Alphabeta Math
CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • 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.
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A Noetherian ring has only finitely many irreducible components in its spectrum

Statement

Assume the Axiom of Choice.

If R is a Noetherian commutative ring, then Spec⁡(R) has only finitely many irreducible components.

Facts & Assumptions

Given: A Noetherian commutative ring R and the Axiom of Choice.

[L1]

Irreducible components of Spec⁡(R) correspond exactly to minimal prime ideals (Irreducible components of the spectrum correspond to minimal prime ideals).

[L2]

A Noetherian ring has only finitely many minimal prime ideals (A Noetherian ring has finitely many minimal prime ideals).

Proof

technique · direct
1.1

By [L2], the ring R has only finitely many minimal prime ideals.

L2
2.1

By [L1], each irreducible component is V(p) for a unique minimal prime p, and each minimal prime gives an irreducible component. Therefore the set of irreducible components is finite.

L1step 1.1
3.1

Hence Spec⁡(R) has only finitely many irreducible components.

step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources