Fix an algebraically closed field $K$, and let $G$ be a unipotent linear algebraic group over $K$ acting morphically on an affine variety $X$. According to [1, Prop. 2.5] we have the following result:
Theorem. Every $G$-orbit on $X$ is (Zariski) closed.
I am looking for a rational version of the above result. More precisely, let $k$ be an arbitrary subfield of $K$ and suppose that $G,X$ and the $G$-action on $X$ are all defined over $k$. The group of $k$-rational points $G(k)$ still acts morphically on $X$.
Is it true that $G(k)$-orbits are closed relative to the Zariski $k$-topology on $X$?
The discussion in [2, Sections 34.1 and 34.2] suggests that the proof of [1, Prop. 2.5] adapts fairly well when $k$ is either perfect or (more generally) $k$ is replaced by some purely inseparable extension in $K$. Unfortunately, I am interested in the case where $k$ is a non-archimedean local field of positive characteristic and $K$ is its algebraic closure.
References would be appreciated in case there is a positive answer to the above question. Thank you in advance.
BIBLIOGRAPHY
[1] R. Steinberg, "Conjugacy Classes in Algebraic Groups", Lecture Notes in Mathematics, Vol. 366, Springer, New York/Berlin, 1974;
[2] J. E. Humphreys, "Linear Algebraic Groups", Graduate Texts in Mathematics, Vol. 21, Springer, New York/Berlin, 1975.