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Let $G$ be an algebraic group over a field $k$. Let $k^s$ be the separable closure of $k$.

I can't seem to figure out why isomorphism classes of twists of $G$ correspond to $k$-torsors over $G$.

It's easy to see that a $k$-torsor over $G$ gives a twist of $G$, but the other way around isn't clear to me. It is bound to involve something from descent theory that I don't know.

In fact, my problem is that I can't seem to define an action of $G$ on $X$. I can only seem to get an action of $G_{k^s}$ on $X_{k^s}$ via the isomorphism of $X_{k^s}$ with $G_{k^s}$. Why does this action descend to $k$?

Let me precise that a twist of $G$ is a variety $X$ over $k$ such that $X_{k^s}$ is isomorphic to $G_{k^s}$ as a variety.

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This isn't true. For example a quadratic twist of an elliptic curve is a twist in your sense, but not a torsor under the original elliptic curve. Even worse a twist of $G$ in your sense need not be a torsor for any group whatsoever. Here is a silly example: $G=\mathbf{Z}/5$, $X=\text{spec}(k\times k\times k')$ where $k'/k$ is a cubic separable extension. –  user18237 Mar 17 '12 at 20:47
    
On page 98 of B. Poonen's notes www-math.mit.edu/~poonen/papers/Qpoints.pdf it clearly states that the set of $k$-torsors is equal to the set of twists of $\mathbf{G}$, where $\mathbf{G}$ is the algebraic group $G$ endowed with its right action of $G$ given by translation. Aren't the twists of $\mathbf{G}$ the same as the twists of $G$? –  Harry Mar 17 '12 at 21:11
    
He means twists of $\mathbf{G}$ as a $G$-torsor, not twists of $G$ as a variety. –  user18237 Mar 17 '12 at 21:18
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Ow I see. Well then it's clear. –  Harry Mar 17 '12 at 21:20
    
Just a minor nitpick: the bijection between the set of $k$-twists and $H^1$ does not follow from the references he gives between parentheses...(That's what had me going.) You need the analogues of those "statements for varieties" in the setting of "torsors" which he sort of explains in Remark 4.5.6. –  Harry Mar 17 '12 at 21:25
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