LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Finite-dimensional commutative algebras decompose into local factors

Statement

Every finite-dimensional commutative unital $k$ -algebra $A$ is a finite product of local $k$ -algebras. For every ideal $I$ , writing $A = \prod A_{i}$ gives $I = \prod I_{i}$ and $A / I ≅ \prod A_{i} / I_{i}$ , and every nonzero factor quotient is local. The zero algebra is the empty product.

Facts & Assumptions

Given: A finite-dimensional commutative unital algebra over any field.

Proof

technique · direct

1.1

A nontrivial idempotent $e$ gives $A ≅ e A \times (1 - e) A$ by $a \mapsto (e a, (1 - e) a)$ , with inverse addition. Both factors have strictly smaller positive dimension. Repeated splitting therefore terminates with finitely many nonzero factors having no nontrivial idempotents. Zero requires no factors.

given

2.1

In such a factor $B$ , let $m_{a}$ be multiplication by $a$ . Choose $n$ with both kernels and images stabilized through $2 n$ , possible by finite dimension. If $v = a^{n} w$ is also in $\ker m_{a}^{n}$ , then $a^{2 n} w = 0$ implies $a^{n} w = 0$ , so $v = 0$ . Rank-nullity gives $B = \ker m_{a}^{n} \oplus im m_{a}^{n}$ . Both spaces are ideals, whose cross-products lie in their zero intersection. Decomposing $1 = u + v$ in them gives $u^{2} = u, v^{2} = v, u v = 0$ . Indecomposability forces one summand to be zero. A zero kernel makes $m_{a}$ bijective and $a$ a unit; a zero image gives $a^{n} = 0$ . Thus every nonunit is nilpotent.

step 1.1

3.1

In a commutative ring the nilpotents form an ideal: if $a^{r} = b^{s} = 0$ , every term of $(a + b)^{r + s}$ vanishes, and $(c a)^{r} = 0$ . This ideal $m$ is proper since $1$ is not nilpotent, and every element outside it is a unit by step 2.1. Consequently $B / m$ is a field, and every proper ideal lies in $m$ ; this proves uniqueness of the maximal ideal without an existence theorem.

step 2.1

4.1

Multiplication by the coordinate idempotents shows that any ideal $I$ contains every coordinate projection of each of its elements. Finite addition then gives $I = \prod I_{i}$ and the displayed quotient isomorphism. If $I_{i} \neq A_{i}$ , it lies in the unique maximal ideal $m_{i}$ ; the quotient has proper ideal $m_{i} / I_{i}$ and every class outside it has a unit representative. This ideal is its unique maximal ideal. If $I_{i} = A_{i}$ , that quotient is zero and may be omitted.

step 1.1step 3.1∎

Sources

Jacobsen, Block fusion systems and the center of the group ring, Lemma 2.32 and Theorem 2.33, pp.18–19; general-field lifting proved locally. Local argument and conventions as displayed above.

Used by

Idempotents lift through finite commutative algebra quotients Lemma

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Jacobsen, Block fusion systems and the center of the group ring, Lemma 2.32 and Theorem 2.33, pp.18–19; general-field lifting proved locally (standard reference, not scraped)