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Every algebra of finite type over a Noetherian ring is finitely presented

Statement

Let $R$ be a Noetherian commutative ring and let $A$ be a commutative $R$ -algebra of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then $A$ is finitely presented as an $R$ -algebra (Finitely presented modules and finitely presented algebras).

Facts & Assumptions

Given: A Noetherian commutative ring $R$ and a commutative $R$ -algebra $A$ of finite type.

[L1]

An $R$ -algebra is of finite type over $R$ exactly when it is isomorphic as an $R$ -algebra to a quotient $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some ideal $a$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

If $R$ is a Noetherian commutative ring then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ).

[L3]

For a commutative ring, being Noetherian is equivalent to every ideal being finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

[L4]

A commutative $R$ -algebra is finitely presented when it is isomorphic as an $R$ -algebra to $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some finitely generated ideal $a$ (Finitely presented modules and finitely presented algebras).

Proof

technique · direct

1.1L1given

Being of finite type, $A$ is isomorphic as an $R$ -algebra to $R [x_{1}, \dots, x_{n}] / a$ for some $n \in N$ and some ideal $a$ of $R [x_{1}, \dots, x_{n}]$ .

2.1L2L3step 1.1

The ring $R [x_{1}, \dots, x_{n}]$ is Noetherian because $R$ is, so its ideal $a$ is finitely generated.

3.1L4step 1.1step 2.1∎

A presentation by a polynomial ring in finitely many variables modulo a finitely generated ideal is exactly what finite presentation as an algebra asks for, so $A$ is finitely presented over $R$ .

Remarks

Over a Noetherian base the two algebra conditions coincide. Finite presentation always implies finite type; this corollary supplies the converse when the base ring is Noetherian, so no relation-finiteness hypothesis needs to be carried in that setting.

Every algebra of finite type over a Noetherian ring is finitely presented

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources