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

Assume the Axiom of Choice.

For a commutative ring $R$ , $Nil (R) = ⋂_{p \in Spec R} p$ , with the empty-intersection convention in force for the zero ring.

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

[L1]

The nilradical of $R$ is $\sqrt{(0)}$ (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).

technique · direct

1.1L1

By [L1], $Nil (R) = \sqrt{(0)}$ .

2.1L2step 1.1

Applying [L2] to the zero ideal gives $\sqrt{(0)} = ⋂_{p \in Spec R} p$ . 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