Chow ring of an affine line with a double origin

Question

Let $k$ be a field. Consider two affine lines $$U_1 = \text{Spec}\ k[s] \ \ \text{and} \ \ U_2 = \text{Spec}\ k[t]$$ over $k$. Let $p_1$ and $p_2$ be the origins on $U_1$ and $U_2$ respectively. Construct a scheme $X$ by gluing $U_1$ and $U_2$ along $$V_1 = U_1 - \{p_1\} \ \ \text{and} \ \ V_2 = U_2 - \{p_2\}$$ via the isomorphism $V_1 \to V_2$ induced by the ring map $$k[t,t^{-1}] \to k[s,s^{-1}], \quad t \mapsto s.$$ Then $X$ is a non-separated non-affine smooth scheme over $k$, and is known as an affine line with a double origin.

What is the Chow ring $A(X)$ of $X$?

PS1: To be precise, the Chow ring $A(Y)$ for a smooth scheme $Y$ over $k$ is the one defined in Chapter 8 in Fulton's Intersection Theory.

PS2: The description of the question above is made more explicit thanks to a comment of Evgeny Shinder.

Answer 1

I apologize for writing something wrong in my comment. Here is a quick amplification of the valid part of my comment. Let $X$ be a scheme that is finitely presented over $\text{Spec}\ k$ for a field $k$. Then Grothendieck's definition of $\text{CH}^1(X)$ as the first graded piece of the gamma filtration is equal to $\text{Pic}(X)$. This is explained, for instance, in Manin's "Lectures on the K-functor in algebraic geometry." However, Fulton's definition is insensitive to nonreduced structure on $X$ and to seminormalization. Thus, Fulton's definition equals $\text{CH}^1$ of the seminormalization of the reduced scheme of $X$.

However, as noted by others, the two definitions do appear to agree for the line with doubled origin. The Picard group is a free cyclic group. I do not know if the two definitions always agree for smooth, finitely presented, but possibly non-separated $k$-schemes. You might check SGA 6, since the seminar participants worked there in great generality (and later authors, such as Thomason, worked in even greater generality).

Answer 2

Let $q_i$ be the image of $p_i$ in $X$ for $i=1,2.$ Then the Chow ring $$A(X) = \mathbb{Z}[X] \oplus \mathbb{Z}[q_1] \cong \mathbb{Z}^2.$$

Since the scheme $X$ is one-dimensional, it suffices to compute two Chow groups: $A^0(X) = A_1(X)$ and $A^1(X) = A_0(X)$. Let $k[x]$ denote the ring $\mathcal{O}_X(X)$ of global sections on $X.$

First, we show $A_1(X) = \mathbb{Z}[X]$. Note that $X$ is a connected scheme. Let $X_1$ and $X_2$ be the images of $U_1$ and $U_2$ in $X$ respectively. Then $X_1$ and $X_2$ are open non-closed subschemes of $X.$ Thus the only closed one-dimensional subscheme of $X$ is itself, which implies that $X$ is irreducible and $A_1(X) = \mathbb{Z}[X]$.

Next, we show $A_0(X) = \mathbb{Z}[q_1]$. Take a closed point $q$ of $X.$ Suppose $q$ is not $q_1$ or $q_2.$ Then it is a closed point of $X_1 \cap X_2$, i.e., a maximal ideal $(f(x)) \neq (x)$ for some irreducible polynomial $f(x)$ in $k[x]$. Therefore, $q$ is rationally equivalent to zero as it is the divisor of the rational function $f(x)$ on $X.$ On the other hand, the zero-cycle $q_1+q_2$ is the divisor of the rational function $x$ on $X$, and hence is rationally equivalent to zero. But $q_1$ is not as it is not the divisor of any rational function on $X$. Therefore, $A_0(X) = \mathbb{Z}[q_1]$.

PS: The answer above corrects a previous incorrect computation $A_0(X) = 0$: the double origin cannot be separated and hence needs a special attention, thanks to naf's comment.