PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

Elements congruent to $1$ modulo a defining ideal are units

Statement

Let $R$ be a commutative ring and let $I \subseteq R$ be an ideal. Assume that $R$ is $I$ -adically complete. If $u \in R$ satisfies

u \equiv 1 (m o d I),

then $u$ is a unit of $R$ .

Consequently, every element of $I$ lies in the Jacobson radical of $R$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I \subseteq R$ , and an $I$ -adically complete ring element $u = 1 - a$ with $a \in I$ .

[L1]

The completion map $R \to \underset{\leftarrow}{l i m} R / I^{n}$ is an isomorphism because $R$ is $I$ -adically complete (Separated and complete filtered modules, The $I$ -adic completion of a module).

Proof

technique · direct

1.1

For each $n \geq 1$ , the finite geometric sum $v_{n} : = 1 + a + a^{2} + \dots + a^{n - 1}$ satisfies $(1 - a) v_{n} = 1 - a^{n} .$ Since $a \in I$ , one has $a^{n} \in I^{n}$ , so the image of $v_{n}$ in $R / I^{n}$ is an inverse to the image of $u = 1 - a$ .

givenalgebra

2.1

The residue classes $(v_{n} m o d I^{n})_{n}$ are compatible: the image of $v_{n + 1}$ in $R / I^{n}$ equals the image of $v_{n}$ because $v_{n + 1} - v_{n} = a^{n} \in I^{n}$ . Therefore they define an element $v \in \underset{\leftarrow}{l i m} R / I^{n} .$

step 1.1construct

3.1

By [L1], there is a unique element $w \in R$ corresponding to $v$ . Since each component of $v$ is an inverse to the image of $u$ , the products $u w$ and $w u$ map to $1$ in every quotient $R / I^{n}$ . Completeness includes separatedness, so the kernel of $R \to \underset{\leftarrow}{l i m} R / I^{n}$ is $0$ ; hence $u w = w u = 1.$ Thus $u$ is a unit.

L1step 2.1algebra

4.1

Let $x \in I$ and $r \in R$ . Then $1 - r x \equiv 1 (m o d I)$ , so step 3.1 shows $1 - r x$ is a unit. The elementary ideal characterization of the Jacobson radical now gives $x \in J (R)$ : if a maximal ideal omitted $x$ , its image would generate the residue field, contradicting invertibility of every $1 - r x$ . Therefore $I \subseteq J (R)$ .

step 3.1algebra∎

Depends on

Used by

Completion of a Noetherian local ring is local with the same residue field Theorem

Dependency tree · two levels

5 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §22.16 (standard reference, not scraped)
The Stacks Project, Lemma 10.96.6 (standard reference, not scraped)

Elements congruent to 11 modulo a defining ideal are units