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A finite-type domain over a field has finite normalization

Statement

Let $k$ be a field and let $A$ be a finite-type integral domain over $k$ . Then the integral closure of $A$ in $Frac (A)$ is a finite $A$ -module. The proof is a Noether-normalisation reduction to the polynomial theorem and uses no choice principle.

Facts & Assumptions

Given: a field $k$ and a finite-type integral domain $A$ over $k$ .

[L1]

Let $k$ be a field and $A$ a nonzero finite-type $k$ -algebra; then there are algebraically independent elements $z_{1}, \dots, z_{d} \in A$ such that $A$ is a module-finite algebra over the polynomial ring $k [z_{1}, \dots, z_{d}]$ , that is, $A$ is generated as a $k [z_{1}, \dots, z_{d}]$ -module by finitely many elements (Noether normalisation yields module finiteness over a polynomial subring, Subalgebra generated by a subset, algebras of finite type, and module-finite algebras).

[L2]

For commutative rings $A \subseteq B$ with $A \neq 0$ and $b \in B$ , the element $b$ is integral over $A$ if and only if there exists a faithful $A [b]$ -module that is finitely generated over $A$ ; in particular a finite-module generation statement of this shape certifies integrality (Integrality and finite-module characterizations for one element, Integral elements over a commutative ring and algebraic integers, Integral ring maps and integral extensions).

[L3]

Let $K$ be a field and $d \geq 0$ , and let $L / K (x_{1}, \dots, x_{d})$ be a finite extension of the rational function field; then the integral closure of $K [x_{1}, \dots, x_{d}]$ in $L$ is a finite module over $K [x_{1}, \dots, x_{d}]$ (Polynomial algebras over fields have finite integral closures).

[L4]

If $A \subseteq B \subseteq N$ are domains with $B$ integral over $A$ , then an element of $N$ is integral over $A$ if and only if it is integral over $B$ ; the integral closure of $A$ in a field extension of $Frac (A)$ is a subring containing $A$ (Integral closure is unchanged across an integral intermediate domain, Integral extensions are transitive, Integral closure in an extension ring and integrally closed domains, Integral elements over a nonzero base ring form a subring).

[L5]

A domain is a nonzero commutative ring without zero divisors, and $Frac (A)$ is its field of fractions, the smallest field containing $A$ (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain, Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors, Field).

[L6]

If $a_{1}, \dots, a_{r}$ are algebraic over a field $F$ , then $F (a_{1}, \dots, a_{r}) / F$ is finite, where $F (a_{1}, \dots, a_{r})$ is the smallest subfield containing $F$ and the $a_{i}$ (An extension generated by finitely many algebraic elements is finite, Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension, Algebraic and transcendental elements and algebraic extensions).

Proof

technique · direct

1.1L1L2L5given

By [L5] the finite-type domain $A$ over $k$ is nonzero, so [L1] applies: fix algebraically independent elements $z_{1}, \dots, z_{d} \in A$ such that $A$ is module-finite over $R : = k [z_{1}, \dots, z_{d}]$ . Every $a \in A$ is integral over $R$ : the ring $A$ is a faithful $R [a]$ -module, because $r \cdot 1_{A} = r \neq 0$ for every nonzero $r \in R [a] \subseteq A$ (evaluate at $1_{A}$ in the domain $A$ ), and it is a finite $R$ -module, so [L2] applies to $b = a$ inside $R \subseteq A$ . Hence $R \subseteq A \subseteq Frac (A)$ with $A$ integral over $R$ .

2.1L1L5L6step 1.1

The extension $F : = Frac (R)$ is a rational function field and $Frac (A) / F$ is finite. Write $A = R a_{1} + \dots + R a_{n}$ with $a_{i} \in A$ , using the module finiteness of step 1.1. Every element of $A$ lies in $F [a_{1}, \dots, a_{n}] \subseteq F (a_{1}, \dots, a_{n})$ , and $F (a_{1}, \dots, a_{n})$ is a field containing $A$ , so $Frac (A) = F (a_{1}, \dots, a_{n})$ by [L5]; each $a_{i}$ is integral over $R$ by step 1.1, hence algebraic over $F$ ; therefore $Frac (A) / F$ is finite by [L6].

3.1L3step 1.1step 2.1

Apply [L3] with the base field $k$ , the algebraically independent elements $z_{1}, \dots, z_{d}$ (so that $R = k [z_{1}, \dots, z_{d}]$ and $F = k (z_{1}, \dots, z_{d})$ ) and the finite extension $L : = Frac (A)$ of step 2.1: the integral closure $B$ of $R$ in $Frac (A)$ is a finite $R$ -module.

4.1L4step 1.1step 3.1∎

Since $A$ is integral over $R$ by step 1.1 and $R \subseteq A \subseteq Frac (A)$ , [L4] shows that an element of $Frac (A)$ is integral over $R$ exactly when it is integral over $A$ ; hence $B$ is exactly the integral closure of $A$ in $Frac (A)$ , and $B$ is a ring with $R \subseteq A \subseteq B$ . If $B = R b_{1} + \dots + R b_{r}$ , then every $A b_{i}$ lies in $B$ , and every $R b_{i}$ lies in $A b_{i}$ because $R \subseteq A$ ; hence $B = A b_{1} + \dots + A b_{r}$ is generated as an $A$ -module by the same finitely many elements. Therefore the integral closure of $A$ in $Frac (A)$ is a finite $A$ -module.

A finite-type domain over a field has finite normalization

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources