Corollary 4

Let $\mathfrak{g}$ be a decomposable Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$. Let $\mathscr{T}$ be the set of commutative subalgebras of $\mathfrak{g}$ consisting of semi-simple elements (cf. Prop. 6). The maximal elements of $\mathscr{T}$ all have the same dimension.

Let $k'$ be an algebraically closed extension of $k$ and $V'= V\otimes_kk',\mathfrak{g}'=$ $\mathfrak{g}\otimes_kk'$. Let $\mathfrak{t}_1,\mathfrak{t}_2$ be maximal elements of $\mathscr{T},\mathfrak{t}'_i=\mathfrak{t}_i\otimes_kk',\mathfrak{h}_i$ the commutant of $\mathfrak{t}_i$ in $\mathfrak{g},\mathfrak{h}'_i=\mathfrak{h}_i\otimes_kk'$. Then $\mathfrak{h}_i$ is a Cartan subalgebra of $\mathfrak{g}$ (Prop. 6) so $\mathfrak{h}'_i$ is a Cartan subalgebra of $\mathfrak{g}'$. Then $\mathfrak{h}_i=\mathfrak{t}_i\times \mathfrak{n}_V(\mathfrak{h}_i)$, hence $\mathfrak{h}'_i=\mathfrak{t}'_i\times \mathfrak{n}_{V'}(\mathfrak{h}'_i)$, so that $\mathfrak{t}'_i$ is the set of semi-simple elements of $\mathfrak{h}'_i$. Since $\mathfrak{g}'$ is decomposable (Cor. $3$)$,\mathfrak{t}'_1$ and $\mathfrak{t}'_2$ are conjugate under Aut$_e(\mathfrak{g}')$ (Prop. 6), so dim$\mathfrak{t}_1=$ dim$\mathfrak{t}_2$.