Let $G$ be a linear algebraic group over an algebraically closed field $\mathbb C$ of characteristic zero and $U$ its unipotent radical, then $H:=G/U$ is a reductive group. Assume that I have a one parameter subgroup $\lambda:\mathbb C^\times\to H$. I would like to lift this to a one parameter subgroup $\tilde\lambda:\mathbb C^\times\to G$ such that $\lambda = \pi\circ \tilde\lambda$, where $\pi:G\to H$ is the canonical projection. If $\mathbb C$ are the complex numbers, then this is always possible by the [two theorems in this paper](http://arxiv.org/pdf/0709.2371v3.pdf).


My question is: Is it true for any algebraically closed field? Do I even need the characteristic zero assumption? I am basically looking for references of such a lifting statement in the context of linear algebraic groups instead of lie groups. 



Lifting one parameter subgroups of algebraic groups
This is true for quotients by smooth unipotent normal subgroups over any field whatsoever: if $f:G \rightarrow G'$ is a surjective homomorphism between group schemes of finite type over a field $k$ such that $\ker f$ is a unipotent smooth group (in particular, affine) then for any $k$-torus $T' \subset G'$ there exists a $k$-torus $T \subset f^{-1}(T')$ such that $f:T \rightarrow T'$ is an isomorphism. The main content of the proof is the structure of connected solvable groups over algebraically closed fields.


For a proof, we can replace $G'$ with $T'$ and replace $G$ with $f^{-1}(T')$ so that $G'=T'$ is a torus. Since $\ker f$ is smooth and affine over $k$, it follows by descent theory that $f$ is smooth and affine (as $G \times\_{G'} G = G \times (\ker f)$). Thus, since $G'$ is now smooth and affine, so is $G$. We can even replace $G$ with $G^0$, so $G$ is a connected solvable linear algebraic group. If $T$ is a maximal $k$-torus in $G$ then $G\_{\overline{k}} = T\_{\overline{k}} \ltimes R\_u(G\_{\overline{k}})$ by the structure theorem for connected solvable groups over algebraically closed fields. 


Under the quotient map $f:G \twoheadrightarrow G'=T'$ necessarily $R\_u(G\_{\overline{k}})$ has to be killed by $f\_{\overline{k}}$. But $G\_{\overline{k}}/R\_u(G\_{\overline{k}})$ is a torus (in fact naturally isomorphic to $T\_{\overline{k}}$), so the image of the smooth unipotent $(\ker f)\_{\overline{k}} = \ker(f\_{\overline{k}})$ in this quotient of $G\_{\overline{k}}$ is trivial. Hence, $\ker(f\_{\overline{k}}) = R\_u(G\_{\overline{k}})$. Thus, $f\_{\overline{k}}$ induces an isomorphism $T\_{\overline{k}} \rightarrow T'\_{\overline{k}}$, so likewise $f$ induces an isomorphism $T \rightarrow T'$. QED


The smoothness of the kernel is essential; e.g., if $\ker f$ is a non-smooth unipotent group scheme then the conclusion is false. The most famous class of examples are the so-called unipotent isogenies between connected semisimple groups, associated to non-degenerate quadratic forms in characteristic 2. More specifically, if $(V,q)$ is any non-degenerate quadratic space of dimension $2n+1 \ge 3$ over a field $k$ of characteristic 2 then the defect space $V^{\perp}$ for the associated symmetric bilinear form $B\_q$ on $V$ is a line and $B\_q$ induces a non-degenerate bilinear form $\overline{B}\_q$ on $\overline{V} = V/V^{\perp}$ that is symplectic (since in characteristic 2), with ${\rm{SO}}(q) \rightarrow {\rm{Sp}}(\overline{B}\_q)$ an isogeny from adjoint type B$\_n$ to *split* simply connected type C$\_n$ whose kernel is a unipotent group scheme (in fact a non-central but commutative direct product of copies of $\alpha\_2$). If $q$ is not homothetic to a split form (e.g., if it is anisotropic) then ${\rm{SO}}(q)$ is $k$-anisotropic, so none of the non-trivial split maximal $k$-tori in the quotient can be lifted (not even up to isogeny).


