nilpotent algebraic groups in terms of extensions Let $N$ be a nilpotent linear algebraic group over a field $k$.

If $k = \mathbb{C}$ and $N$ is connected, one can write $N = U \times T$, where $U$ is a unipotent algebraic group and $T$ is a torus. Can we do the same for any $k$ (with char = 0)?
Is there a characterization of a general $N$ (not necessarily connected) in terms of extensions of algebraic groups, for a general $k$ (with char = 0)?

 A: This holds in characteristic zero in general: any nilpotent linear algebraic $k$-group is the direct product of a (connected) unipotent group and a virtual torus (an algebraic group whose identity component is a torus).
First, let us prove it in the connected case. Let $U$ be the unipotent radical and $D$ a $k$-defined Levi factor. So $D$ is reductive and $N=U\rtimes D$. Since $N$ is solvable, $D$ is abelian. Since $N$ is nilpotent, we have $[N,D]=1$ ($*$), hence $N=D\times U$. This thus characterizes $D$ as the unique maximal torus in $N$, since for any torus in $N$, its projection in $U$ is a torus and thus is trivial.
Now assume $N$ arbitrary. By the previous case, we can write $N^\circ=D\times U$ with $D,U$ defined over $k$. On the other hand, there exists a finite algebraic subgroup $F$ such that $FN^\circ=N$. Replacing $F$ with its $k$-closure, we can suppose that it is $k$-defined. Then since $N$ is nilpotent, we have $[F,N^\circ]=1$. Hence $[FD,U]=1$ and $FDU=N$. We have to check that $FD\cap U=1$: indeed $FD\cap U$ is a unipotent subgroup on the one hand, and has a finite index subgroup contained in $D$ on the other hand, and hence $FU$ is finite; being contained in $U$ it is therefore trivial. Hence $N=FD\times U$, where $U$ is connected (and unipotent) and $FD$ is a virtual torus, defined over $k$.

Edit: I was asked by email why $[F,N^\circ]=1$. Otherwise, $F$ acts non-trivially on the Lie algebra $\mathfrak{n}$ of $N^\circ$ (indeed, the exponential intertwines the actions of $F$ on $N^\circ$ and on $\mathfrak{n}$). Using complete reducibility of actions of finite groups in characteristic zero, and by nilpotency of $\mathfrak{n}$, there exists $i$ such that $F$ acts non-trivially on $\mathfrak{n}^i/\mathfrak{n}^{i+1}$ (where $(\mathfrak{n}^i)$ is the lower central series). Modding out by $(N^\circ)^{i+1}$, one can suppose that $F$ acts non-trivially on the center $Z$ of $N^\circ$. Then $[F,Z]=[F,[F,Z]]$ is non-trivial (again, using complete reducibility). This contradicts nilpotency of $N$.
