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The Artin–Tate lemma with integrality in place of module finiteness

Statement

Let $A \subseteq B \subseteq C$ be commutative rings, each a subring of the next (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication), with $A$ Noetherian. Suppose $C$ is of finite type over $A$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras) and every element of $C$ is integral over $B$ (Integral elements over a commutative ring and algebraic integers). Then $C$ is module-finite over $B$ , and $B$ is of finite type over $A$ .

Facts & Assumptions

Given: Commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C$ of finite type over $A$ , and every element of $C$ integral over $B$ .

[L1]

$R [a_{1}, \dots, a_{n}]$ is the smallest subring of $A$ containing the image of $R$ and $a_{1}, \dots, a_{n}$ ; an algebra is of finite type over $R$ when it equals $R [a_{1}, \dots, a_{n}]$ for some finite list, and 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).

[L2]

For a homomorphism of commutative rings $A \to B$ , an element $b \in B$ is integral over $A$ when it is a root of a monic polynomial in $A [X]$ (Integral elements over a commutative ring and algebraic integers).

[L3]

For commutative rings $A \subseteq B$ with $A$ a subring of $B$ and $b_{1}, \dots, b_{n} \in B$ integral over $A$ , the subalgebra $A [b_{1}, \dots, b_{n}]$ is module-finite over $A$ (A subalgebra generated by finitely many integral elements is module-finite).

[L4]

For commutative rings $A \subseteq B \subseteq C$ , each a subring of the next, with $A$ Noetherian, $C$ of finite type over $A$ and $C$ module-finite over $B$ , the ring $B$ is of finite type over $A$ (Artin–Tate lemma: an intermediate ring over which a finite-type algebra is module-finite is itself of finite type).

[L5]

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

Proof

technique · direct

1.1

Fix $r \in N$ and $t_{1}, \dots, t_{r} \in C$ with $C = A [t_{1}, \dots, t_{r}]$ . The subring $B [t_{1}, \dots, t_{r}]$ of $C$ contains $B$ , hence contains $A$ , and contains every $t_{i}$ ; since $A [t_{1}, \dots, t_{r}]$ is the smallest subring of $C$ with those two properties, $C = A [t_{1}, \dots, t_{r}] \subseteq B [t_{1}, \dots, t_{r}] \subseteq C$ . So $C = B [t_{1}, \dots, t_{r}]$ is of finite type over $B$ , generated by the same elements.

L1L5given

2.1

Every element of $C$ is integral over $B$ , so in particular each $t_{i}$ is; $B$ is a subring of $C$ ; hence $B [t_{1}, \dots, t_{r}]$ is module-finite over $B$ . By step 1.1 that ring is $C$ , so $C$ is module-finite over $B$ .

L2L3step 1.1

3.1

The hypotheses of the Artin–Tate lemma are now all in place for $A \subseteq B \subseteq C$ : $A$ is Noetherian, $C$ is of finite type over $A$ , and $C$ is module-finite over $B$ . Therefore $B$ is of finite type over $A$ .

L4step 2.1∎

Remarks

Finite type over $A$ gives finite type over $B$ for free. Step 1.1 uses only that $A \subseteq B$ : the same algebra generators work over the larger base ring. That is what lets the integrality hypothesis be applied to the finitely many $t_{i}$ rather than to all of $C$ at once.
Integrality of every element is more than integrality of the generators, and only the generators are used. The hypothesis as stated is the usual one and is what the applications supply; the proof needs it only at $t_{1}, \dots, t_{r}$ .

Depends on

Used by

Noether's finiteness theorem: the invariants of a finite group acting on a finite-type algebra over a Noetherian ring form an algebra of finite type Theorem

Dependency tree · two levels

21 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (16.21) (standard reference, not scraped)
M. Hochster, Introduction to Commutative Algebra, Math 614, Theorem 5.8 (standard reference, not scraped)