Consider the group scheme $\mathcal T$ over $\mathbf P^1$ given locally (variable $t$) by the equation


$x^2 - f(t)y^2 = 1$


where $f(t)$ is a polynomial of degree $r$ with distinct roots (assume that $r$ is even and the field is algebraically closed). What is the group of torsors for $\mathcal T$, i.e. $\mathrm H^1(\mathbf P^1, \mathcal T)$?


Using the machinery of spectral sequences I was able to deduce that my group is an extension of the Jacobian $J(C)$, where $C$ is the double cover that splits $\mathcal T$:


$1 \to (\mathbb Z/2\mathbb Z)^{r-1} \to \mathrm H^1(\mathbf P^1, \mathcal T) \to J(C) \to 1$


Unfortunately, I cannot confirm whether my answer is correct. It is difficult to believe that such example hasn't been computed anywhere in the literature, but I couldn't find any reference. Would be happy if someone could give a hint or a reference.



Torsors under the group scheme over projective line
Taking cohomology of the short exact sequence $\mathcal T \to f\_\* \mathbb G\_m \to \mathbb G\_m$ gives a long exact sequence:


$H^0(C, \mathbb G\_m) \to H^0 (\mathbb P^1, \mathbb G\_m) \to H^1(\mathbb P^1, \mathcal T) \to H^1(C, \mathbb G\_m) \to H^1(\mathbb P^1, \mathbb G\_m) $


We can identify most of the terms:


$k^\times \to k^\times \to H^1(\mathbb P^1, \mathcal T) \to \mathcal J \times \mathbb Z \to \mathbb Z $


Assuming $k$ is algebraically closed, the arrow $k^\times \to k^\times$, which is squaring, is an isomorphism. The map $\mathcal J \times \mathbb Z \to \mathbb Z$ is the norm map, which sends $\mathcal J$ to $0$ and doubles $\mathbb Z$, so the kernel is $\mathcal J$.


Hence I think your group is just $\mathcal J$. $\mathcal J$ is an extension of itself by $(\mathbb Z/2)^{r-2}$, but I don't know where the $-1$ comes from. It may depend on how you extend your group scheme to $\infty$, which you didn't really specify.


