PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Primitive ideals are prime in the noncommutative sense

Statement

Let $I$ be a primitive ideal of $U (g)$ and let $A, B$ be two-sided ideals of $U (g)$ with $A B \subseteq I$ . Then $A \subseteq I$ or $B \subseteq I$ . The claim holds for every complex Lie algebra $g$ , and equivalently the quotient ring $U (g) / I$ is prime.

Facts & Assumptions

Given: A complex Lie algebra $g$ , a primitive ideal $I$ , a simple left $U (g)$ -module $M$ with $I = {Ann}_{U (g)} (M)$ , and two-sided ideals $A, B ⊴ U (g)$ with $A B \subseteq I$ .

[F1]

$I$ is the annihilator of the simple module $M$ ; the annihilator of a module is a two-sided ideal (Primitive ideals of an enveloping algebra, The annihilator of a module over an enveloping algebra).

[F2]

The product $A B$ consists of finite sums $\sum_{k} a_{k} b_{k}$ and is a two-sided ideal (The sum $I + J$ and product $I J$ of two-sided ideals, The sum and product of two-sided ideals are two-sided ideals). For an ideal $C$ and submodule $N$ , write $C N$ for the finite sums $\sum_{k} c_{k} n_{k}$ ; these form a submodule since $u (c_{k} n_{k}) = (u c_{k}) n_{k}$ and $u c_{k} \in C$ (Left, right and two-sided ideals). Distributing finite sums and using associativity gives $A (B M) = (A B) M$ .

[F3]

$M$ is nonzero and its only submodules are $0$ and $M$ ; in particular a submodule $N \subseteq M$ with $N \neq 0$ equals $M$ (Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct

1.1F1F2F3given

Suppose $B \subseteq̸ I$ , i.e. $B \subseteq̸ {Ann}_{U (g)} (M)$ . Then some $b \in B$ has $b M \neq 0$ , and $B M$ is a nonzero submodule of $M$ : for $u \in U (g)$ , one has $u \sum_{k} b_{k} m_{k} = \sum_{k} (u b_{k}) m_{k} \in B M$ because $u b_{k} \in B$ . By [F3], $B M = M$ .

2.1step 1.1F1F2algebra

Using the associativity of the action and the containment $A B \subseteq I$ : $A M = A (B M) = (A B) M \subseteq I M = 0$ , so every $a \in A$ annihilates $M$ and therefore $A \subseteq {Ann}_{U (g)} (M) = I$ .

3.1step 1.1step 2.1F2algebra∎

The argument shows that $B \subseteq̸ I$ forces $A \subseteq I$ ; contrapositively, if $A \subseteq̸ I$ then $B \subseteq I$ . Hence $A \subseteq I$ or $B \subseteq I$ , and no finite-dimensionality of $g$ was used. Passing to $U (g) / I$ , two-sided ideals of the quotient correspond to two-sided ideals of $U (g)$ containing $I$ , and the product condition becomes $\overset{ˉ}{A} \overset{ˉ}{B} = 0$ ; the displayed alternative is exactly the primeness of $U (g) / I$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. Etingof, Representations of Lie Groups (18.757, MIT OCW 2023 full notes) (standard reference, not scraped)
D. Barbasch, Cells in Weyl groups and primitive ideals (AIM workshop notes, 2006) (standard reference, not scraped)