Counterexample to Lie's theorem over a field, that is not algebraically closed What would be a counterexample to Lie's Theorem over a field (of characterictic $0$) that is not algebraically closed ?
 A: Take a $3$-dimensional real solvable Lie algebra $L$, with basis $(x,y,z)$ and Lie brackets
$$
[z,x]=\alpha y,\; [z,y]=\beta x
$$
with real numbers $\alpha,\beta$ such that $\alpha\beta<0$. Then $L$ cannot embedded into a Lie algebra of upper-triangular matrices of any degree. Hence Lie's Theorem does not hold. For a proof, see page $21$ here.
In characteristic $p>0$, the standard counterexample is for the non-abelian Lie algebra $\mathfrak{r}_2(K)$ of dimension $2$ over a field $K$ of characteristic $p>0$, with basis $E,F$ and Lie bracket $[E,F]=E$, and a $p$-dimensional representation $\rho$ given by 
$$
\rho (E)(e_i)=e_{i+1}, i\le p-1,\; \rho(E)(e_p)=e_1, \rho(F)e_i=(i-1)e_i.
$$
However, Lie's Theorem does hold in characteristic $p>0$ provided the dimension of the vector space is less than the characteristic. For details see my MO-answer.
A: I guess that you mean the theorem that every solvable subalgebra of $\mathfrak{gl}_n$ is conjugate to a subalgebra of the subalgebra of upper triangular matrices (when the field is algebraically closed of characteristic zero).
If $K$ is not algebraically closed, and $L$ is an extension of degree $n\ge 1$, view $L$ as an abelian subalgebra of $\mathfrak{gl}_n$. Then $L$ is not conjugate into upper triangular matrices.
Dietrich's answer yields something a little stronger: a finite-dimensional solvable Lie algebra that is not embeddable into any finite-dimensional upper triangular Lie algebra. He can also get it for an arbitrary non-algebraically closed field, choosing a semidirect product $K^n\rtimes K$ where the action is by a companion matrix of an irreducible polynomial of degree $n\ge 2$.
For a related (more subtle) question, see 
https://mathoverflow.net/questions/207154/about-supersolvable-lie-algebras
