# The Picard group of a curve contained in $\mathbb{P}_F^1$

Let $$F$$ be an algebraically closed field and $$X$$ be a curve obtained by removing at least two points on the projective line $$\mathbb{P}^1_F$$, i.e., $$\mathbb{P}^1_F-X$$ is a reduced separated divisor on $$\mathbb{P}^1_F$$ of degree $$\geq 2$$.

We know that the Picard group of $$\mathbb{P}^1_F$$ is $$\mathbb{Z}$$. How does one deduce that the Picard group of $$X$$ is $$0$$?

In this case we can view $$X$$ as $$\mathbb{A}^1_F-\{p_1,...,p_r\}$$, where $$r \geq 1$$, since the projective line minus a point is the affine line as seen in Is the projective line minus one point always isomorphic to the affine space?. Since $$\mathbb{A}^1_F = \mathrm{Spec}\,F[x]$$, we know that $$\mathrm{Pic}(\mathbb{A}_F^1) = 0$$. Does the removal of points have any relation to the localization of the UFD $$F[x]$$?

• There are problems with what you write: it is not true that the projective line minus some closed point is always the affine line. One must assume that $F$ is algebraically closed or the point you remove is $F$-rational in order to have that result. On the other hand, it is always true that $X$ is affine, it just might not be $\Bbb A^1_F$. (For an example, consider removing $(x^2+y^2)$ from $\Bbb P^1_{\Bbb R}$.) Jun 29, 2021 at 6:48
• @KReiser Yes I forgot to add that $F = \bar{F}$, this was what's mentioned in the paper I'm reading. However in the linked Stackexchange post, there is no mention of algebraic closure. No I'm not saying that $X$ in our case is the affine line, I meant that since we take the complement of a reduced divisor a degree at least 2, $X$ can be seen as the affine line minus at least one other point. Jun 29, 2021 at 6:56
• The linked post is implicitly assuming $F=\overline{F}$ or the point removed is $F$-rational. As far as the other part of your comment, I think we're in agreement - I phrased the final non-parenthetical sentence in my previous comment a little poorly, and it's too late to edit it. Jun 29, 2021 at 7:05

Corollary 6.16: If $$X$$ is a noetherian, integral, separated locally factorial scheme, then there is a natural isomorphism $$\operatorname{Cl} X\cong\operatorname{Pic} X$$.
Proposition 6.5(c): Suppose $$X$$ is a noetherian integral separated scheme which is regular in codimension one. Let $$Z$$ be an irreducible closed subset of codimension one, and let $$U$$ be it's open complement. Then there is an exact sequence $$\Bbb Z\to\operatorname{Cl} X\to \operatorname{Cl} U\to 0$$ where $$1\in\Bbb Z$$ maps to $$[Z]\in \operatorname{Cl} X$$.
Combining these statements with the result that $$\operatorname{Cl} \Bbb P^1_F=\Bbb Z$$ generated by a closed $$F$$-rational point, we see that $$\operatorname{Cl} \Bbb A^1_F=0$$, and therefore every open subscheme also has vanishing class (and therefore Picard) group.
Alternately, one may use the homotopy invariance of the class/Picard group - see proposition 6.6 which states that if $$X$$ satisfies the assumptions of proposition 6.5 above, then $$\operatorname{Cl} X\cong \operatorname{Cl} X\times \Bbb A^1$$; or here for the statement that if $$X$$ is normal, then $$\operatorname{Pic} X\cong \operatorname{Pic} X\times\Bbb A^1$$. As the class/Picard group of $$\operatorname{Spec} F$$ is trivial, the class/Picard group of $$\Bbb A^1_F$$ is also trivial, and then we may continue on with proposition 6.5(c) as above.
• Thanks, just one question: "and therefore every open subscheme also has vanishing class group", here I suppose it's because we have the exact sequence $$\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow \mathrm{Cl}(U) \rightarrow 0$$ and so there're no other choices for $\mathrm{Cl}(U)$. But the complement of $U$ is a finite subset of closed points of size greater than 1, which isn't irreducible. So $Z$ does not satisfy the hypothesis of 6.5(c), no? Jun 30, 2021 at 3:27