The Radical of $SL(n,k)$ For an algebraically closed field $k$, I'd like to show that the algebraic group $G=SL(n,k)$ is semisimple.  Since $G$ is connected and nontrivial, this amounts to showing that the radical of $G$, denoted $R(G)$, is trivial.  $R(G)$ can be defined as the unique largest normal, solvable, connected subgroup of $G$.
I know that the group of $n$th roots of unity of $k$ is inside of $G$, and it is normal and solvable (being in the center of $G$) but not connected, having one irreducible component for each root of unity.  What are other normal subgroups in $G$?  How can I show that $R(G)=e?$
 A: The fact is that the quotient $\mathrm{PSL}_n(k)$ of $\mathrm{SL}_n(k)$ by its center is simple. Since the center of $\mathrm{SL}_n(k)$ consists, as you say, of the $n$th roots of unity, this shows that there are no nontrivial connected normal subgroups of $\mathrm{SL}_n(k)$.
The fact that the projective special linear group is simple is not entirely trivial. There is  a proof using Tits systems in the famous Bourbaki book on Lie groups and Lie algebras (chapter 4), which is of course more general. A more elementary approach, just using linear algebra, can be found in Grove's book "Classical groups and geometric algebra".
A: Here is a completely different approach, which however works only over a field of char. $0$: it is enough to prove that the Lie algebra is simple. For $\mathfrak{sl}_n$, this follows from the observation that any non-zero ideal must be contain a simultaneous eigenvector for the action by Lie bracket of the diagonal trace $0$ matrices. Such an eigenvector is either a diagonal trace $0$ matrix or (up to scalars) a matrix unit $e_{ij}$ with $i \neq j$. In the latter case the formula $[e_{ij},e_{ji}]=e_{ii}-e_{jj}$ implies that our ideal contains a non-zero diagonal trace zero matrix. Using this matrix it's now easy to show that our ideal is actually everything.
