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Let $X$ be a Brauer–Severi variety over $k$.

I understood that the automorphism scheme $\mathrm{Aut}_{X/k}=G$ as a group scheme acting on $X$ via $$(m,pr_2):G\times X \rightarrow X\times X$$ (the multiplication morphism), and that this morphism is surjective.


What about the action of $\mathrm{Aut}_{X/k}(k)$ on $X$?

Is there for example a transitive action of $\mathrm{Aut}_{X/k}(k)$ on the closed points of $X$? If not, 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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