L}$ denotes the pullback of $s$ to $X$. On the other hand we have

$f^*\text{div}_\mathcal {L}(s) = f^*[Z(s)]_{k - 1} = [f^{-1}(Z(s))]_{k + r - 1},$

by Lemmas 42.23.2 and 42.14.4. Since $Z(f^*(s)) = f^{-1}(Z(s))$ we win. $\square$

Lemma 42.25.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Let $\alpha$ be a $k$-cycle on $Y$. Then

$f^*(c_1(\mathcal{L}) \cap \alpha ) = c_1(f^*\mathcal{L}) \cap f^*\alpha$

in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$.

Proof. Write $\alpha = \sum n_ i[W_ i]$. We will show that

$f^*(c_1(\mathcal{L}) \cap [W_ i]) = c_1(f^*\mathcal{L}) \cap f^*[W_ i]$

in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$ by producing a rational equivalence on the closed subscheme $f^{-1}(W_ i)$ of $X$. By the discussion in Remark 42.19.5 this will prove the equality of the lemma is true.

Let $W \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$. Consider the closed subscheme $W' = f^{-1}(W) = W \times _ Y X$ so that we have the fibre product diagram

$\xymatrix{ W' \ar[r] \ar[d]_ h & X \ar[d]^ f \\ W \ar[r] & Y }$

We have to show that $f^*(c_1(\mathcal{L}) \cap [W]) = c_1(f^*\mathcal{L}) \cap f^*[W]$. Choose a nonzero meromorphic section $s$ of $\mathcal{L}|_ W$. Let $W'_ i \subset W'$ be the irreducible components of $\delta$-dimension $k + r$. Write $[W']_{k + r} = \sum n_ i[W'_ i]$ with $n_ i$ the multiplicity of $W'_ i$ in $W'$ as per definition. So $f^*[W] = \sum n_ i[W'_ i]$ in $Z_{k + r}(X)$. Since each $W'_ i \to W$ is dominant we see that $s_ i = s|_{W'_ i}$ is a nonzero meromorphic section for each $i$. By Lemma 42.25.1 we have the following equality of cycles

$h^*\text{div}_{\mathcal{L}|_ W}(s) = \sum n_ i\text{div}_{f^*\mathcal{L}|_{W'_ i}}(s_ i)$

in $Z_{k + r - 1}(W')$. This finishes the proof since the left hand side is a cycle on $W'$ which pushes to $f^*(c_1(\mathcal{L}) \cap [W])$ in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$ and the right hand side is a cycle on $W'$ which pushes to $c_1(f^*\mathcal{L}) \cap f^*[W]$ in $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X)$. $\square$

Lemma 42.25.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a proper morphism. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Let $s$ be a nonzero meromorphic section $s$ of $\mathcal{L}$ on $Y$. Assume $X$, $Y$ integral, $f$ dominant, and $\dim _\delta (X) = \dim _\delta (Y)$. Then

$f_*\left(\text{div}_{f^*\mathcal{L}}(f^*s)\right) = [R(X) : R(Y)]\text{div}_\mathcal {L}(s).$

as cycles on $Y$. In particular

$f_*(c_1(f^*\mathcal{L}) \cap [X]) = c_1(\mathcal{L}) \cap [Y].$

Proof. The last equation follows from the first since $f_*[X] = [R(X) : R(Y)][Y]$ by definition. It turns out that we can re-use Lemma 42.18.1 to prove this. Namely, since we are trying to prove an equality of cycles, we may work locally on $Y$. Hence we may assume that $\mathcal{L} = \mathcal{O}_ Y$. In this case $s$ corresponds to a rational function $g \in R(Y)$, and we are simply trying to prove

$f_*\left(\text{div}_ X(g)\right) = [R(X) : R(Y)]\text{div}_ Y(g).$

Comparing with the result of the aforementioned Lemma 42.18.1 we see this true since $\text{Nm}_{R(X)/R(Y)}(g) = g^{[R(X) : R(Y)]}$ as $g \in R(Y)^*$. $\square$

Lemma 42.25.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $p : X \to Y$ be a proper morphism. Let $\alpha \in Z_{k + 1}(X)$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Then

$p_*(c_1(p^*\mathcal{L}) \cap \alpha ) = c_1(\mathcal{L}) \cap p_*\alpha$

in $\mathop{\mathrm{CH}}\nolimits _ k(Y)$.

Proof. Suppose that $p$ has the property that for every integral closed subscheme $W \subset X$ the map $p|_ W : W \to Y$ is a closed immersion. Then, by definition of capping with $c_1(\mathcal{L})$ the lemma holds.

We will use this remark to reduce to a special case. Namely, write $\alpha = \sum n_ i[W_ i]$ with $n_ i \not= 0$ and $W_ i$ pairwise distinct. Let $W'_ i \subset Y$ be the image of $W_ i$ (as an integral closed subscheme). Consider the diagram

$\xymatrix{ X' = \coprod W_ i \ar[r]_-q \ar[d]_{p'} & X \ar[d]^ p \\ Y' = \coprod W'_ i \ar[r]^-{q'} & Y. }$

Since $\{ W_ i\}$ is locally finite on $X$, and $p$ is proper we see that $\{ W'_ i\}$ is locally finite on $Y$ and that $q, q', p'$ are also proper morphisms. We may think of $\sum n_ i[W_ i]$ also as a $k$-cycle $\alpha ' \in Z_ k(X')$. Clearly $q_*\alpha ' = \alpha$. We have $q_*(c_1(q^*p^*\mathcal{L}) \cap \alpha ') = c_1(p^*\mathcal{L}) \cap q_*\alpha '$ and $(q')_*(c_1((q')^*\mathcal{L}) \cap p'_*\alpha ') = c_1(\mathcal{L}) \cap q'_*p'_*\alpha '$ by the initial remark of the proof. Hence it suffices to prove the lemma for the morphism $p'$ and the cycle $\sum n_ i[W_ i]$. Clearly, this means we may assume $X$, $Y$ integral, $f : X \to Y$ dominant and $\alpha = [X]$. In this case the result follows from Lemma 42.25.3. $\square$

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