LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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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

1.1

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 $\sqrt{I_{1} \cap \dots \cap I_{n}} \subseteq \sqrt{I_{1}} \cap \dots \cap \sqrt{I_{n}}$ .

L1given

1.2

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 $x \in \sqrt{I_{1} \cap \dots \cap I_{n}}$ .

choosegivenalgebra

2.1

The two inclusions prove the stated equality.

step 1.1step 1.2∎

Depends on

The radical of an ideal is an ideal

Used by

A radical of a product and intersection computation Example
The radical of a product of ideals Lemma

Dependency tree · two levels

3 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

M. Hochster, Introduction to Commutative Algebra, Math 614 notes (2020) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §2 Ideals (standard reference, not scraped)