0
$\begingroup$

I know that $\text{Pic}(\mathbb{A}^n)=0$, but it turns out that $\text{Pic}(\mathbb{A}^1-\{0\})=0$ also. Is there a simple explanation for this?

$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

$\mathbb{A}^1_k - (0) = Spec(k[x]_x)$, which is the spec of a Noetherian UFD, see here .

Hartshorne proposition II.6.2 : Let $A$ be a Noetherian domain. Then, $A$ is a unique factorization domain if and only if $X = Spec(A)$ is normal and $Cl(X) = 0$.

Then just use that $Cl(X) = Pic(X)$ for schemes that are regular at all points and satisfy $(*)$ (Noetherian, integral, separated, regular in codimension 1). This is proposition II.6.11 in Hartshorne.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.