Let $g$ be a Lie algebra, $g'$ a subalgebra, $h$ an ideal of $g$, $t = g/h$ and $\phi$ the canonical mapping of $g$ onto $t$. If $g$ is solvable then $D^k g = \{0\}$ for some integer $k$, hence $D^k g' \subset D^k g = \{0\}$, $D^k t = \phi(D^k g) = \{0\}$ and hence $g'$ and $t$ are solvable. If $h$ and $t$ are solvable there exist integers $s, t$ such that

then $D^t g \subset h$, hence $D^{s+t} g = D^s(D^t g) \subset D^s h = \{0\}$ and $g$ is solvable. The last assertion follows from the second by induction on the number of factors.

(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.

Let $a$ be an ideal of $g$ and $\phi$ the canonical mapping of $g$ onto $g/a$. If the radical of $g/a$ is zero, then $\phi(r)$, which is a solvable ideal of $g/a$, is zero; hence $r \subset a$.

On the other hand, the inverse image $\phi^{-1}(r')$ of the radical $r'$ of $g/r$ is an ideal of $g$ which is solvable by Proposition 1 and hence is equal to $r$; therefore $r' = \{0\}$.

The product $r'$ of the $r_i$ is a solvable ideal (Proposition 1) and hence $r' \subset r$. The canonical image of $r$ in $g_i$ is a solvable ideal of $g_i$ and hence is contained in $r_i$; hence $r \subset r'$.

If b is an ideal of g contained in a such that $\operatorname{Tr} bs = 0$ for all $b \in b$ and all $s \in S$, then in particular, by definition of S, $\operatorname{Tr}(b^n) = 0$ for every integer $n > 0$ and hence b is nilpotent (Algebra, Chapter VII, § 5, no. 5, Corollary 4 to Proposition 13); as the elements of b are all nilpotent, $b = \{0\}$ (§ 4, no. 3, Lemma 2). We first apply this to the ideal $[g, a]$ of g. If $x \in g, a \in a, s \in S$, then $\operatorname{Tr}[x, a]s = \operatorname{Tr}(xas - axs) = \operatorname{Tr} x(as - sa) = 0$ since $as = sa$; hence $[g, a] = \{0\}$. Hence the elements of g commute with those of a and hence also with those of S. If $x, y$ belong to g and $s \in S$, then

since $ys = sy$; then taking b to be the ideal $\mathscr{D}g \cap a$, it follows that $\mathscr{D}g \cap a = \{0\}$.

We already know that $s \subset \mathscr{D}g \cap r$. Hence it will suffice to show that if $\rho$ is a finite-dimensional simple representation of g then $\rho(\mathscr{D}g \cap r) = \{0\}$. Let k be the least integer $\geq 0$ such that $\rho(\mathscr{D}^{k+1}r) = \{0\}$; we write $g' = \rho(g)$, $a' = \rho(D^k r)$; as $\mathscr{D}^k r$ is an ideal of g, $a'$ is an ideal of $g'$; this ideal is commutative since $\rho(\mathscr{D}^{k+1}r) = \{0\}$. If V is the space of $\rho$, $g' \subset \operatorname{gl}(V)$ and V is a simple $g'$-module. Then $\rho(\mathscr{D}g \cap \mathscr{D}^k r) \subset \mathscr{D}g' \cap a' = \{0\}$. If $k > 0$, then $\mathscr{D}^k r \subset \mathscr{D}g$ and $\rho(\mathscr{D}^k r) = \{0\}$, contrary to the definition of k. Hence $k = 0$, that is

Here $r = g$, whence $s = \mathscr{D}g$. Hence $\varphi(\mathscr{D}g) = \{0\}$, which shows that $g' = \rho(g)$ is commutative. Every element $\neq 0$ of $L$ is invertible by Schur’s Lemma; hence $L$ is a field.

Corollary 2 (Lie’s Theorem). Let $g$ be a solvable Lie algebra; suppose that $K$ is algebraically closed. Let $M$ be a $g$-module of finite dimension over $K$ and let $(M_i)_{0 \leq i \leq r}$ be a Jordan-Hölder series of $M$. Then $M_{i-1}/M_i$ is of dimension 1 over $K$ for $1 \leq i \leq r$ and, for all $x \in g$, $x_{M_{i-1}/M_i} = \lambda_i(x) \cdot 1$, where $\lambda_i$ is a linear form on $g$ which is zero on $\mathscr{D}g$. In particular, every simple $g$-module of finite dimension over $K$ is in fact of dimension 1.

Let $\rho_i$ be the representation of $g$ on $M_{i-1}/M_i$. The associative algebra $L_i$ generated by 1 and $\rho_i(g)$ is a field, a finite extension of $K$ and therefore equal to $K$; and $M_{i-1}/M_i$ is a simple $L_i$-module, whence $\dim M_{i-1}/M_i = 1$. The rest of the corollary is obvious.

Every ideal is part of a Jordan-Hölder series of $g$, considered as the space of the adjoint representation (Algebra, Chapter I, § 6, no. 14, Corollary to Theorem 8); then it suffices to apply Corollary 2.

The condition is necessary by Corollary 1. It is sufficient since $g/\mathscr{D}g$ is commutative.

