LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 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.

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The radical of a finite intersection

Statement

Let $R$ be a commutative ring, and let $I_{1}, \dots, I_{n} ⊴ R$ with $n \geq 1$ . Then $\sqrt{I_{1} \cap \dots \cap I_{n}} = \sqrt{I_{1}} \cap \dots \cap \sqrt{I_{n}} .$

Facts & Assumptions

Given: A commutative ring $R$ , ideals $I_{1}, \dots, I_{n} ⊴ R$ , and an integer $n \geq 1$ .

[L1]

Radical is order-preserving on ideals (The radical of an ideal is an ideal).

Proof

technique · direct

1.1L1given

Since $I_{1} \cap \dots \cap I_{n} \subseteq I_{j}$ for every $j$ , [L1] gives $\sqrt{I_{1} \cap \dots \cap I_{n}} \subseteq \sqrt{I_{j}}$ for every $j$ , hence $\sqrt{I_{1} \cap \dots \cap I_{n}} \subseteq \sqrt{I_{1}} \cap \dots \cap \sqrt{I_{n}}$ .

1.2choosegivenalgebra

Conversely, let $x \in \sqrt{I_{1}} \cap \dots \cap \sqrt{I_{n}}$ . For each $j$ , choose $m_{j} \geq 1$ with $x^{m_{j}} \in I_{j}$ , and set $N = m_{1} + \dots + m_{n}$ . Then $x^{N} = x^{N - m_{j}} x^{m_{j}} \in I_{j}$ for every $j$ , so $x^{N} \in I_{1} \cap \dots \cap I_{n}$ . Hence $x \in \sqrt{I_{1} \cap \dots \cap I_{n}}$ .

2.1step 1.1step 1.2∎

The two inclusions prove the stated equality.

Depends on

The radical of an ideal is an ideal

The radical of a finite intersection

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources