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Let $G$ be a smooth algebraic group over a field $k$. Then $Aut(G)=\mathbf{G}(k)$, where $\mathbf{G}$ is the trivial right $G$-torsor.

My attempt:

Given a $k$-rational point $\operatorname{Spec}(k) \xrightarrow{g} G$, then one obtains an automorphism of $G$ by composing the structure morphism $G \rightarrow \operatorname{Spec}(k)$ with $g$.

For the other direction I am getting confused, and haven't been able to come up with a decent approach.

Any help please?

proofromthebook
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    Automorphism as what? And how does the composed map become invertible, when it factors through something which is generally much smaller than $G$? – Tobias Kildetoft Jun 11 '18 at 17:58
  • I'm confused : for $G = \Bbb G_m$, don't we have $Aut_{AlgGrp}(G) = { \pm 1 }$ ? See here. And it seems to me that $Aut_{Schemes}(G) = { \pm 1 } \times k^*$. – Watson Jun 11 '18 at 18:02
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    You are right, I meant the automorphisms of the trivial right $G$-torsor (edited). This is exercise 5.15 in Poonen rational points on varieties lecture notes. – proofromthebook Jun 11 '18 at 20:45

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That is not true. For example $\textrm{Aut}(SL_2) = PGL_2(k)$. In any linear algebraic group, $\textrm{Inn}(G) \cong (G/Z)(k)$, where $Z$ is the center, and the full automorphism group is an extension of that by it's outer automorphisms.

In view of the edit:

An automorphism or torsors $ \varphi: \mathbf{G} \to \mathbf{G}$ is $G$-equivariant, hence $\varphi(g)=\varphi(e\cdot g)=\varphi(e)\cdot g$ where "$\cdot$" denotes right multiplication. Following the identity element, what can you conclude?

wskrsk
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