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Let $X$ be a variety -- one can compute $\text{Pic}(X) = H^1(X, \mathcal{O}^*_X)$ by choosing a Cech cover which is acyclic with respect to $H^\bullet(-, \mathcal{O}^*)$.

Can one always do this? It seems to me that the answer is no. For example, take the quadric cone $X = \text{Spec} k[x,y,z]/xy-z^2$. $\text{Pic}(X)$ is generated by the two rulings, and it seems like there isn't a distinguished open containing zero which cuts out a ruling. Does this sound correct?

Are there known characterizations for when it's possible to find such a Cech cover?

user148177
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1 Answers1

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1) Contrary to what you believe, your cone has zero Picard group: Hartshorne Example 6.11.3, page 142.
2) However your basic intuition is correct, you can't find Čech covers acyclic for $\mathcal O^*$ in general.

For example, take a complete projective smooth curve $\bar C$ of genus $\gt 0$ and delete a point, obtaining the affine curve $C$.
Any open subset $U\subset C$ is obtained by deleting finitely many points from $C$ and such a $U$ has a huge Picard group, very close to $Pic(\bar C)$, which is a $g$-dimensional abelian variety.
In conclusion $C$ has no cover consisting of open subsets acyclic for $\mathcal O^*$.

Georges Elencwajg
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