Exercise 12, § 3

Let $k'$ be a finite extension of $k,\mathfrak{a}'$ a semi-simple $k'$-Lie algebra, and $\mathfrak{a}$ the underlying $k$-Lie algebra. Show that the parabolic (resp. Borel) subalgebras of $\mathfrak{a}$ are the same as those of $\mathfrak{a}'$. (Extend scalars to an algebraic closure of $k$, and use Exerc. 11.)