LemmaStatement: AI-generatedProof: AI-generated

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.

Determinant on the stable general linear group

PDF p. 9: Surjectivity of the determinant on K1

Statement

For a commutative unital ring $A$ , let ${GL}_{n} (A)$ be the group of invertible $n$ by $n$ matrices and $GL (A)$ their union under $M \mapsto diag (M, 1)$ . The ordinary determinant defines a homomorphism $GL (A) \to A^{\times}$ , where $A^{\times}$ is the group of units of $A$ . It factors through the abelianization $GL (A)^{a b}$ and sends the class of the one by one matrix $(u)$ to $u$ .

The determinant of an $n$ by $n$ matrix is the sum over permutations $σ$ of $sign (σ)$ times the product of the entries in positions $(i, σ (i))$ . The abelianization of a group is its quotient by the subgroup generated by commutators.

Facts & Assumptions

Given: A commutative unital ring $A$ .

Goal: Construct the indicated determinant homomorphism and calculate it on one by one matrices.

Proof

technique · Use the permutation formula and the universal property of abelianization

1.1givenalgebra

The permutation formula gives $\det (M N) = \det (M) \det (N)$ : on expanding, terms with repeated intermediate indices cancel in pairs, and the remaining terms are indexed by permutations. Also $\det (I) = 1$ and $\det (diag (M, 1)) = \det (M)$ . If $M$ is invertible, multiplicativity gives $\det (M) \det (M^{- 1}) = 1$ . Thus determinant defines a homomorphism $GL (A) \to A^{\times}$ independent of the chosen matrix size.

2.1step 1.1algebra

The unit group $A^{\times}$ is abelian because $A$ is commutative. Hence every commutator in $GL (A)$ has determinant $1$ , so the homomorphism factors uniquely through $GL (A)^{a b}$ . The determinant of $(u)$ is $u$ by the permutation formula.

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the statement.

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