Let $U$ be an open subset of $\mathbb P^1$ without two points (say $t=0$ and $t=\infty$) and $j: U\to \mathbb P^1$ be an open immersion. Ground field $k$ is algeraically closed. Let $G$ be the group scheme over $\mathbb P^1$ given by


$x^2 - ty^2 = 1$


I am interested in computing the stalks of the skyscraper $R^1j\_\*G$. It looks that the stalk at point $t=0$ must be isomorphic to the group


$\mathrm H^1\_{et}(k((t)), G) \simeq \mathbb Z/2$


but cannot confirm this result.


My motivation comes from [this](https://mathoverflow.net/questions/181239/compute-higher-direct-image-for-gm-under-open-embedding) example, but I want to apply it for the group scheme $G$.


P.S. Moreover, it looks that $j\_\*G \simeq G$, just by computing the stalks of the quotient for a natural map $G \to j\_\*G$ on $\mathbb P^1$. Cannot confirm this either.



Stalks of higher direct image under open embedding
$R^1 j\_\*G$ is defined as a sheafification of the presheaf $U \to H^1 ( j^{-1} U, G)$. The stalk is defined as a certain forward limit. The sheafification of a presheaf has the same stalk as the presheaf. Hence the stalk is equal to the forward limit over all neighborhoods $U$ of $0$ of $H^1( j^{-1} U, G)$.


You can replace the forward limit of cohomology groups by an inverse limit of schemes: The inverse limit over neighborhoods $U$ of $0$ of $j^{-1} U$. This scheme is precisely the spectrum of the field of fractions of the strict henselization of $k[x]$ at $0$. Most likely the etale cohomology computation you did for the completion works for the strict henselization as well. If not, then one can check that etale cohomology of algebraic groups is the same for each: [See this answer.](https://mathoverflow.net/a/199084/18060)


For $R^0$, there is clearly a map from sections of $G$ to sections of the pushforward, because each section of $G$ restricts to a section of $G$ on the open set. (This is the push-pull adjunction). To check that this map is an isomorphism, it is enough to check it on stalks. So you must show that for $(x,y) \in k((t))$ such that $x^2-ty^2=1$, $x$ and $y$ are in fact in $k[[t]]$. This is pretty easy to do using leading terms - then oberseve that your proof works equally well with strict henselization instead of completion.


