Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 dimkA=n with nN (Finite-dimensional vector space, and its dimension dimFV; 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 10; it is an integral domain, and it is a commutative division ring), and a commutative k-algebra A with structure map ηA ⁣:kA 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 ⁣:RA 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×VV 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 dimFV is the unique nN with a basis B satisfying Bn (Finite-dimensional vector space, and its dimension dimFV; infinite-dimensional means having no finite basis).

[L5]

If V is finite dimensional over F with dimFV=n and U is a linear subspace of V, then U is finite dimensional over F and dimFUn (If dimFV=n and U is a linear subspace of V, then U is finite-dimensional, dimFUn, and dimFU=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
1.1

The algebra action λa=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 dimkA=n refers to.

L1L2given
2.1

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 λa=ηA(λ)a is a product of an element of A with an element of I.

L1L3step 1.1
3.1

By the subspace theorem I is finite dimensional over k with dimkIn; fix a basis b1,,br of I with rn. Every element of I is then a finite k-linear combination i=1rλibi with λik.

L4L5L6L7step 2.1
4.1

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.

L8L9step 3.1
5.1

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 rn on the number of generators, which the general theorem does not.

L10L11L12step 1.1step 4.1

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

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources