Let $k$ be an algebraically closed field, and $X$ a connected smooth projective curve over $X$. Let $F$ be the function field of $k$. Let $G$ be an algebraic group over $k$ (assume that it is smooth, or reductive, if needed). Can you explain why $H^1 (Gal(F^{sep}/F) , G(F^{sep}))$ is zero, or provide a simple reference?


Thank you,
Sasha



Galois cohomology of a non-abelian group over a function field
You need to assume that $G$ is connected or else that is false. Assuming that $G$ is connected, this follows immediately from Steinberg's Theorem, i.e., the proof of Serre's Conjecture I. There is a copy of Steinberg's paper as an appendix of Serre's book on Galois Cohomology. Serre explains in the book how to deduce your statement from Steinberg's Theorem. Also Qiaochu Yuan is correct; you should use $G(\overline{F})$, not $G(F)$. Finally, since Galois cohomology is insensitive to nilpotents in $G$, you immediately reduce to the case that $G$ is smooth and connected.


