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A primitive ideal determines a central character

Statement

Let g be a finite-dimensional complex Lie algebra and let I be a primitive ideal of U(g), say I=Ann⁡U(g)(M) with M simple. Then Z(U(g)) acts on M by a central character χI ⁣:Z(U(g))→C, and

I∩Z(U(g))=ker⁡χI.

In particular I∩Z(U(g)) is a maximal ideal of the commutative algebra Z(U(g)), and U(g)ker⁡χI⊆I.

Facts & Assumptions

Given: A finite-dimensional complex Lie algebra g, a simple left U(g)-module M, and the primitive ideal I=Ann⁡U(g)(M).

[F1]

U(g) is countable-dimensional for finite-dimensional g, so Dixmier's lemma applies: End⁡U(g)(M)=C, every central element acts on M by a scalar, and these scalars form a unital C-algebra homomorphism χI ⁣:Z(U(g))→C with zm=χI(z)m (Dixmier's lemma: endomorphisms of a simple module over a countable-dimensional algebra, Central character of a Lie algebra module).

[F2]

A unital homomorphism χ ⁣:Z(U(g))→C has image C, so Z(U(g))/ker⁡χ≅C is a field and ker⁡χ is a maximal ideal; maximal means maximal among proper ideals (R/M is a field if and only if M is a maximal ideal, Prime ideals and maximal ideals in a commutative ring).

[F3]

I is a two-sided ideal equal to the kernel of the action; M≠0 and 1 acts as the identity, so I is proper (The annihilator of a module over an enveloping algebra, Primitive ideals of an enveloping algebra, Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct
1.1F1F3given

By [F1] there is a central character χI ⁣:Z(U(g))→C with zm=χI(z)m for all z∈Z(U(g)), m∈M. If z∈I∩Z(U(g)), then z acts on M as 0 and as the scalar χI(z); since M≠0 this forces χI(z)=0, so I∩Z(U(g))⊆ker⁡χI.

1.2F1F3given

Conversely, if z∈ker⁡χI, then zm=χI(z)m=0 for every m∈M, so z∈Ann⁡U(g)(M)=I, hence z∈I∩Z(U(g)). Thus ker⁡χI⊆I∩Z(U(g)).

2.1step 1.1step 1.2F2

Steps 1.1 and 1.2 give I∩Z(U(g))=ker⁡χI, which is a maximal ideal of Z(U(g)) by [F2] because χI is a unital homomorphism onto C.

3.1step 2.1F3algebra∎

Every z∈ker⁡χI lies in the two-sided ideal I, so every product uz with u∈U(g) lies in I; as these products generate the two-sided ideal U(g)ker⁡χI, one has U(g)ker⁡χI⊆I.

Depends on

Used by

Dependency tree · two levels

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Sources