# Flat cohomology for finite infinitesimal group scheme over a perfect field

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

-
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.
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 journals.cambridge.org/action/… or Lemma 2.7 (a) in arxiv.org/abs/1410.2621 – 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