The Stacks project

42.28 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 42.28.1. Let $(S, \delta )$ be as in Situation 42.7.1. 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 $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$.

Proof. Immediate from the key Lemma 42.27.1 and the discussion preceding it. $\square$

Lemma 42.28.2. Let $(S, \delta )$ be as in Situation 42.7.1. 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

\[ c_1(\mathcal{L}) \cap - : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X) \]

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 42.25.1 is zero. By Definition 42.19.1 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 42.26.4 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 42.28.1 we conclude that

\[ c_1(\mathcal{L}) \cap \alpha = c_1(\mathcal{O}_ X) \cap \beta . \]

However, by Lemma 42.25.2 we see that the right hand side is zero in $\mathop{\mathrm{CH}}\nolimits _ k(X)$ as desired. $\square$

Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$ be invertible on $X$. We will denote

\[ c_1(\mathcal{L}) \cap - : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(X) \]

the operation $c_1(\mathcal{L}) \cap - $. This makes sense by Lemma 42.28.2. We will denote $c_1(\mathcal{L})^ s \cap -$ the $s$-fold iterate of this operation for all $s \geq 0$.

Lemma 42.28.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible on $X$. For any $\alpha \in \mathop{\mathrm{CH}}\nolimits _{k + 2}(X)$ 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 $\mathop{\mathrm{CH}}\nolimits _ k(X)$.

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 42.26.4 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 42.28.1 and the definitions. $\square$

Comments (2)

Comment #6291 by Yi Shan on

In the statement before Lemma 02TJ, why the operation maps the Chow group of -cycles to that of -cycles? Should the here be replaced by ?

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