Let $\mathfrak{g}$ be a real (fin. dim.) Lie algebra.
The complexification of $\mathfrak g$, denoted by $\mathfrak g_{\mathbb C}$, is simply the complexification of $\mathfrak g$ as a vector space (i.e., $\mathfrak g_{\mathbb C}=\mathbb C\otimes \mathfrak g$), endowed with a natural Lie bracket. Let $\mathfrak a$ be an ideal of $\mathfrak g$. We then also have the ideal $\mathfrak a_{\mathbb C}\subset\mathfrak g_{\mathbb C}$.
Let $\mathfrak{rad}(\mathfrak g)$ be the radical of $\mathfrak g$, i.e., the largest solvable ideal. I want to show that
$$
\mathfrak{rad}(\mathfrak g_{\mathbb C})=\mathfrak{rad}(\mathfrak g)_{\mathbb C}.
$$
I can show the $\supset$-inclusion: For any ideal $\mathfrak{a}\subset\mathfrak{g}$, we define recursively
$$
D(\mathfrak a)=[\mathfrak a,\mathfrak a],\quad D^{n+1}(\mathfrak a)=[D^n(\mathfrak a),D^n(\mathfrak a)].
$$
One can then show that
$$
D^n(\mathfrak a)_\mathbb C = D^n(\mathfrak a_\mathbb C),
$$
from which the inclusion $\supset$ follows.
I'm stuck at the other inclusion. Here are my thoughts:
It's not true that each (complex) ideal $\mathfrak a\subset\mathfrak g_\mathbb C$ is of the form $\mathfrak a=\mathfrak b_\mathbb C$ where $\mathfrak b\subset\mathfrak g$ is an ideal of $\mathfrak g$ (an easy example is achieved in dimension 2 with the trivial Lie bracket), so we can't go that route.
It is known that $[\mathfrak g,\mathfrak g]^\perp=\mathfrak{rad}(\mathfrak g)$ (where the orthogonality is w.r.t. the Cartan killing form). I wonder then if it's generally true that for any ideal $\mathfrak a\subset\mathfrak g$ we have $(a^\perp)_\mathbb C=(\mathfrak a_\mathbb C)^\perp$, in which case we would be done. Or if this doesn't hold generally, then at least $[\mathfrak g_\mathbb C,\mathfrak g_\mathbb C]^\perp\subset [\mathfrak g,\mathfrak g]^\perp_\mathbb C$ might maybe be easier to show. I was thinking of maybe working with a basis $(x_i)$ for $[\mathfrak g,\mathfrak g]^\perp$, which can be extended to a basis $(x_i,y_j)$ for $\mathfrak g$, and this would also be a basis for $\mathfrak g_\mathbb C$. So we could write and $x\in [\mathfrak g_\mathbb C,\mathfrak g_\mathbb C)]^\perp$ as $x=\sum \lambda_i x_i +\sum \mu_j y_j$. For any $[z,w]\in[\mathfrak g,\mathfrak g]$ we would then have $[x,[z,w]]=\sum \mu_j [y_j,[z,w]]$, but I don't see how this would yield $\mu_j=0$.
Best Answer
I think you are on the right track with the idea in your final paragraph. Show that the Killing form of the complexification is the natural $\mathbb C$-bilinear extension of the Killing form of $\mathfrak{g}$; and that $[\mathfrak g, \mathfrak g]_{\mathbb C}$ can be naturally identified with $[\mathfrak g_{\mathbb C}, \mathfrak g_{\mathbb C}]$. Finally, the orthogonal complement to $[\mathfrak g, \mathfrak g]_{\mathbb C}$ w.r.t. the extended Killing form should be the span of the orthogonal to $[\mathfrak g, \mathfrak g]$ under the original Killing form.
As a check, none of this should rely on any special properties of the field extension $\mathbb C |\mathbb R$ as the result holds true for any field extension $L|K$, at least in characteristic zero.
Added: The step that was rightly pointed out as not obvious in the comment holds very generally:
Let $L|K$ be a field extension, $V$ a $K$-vector space, $b:V\times V \rightarrow K$ a symmetric bilinear form. Denote $V_L :=L\otimes_K V$ and let $b_L$ be the unique extended form $V_L \times V_L \rightarrow L$. Then for any subspace $U \subseteq V$, under the obvious identifications we have: $$(U^\perp)_L = (U_L)^\perp$$ Proof: As you say, $\subseteq$ is easy. To see $\supseteq$, write an element $y\in (U_L)^\perp$ as $\sum_i l_i \otimes v_i$ with the $l_i$ linearly independent over $K$. Then for every $u \in U$: $$0 \stackrel{y\in (U_L)^\perp}= b_L(1\otimes u, y)= \sum_i \underbrace{b(u,v_i)}_{\in K} \cdot l_i $$ By linear independence of the $l_i$, all $b(u,v_i)=0$. So each $v_i \in U^\perp$, q.e.d.
(A ridiculously more general version of this is stated and proved in Bourbaki's Algèbre chap. IX §1 no. 4 prop. 3.)