why is Borel subgroup not nilpotent?

Let $G$ be a simple linear group group over an algebraically closed field $k$, and let $B$ be a maximal solvable subgroup.

If things are happening over $\mathbb{C}$ then I know how to show that $B$ is not nilpotent: elements of torus lie subalgebra of $B$ act in a non-nilpotent way via adjoint representation.

But is there an equivalence between Lie algebra being nilpotent and Lie group being nilpotent in general, when there is no exponential map? what does one do in positive characteristic?

So the question is: how to show that $B$ is not nilpotent when $k \neq \mathbb{C}$?

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Over any algebraically closed field, in a connected algebraic group a Borel being nilpotent implies the Borel is the whole group. –  Jason Polak Dec 4 '13 at 21:48
could you explain how one can see that, please? –  Dima Sustretov Dec 4 '13 at 21:56
Check Humphreys's 'Linear Algebraic Groups', Proposition 21.4B(a). Use induction on the dimension of $G$, the dimension zero case being clear. If $\mathrm{dim}(G)>0$ then $\mathrm{dim}(B) > 0$ because Borel subgroups cover the whole group. Now pass to $G/Z(G)^\circ$ which has smaller dimension because $Z(G)^\circ \subseteq Z(B)$ where $Z(B)$ is the center of $B$, which is nontrivial because $B$ is nilpotent, and use the induction hypothesis. –  Jason Polak Dec 4 '13 at 22:04
great! thanks for the reference. –  Dima Sustretov Dec 4 '13 at 22:20