LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 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 an ideal is an ideal

Statement

Let $R$ be a commutative ring and let $I ⊴ R$ be an ideal. Then $\sqrt{I}$ is an ideal of $R$ containing $I$ . If $J ⊴ R$ is another ideal with $I \subseteq J$ , then $\sqrt{I} \subseteq \sqrt{J}$ . Moreover,

\sqrt{\sqrt{I}} = \sqrt{I} .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , and, for the order-preservation clause, an ideal $J ⊴ R$ with $I \subseteq J$ .

[L1]

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

[L2]

In a commutative ring, $(x + y)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) x^{k} y^{n - k}$ for every natural number $n$ (The binomial theorem over an arbitrary commutative ring).

Proof

technique · direct

1.1

If $a \in I$ , then $a^{1} \in I$ , so $a \in \sqrt{I}$ by [L1]. Thus $I \subseteq \sqrt{I}$ . If $r \in R$ and $x \in \sqrt{I}$ , choose $n \geq 1$ with $x^{n} \in I$ . Then $(r x)^{n} = r^{n} x^{n} \in I$ , so $r x \in \sqrt{I}$ .

L1givenalgebra

1.2

Let $x, y \in \sqrt{I}$ . Choose $m, n \geq 1$ with $x^{m} \in I$ and $y^{n} \in I$ , and set $N = m + n$ . By [L2], every term of $(x + y)^{N}$ has the form $(\binom{N}{k}) x^{k} y^{N - k}$ . For each $k$ , either $k \geq m$ or $N - k \geq n$ ; otherwise $k \leq m - 1$ and $N - k \leq n - 1$ , which would force $N \leq m + n - 2$ . Hence each term lies in $I$ , so $(x + y)^{N} \in I$ and $x + y \in \sqrt{I}$ .

L1L2choosealgebra

2.1

Steps 1.1 and 1.2 show that $\sqrt{I}$ is an ideal containing $I$ . If $I \subseteq J$ and $x \in \sqrt{I}$ , any power of $x$ lying in $I$ also lies in $J$ , so $x \in \sqrt{J}$ . Thus radical is order-preserving.

step 1.1step 1.2L1

3.1

If $x \in \sqrt{\sqrt{I}}$ , choose $m \geq 1$ with $x^{m} \in \sqrt{I}$ , and then choose $n \geq 1$ with $x^{m n} \in I$ . By [L1], this means $x \in \sqrt{I}$ . Together with step 2.1 applied to $I \subseteq \sqrt{I}$ , this proves $\sqrt{\sqrt{I}} = \sqrt{I}$ .

L1step 2.1choosealgebra

4.1

The radical construction therefore sends ideals to radical ideals, contains the original ideal, and is order-preserving and idempotent.

step 2.1step 3.1∎

Depends on

Used by

The radical of a finite intersection Lemma
Vanishing sets detect radicals Lemma
A Noetherian ring has finitely many minimal prime ideals Theorem

Dependency tree · two levels

7 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, §2 Ideals (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §2 Ideals (standard reference, not scraped)