[Intersection Theory I] Chow's Groups with Examples

Intersection theory plays an important role in the setting of the Grothendieck-Riemann-Roch's theorem. One wants to replace the homology ring in complex geometry, so that one can have the theory of Chern classes in algebraic geometry. And the Chow's ring plays such a role.

Let $X$ be a scheme of dimension $n$. A $k$-cycle of $X$ is of the form $\sum_{V}n_V[V]$ where $V$ is the $k$-dimensional subscheme of $X$, and $n_V\in \mathbb{Z}$ with finitely many $n_V\ne 0$. One can see that the set of $k$-cycle forms a group with additional law induced from $\mathbb{Z}$. Let $W$ be a $k+1$-dimensional subscheme of $X$, and $\phi\in k(W)$. One can see zeros and poles of $\phi$ are $k$-dimensional subschemes of $X$, with scheme theoretic multiplicities. We denote $div(\varphi):=\sum_{V}n_V[V]$, where $V$ are zeroes or poles of $\varphi$, and $n_V$ are multiplicities of $\phi$ at $V$. Sums of $k$-cycle comes from this way also forms a group, denoted $B_k(X)$, which is a subgroup of $Z_k(X)$. The $k$-th Chow's group is the quotient $A_k(X):=Z_k(X)/B_k(X)$. We will compute some Chow's groups as examples.

Example 1. Let $X$ be a scheme with pure dimension $n$, one can see that $A_n(X)$, by definition, is just $Z_n(X)$, and $Z_n(X)\cong \mathbb{Z}^n$, where $n$ is the number of irreducible components of $X$. In particular, if $X=\{pt\}$ be a point, then its dimension is $0$. By the example above, one can see $A_0(X)\cong\mathbb{Z}$.

Example 2. Let $H$ be a hyperplane in $\mathbb{P}^{n}$ given by a linear equation $f=0$, and $V$ be a hypersurface in $\mathbb{P}^n$ given by an equation $g=0$, where $g$ is a homogeneous polynomial of degree $d$, then one can show that $d[H]\sim[V]$ in $Z_{n-1}(\mathbb{P}^n)$, since one can choose $\phi:=g/f^d\in k(\mathbb{P}^n)$, and $div(\phi)=[V]-d[H]$, which is zero in $A_{n-1}(\mathbb{P}^n)$.

Example 3. Let $X=\mathbb{A}^n$, we will prove that $A_0(\mathbb{A}^n)$ is trivial. If we take any point $P\in\mathbb{A}^n$, and take any line $L\subset\mathbb{A}^n$ going through $P$. In $k(L)$, we can choose a linear function, that intersects $L$ at a unique point $P$. This yields $[P]\in B_0(X)$. Hence, $A_0(\mathbb{A}^n)$ is just trivial.

Example 4. Let $X=\mathbb{P}^n$, we will prove that $A_0(\mathbb{P}^n)\cong\mathbb{Z}$. Take any point $P,Q$ in $X$, and a line $L$ going through $P,Q$. Because the genus of $L$ is just zero, there exists a rational function $f\in k(L)$, such that $div(f)=P-Q$. Hence $P\sim Q\in A_0(\mathbb{P}^n)$. This yields $A_0(\mathbb{P}^n)\cong \mathbb{Z}$.

If $V\subset X$ the closed subscheme of $X$, then $i: V\to X$ the embedding. This induces the push-forward map $i_*: A_k(V)\to A_k(X)$. Note that the push-forward map is not injective in general. If $X=\mathbb{P}^2$, and $E$ is an elliptic curve on $X$, then consider $j_*: A_0(E)\to A_0(X)$. If $P\ne Q$ are two points on $E$, then $j_*(P-Q)=P-Q$ are zero on $A_0(X)$ by Example 3, but we know that $P-Q$ is not zero in $Pic(E)$.

