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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 $F[x]$?

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  • $\begingroup$ 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 $\Bbb P^1_{\Bbb R}$.) $\endgroup$
    – KReiser
    Jun 29, 2021 at 6:48
  • $\begingroup$ @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. $\endgroup$ Jun 29, 2021 at 6:56
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Jun 29, 2021 at 7:05

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All the material you need to attack this problem will be located in the chapter on divisors in your favorite algebraic geometry book. For instance, here's the relevant material from Hartshorne chapter II section 6:

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.

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  • $\begingroup$ 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? $\endgroup$ Jun 30, 2021 at 3:27
  • $\begingroup$ Use induction and remove one point at a time. $\endgroup$
    – KReiser
    Jun 30, 2021 at 3:29

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