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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

1.1

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].

L1L2given

1.2

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 $x \in \sqrt{I J}$ .

L2choosealgebra

2.1

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 $\sqrt{I \cap J}$ .

step 1.1step 1.2L1∎

Depends on

Used by

A radical ideal in a Noetherian ring is a finite intersection of minimal primes Corollary

Dependency tree · two levels

4 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)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §2 Ideals (standard reference, not scraped)