Let $\mathbb{K}$ be an arbitrary field, and $G$ an algebraic group (or group variety?) over this field. A Borel subgroup of $G$ is a connected solvable subgroup variety $B$ of $G$ such that $G/B$ is complete. A parabolic subgroup of $G$ is a subgroup variety $P$ such that $G/P$ is complete.
From section 29.4 of Tauvel and Yu, I know that if $\mathbb{K}$ is algebraically closed and has characteristic zero, then a Borel subalgebra of a finite-dimensional Lie $\mathbb{K}$-algebra $\mathfrak{g}$ (not necessarily semisimple) is defined to be a maximal solvable subalgebra $\mathfrak{b} \leq \mathfrak{g}$. A parabolic subalgebra $\mathfrak{p} \leq \mathfrak{g}$ is any subalgebra containing a Borel subalgebra.
Tauvel and Yu prove that under these hypotheses on the field ($\mathbb{K}$ is algebraically closed and has characteristic zero), if $G$ is a connected algebraic group with Lie algebra $\mathfrak{g}$, then parabolic subalgebras of $\mathfrak{g}$ are precisely the Lie algebras of parabolic subgroups of $G$.
Now if we generalize so that $\mathbb{K}$ is not algebraically closed (or more generally, but less importantly, when the characteristic is nonzero), I would like to know to what extent these definitions and correspondences carry over? Moreover, if $\overline{\mathbb{K}}$ is the algebraic closure of $\mathbb{K}$, what properties are preserved and reflected by "extending" the scalars?
For my particular purposes, I am most interested in the case $\mathbb{K}=\mathbb{R}$. However, it seems that everywhere I look the complex case is emphasized. I like to think very categorically, and in a "top-down" sort of way, so I am hoping to get a good overview of where I am going and what to expect before I slog through the complex theory.
If it helps at all, I trying to get a handle on parabolic subgroups and subalgebras so that I can reach a more comprehensive understanding of parabolic Cartan geometries and (eventually) BGG resolutions.
Unfortunately for me, my knowledge of algebraic geometry and algebraic groups is very limited, so I apologize if the answer to my question is very obvious.