Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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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.

For every square matrix over a commutative ring, det⁡(AT)=det⁡(A)

Statement

For every n≥1, every commutative ring R, and every A∈Mn(R), det⁡(AT)=det⁡(A).

Facts & Assumptions

Given: A square matrix A=(aij) over a commutative ring.

[L3]

Sign is a homomorphism into {1,−1}, so sgn⁡(σ−1)=sgn⁡(σ) (The sign is a homomorphism Sn→{+1,−1}, surjective exactly when n≥2).

Proof

technique · direct
1.1

Substituting [L2] into [L1] gives det⁡(AT)=∑σ∈Snsgn⁡(σ)∏i<nai,σ(i).

L1L2
2.1

Reindex the sum by τ=σ−1 and the product by j=σ(i). Commutativity and [L3] turn the expression into ∑τ∈Snsgn⁡(τ)∏j<naτ(j),j=det⁡(A).

step 1.1L3L4algebra∎

Depends on

Used by

Dependency tree · two levels

22 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