Lie's theorem is field- and characteristic-free

Over $\mathbb{R}$, let the one-dimensional abelian Lie algebra $\mathbb{R}t$ act on $V={\mathbb{R}}^2$ by $\rho (t)=R=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. This is a representation by [L3] and the acting algebra is solvable by [L2]. But $R$ has characteristic polynomial $T^2+1$, so it has no real eigenvector and hence no common invariant eigenline. Indeed this two-dimensional real module is irreducible, because every nonzero proper subspace would be a line. Thus Lie's theorem fails without a splitting-field hypothesis even in characteristic zero.

For the characteristic obstruction, let $k$ have characteristic $p>0$, let $\mathfrak{a}=kx\oplus ky$ with $[x,y]=y$, and let $V$ have basis $v_0,\dots ,v_{p-1}$. Define $X{v}_i=i{v}_i$ and $Y{v}_i=v_{i+1}$ with indices modulo $p$. For $i<p-1$, $(XY-YX)v_i=(i+1-i)v_{i+1}=Y{v}_i$, while for $i=p-1$ it is $(0-(p-1))v_0=v_0=Y{v}_{p-1}$. Hence $[X,Y]=Y$, so $x\mapsto X$, $y\mapsto Y$ is a representation by [L3]. Also ${\mathfrak{a}}^{(1)}=ky$ and ${\mathfrak{a}}^{(2)}=0$, so the acting algebra is solvable by [L2].

The $p$ eigenvalues $0,1,\dots ,p-1$ of $X$ are distinct and its eigenspaces are exactly the lines $k{v}_i$, but $Y$ cyclically moves each such line to the next, so $X$ and $Y$ have no common eigenvector. More strongly, if $0\mathrm{\ne }W\subseteq V$ is invariant and a vector of $W$ has nonzero $v_i$-coordinate, the Lagrange polynomial $P_i(T)={\prod }_{j\mathrm{\ne }i}(T-j)\mathrm{/}(i-j)$ gives $P_i(X)W\subseteq W$ and extracts a nonzero multiple of $v_i$; repeated application of $Y$ then puts every $v_j$ in $W$. Thus the module is irreducible of dimension $p>1$.

Step 1.1 violates the common-eigenvector conclusion over a non-algebraically-closed characteristic-zero field, and steps 1.2–2.1 violate it over a field of positive characteristic. These independent witnesses show that neither omitted hypothesis is cosmetic. Both constructions are finite and use no choice.