Let $\mathfrak{g}$ be a finite-dimensional Lie algebra. The radical of a Lie algebra $\mathfrak{g}$, denoted as $\mathrm{rad}(\mathfrak{g})$, is the unique maximal solvable ideal of $\mathfrak{g}$.
Let $\mathfrak{g}$ be a finite-dimensional complex Lie algebra and $\mathfrak{g}_{\mathbb{R}}$ its realification, which means that $\mathfrak{g}_{\mathbb{R}}$ is $\mathfrak{g}$ considered as a real Lie algebra. Of course we have $\mathrm{rad}(\mathfrak{g})\subset \mathrm{rad}(\mathfrak{g}_{\mathbb{R}})$ since solvability does not rely on the ground field. But can this inclusion be strict?
Thank you in advance for any help.
(Note: From this question we know that for finite-dimensional complex Lie algebra $\mathfrak{g}$, if $\mathrm{rad}(\mathfrak{g}) = \{0\}$, then $\mathrm{rad}(\mathfrak{g}_{\mathbb{R}}) = \{0\}$.)
Best Answer
$\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}\DeclareMathOperator\rad{rad}$Let $K\subset L$ be fields. Let $\g$ be a finite-dimensional Lie algebra over $L$. I claim that $\rad(\g)=\rad(\g|_K)$. The latter means the sum of all solvable ideals ($\g$ might be infinite-dimensional over $K$).
As already mentioned, $\rad(\g)\subset\rad(\g|_K)$ is clear (since $\rad(\g)$ is an ideal of $\g|_K$). Assuming by contradiction the inclusion proper, there exists a solvable ideal of $\g|_K$ not contained in $\rad(\g)$. Projecting, we deduce that the Lie $K$-algebra $\g/\rad(\g)$ contains a nonzero solvable ideal. Hence passing to an element in its derived series, we deduce that the Lie $K$-algebra $\g/\rad(\g)$ contains a nonzero abelian ideal. By the next lemma, $\g/\rad(\g)$ also contains a nonzero abelian ideal (as $L$-algebra).
Lemma: Let $\h$ be a Lie algebra over $L$ and $V$ an abelian ideal of $\h|_K$. Then the $L$-span $W$ of $\h$ is an abelian ideal of $\h$.
Proof: we have $[\sum_i t_iv_i,\sum_i u_jv'_j]=\sum_{i,j}t_iu_i[v_i,v'_j]=0$, so $W$ is abelian. For $h\in\h$ we have $[h,\sum_i t_iv_i]=\sum_i t_i[h,v_i]$ so $W$ is an ideal.