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Let X be a Brauer-Severi variety over k. I have understand that the automorphism scheme Aut_{X/k}=G as a group scheme acts on X via (m,pr_2):GxX /rightarrow XxX (multiplication morphism) and that this morphism is surjectiv.

But what is about the action of Aut_{X/k}(k) on X?

Is there for example a transitiv action of Aut_{X/k}(k) on the closed points of X?

Or what kind of assumption is needed to get this?

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It is clear that there cannot be a transitive action on the closed points in general: the isomorphism class of the residue field, in particular also degree, of a closed point must be preserved. –  ulrich Nov 19 '12 at 8:38
    
I see, but supposing this is the case what can we say? –  Peter Nov 19 '12 at 8:56
    
So suppose there are closed points x and y such that the residue fields are isomorphis as k-algebras does there exsist an element of Aut_{X/k}(k) mapping x to y? –  Peter Nov 19 '12 at 9:13
    
@Peter: Have you tried looking at what happens for the simplest non-trivial Brauer-Severi varieties, namely conics which do not contain a rational point? –  Daniel Loughran Nov 19 '12 at 9:29
    
Or you could even consider the action on $\mathbb{P}^1$, say over a finite field, and then counting will show that the action is not transitive. –  ulrich Nov 19 '12 at 14:33

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