The Chow ring of affine space

Question

I want to show that $\operatorname{CH}(\mathbb{A}^n) = \mathbb{Z}$ from first principles. There's a proof of this in 3264 by Eisenbud and Harris that utilizes the definition of rational equivalence given by differences of cycles on $\mathbb{P}^1 \times \mathbb{A}^n$; I would like to show this using the divisors on subvarieties formulation.

The key part of the proof is to show that any proper subvariety $Y \subset \mathbb{A}^n$ is rationally equivalent to zero. The idea I have is to find a subvariety $Z$ such that $Y \subset Z \subset \mathbb{A}^n$ and $Z$ is one dimension larger than $Y$; then $Y$ is a divisor on $Z$ since it is cut out by equations, and therefore rationally equivalent to zero. Does this work? How can I show rigorously that we can always find such a $Z$?

Answer 1

This proof follows from the suggestion of Georges Elencwajg that I study the Bible.

Proposition: Let $X$ be a scheme, and $\pi:E \rightarrow X$ a locally free sheaf of rank $r$ on $X$. Then for all $k \geq 0$, the pull back homomorphism $\pi^\ast :\operatorname{CH}_k(X) \rightarrow \operatorname{CH}_{k+r}(E)$ is surjective.

Proof: Note that $\pi^\ast$ maps $\operatorname{CH}_k(X)$ to $\operatorname{CH}_{k+r}(E)$ since a cycle will gain a factor of $\mathbb{A}^r$ under preimage. We proceed via induction on $\dim X$; let $U$ be an open affine on which $E$ is trivial, e.g, such that $E|_U \cong U \times \mathbb{A}^r$, and set $Y = U^c$. By excision, we have the following diagram:

By the four lemma, if we can show that $\operatorname{CH}_k(Y) \rightarrow \operatorname{CH}_{k+r}(E|_Y)$ and $\operatorname{CH}_k(U) \rightarrow \operatorname{CH}_K(U \times \mathbb{A}^r)$ are surjections, it follows that $\pi^\ast$ is a surjection, and the former map is a surjection by the inductive hypothesis. Therefore, we need only show that the rightmost arrow factors, e.g, we have reduced the proof the assertion to the case of $X = \operatorname{Spec} R$ affine, and $E = X \times \mathbb{A}^r$ is the trivial bundle (by choice of $U$ above). Moreover, we can factor $\pi$ as a series of projections: $$\pi:E = X \times \mathbb{A}^r \rightarrow X \times \mathbb{A}^{r-1} \rightarrow \cdots \rightarrow X \times \mathbb{A}^1 \rightarrow X$$ Therefore, we may assume that $r=1$ by treating each locally free sheaf in the chain as a rank $1$ free sheaf on the scheme below, and we have that $$E= X \times \mathbb{A}^1 = \operatorname{Spec} R \otimes_k \operatorname{Spec} k[t] = \operatorname{Spec} R[t]$$

We are then tasked with showing that $\pi^\ast: \operatorname{CH}_k(X) \rightarrow \operatorname{CH}_{k+1}(X \times \mathbb{A}^1)$ is surjective; to that end, let $V \subset X \times \mathbb{A}^1$ be a $(k+1)$-dimensional subvariety, and let $W = \overline{\pi(V)}$. If $\dim W = k$, then $V = W \times \mathbb{A}^1$ since $\pi$ is a projection, so $[V] = \pi^\ast[W]$. If $\dim W = k+1$ (this is the only other option) we must show that $[V]$ is in the image of the induced pullback map $\operatorname{CH}_k(W) \rightarrow \operatorname{CH}_{k+1}(W \times \mathbb{A}^1)$. We can therefore assume that $W = X$, let $I(V)$ be the vanishing ideal of $V$, and consider the related ideal $$I(V) \otimes_R K \subset K[t]$$ where $K$ is the fraction field of $R$. This ideal is not the unit ideal, as otherwise $V= W \times \mathbb{A}^1$ which we handled above. Since $K[t]$ is a PID, $I(V) \otimes_R K[t] = (\varphi)$ for some $\varphi \in K[t]$. Then the divisor of $\varphi$ (taken as a function on $X \times \mathbb{A}^1$) is $[V]$ by construction, up to terms of the form $\pi^\ast [W_i]$ for $W_i \subset X$ corresponding to tensoring with $K$, from which the result follows. $\blacksquare$

This result gives an easy alternative derivation of the Chow ring of affine space by showing that all subvarieties are rationally equivalent to zero. First, we have that $\operatorname{CH}_0(\mathbb{A}^n) =0$ for all $n$; to see this, for any $x \in \mathbb{A}^n$, pick a line $L \cong \mathbb{A}^1 \subseteq \mathbb{A}^n$ through $x$ and a function on $L$ vanishing (only) on $x$. Then, by the above result, $\operatorname{CH}_0(\mathbb{A}^{n-k}) \rightarrow \operatorname{CH}_k(\mathbb{A}^n)$ is surjective for all $k< n$, from which the result follows.