Corollary

Tag 0A01, permanent. Cite it as [Bourbaki, Tag 0A01].

Lie Groups and Lie Algebras, chapter I, § 6. Semi-simple Lie algebras

Let $g, g'$ be Lie algebras, $s$ (resp. $s'$ ) the nilpotent radical of $g$ (resp. $g'$ ) and $f$ a homomorphism of $g$ onto $g'$ . (a) Then $s' = f(s)$ . (b) $g'$ is reductive if and only if the kernel of $f$ contains $s$ .

If $r, r'$ are the radicals of $g, g'$ , then $s' = [g', r'] = [f(g), f(r)] = f([g, r]) = f(s)$ . Assertion (b) is an immediate consequence of (a).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Theorem 4, § 6

Cites

Nothing.