LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-27

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The radical of a product of ideals

Statement

Let $R$ be a commutative ring and let $I, J ⊴ R$ be ideals. Then $\sqrt{I J} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J} .$

Facts & Assumptions

Given: A commutative ring $R$ and ideals $I, J ⊴ R$ .

[L1]

The radical of a finite intersection is the intersection of the radicals (The radical of a finite intersection).

[L2]

The product ideal $I J$ is generated by finite sums of products $i j$ with $i \in I$ and $j \in J$ (The sum $I + J$ and product $I J$ of two-sided ideals).

Proof

technique · direct

1.1L1L2given

Every generator $i j$ of $I J$ lies in both $I$ and $J$ , so $I J \subseteq I \cap J$ . Therefore $\sqrt{I J} \subseteq \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$ by [L1].

1.2L2choosealgebra

If $x \in \sqrt{I} \cap \sqrt{J}$ , choose $m, n \geq 1$ with $x^{m} \in I$ and $x^{n} \in J$ . Then $x^{m + n} = x^{m} x^{n} \in I J$ , so $x \in \sqrt{I J}$ .

2.1step 1.1step 1.2L1∎

Step 1.1 gives $\sqrt{I J} \subseteq \sqrt{I} \cap \sqrt{J}$ , step 1.2 gives the reverse inclusion, and [L1] identifies that common ideal with $\sqrt{I \cap J}$ .

The radical of a product of ideals

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources