CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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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

1.1L2

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

2.1L1step 1.1

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.

3.1step 2.1∎

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

Depends on

Used by

Locally factorial scheme Definition
Weil divisor normal noetherian scheme Definition
Dense relative-dimension strata in flat finitely presented fibres Lemma

Dependency tree · two levels

9 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Corollary 14.9 (standard reference, not scraped)
The Stacks Project, Section 10.26: Irreducible components of spectra (standard reference, not scraped)