Lie's theorem say that
Let $L$ be a solvable, nonzero subalgebra of $\mathfrak{gl}(V)$, with $V$ a finite dimensional vector space over an algebraically closed field $F$ of characteristic zero. Then $V$ contains a common eigenvector for all the endomorphisms in $L$.
It is true for field with zero characteristic. I am unable to find a counter example when $\mathrm{char}~F= p$, a prime. Can anybody explain ?
Take the $3$-dimensional lie algebra spanned by $1, x, d/dx$ acting on the vector space $k[x]/(x^p)$ in the obvious way. The point is that both $x$ and $d/dx$ are nilpotent with one dimensional kernels, but these kernels are different, so there can't possibly be a common eigenvector.
Whereas in characteristic 0, applying trace to the commutation relation $[d/dx,x]=1$, we get that 1 must act by zero on any finite dimensional representation and therefore the same relation tells us that any eigenvector of $x$ is a zero eigenvector of $d/dx$. So there is no hope of this sort of counter-example working in characteristic 0.