Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / A / 1

We prove Lie's theorem in the required form by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise solvability gives $[L,L]\ne L$, so choose a codimension-one ideal $K\supseteq [L,L]$ and $x\in L\setminus K$. By induction there are $v\ne 0$ and a linear functional $\lambda :K\to \mathbb{C}$ with $av=\lambda (a)v$ for all $a\in K$.

Let $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Choose the cyclic basis $v,xv,\dots ,x^{r-1}v$ up to its first linear dependence. The commutator identity $a{x}^j=xa{x}^{j-1}+[a,x]x^{j-1}$, with $[a,x]\in K$, proves inductively thatHence $W$ is invariant under $K$ and $x$, and every $a\in K$ has trace $r\lambda (a)$ on $W$. In particular ${\mathrm{tr}}_W([x,a])=0$ implies $r\lambda ([x,a])=0$. Characteristic zero gives $\lambda ([x,a])=0$.

The simultaneous weight space $V_{\lambda }=\{w:aw=\lambda (a)w\forall a\in K\}$ is nonzero and stable under $x$, becauseOver $\mathbb{C}$, the restriction of $x$ to this nonzero finite-dimensional space has an eigenvector. That vector is an eigenvector of both $x$ and every $a\in K$, hence of every element of $L$.

Let $\pi :L\to L/I$ be the quotient map. A full flag of ideals in $L/I$ lifts by inverse images to a flag from $I$ to $L$, with dimensions increasing by one. Prepending $0<I$ supplies the required full flag in $L$. Thus supersolubility lifts across a one-dimensional ideal.

Over $\mathbb{C}$, a nonzero soluble $L$ has a one-dimensional ideal: apply part (a) to its adjoint image acting on $L$. A common eigenvector spans an adjoint-invariant line, which is an ideal. The quotient is soluble and has smaller dimension, so induction and the lifting result give a full flag of ideals. Therefore a soluble complex Lie algebra is supersoluble.

Conversely this implication needs no field restriction. If $L$ has a full flag, its first nonzero ideal $I$ is one-dimensional and abelian, and $L/I$ has a full flag. By induction $L/I$ is soluble, so some term of the derived series of $L$ lies in $I$; its next derived term is zero. Hence

Let $A$ be the two-dimensional affine Lie algebra with basis $u,v$ and bracket $[u,v]=v$, and putBoth factors have zero centre, so $Z(L)=0$. The line $\mathbb{C}v$ in the second factor is an ideal, making $L$ PSS. But its quotient onto the nonabelian simple algebra ${\mathfrak{sl}}_2$ is not soluble. A quotient of a soluble algebra would be soluble, so $L$ is not soluble.

If $L$ is abelian and nonzero, $L^{\prime }=0<L$. Otherwise choose $x\notin Z=Z(L)$ such that $Z+Fx$ is an ideal. For each $a\in L$ writeThe coefficient $\lambda (a)$ is uniquely defined and linear. The Jacobi identity givesso $\lambda ([a,b])=0$. If $\lambda \ne 0$, then $L^{\prime }\subseteq \ker \lambda <L$.

If $\lambda =0$, the same equation says $[[a,b],x]=0$ for all $a,b$. Hence $L^{\prime }\subseteq C_L(x)$. Since $x$ is not central, its centralizer is proper. Thus in both casesNo algebraic closure or characteristic restriction was used.