trivial Picard group

let $S=\operatorname{Spec}(A)$ be an affine scheme. For which ring $A$, not field is it known that $H^1(S,\mathcal{O}_S^{*})$ is trivial?

If $X\to S$ is a finite map and $H^1(S,\mathcal{O}_S^{*})$ is trivial, is it true that also $H^1(X,\mathcal{O}_X^{*})$ is trivial?

Thanks

1 Answer

Sample answers to your first question: If $S$ is Spec of a local ring, or of a UFD, then $H^1(S, \mathcal O_S^{\times})$ is trivial.

The answer to your second question is no: $X =$Spec $\mathbb C[x,y]/(y^2 - x^3 +x) \to S =$ Spec $\mathbb C[x]$ gives a counterexample of a geometric nature, and $X =$Spec $\mathbb Z[\sqrt{-5}] \to S =$ Spec $\mathbb Z$ gives a counterexample of an arithmetic nature.

• is this true for coherent cohomology? Namely if $S$ is affine and $\mathcal{F}$ is coherent on $X$, $X\to S$ finite, is it true that $H^i(X,\mathcal{F})=0$ for $i>0$? – ullo Jul 6 '12 at 14:31
• Yes, because $X$ is affine under your hypotheses! – Georges Elencwajg Jul 6 '12 at 15:31
• @Matt E: can you explain how to compute $H^1(X,\mathcal{O}_X^\times)$ in your geometric example? – adrido Jun 5 '14 at 19:19
• @adrido: Dear adrido, Recall that $H^1(X,\mathcal O_X^{\times})$ is the Picard group of $X$. In the geometric example, the curve $X$ is a once-punctured geometric curve, so its Pic is equal to Pic of an elliptic curve modulo the copy of $\mathbb Z$ generated by a single point (i.e. the point we punctured), which is naturally isomorphic to Pic${}^0$ of the elliptic curve. But this is isomorphic to the elliptic curve itself. In other words, in the geometric example, we have $H^1(X,\mathcal O_X^{\times})$ is isomorphic to an elliptic curve. Regards, – Matt E Jun 6 '14 at 5:55
• @MattE: Dear Matt, that's brilliant, thank you! so the projective curve defined by $wy^2-x^3+wx^2$ is elliptic and once you take out a point (w=0) you get the affine curve above. I wonder if you can generalize this trick to compute the Picard group of affine varieties that are cones over a nice projective variety? e.g. V(xy-z^2)? – adrido Jun 6 '14 at 16:26