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Finite normalization commutes with principal localization

Statement

Let $k$ be a field, let $A$ be a finite-type integral domain over $k$ , let $B$ be the integral closure of $A$ in $Frac (A)$ , and let $0 \neq f \in A$ . Then the integral closure of the principal localisation $A_{f}$ inside the field $Frac (A)$ is exactly the localisation $B_{f}$ , and $B_{f}$ is a finite $A_{f}$ -module.

Facts & Assumptions

Given: a field $k$ , a finite-type integral domain $A$ over $k$ , its integral closure $B$ in $Frac (A)$ , and an element $0 \neq f \in A$ .

[L1]

Let $A \to B$ be a homomorphism of commutative rings, $S \subseteq A$ multiplicative and $b \in B$ : if $b$ is integral over $A$ then $b / 1$ is integral over $S^{- 1} A$ . Moreover, if $A$ is a domain, $S \subseteq A ∖ {0}$ , $K$ is a field extension of $Frac (A)$ and $\overline{A}$ is the integral closure of $A$ in $K$ , then the integral closure of $S^{- 1} A$ in $K$ is exactly $S^{- 1} \overline{A}$ (Integrality and integral closure commute with localisation, Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L2]

For a commutative ring $R$ and $f \in R$ the powers $S_{f} = {f^{n} : n \in N}$ form a multiplicative subset, and the principal localisation is $R_{f} = S_{f}^{- 1} R$ with elements written $r / f^{n}$ ; the element $f$ becomes a unit, and an element $r / 1$ is zero exactly when $f^{n} r = 0$ for some $n$ (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ , Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L3]

The integral closure of a finite-type domain over a field $k$ in its fraction field is a finite module over the domain (A finite-type domain over a field has finite normalization, Subalgebra generated by a subset, algebras of finite type, and module-finite algebras, Integral closure in an extension ring and integrally closed domains).

[L4]

If $A$ is a domain and $0 \neq f \in A$ , then $A_{f}$ is a domain containing $A$ and the identity on $A$ exhibits $Frac (A)$ as a field containing $A_{f}$ , and $Frac (A_{f}) = Frac (A)$ : each element of $Frac (A_{f})$ is a fraction of elements of $A_{f}$ , hence lies in $Frac (A)$ , while each $a / b$ with $a, b \in A$ , $b \neq 0$ , equals $(a / 1) / (b / 1)$ with $a / 1, b / 1 \in A_{f}$ (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, Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

[L5]

If $B$ is a finite $A$ -module with generators $b_{1}, \dots, b_{m}$ and $A \subseteq A_{f}$ , then $B_{f} = A_{f} (b_{1} / 1) + \dots + A_{f} (b_{m} / 1)$ is a finite $A_{f}$ -module: every element of $B_{f}$ is $b / f^{n}$ with $b = \sum_{i} a_{i} b_{i}$ , and $b / f^{n} = \sum_{i} (a_{i} / f^{n}) (b_{i} / 1)$ (Module finiteness is transitive along a tower of algebras, Generated submodule, cyclic and finitely generated modules, module basis and free module, Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

Proof

technique · direct

1.1L2L3L4given

The element $f$ is nonzero in the domain $A$ , so the multiplicative subset $S_{f} = {f^{n}}$ is contained in $A ∖ {0}$ by [L2] and [L4]. By [L3] the integral closure $B$ of $A$ in $Frac (A)$ is a finite $A$ -module. The localisation $A_{f}$ is a domain containing $A$ , and by [L4] the field $Frac (A)$ contains $A_{f}$ and equals its fraction field, so the phrase "the integral closure of $A_{f}$ inside $Frac (A)$ " is computed inside a field extension of $Frac (A_{f})$ .

2.1L1L2step 1.1

Applying the third clause of [L1] with the domain $A$ , the multiplicative subset $S = S_{f} \subseteq A ∖ {0}$ , the field $K = Frac (A)$ and $\overline{A} = B$ : the integral closure of $S_{f}^{- 1} A = A_{f}$ in $K = Frac (A)$ is exactly $S_{f}^{- 1} B = B_{f}$ . This is a genuine equality inside the fixed field $Frac (A)$ , not an isomorphism chosen afterwards, so it is canonical.

3.1L2L5step 1.1step 2.1∎

By [L5], applied to the finite generating list $b_{1}, \dots, b_{m}$ of the $A$ -module $B$ from step 1.1, the localisation $B_{f}$ is generated as an $A_{f}$ -module by the finitely many elements $b_{1} / 1, \dots, b_{m} / 1$ , so $B_{f}$ is a finite $A_{f}$ -module. Combined with step 2.1, the integral closure of $A_{f}$ inside $Frac (A)$ is $B_{f}$ , a finite $A_{f}$ -module. If $f$ is a unit of $A$ , then $S_{f}$ contains $1$ and $A_{f} = A$ , so the statement reduces to $B_{f} = B$ ; the hypothesis $f \neq 0$ is exactly what makes $S_{f}$ a subset of $A ∖ {0}$ , and it is used nowhere else.

Finite normalization commutes with principal localization

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources