Lie's theorem

Every finite-dimensional complex representation of a solvable Lie algebra has a common eigenvector. Equivalently, an irreducible finite-dimensional complex representation is one-dimensional; iterating gives simultaneous upper triangularization.

The solvable Lie algebra $\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, has an infinite-dimensional Irreducible Lie algebra representation on $\mathbb{C}[t]$: $xf(t)=tf(t)$ and $yf(t)=f(t-1)$. Indeed $[x,y]f=yf$. An invariant subspace is an ideal $p(t)\mathbb{C}[t]$ because it is invariant under multiplication by $t$. Shift invariance gives $p(t)\mid p(t-1)$, hence $p(t-1)=p(t)$ and $p$ is constant. This shows why the Lie theorem needs finite-dimensional representations, even when the algebra itself is finite-dimensional.

A Lie algebra representation is simultaneously upper triangular precisely when it preserves a complete flag, with $\dim V_j=j$. A basis adapted to the invariant flag makes every representing matrix upper triangular. For a complex finite-dimensional representation of a solvable Lie algebra, the Lie theorem provides a common eigenvector; iteration on the quotient representations constructs the flag.

In a field of characteristic $p$, take a basis $e_0,\dots ,e_{p-1}$, and let $x{e}_j=e_{j-1}$ with indices modulo $p$, while $y{e}_j=j{e}_j$. Then $[x,y]=x$, so $kx+ky$ is a solvable Lie algebra, but the distinct eigenvalues of $y$ force any common eigenvector to be a coordinate vector, and $x$ cyclically permutes those vectors. There is consequently no common eigenvector, even over an algebraically closed field.