# Chow ring of an affine line with a double origin

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.

• Probably you mean that $A_0(X) = 0$, and $A_1(X) = \mathbf{Z}$ (it's not a zero ring). For your $X$, you need to specify an isomorphism between $V_1$ and $V_2$; if $p_1$ and $p_2$ is the same closed point on both, the same argument as in the algebraically closed case should work. If $p_1$ and $p_2$ are different points (especially, with different residue fields) it's not clear what the gluing between $V_1$ and $V_2$ is. Jul 6, 2022 at 10:26
• I get something else. Jul 6, 2022 at 13:48
• @EvgenyShinder You are right. I edited the question. The two points $p_1$ and $p_2$ are both closed of degree one, but they are not the "same" point as they belong to two different affine lines. Jul 6, 2022 at 14:34
• I suppose by applying the automorphis $x \mapsto x + c$ we can assume that $p_1 = p_2 = 0$. Then it should be the same result as over algebraically closed field with the same proof. Jul 6, 2022 at 16:11
• @Johan What do you get? To the OP, what definition of Chow ring do you use? Be aware: the definition in Fulton’s textbook is different than the definition that Grothendieck used (this is most evident for nonreduced schemes). Jul 6, 2022 at 17:40

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).

• It's fine. I don't remember what you wrote. Thanks for the explanations on the different definitions of Chow ring. Jul 7, 2022 at 20:21

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.

• Your computation is not correct if $p$ is $p_1$ or $p_2$: one only gets that $[p_1] + [p_2]$ is $0$ in $A^1(X)$, so $A^1(X) \cong \mathbb{Z}$.
– naf
Jul 7, 2022 at 5:02