## 81.32 Degrees of zero cycles

This section is the analogue of Chow Homology, Section 42.41. We start with defining the degree of a zero cycle on a proper algebraic space over a field.

Definition 81.32.1. Let $k$ be a field. Let $p : X \to \mathop{\mathrm{Spec}}(k)$ be a proper morphism of algebraic spaces. The degree of a zero cycle on $X$ is given by proper pushforward

$p_* : \mathop{\mathrm{CH}}\nolimits _0(X) \longrightarrow \mathop{\mathrm{CH}}\nolimits _0(\mathop{\mathrm{Spec}}(k)) \longrightarrow \mathbf{Z}$

(Lemma 81.16.3) composed with the natural isomorphism $\mathop{\mathrm{CH}}\nolimits _0(\mathop{\mathrm{Spec}}(k)) \to \mathbf{Z}$ which maps $[\mathop{\mathrm{Spec}}(k)]$ to $1$. Notation: $\deg (\alpha )$.

Let us spell this out further.

Lemma 81.32.2. Let $k$ be a field. Let $X$ be a proper algebraic space over $k$. Let $\alpha = \sum n_ i[Z_ i]$ be in $Z_0(X)$. Then

$\deg (\alpha ) = \sum n_ i\deg (Z_ i)$

where $\deg (Z_ i)$ is the degree of $Z_ i \to \mathop{\mathrm{Spec}}(k)$, i.e., $\deg (Z_ i) = \dim _ k \Gamma (Z_ i, \mathcal{O}_{Z_ i})$.

Proof. This is the definition of proper pushforward (Definition 81.8.1). $\square$

Lemma 81.32.3. Let $k$ be a field. Let $X$ be a proper algebraic space over $k$. Let $Z \subset X$ be a closed subspace of dimension $d$. Let $\mathcal{L}_1, \ldots , \mathcal{L}_ d$ be invertible $\mathcal{O}_ X$-modules. Then

$(\mathcal{L}_1 \cdots \mathcal{L}_ d \cdot Z) = \deg ( c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_1) \cap [Z]_ d)$

where the left hand side is defined in Spaces over Fields, Definition 71.18.3.

Proof. Let $Z_ i \subset Z$, $i = 1, \ldots , t$ be the irreducible components of dimension $d$. Let $m_ i$ be the multiplicity of $Z_ i$ in $Z$. Then $[Z]_ d = \sum m_ i[Z_ i]$ and $c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_ d) \cap [Z]_ d$ is the sum of the cycles $m_ i c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_ d) \cap [Z_ i]$. Since we have a similar decomposition for $(\mathcal{L}_1 \cdots \mathcal{L}_ d \cdot Z)$ by Spaces over Fields, Lemma 71.18.2 it suffices to prove the lemma in case $Z = X$ is a proper integral algebraic space over $k$.

By Chow's lemma there exists a proper morphism $f : X' \to X$ which is an isomorphism over a dense open $U \subset X$ such that $X'$ is a scheme. See More on Morphisms of Spaces, Lemma 75.40.5. Then $X'$ is a proper scheme over $k$. After replacing $X'$ by the scheme theoretic closure of $f^{-1}(U)$ we may assume that $X'$ is integral. Then

$(f^*\mathcal{L}_1 \cdots f^*\mathcal{L}_ d \cdot X') = (\mathcal{L}_1 \cdots \mathcal{L}_ d \cdot X)$

by Spaces over Fields, Lemma 71.18.7 and we have

$f_*(c_1(f^*\mathcal{L}_1) \cap \ldots \cap c_1(f^*\mathcal{L}_ d) \cap [Y]) = c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_ d) \cap [X]$

by Lemma 81.19.4. Thus we may replace $X$ by $X'$ and assume that $X$ is a proper scheme over $k$. This case was proven in Chow Homology, Lemma 42.41.4. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).