TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29

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Integrality and integral closure commute with localisation

Statement

Let $A \to B$ be a homomorphism of commutative rings, let $S \subseteq A$ be multiplicative, and let $b \in B$ .

If $b$ is integral over $A$ , then $b / 1$ is integral over $S^{- 1} A$ in $S^{- 1} B$ .
If $b / 1$ is integral over $S^{- 1} A$ in $S^{- 1} B$ , then some $s \in S$ makes $s b$ integral over $A$ .

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 $S^{- 1} \overline{A}$ .

Facts & Assumptions

Given: A ring map $A \to B$ , a multiplicative subset $S \subseteq A$ , and an element $b \in B$ .

[L1]

An element is integral over a ring exactly when it satisfies a monic polynomial with coefficients in that ring (Integral ring maps and integral extensions).

[L2]

The integral closure of a domain in a field is the set of elements integral over the domain (Integral closure in an extension ring and integrally closed domains).

[L3]

Localisation uses fractions with the usual ring laws (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

[L4]

A fraction $r / s$ is zero in a localisation exactly when some denominator annihilates $r$ (Equality, vanishing, and the kernel of the localisation map).

Proof

technique · direct

1.1L1L3given

If $b$ satisfies $b^{n} + a_{n - 1} b^{n - 1} + \dots + a_{0} = 0$ with $a_{i} \in A$ , then the same identity in $S^{- 1} B$ reads $(b / 1)^{n} + (a_{n - 1} / 1) (b / 1)^{n - 1} + \dots + a_{0} / 1 = 0$ . By [L1], this makes $b / 1$ integral over $S^{- 1} A$ .

1.2L1L3L4givenalgebra

Conversely, assume $b / 1$ is integral over $S^{- 1} A$ . Choose a monic relation $(b / 1)^{n} + (a_{n - 1} / s_{n - 1}) (b / 1)^{n - 1} + \dots + a_{0} / s_{0} = 0$ with $a_{i} \in A$ and $s_{i} \in S$ . Let $t : = s_{0} \dots s_{n - 1}$ and put $y : = t b$ . Multiplying the relation by $t^{n}$ gives $(y / 1)^{n} + c_{n - 1} (y / 1)^{n - 1} + \dots + c_{0} / 1 = 0$ with each $c_{i} \in A$ . Hence [L4] gives some $u \in S$ with $u (y^{n} + c_{n - 1} y^{n - 1} + \dots + c_{0}) = 0$ in $B$ . For $s : = u t$ and $z : = s b = u y$ , multiplying that equation by $u^{n - 1}$ yields $z^{n} + c_{n - 1} u z^{n - 1} + \dots + c_{0} u^{n} = 0$ , a monic equation over $A$ . So $s b$ is integral over $A$ .

2.1L1L2step 1.1

Now assume $A$ is a domain, $S \subseteq A ∖ {0}$ , $K / Frac (A)$ is a field extension, and $\overline{A}$ is the integral closure of $A$ in $K$ . If $x = a / s$ with $a \in \overline{A}$ and $s \in S$ , then $a$ is integral over $A$ , so step 1.1 makes $x$ integral over $S^{- 1} A$ .

3.1L2step 1.2step 2.1∎

Conversely, let $x \in K$ be integral over $S^{- 1} A$ . Step 1.2 gives $s \in S$ with $s x$ integral over $A$ , so [L2] gives $s x \in \overline{A}$ . Therefore $x = (s x) / s \in S^{- 1} \overline{A}$ . Combining this with step 2.1 proves that the integral closure of $S^{- 1} A$ in $K$ is exactly $S^{- 1} \overline{A}$ .

Integrality and integral closure commute with localisation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources