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Tag 02TG

41.26. Intersecting with an invertible sheaf and rational equivalence

Applying the key lemma we obtain the fundamental properties of intersecting with invertible sheaves. In particular, we will see that $c_1(\mathcal{L}) \cap -$ factors through rational equivalence and that these operations for different invertible sheaves commute.

Lemma 41.26.1. Let $(S, \delta)$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ integral and $\dim_\delta(X) = n$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. Choose a nonzero meromorphic section $s$ of $\mathcal{L}$ and a nonzero meromorphic section $t$ of $\mathcal{N}$. Set $\alpha = \text{div}_\mathcal{L}(s)$ and $\beta = \text{div}_\mathcal{N}(t)$. Then $$c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{L}) \cap \beta$$ in $A_{n - 2}(X)$.

Proof. Immediate from the key Lemma 41.25.1 and the discussion preceding it. $\square$

Lemma 41.26.2. Let $(S, \delta)$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be invertible on $X$. The operation $\alpha \mapsto c_1(\mathcal{L}) \cap \alpha$ factors through rational equivalence to give an operation $$c_1(\mathcal{L}) \cap - : A_{k + 1}(X) \to A_k(X)$$

Proof. Let $\alpha \in Z_{k + 1}(X)$, and $\alpha \sim_{rat} 0$. We have to show that $c_1(\mathcal{L}) \cap \alpha$ as defined in Definition 41.23.1 is zero. By Definition 41.19.1 there exists a locally finite family $\{W_j\}$ of integral closed subschemes with $\dim_\delta(W_j) = k + 2$ and rational functions $f_j \in R(W_j)^*$ such that $$\alpha = \sum (i_j)_*\text{div}_{W_j}(f_j)$$ Note that $p : \coprod W_j \to X$ is a proper morphism, and hence $\alpha = p_*\alpha'$ where $\alpha' \in Z_{k + 1}(\coprod W_j)$ is the sum of the principal divisors $\text{div}_{W_j}(f_j)$. By Lemma 41.24.3 we have $c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap \alpha')$. Hence it suffices to show that each $c_1(\mathcal{L}|_{W_j}) \cap \text{div}_{W_j}(f_j)$ is zero. In other words we may assume that $X$ is integral and $\alpha = \text{div}_X(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}_X(f)$ for some $f \in R(X)^*$. We can think of $f$ as a regular meromorphic section of the invertible sheaf $\mathcal{N} = \mathcal{O}_X$. Choose a meromorphic section $s$ of $\mathcal{L}$ and denote $\beta = \text{div}_\mathcal{L}(s)$. By Lemma 41.26.1 we conclude that $$c_1(\mathcal{L}) \cap \alpha = c_1(\mathcal{O}_X) \cap \beta.$$ However, by Lemma 41.23.2 we see that the right hand side is zero in $A_k(X)$ as desired. $\square$

Let $(S, \delta)$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be invertible on $X$. We will denote $$c_1(\mathcal{L})^s \cap - : A_{k + s}(X) \to A_k(X)$$ the operation $c_1(\mathcal{L}) \cap -$. This makes sense by Lemma 41.26.2. We will denote $c_1(\mathcal{L}^s \cap -$ the $s$-fold iterate of this operation for all $s \geq 0$.

Lemma 41.26.3. Let $(S, \delta)$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. For any $\alpha \in A_{k + 2}(X)$ we have $$c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha$$ as elements of $A_k(X)$.

Proof. Write $\alpha = \sum m_j[Z_j]$ for some locally finite collection of integral closed subschemes $Z_j \subset X$ with $\dim_\delta(Z_j) = k + 2$. Consider the proper morphism $p : \coprod Z_j \to X$. Set $\alpha' = \sum m_j[Z_j]$ as a $(k + 2)$-cycle on $\coprod Z_j$. By several applications of Lemma 41.24.3 we see that $c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap c_1(p^*\mathcal{N}) \cap \alpha')$ and $c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{N}) \cap c_1(p^*\mathcal{L}) \cap \alpha')$. Hence it suffices to prove the formula in case $X$ is integral and $\alpha = [X]$. In this case the result follows from Lemma 41.26.1 and the definitions. $\square$

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\section{Intersecting with an invertible sheaf and rational equivalence}
\label{section-commutativity}

\noindent
Applying the key lemma we obtain the fundamental properties of intersecting
with invertible sheaves. In particular, we will see that
$c_1(\mathcal{L}) \cap -$ factors through rational equivalence and
that these operations for different invertible sheaves commute.

