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Let $G$ be a finite infinitesimal group scheme (e.g.$\mu_p,\alpha_p) $ over a perfect field $k$, how much is known about $H^1_{fppf}(k,G)$?

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What precisely do you want to know about it? – Kestutis Cesnavicius Jan 27 '13 at 12:30
I know for $\mu_p$ and $\alpha_p$ these are trivial. Do I have more examples? Are these all trivial for commutative finite groups? – stefan Jan 27 '13 at 14:04
Sorry, I meant for commutative finite infinitesimal groups. – stefan Jan 27 '13 at 14:25
@stefan: Use the connected-etale sequence to see the triviality whenever $G$ is commutative and infinitesimal. – user30180 Jan 27 '13 at 14:34

I think that $H^1(k,G)=1$ for all infinitesimal $G$.

Let us make some preliminary comments on the process of perfection. If $X=Spec(A)$ is an affine $k$-scheme then we can form its perfection $X^{perf}=Spec(A^{perf})$ where $A^{perf}=\varinjlim_{\sigma} A$ is the direct limit of the $\mathbb{N}$-indexed system formed by the Frobenius of $A$. There is a canonical map $X^{perf}\to X$. If $X$ is a $k$-group scheme, then $X^{perf}$ is a $k$-group scheme. If $X$ is a torsor under a $k$-group scheme $G$, then $X^{perf}$ is a torsor under the $k$-group scheme $G^{perf}$.

Let us come back to the question. Let $X$ be a torsor under $G$. Then $X^{perf}$ is a torsor under $G^{perf}$. Since $G$ is infinitesimal we have $G^{perf}=1$. Thus $X^{perf}=Spec(k)$. The map $X^{perf}\to X$ shows that $X$ has a $k$-point, hence is trivial.

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It doesn't seem apparent that the formation of the perfection is compatible with passage to an fppf cover, so some more argument seems to be needed to explain why $X^{\rm{perf}}$ is a $G^{\rm{perf}}$ torsor. – nfdc23 Jan 24 at 21:44
In case it is of interest, here are some references which prove slightly more general statements that imply $H^1(k, G) = 1$: Lemma 5.7 (b) in… or Lemma 2.7 (a) in – Kestutis Cesnavicius Jan 24 at 21:49
If $X$ is a scheme of finite type over a field $K$ then its maximal geometrically reduced closed subscheme $X'$ (Galois descent of Zariski closure of $X(K_s) \subset X_{K_s}$) is compatible with direct products and functorial, so if $G$ is a $K$-group scheme of finite type then $G'$ is the maximal smooth closed $K$-subgroup and if $X$ is a $G$-torsor then $X'$ is a $G'$-torsor provided $X'$ is non-empty. Using pushout along $G' \rightarrow G$, it follows that H$^1(K,G')$ is the subset of H$^1(K,G)$ split over $K_s$. For perfect $K$ we have $G' = G_{\rm{red}}$ and $K_s=\overline{K}$. QED – nfdc23 Jan 24 at 21:52
"some more argument seems to be needed to explain why $X^{perf}$ is a $G^{perf}$-torsor" : perfectization commutes with products, so if the map $G\times X \to X\times X$ is an isomorphism then $G^{perf}\times X^{perf} \to X^{perf}\times X^{perf}$ also. – Matthieu Romagny Jan 24 at 22:00
Ah, good trick (also explains why $G^{\rm{perf}}$ is naturally a $k^{\rm{perf}}$-group). – nfdc23 Jan 24 at 22:06

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