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CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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The nilradical is the intersection of all prime ideals

Statement

Assume the Axiom of Choice.

For a commutative ring R, Nil⁡(R)=⋂p∈Spec⁡Rp, with the empty-intersection convention in force for the zero ring.

Facts & Assumptions

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

[L1]

The nilradical of R is (0) (The nilradical and reduced rings).

[L2]

The radical of any ideal is the intersection of the prime ideals containing it (The radical of an ideal is the intersection of the prime ideals containing it).

Proof

technique · direct
1.1L1

By [L1], Nil⁡(R)=(0).

2.1L2step 1.1

Applying [L2] to the zero ideal gives (0)=⋂p∈Spec⁡Rp. Combining this with step 1.1 yields the claimed formula for the nilradical.

3.1step 2.1∎

Therefore the nilradical is exactly the intersection of all prime ideals of R.

Depends on

Used by

Dependency tree · two levels

5 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