LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Primes containing an ideal contain its radical

Statement

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $p$ be a prime ideal with $I \subseteq p$ . Then $\sqrt{I} \subseteq p$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , and a prime ideal $p$ containing $I$ .

[L1]

An element lies in $\sqrt{I}$ exactly when some positive power of it lies in $I$ (The radical of an ideal).

[L2]

A prime ideal is proper and contains a factor whenever it contains a product (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1L1givenchoose

Let $x \in \sqrt{I}$ . Choose $n \geq 1$ with $x^{n} \in I$ . Since $I \subseteq p$ , one has $x^{n} \in p$ .

2.1L2step 1.1algebra

Repeatedly applying primality from [L2] to the factorization $x^{n} = x \cdot x^{n - 1}$ shows that $x \in p$ : if $x \notin p$ , then $x^{n - 1} \in p$ ; repeating the same argument eventually forces $x \in p$ after all.

3.1step 2.1∎

Every element of $\sqrt{I}$ lies in $p$ , so $\sqrt{I} \subseteq p$ .

Primes containing an ideal contain its radical

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources