Proposition 14

Let $k'$ be an extension of $k$ and $\mathfrak{g}'=\mathfrak{g}\otimes_kk'$. Let $\mathfrak{m}$ be a subalgebra of $\mathfrak{g}$ and $\mathfrak{m}'=\mathfrak{m}\otimes_kk'$. If $\mathfrak{m}'$ is a parabolic (resp. Borel ) subalgebra of $\mathfrak{g}',\mathfrak{m}$ is a parabolic (resp. Borel ) subalgebra of $\mathfrak{g}$.

By Prop. 8 and 11, it suffices to prove that $\mathfrak{m}$ contains a splitting Cartan subalgebra of $\mathfrak{g}$. Let $\mathfrak{b}$ be a Borel subalgebra of $\mathfrak{g}$. Then $\mathfrak{b}'=\mathfrak{b}\otimes_kk'$ is a Borel subalgebra of $\mathfrak{g}'$, so $\mathfrak{m}'\cap \mathfrak{b}'$ contains a Cartan subalgebra of $\mathfrak{g}'$ (Prop. 10). Let $\mathfrak{t}$ be a Cartan subalgebra of $\mathfrak{m}\cap \mathfrak{b}$. Then $\mathfrak{t}\otimes_kk'$ is a Cartan subalgebra of $\mathfrak{m}'\cap \mathfrak{b}'$, and hence of $\mathfrak{g}'$ (Chap. VII, §3, no. 3, Prop. 3). Consequently, $\mathfrak{t}$ is a Cartan subalgebra of $\mathfrak{g}$, and it is splitting since it is contained in $\mathfrak{b}$.