What would be an example of a connected finite group scheme over a field $k$ that is not solvable? Here $k$ is algebraically closed.


Let $\operatorname{GL}\_n$ be the general linear group scheme over a field of characteristic $p>0$. Now consider the $H=\operatorname{ker}{F^r:\operatorname{GL}\_n \to \operatorname{GL}\_n}$, where $F^r$ is the frobenius map which raises each element of the matrix $\operatorname{GL}\_n(R)$ to $p^r$. In this case, is $H$ connected and not solvable?



Example of a connected finite group scheme which is not solvable
The connected finite kernel $H$ is not solvable, provided that $n>2$ or $p>2$, see the edit below.


$\def\eps{\varepsilon} \def\m{\mathfrak{m}}$Suppose by contradiction that $H$ is solvable and the $m$-th derived group of $H$ is trivial (where $[H,H]$ is the *first* derived group). It implies that for any $k$-algebra $R$ the $m$-th derived group of the abstract group $H(R)$ is also trivial (because for any $G$ we have $[G(R), G(R)]\subset D(G)(R)$). Put $l=2^{m+1}$ and consider $R=k[\eps\_1,\dotsc,\eps\_l]/(\eps\_1,\dotsc,\eps\_l)^{(p)}$ where $(p)$ means taking $p$-th powers of all the elements of the ideal. Put also $\m=(\eps\_1,\dotsc,\eps\_l)$ to be the maximal ideal. I claim that $D(D(\dotso D(H(R)))\neq 1$ ($D$ iterated $m$ times). Indeed, $H(R)=1+\m\operatorname{Mat}\_n(R)$. Now note that for any two matrices $A, B\in \operatorname{Mat}\_n(R)$ we have (sums represented by dots are actually finite) $$[1+\eps\_i A,1+\eps\_j B]=(1+\eps\_i A)(1+\eps\_j B)(1-\eps\_i A+\eps^{2}\_iA^2-\dotsb)(1-\eps\_j B+\eps\_j^2 B^2-\dotsb)=1+\eps\_i\eps\_j(AB-BA) \pmod{\m^3}.$$
Now pick a tuple $A\_1,\dotsc A\_{2^{m+1}}$ of elements of the Lie algebra $\mathfrak{gl}\_n$ over $k$ such that the $(m+1)$-times nested commutator of them is $B\neq 0$. The $(m+1)$-times nested commutator of the elements $1+\eps\_iA\_i\in H(R)$ then equals $1+\eps\_1\dotsm\eps\_{2^{m+1}}B$ modulo $\m^{2^{m+1}+1}$ and thus is not equal to $1$.


Edit: This argument uses that the Lie algebra $\mathfrak{gl}\_n$ over a field of characteristic $p$ is not solvable. This is true if $n>2$ or $p>2$, but $\mathfrak{gl}\_2$ in characteristic $2$ is actually solvable. This is because $\mathfrak{sl}\_2=\ker(\mathfrak{gl\_2}\xrightarrow{\operatorname{tr}}k)$ has non-trivial center, spanned by the diagonal matrix $\mathrm{diag}(1,1)$, and its quotient by the center is abelian.


