Let $X$ be a (projective) curve (or more generally a scheme).
I would like to "find" a open cover $\mathcal{U}$ of $X$ such that $H(\mathcal{U},\mathcal{O}_X^*) = H(X,\mathcal{O}_X^*)$, i.e. find a Leray cover, to be able to do calculations (manly, but not exclusivly in the picard group).
Does anyone know if such a cover has to exist, and how to find it? If there is no general (known) solution, special cases would also be very much appreciated.
Edit:
What I have tried: I considered the simple case where $X = Spec(A)$ for some Dedekind domain $A$ with finite class number. If $\mathfrak{p}_1,...,\mathfrak{p}_r$ generate the class group, and $a \in \mathfrak{p}_1 \cap ... \cap \mathfrak{p}_r$, then for $U = Spec(A_a) = D(a)$ we should have $Pic(U) = H^1(U, \mathcal{O}_X^*) = 0$ (since $A_a$ has class number $0$). If we chose different generators, then we should be able to get a similar $b \in A$ with $V = Spec(A_b) = D(b)$ and $Pic(V) = H^1(V, \mathcal{O}_X^*) = 0$ and $X = U \cup V$. But I am not sure if $H^p(U, \mathcal{O}_X^*) = 0$ for $p \geq 2$. (Rephrasing this part: If $R$ is a principal ideal domain, do we know how $H(Spec(R), \mathcal{O}_{Spec(R)}^*)$ looks like? )
The intersection $U \cap V = D(a b)$ would look the same, so there would be no problem (if $H^p(U, \mathcal{O}_X^*) = 0$ works for $U$ and $V$).
From there I was hoping to, for a suitable curve $X$, cover $X$ with affine patches of the form above, and do the same for every affine patch. But I am not sure if this works, and even if it does, it is not as general as I would like.