Let V be the space of $\rho$; let $(V_i)_{0 \leq i \leq r}$ be a Jordan-Hölder series for the r-module structure on V and let $\rho_i$ be the representation of r with space $V_i / V_{i-1}$ ($1 \leq i \leq r$). If $\rho(x)$ is nilpotent, so is $\rho_i(x)$; as for all $i$ the algebra generated by $\rho_i(x)$ is a field, $\rho_i(x) = 0$. Conversely, if $\rho_i(x) = 0$ for all $i$, $\rho(x) = 0$. This shows that the set $a$ of $x \in r$ such that $\rho(x)$ is nilpotent is an ideal of r. On the other hand, $[g, a] \subset \mathscr{D}g \cap r \subset n \cap r \subset a$ and hence $a$ is an ideal of g. This proves that $a \subset n$.

Let these sets be denoted by $a, b, c, d$. The inclusions $a \subset b \subset d \subset c$ are clear. $c \subset a$ by Corollary 6 applied to the adjoint representation of g.

We know that $\mathrm{ad}\, x = \mathrm{ad}\, s + \mathrm{ad}\, n, [\mathrm{ad}\, s, \mathrm{ad}\, n] = 0$ and $\mathrm{ad}\, n$ is nilpotent ($§ 4$, Lemma 1). We show that $\mathrm{ad}\, s$ is semi-simple. It suffices to do this when K is algebraically closed (cf. Algebra, Chapter VIII, § 9, no. 2, Proposition 3). Then let $(e_i)_{1 \leq i \leq n}$ be a basis of V such that $s(e_i) = \lambda_i e_i$ ($\lambda_i \in K$). Let $(E_{ij})$ be the canonical basis of $M_n(K) = \mathfrak{gl}(V)$. By formulae (5) of $§ 1$,

and hence $\mathrm{ad}\, s$ is semi-simple. The last assertion of the lemma follows from Algebra, Chapter VIII, § 9, no. 4, Proposition 8.

It suffices to prove this when K is algebraically closed, which we shall assume henceforth. Let s and n be the semi-simple and nilpotent components of z and let $(e_i)$ be a basis of M such that $s(e_i) = \lambda_i e_i$ ($\lambda_i \in K$). Let $V \subset K$ be the vector space over $\mathbf{Q}$ generated by the $\lambda_i$. We need to prove that $V = \{0\}$. Let f be a $\mathbf{Q}$-linear form on V and let t be the endomorphism of M such that

$te_i = f(\lambda_i)e_i$. If $(E_{ij})$ is the canonical basis of $\mathfrak{gl}(M)$ defined by $E_{ij}e_k = \delta_{jk}e_i$, then

There exists a polynomial P with no constant term and with coefficients in K such that $P(\lambda_i - \lambda_j) = f(\lambda_i) - f(\lambda_j)$ for all i and j (for if $\lambda_i - \lambda_j = \lambda_h - \lambda_k$, then $f(\lambda_i) - f(\lambda_j) = f(\lambda_h) - f(\lambda_k)$ and, if $\lambda_i - \lambda_j = 0$, $f(\lambda_i) - f(\lambda_j) = 0$). Then $\mathrm{ad}\ t = P(\mathrm{ad}\ s)$. On the other hand, $\mathrm{ad}\ s$ is a polynomial with no constant term in $\mathrm{ad}\ z$. Now $(\mathrm{ad}\ z)(A) \subset B$, whence also $(\mathrm{ad}\ t)(A) \subset B$. By the hypothesis $0 = \mathrm{Tr}(zt) = \sum \lambda_i f(\lambda_i)$, whence $0 = f(\mathrm{Tr}(zt)) = \sum f(\lambda_i)^2$. Since the $f(\lambda_i)$ are rational numbers, $f = 0$, which completes the proof.

It can obviously be reduced to the case where g is a Lie subalgebra of $\mathfrak{gl}(M)$ and ρ is the identity mapping. If g is solvable, $\mathscr{D}$g is contained in the largest nilpotency ideal of the identity representation of g (Theorem 1) and hence is orthogonal to g with respect to β (§ 4, Proposition 4 (d)). Suppose that $\mathscr{D}$g is orthogonal to g with respect to β. We prove that g is solvable. Let T be the set of $t \in \mathfrak{gl}(M)$ such that $[t, g] \subset \mathscr{D}g$. If $t \in T$ and x, y belong to g, then $[t, x] \in \mathscr{D}g$ and hence

This is an immediate consequence of Proposition 5 (b).

$\mathscr{D}g$ is a characteristic ideal and the Killing form is completely invariant (§ 3, no. 6, Proposition 10). Hence the orthogonal of $\mathscr{D}g$ with respect to the Killing form is a characteristic ideal (§ 3, no. 6, Proposition 7 (b)).

r ∩ a is a solvable ideal of a and hence is contained in the radical r' of a. Conversely, r' is an ideal of g (Corollary 2 and § 1, no. 4, Proposition 2) and hence r' ⊂ r.

Corollary 2 can be made more precise as follows:

Let D be a derivation of g. Let g' = g + Kx_0 be a Lie algebra in which g is an ideal of codimension 1 such that Dx = [x_0, x] for all x ∈ g (§ 1, no. 8, Example 1). By Corollary 3 to Proposition 5, r is contained in the radical r' of g'. Then D(r) = [x_0, r] ⊂ [g', g'] ∩ r' = s'. For all x ∈ s', ad_{g'} x is nilpotent (Theorem 1). Hence, for all x ∈ s' ∩ g, ad_g x is nilpotent. Hence D(r) is contained in the nilpotent ideal s' ∩ g of g.