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

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The integral closure of a domain in a field extension is integrally closed

Statement

Let $A$ be a domain, let $K$ be a field extension of $Frac (A)$ , and let $\overline{A}$ be the integral closure of $A$ in $K$ . Then $\overline{A}$ is an integrally closed domain.

Facts & Assumptions

Given: A domain $A$ , a field extension $K / Frac (A)$ , and the integral closure $\overline{A}$ of $A$ in $K$ .

[L1]

The integral closure of $A$ in $K$ is the set of elements of $K$ integral over $A$ , and a domain is integrally closed when every element of its field of fractions integral over it already lies in the domain (Integral closure in an extension ring and integrally closed domains).

[L2]

In a nonzero integral extension, the elements integral over the base form a subring (Integral elements over a nonzero base ring form a subring).

[L3]

Integral maps are transitive (Integral extensions are transitive).

[A1]

Any subring of a field is a domain, so its field of fractions embeds into that field.

Proof

technique · direct

1.1L1L2A1given

Because $A$ is a domain, it is nonzero, so [L2] applies to the inclusion $A \subseteq K$ . Therefore [L1] implies that $\overline{A}$ is a subring of the field $K$ containing $A$ , and [A1] makes $\overline{A}$ a domain.

1.2L3A1given

Let $x \in Frac (\overline{A})$ be integral over $\overline{A}$ . By [A1] we may regard $x$ as an element of $K$ . Then [L3] shows that $x$ is integral over $A$ .

2.1L1step 1.1step 1.2∎

Since $\overline{A}$ is, by [L1], exactly the set of elements of $K$ integral over $A$ , step 1.2 gives $x \in \overline{A}$ . Thus every element of $Frac (\overline{A})$ integral over $\overline{A}$ already lies in $\overline{A}$ , so $\overline{A}$ is integrally closed.

Depends on

Used by

Dependency tree · two levels

10 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (10.32) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Corollary 6.8 and Definition 6.9 (standard reference, not scraped)