CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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.

$M_{n} (F)$ is noncommutative for every $n \geq 2$

Statement

For every field $F$ and every natural $n \geq 2$ , the ring $M_{n} (F)$ is not commutative.

Facts & Assumptions

Given: A field $F$ and a natural $n \geq 2$ .

[L1]

Matrix units satisfy $E_{i j} E_{k ℓ} = δ_{j k} E_{i ℓ}$ ( $E_{i j} E_{k ℓ} = δ_{j k} E_{i ℓ}$ ).

Proof

technique · direct

1.1

By [L1], $E_{01} E_{10} = E_{00}$ while $E_{10} E_{01} = E_{11}$ ; these products differ at entry $(0, 0)$ because $1 \neq 0$ in a field.

givenL1

2.1

Thus two elements of $M_{n} (F)$ fail to commute, so the matrix ring is noncommutative.

step 1.1L1∎

Depends on

Used by

Two explicit 2 by 2 matrices do not commute Example
False statement: every semisimple ring is commutative False statement

Dependency tree · two levels

7 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

S. Axler, Linear Algebra Done Right, 4th ed., §3C (standard reference, not scraped)

Mn(F)Mn​(F) is noncommutative for every n≥2n≥2