Let $k$ be a field and $G$ a connected semisimple algebraic group over $k$. 


If $k$ is algebraically closed, then it is well known that all Borel subgroups of $G$ are conjugate by the action of $G(k)$. I would like to know whether this is also true over non-closed fields.



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Recall that a Borel subgroup $B$ of $G$ over $k$ is a closed subgroup of $G$ over $k$ such that the base change of $B$ to the algebraic closure $\bar{k}$ is a Borel subgroup of $G \times\_k \bar{k}$. 


Of course Borel subgroups need not exist in general; when they do one says that $G$ is quasi-split. If $G$ is not quasi-split then the statement is vacuously true.



Conjugacy of Borel subgroups over arbitrary fields
Yes. This follows directly from Theorem 20.9 (i) in "Armand Borel, Linear Algebraic Groups, Second enlarged edition, 1991" which goes like this:


**Theorem**: Let $ G $ be a connected reductive group over a field $ k $. The minimal parabolic $k$-subgroups of $G$ are conjugate under $G(k)$.


