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.

  • 4
    $\begingroup$ To be more precise, if $f:C \rightarrow \mathbf{P}^1$ is the finite flat degree-2 covering map then it seems you really mean for $\mathscr{T}$ to be the relative identity component of the kernel of the norm homomorphism $N:f_{\ast}({\rm{GL}}_1)\rightarrow {\rm{GL}}_1$ (perhaps also assume ${\rm{char}}(k)\ne 2$); is that right? Then low-degree fppf cohomology sequences seem to do the job (no need for spectral sequences); use that to double-check your calculations. $\endgroup$ – user74230 Feb 5 '15 at 5:25
  • $\begingroup$ Thank you for suggestion. Unfortunately, for this approach we need to know $H^1(\mathbb P^1, f_*(\mathbb G_m))$ which is even more difficult to compute, isn't it? $\endgroup$ – lime Mar 5 '15 at 6:30
  • $\begingroup$ Since $f$ is finite, ${\rm{H}}^i(\mathbf{P}^1, f_{\ast}(\mathscr{F})) = {\rm{H}}^i(C, \mathscr{F})$ for any $i$ and any sheaf $\mathscr{F}$ on the etale site of $C$. Thus, the H$^1$ you mention is Pic($C$) (extension of $\mathbf{Z}$ by $J(C)(k)$). $\endgroup$ – user74230 Mar 5 '15 at 12:36
  • $\begingroup$ Apparently I missed completely the fact that the cover was finite and hence the spectral sequence for push-forward degenerated. Thank you, this solved the problem. $\endgroup$ – lime Mar 6 '15 at 2:41

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.

  • $\begingroup$ I believe what you mean is that the squaring map is surjective. My original computation was incorrect. According to $\infty$ point, I believe it will extend uniquely, but depending on parity of the degree of $f(t)$ will give $\mathbb G_m$ or $\mathbb G_a\times \mathbb Z/2$ in fiber. Everything fits now. Thank you Will. $\endgroup$ – lime Mar 6 '15 at 2:53

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