Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 A be the integral closure of A in K. Then A is an integrally closed domain.

Facts & Assumptions

Given: A domain A, a field extension K/Frac(A), and the integral closure 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
1.1

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

L1L2A1given
1.2

Let xFrac(A) be integral over A. By [A1] we may regard x as an element of K. Then [L3] shows that x is integral over A.

L3A1given
2.1

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

L1step 1.1step 1.2

Depends on

Used by

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