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Unformatted text preview: a ∈ L then ϕ[ax] = [ϕ(a)ϕ(x)] = [ϕ(a)0] = 0. So, [ax] ∈ ker ϕ.
Conversely, if I ⊂ L is an ideal, a ∈ L, x ∈ I then
[(a + I )(x + I )] ⊆ [ax] + [aI ] + [Ix] + [II ] ⊆ [ax] + I So, the bracket is welldeﬁned in L/I , [ϕ(a)ϕ(x)] = ϕ[ax] and I = ker ϕ. Proposition 2.1.5. Any epimorphism (surjective homomorphism) of Lie algebras ϕ :
L → L gives a 11 correspondence between ideals I in L and ideals I of L containing
ker ϕ.
Proof. The correspondence is given by I = ϕ(I ). This is an ideal since [L , I ] =
[ϕ(L), ϕ(I )] = ϕ[LI ] ⊆ ϕ(I ) = I and I = ϕ−1 (I ) which is an ideal since it is the
kernel of L → L → L /I .
Example 2.1.6. The center of a Lie algebra L is deﬁned to be
Z (L) = {x ∈ L  [xL] = 0}. Then Z (L) is an ideal in L. Z (L) also the kernel of the adjoint representation ad : L →
gl(L). If I ⊆ L is an ideal, then the adjoint representation restricts to a representation
on I : ad : L ...
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.
 Fall '11
 K.Igusa
 Algebra

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