Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-26
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.

An algebra that is finite dimensional as a vector space over a field is a Noetherian ring

Example

Let k be a field and let A be a commutative k-algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) whose underlying k-vector space is finite dimensional, say dim⁡kA=n with n∈N (Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis). Then A is a Noetherian ring, and every ideal of A is generated by at most n elements.

Facts & Assumptions

Given: A field k, which is in particular a commutative ring (Every field is a commutative ring with 1≠0; it is an integral domain, and it is a commutative division ring), and a commutative k-algebra A with structure map ηA ⁣:k→A whose underlying k-vector space is finite dimensional of dimension n.

[L1]

An R-algebra is a unital ring A with a unital ring homomorphism ηA ⁣:R→A of central image; the induced scalar action ra:=ηA(r)a makes A an R-module (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[L2]

A vector space over a field F is a set V with an addition making (V,+,0V) an abelian group and a scalar multiplication F×V→V satisfying λ(u+v)=λu+λv, (λ+μ)v=λv+μv, (λμ)v=λ(μv) and 1Fv=v (Vector space over a field).

[L3]

A linear subspace of a vector space V over F is a subset containing 0V and closed under addition and under scalar multiplication, and it is itself a vector space over F under the restricted operations (Linear subspace of a vector space).

[L4]

V is finite-dimensional over F when it has a finite basis, and dim⁡FV is the unique n∈N with a basis B satisfying B≈n (Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis).

[L5]

If V is finite dimensional over F with dim⁡FV=n and U is a linear subspace of V, then U is finite dimensional over F and dim⁡FU≤n (If dim⁡FV=n and U is a linear subspace of V, then U is finite-dimensional, dim⁡FU≤n, and dim⁡FU=n if and only if U=V).

[L7]

The span of a subset is the set of its finite linear combinations, and a subset spans V when its span is V (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S).

[L8]

In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra (In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra).

[L10]

An algebra is module-finite over R when it is finitely generated as an R-module (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L11]

A module-finite commutative algebra over a Noetherian commutative ring is a Noetherian ring (A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two).

Verification

technique · direct
1.1L1L2given

The algebra action λ⋅a=ηA(λ)a makes A a k-vector space: (A,+,0A) is an abelian group, and the four displayed scalar identities are exactly the module axioms that the algebra action satisfies. This is the vector-space structure the hypothesis dim⁡kA=n refers to.

2.1L1L3step 1.1

Every ideal I of A is a linear subspace of that vector space: it contains 0A, is closed under addition, and is closed under the scalar action because λ⋅a=ηA(λ)a is a product of an element of A with an element of I.

3.1L4L5L6L7step 2.1

By the subspace theorem I is finite dimensional over k with dim⁡kI≤n; fix a basis b1,…,br of I with r≤n. Every element of I is then a finite k-linear combination ∑i=1rλibi with λi∈k.

4.1L8L9step 3.1

Hence I=(b1,…,br) as an ideal of A: each element ∑iλibi of I equals ∑iηA(λi)bi, which lies in the ideal generated by b1,…,br, and conversely that ideal is contained in I because every bi lies in I. So every ideal of A is generated by at most n elements, and A is Noetherian.

5.1L10L11L12step 1.1step 4.1∎

The same conclusion follows from the module-finite theorem, and the two agree. A finite basis of A generates A as a k-module, so A is module-finite over k; k is a Noetherian ring; and a module-finite commutative algebra over a Noetherian ring is Noetherian. The direct argument above is recorded because it also produces the bound r≤n on the number of generators, which the general theorem does not.

Remarks

  • Finite dimension over k is much stronger than finite type over k. A polynomial ring k[x] is of finite type over k and is Noetherian, but is not finite dimensional as a k-vector space; the bound on the number of generators of an ideal disappears there, as the companion false-statement item on this page records for k[x,y].

  • Commutativity of A is assumed only because this page works with commutative rings. The same argument applies verbatim to a left ideal of a finite-dimensional algebra that is not commutative.

Depends on

Used by

Dependency tree · two levels

71 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