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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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In the Artin–Tate setup the intermediate ring is module-finite over the coefficient subalgebra

Statement

Keep the data and the notation of The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type: commutative rings A⊆B⊆C, each a subring of the next, with A Noetherian, C=A[t1,…,tr], a B-module generating list y1,…,yn of C with y1=1C, coefficients zij,zijk∈B as displayed there, and A′ the A-subalgebra of B they generate.

Then C is generated as an A′-module by y1,…,yn, so C is module-finite over A′; and B is module-finite over A′ as well (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

Facts & Assumptions

Given: The data of The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type, and the set M:=∑j=1nA′yj={ ∑j=1najyj:aj∈A′ }⊆C.

[L1]

In that setup A⊆A′⊆B, the algebra A′ is of finite type over A, and A′ is a Noetherian ring (The Artin–Tate coefficient subalgebra is a Noetherian algebra of finite type).

[L2]

R[a1,…,an] is the smallest subring of A containing the image of R and a1,…,an; 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).

[L3]

For a ring R, a left R-module M and S⊆M, the submodule ⟨S⟩R is the set of finite sums ∑i=1krisi with k∈N, ri∈R and si∈S (The submodule generated by a subset consists of the finite R-linear combinations of that subset).

[L4]

Let A be a Noetherian commutative ring and B a commutative A-algebra module-finite over A with A a subring of B; then every ring C with A⊆C⊆B is module-finite over A and is a Noetherian ring (A module-finite algebra over a Noetherian ring is a Noetherian ring, and so is every ring between the two).

[L5]

A subring contains the identity of the ambient ring and is closed under sums, additive inverses and products (Subring: a subset containing 1R and closed under addition, additive inverses and multiplication).

Proof

technique · direct
1.1L1L3L5given

The set M=∑j=1nA′yj is the A′-submodule of C generated by y1,…,yn, by the finite-sum description of a generated submodule; it is in particular an additive subgroup of C closed under multiplication by elements of A′. Since y1=1C we have A′=A′y1⊆M, and therefore A⊆M and 1C∈M.

2.1L1L3L5step 1.1

M is closed under multiplication, hence is a subring of C. It suffices to multiply two generators with coefficients: for a,a′∈A′, (ayi)(a′yj)=aa′(yiyj)=aa′∑k=1nzijkyk=∑k=1n(aa′zijk)yk, and each aa′zijk lies in A′ because A′ is a subring of B containing every zijk. A product of two elements of M expands by distributivity into a finite sum of such terms, so it lies in M; with the additive subgroup property and 1C∈M from step 1.1, M is a subring of C.

3.1L2L3step 1.1step 2.1

M contains each ti, since ti=∑j=1nzijyj and every zij lies in A′. So M is a subring of C containing A and t1,…,tr. But C=A[t1,…,tr] is the smallest such subring, so C⊆M; and M⊆C by construction. Hence C=M, and y1,…,yn generate C as an A′-module, so C is module-finite over A′.

4.1L1L4step 3.1∎

Now A′ is a Noetherian commutative ring, A′ is a subring of C, and C is module-finite over A′; the intermediate-ring clause applied to A′⊆B⊆C gives that B is module-finite over A′.

Remarks

  • No induction on total degree is needed. Showing directly that ∑jA′yj is a subring containing A and the algebra generators, and then invoking minimality of A[t1,…,tr], replaces the degreewise reduction of a polynomial expression; the multiplication table yiyj=∑kzijkyk is exactly what makes the closure argument work in one line.

  • y1=1C is used, and used here. It is what puts A′ itself inside M, without which M need not contain A and the minimality argument would not apply.

  • Only A′ is known to be Noetherian. The final step is applied with A′ as the Noetherian base, never with B, which is exactly the point of the Artin–Tate argument.

Depends on

Used by

Dependency tree · two levels

23 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