Also, if $U\subset X$ an open subset, and $j: U\to X$ the embedding. This induces the pull-back map $j^*: A_k(X)\to A_k(U)$ sending $[V]\mapsto [V\cap U]$. And the pull-back map in this case is always surjective. We have the following important

Proposition 5. Let $V\subset X$ be a closed subscheme of $X$, and $U:=X\setminus V$, then we have the following exact sequence

$$A_k(V)\xrightarrow{i_*} A_k(X)\xrightarrow{j^*}A_k(U)\to 0$$


The proof can be found in the note of Gathmann, Chapter IX.

Proposition 6. Let $\pi: E\to X$ be a vector bundle of rank $r$ on $X$. Then the pull-back map $\pi^*: A_k(X)\to A_{k+r}(E)$ sending $[V]\mapsto [\pi^{-1}(V)]$ is well-defined and it is a group isomorphism.

Using two propositions, we will compute the Chow's ring of $\mathbb{A}^n$, and $\mathbb{P}^n$.

Proposition 7. We have $A_n(\mathbb{A}^n)\cong \mathbb{Z}$ and $A_k(\mathbb{A}^n)=0$ for all $k\le n-1$.

Proof. The first statement just follows from Example 1. For the second, we use induction and use Proposition 6. There is nothing to prove when $n=0$. When $n\ge 1$, it can be seen that $\mathbb{A}^{n}$ is a line bundle over $\mathbb{A}^{n-1}$, and using Proposition 6, we have $A_{k}(\mathbb{A}^{n-1})\cong A_{k+1}(\mathbb{A}^{n})$. And the second statement follows from the induction hypothesis (Q.E.D)

Proposition 8. We have $A_k(\mathbb{P}^n)\cong \mathbb{Z}$, for all $0\le k\le n$, and $A_k(\mathbb{P}^n)$ is generated by the class of $k$-dimensional linear subspace of $\mathbb{P}^n$.

Proof. We will prove this by induction. When $n=0$, then there is nothing to prove. When $n\ge 1$, using Proposition 6, we have the following exact sequence

$$A_k(\mathbb{P}^{n-1})\to A_k(\mathbb{P}^n)\to A_k(\mathbb{A}^{n})\to 0$$

For $k\le n-1$, Proposition 8 yields $A_k(\mathbb{A}^{n})=0$. So, the push-forward map $A_k(\mathbb{P}^{n-1})\to A_k(\mathbb{P}^n)$ is surjective. By duplicating this process, one can see the push-forward map $A_k(\mathbb{P}^{k+1})$ to $A_{k}(\mathbb{P}^n)$ is surjective, and it maps a class of $k$-dimensional hyperplane $[H]$ in $\mathbb{P}^{n-1}$ to a class of $k$-dimensional hyperplane in $\mathbb{P}^{n}$. And by Example 2 (or induction hypothesis), $A_k(\mathbb{P}^{k+1})$ is generated by $[H]$. This yields $[H]$ is also the generator for $A_k(\mathbb{P}^n)$.

We will prove next that the push-forward map is injective. By this, we mean, if $\sum_Vn_V[V]\in A_k(\mathbb{P}^{k+1}$) maps to $0\in A_k(\mathbb{P}^n)$, then there exists finitely many $W_s\in Z_{k+1}(X), \phi_s\in k(W_s)$ such that $\sum_sdiv(\phi_s)=\sum_Vn_V[V]$, then $\sum_Vn_V[V]$ is zero in $A_k(\mathbb{P}^{k+1})$. And we already proved in Example 2 that $[V]\sim \deg V[H]$ in $Z_k(\mathbb{P}^{k+1})$. So $\sum_{V}n_V[V]$ is zero in $A_k(\mathbb{P}^{k+1})$ iff $\sum_Vn_V\deg V=0$.

But then, because $\sum_Vn_V\deg V=\sum_s\deg div(\phi_s|_{\mathbb{P}^{k+1}})$, it is sufficient for us to prove that $\deg div(\phi_s|_{\mathbb{P}^{n+1}})=0$, and this follows from Hartshorne I.7.7. (Q.E.D)