Proposition 2

Let $g$ be a Lie algebra. The following conditions are equivalent:

(a) $g$ is solvable; (b) there exists a decreasing sequence $g = g_0 \supset g_1 \supset \cdots \supset g_n = \{0\}$ of ideals of $g$ such that the algebras $g_i/g_{i+1}$ are commutative ($i = 0, 1, \ldots, n-1$); (c) there exists a decreasing sequence $g = g'_0 \supset g'_1 \supset \cdots \supset g'_p = \{0\}$ of subalgebras of $g$ such that $g'_{i+1}$ is an ideal of $g'_i$ and $g'_i/g'_{i+1}$ is commutative ($i = 0, 1, \ldots, p-1$); (d) there exists a decreasing sequence $g = g''_0 \supset g''_1 \supset \cdots \supset g''_q = \{0\}$ of subalgebras of $g$ such that $g''_{i+1}$ is an ideal of $g''_i$ of codimension 1 ($i = 0, 1, \ldots, q-1$).

(a) $\Rightarrow$ (b): it suffices to consider the sequence of derived ideals of $g$. (b) $\Rightarrow$ (c): this is obvious. (c) $\Rightarrow$ (d): suppose that condition (c) holds; every vector subspace of $g'_i$ containing $g'_{i+1}$ is an ideal of $g'_i$, whence immediately (d). (d) $\Rightarrow$ (a): this follows immediately from the fact that an extension of a solvable algebra by a solvable algebra is solvable.

† The reader will note that the hypothesis on the characteristic of $K$ is not used in nos. 1 and 2 of this paragraph.

Examples of solvable Lie algebras

I. Let $g$ be a 2-dimensional vector space over $K$ and $(e_1, e_2)$ a basis of $g$. There exists one and only one alternating bilinear multiplication $(x, y) \mapsto [x, y]$ on $g$ such that $[e_1, e_2] = e_2$. It is easily verified that $g$ is thus given a solvable Lie algebra structure. Now let $h$ be a non-commutative Lie algebra of dimension 2 over $K$. We show that $h$ is isomorphic to $g$. Let $(f_1, f_2)$ be a basis of $h$. The element $[f_1, f_2]$ is non-zero (otherwise $h$ would be commutative) and hence it generates a 1-dimensional subspace $t$ of $h$. Then $[h, h] = t$. Let $(e'_1, e'_2)$ be a basis of $h$ such that $e'_2 \in t$. Then $[e'_1, e'_2] = \lambda e'_2$ with $\lambda \neq 0$. Replacing $e'_1$ by $\lambda^{-1} e_1$, it can be assumed that $\lambda = 1$, whence our assertion.

II. Formulae (5) of § 1 prove that $\mathscr{D}t(n, K) = n(n, K)$. As $n(n, K)$ is nilpotent and hence solvable, $t(n, K)$ is solvable. Therefore $st(n, K)$ is solvable. In particular, $st(2, K)$ is isomorphic to the algebra of Example I.