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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Radicals commute with localization

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be a multiplicative subset, and let $I ⊴ R$ be an ideal. Then

S^{- 1} ⁣ \sqrt{I} = \sqrt{S^{- 1} I}

as ideals of $S^{- 1} R$ .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and an ideal $I ⊴ R$ .

[L1]

An element lies in the radical of an ideal exactly when some positive power lies in that ideal (The radical of an ideal).

[L2]

In $S^{- 1} R$ , one has $r / s = r^{'} / s^{'}$ exactly when $u (r s^{'} - r^{'} s) = 0$ for some $u \in S$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L3]

The extended ideal is $S^{- 1} I = {a / u : a \in I, u \in S}$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ).

Proof

technique · direct

1.1

If $a / s \in S^{- 1} ⁣ \sqrt{I}$ , choose $n \geq 1$ with $a^{n} \in I$ . Then $(a / s)^{n} = a^{n} / s^{n} \in S^{- 1} I$ , so $a / s \in \sqrt{S^{- 1} I}$ by [L1]. This proves $S^{- 1} ⁣ \sqrt{I} \subseteq \sqrt{S^{- 1} I}$ .

L1L3givenalgebra

1.2

Conversely, let $r / s \in \sqrt{S^{- 1} I}$ . Choose $n \geq 1$ with $r^{n} / s^{n} \in S^{- 1} I$ , and then choose $a \in I$ and $u \in S$ with $r^{n} / s^{n} = a / u$ . By [L2], some $t \in S$ satisfies $t (u r^{n} - a s^{n}) = 0$ . Hence $(t u) r^{n} = t a s^{n} \in I$ , so $((t u) r)^{n} = (t u)^{n - 1} ((t u) r^{n}) \in I$ . Thus $(t u) r \in \sqrt{I}$ by [L1], and $r / s = ((t u) r) / ((t u) s)$ lies in $S^{- 1} ⁣ \sqrt{I}$ .

L1L2L3choosealgebra

2.1

Steps 1.1 and 1.2 prove the equality $S^{- 1} ⁣ \sqrt{I} = \sqrt{S^{- 1} I}$ .

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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)
The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)