CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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.

A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit

Let $R$ be a commutative ring, $n \geq 1$ , and $A \in M_{n} (R)$ . Then $A$ is invertible if and only if $\det (A)$ is a unit of $R$ .

Given: $R, n, A$ as in the statement.

[L1]

If a positive-sized square matrix is invertible, its determinant is a unit (An invertible square matrix over a commutative ring has unit determinant).

[L2]

[F1]

A square matrix is invertible when it has a two-sided multiplicative inverse (Invertible square matrices and similarity over a commutative ring).

[L3]

technique · direct

1.1L1

If $A$ is invertible, [L1] says directly that $\det (A)$ is a unit.

1.2L3

Conversely, suppose $u : = \det (A)$ is a unit, and let $u^{- 1}$ be its inverse.

2.1step 1.2L2algebra

Multiplying both identities in [L2] by the scalar $u^{- 1}$ gives $A (u^{- 1} adj (A)) = (u^{- 1} adj (A)) A = I_{n}$ .

3.1step 2.1F1step 1.1∎

Thus $u^{- 1} adj (A)$ is a two-sided inverse, so $A$ is invertible. Together with step 1.1, this proves both directions.

Depends on