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A Noetherian ring has only finitely many irreducible components in its spectrum
Statement
Assume the Axiom of Choice.
If is a Noetherian commutative ring, then has only finitely many irreducible components.
Facts & Assumptions
Given: A Noetherian commutative ring and the Axiom of Choice.
Irreducible components of correspond exactly to minimal prime ideals (Irreducible components of the spectrum correspond to minimal prime ideals).
A Noetherian ring has only finitely many minimal prime ideals (A Noetherian ring has finitely many minimal prime ideals).
Proof
By [L2], the ring has only finitely many minimal prime ideals.
By [L1], each irreducible component is for a unique minimal prime , and each minimal prime gives an irreducible component. Therefore the set of irreducible components is finite.
Hence has only finitely many irreducible components.