Radical of $\mathfrak{gl}_n$

If $I$ is a solvable ideal of $\mathfrak{gl}_n(\mathbb{F})$ then $I\cap \mathfrak{sl}_n(\mathbb{F})$ is a solvable ideal of $\mathfrak{sl}_n(\mathbb{F})$ which is semisimple (in fact simple) as long as the characteristic of $\mathbb{F}$ does not divide $n$, so the radical of $\mathfrak{gl}_n(\mathbb{F})$ intersects trivially with $\mathfrak{sl}_n(\mathbb{F})$, and this forces it to have dimension $1$.

On the other hand, if we take the example $\mathfrak{gl}_2(\mathbb{F}_2)$ then one can check that this is actually solvable, so in this case the radical does not just consist of the scalar matrices (the derived subalgebra is contained in $\mathfrak{sl}_2(\mathbb{F}_2)$ which contains the scalar matrices. The quotient of $\mathfrak{sl}_2(\mathbb{F}_2)$ by the scalar matrices has dimension $2$ and is thus solvable).