\begin{lemma}
\label{lemma-commutativity-on-integral}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$ be locally of finite type over $S$.
Assume $X$ integral and $\dim_\delta(X) = n$.
Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$.
Choose a nonzero meromorphic section $s$ of $\mathcal{L}$
and a nonzero meromorphic section $t$ of $\mathcal{N}$.
Set $\alpha = \text{div}_\mathcal{L}(s)$ and
$\beta = \text{div}_\mathcal{N}(t)$.
Then
$$c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{L}) \cap \beta$$
in $A_{n - 2}(X)$.
\end{lemma}

\begin{proof}
Immediate from the key Lemma \ref{lemma-key-formula}
and the discussion preceding it.
\end{proof}

\begin{lemma}
\label{lemma-factors}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$ be locally of finite type over $S$.
Let $\mathcal{L}$ be invertible on $X$.
The operation $\alpha \mapsto c_1(\mathcal{L}) \cap \alpha$
factors through rational equivalence to give an operation
$$c_1(\mathcal{L}) \cap - : A_{k + 1}(X) \to A_k(X)$$
\end{lemma}

\begin{proof}
Let $\alpha \in Z_{k + 1}(X)$, and $\alpha \sim_{rat} 0$.
We have to show that $c_1(\mathcal{L}) \cap \alpha$
as defined in Definition \ref{definition-cap-c1} is zero.
By Definition \ref{definition-rational-equivalence} there
exists a locally finite family $\{W_j\}$ of integral closed
subschemes with $\dim_\delta(W_j) = k + 2$ and rational functions
$f_j \in R(W_j)^*$ such that
$$\alpha = \sum (i_j)_*\text{div}_{W_j}(f_j)$$
Note that $p : \coprod W_j \to X$ is a proper morphism,
and hence $\alpha = p_*\alpha'$ where $\alpha' \in Z_{k + 1}(\coprod W_j)$
is the sum of the principal divisors $\text{div}_{W_j}(f_j)$.
By Lemma \ref{lemma-pushforward-cap-c1} we have
$c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap \alpha')$.
Hence it suffices to show that each
$c_1(\mathcal{L}|_{W_j}) \cap \text{div}_{W_j}(f_j)$ is zero.
In other words we may assume that $X$ is integral and
$\alpha = \text{div}_X(f)$ for some $f \in R(X)^*$.

\medskip\noindent
Assume $X$ is integral and $\alpha = \text{div}_X(f)$ for some $f \in R(X)^*$.
We can think of $f$ as a regular meromorphic section of the invertible
sheaf $\mathcal{N} = \mathcal{O}_X$. Choose a meromorphic section
$s$ of $\mathcal{L}$ and denote $\beta = \text{div}_\mathcal{L}(s)$.
By Lemma \ref{lemma-commutativity-on-integral}
we conclude that
$$c_1(\mathcal{L}) \cap \alpha = c_1(\mathcal{O}_X) \cap \beta.$$
However, by Lemma \ref{lemma-c1-cap-additive} we see that the right hand side
is zero in $A_k(X)$ as desired.
\end{proof}

\noindent
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$ be locally of finite type over $S$.
Let $\mathcal{L}$ be invertible on $X$.
We will denote
$$c_1(\mathcal{L})^s \cap - : A_{k + s}(X) \to A_k(X)$$
the operation $c_1(\mathcal{L}) \cap -$. This makes sense by
Lemma \ref{lemma-factors}. We will denote $c_1(\mathcal{L}^s \cap -$
the $s$-fold iterate of this operation for all $s \geq 0$.

\begin{lemma}
\label{lemma-cap-commutative}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$ be locally of finite type over $S$.
Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$.
For any $\alpha \in A_{k + 2}(X)$ we have
$$c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha$$
as elements of $A_k(X)$.
\end{lemma}

\begin{proof}
Write $\alpha = \sum m_j[Z_j]$ for some locally finite
collection of integral closed subschemes $Z_j \subset X$
with $\dim_\delta(Z_j) = k + 2$.
Consider the proper morphism $p : \coprod Z_j \to X$.
Set $\alpha' = \sum m_j[Z_j]$ as a $(k + 2)$-cycle on
$\coprod Z_j$. By several applications of
Lemma \ref{lemma-pushforward-cap-c1} we see that
$c_1(\mathcal{L}) \cap c_1(\mathcal{N}) \cap \alpha = p_*(c_1(p^*\mathcal{L}) \cap c_1(p^*\mathcal{N}) \cap \alpha')$
and
$c_1(\mathcal{N}) \cap c_1(\mathcal{L}) \cap \alpha = p_*(c_1(p^*\mathcal{N}) \cap c_1(p^*\mathcal{L}) \cap \alpha')$.
Hence it suffices to prove the formula in case $X$ is integral
and $\alpha = [X]$. In this case the result follows
from Lemma \ref{lemma-commutativity-on-integral} and the definitions.
\end{proof}

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