CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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The algebraic numbers in $C$ form an algebraic closure of $Q$

Statement

Let $A : = {z \in C : z is algebraic over Q} .$ Then $A$ is an algebraic closure of $Q$ .

Facts & Assumptions

Given: The subset $A \subseteq C$ of elements algebraic over $Q$ .

[L1]

The algebraic elements in an extension form a subfield (The elements of an extension algebraic over the base field form a subfield).

[L2]

The field $C$ is algebraically closed (The complex numbers are algebraically closed).

[L3]

A field generated by finitely many algebraic elements over the base field is finite over that base (An extension generated by finitely many algebraic elements is finite).

[L4]

An element is algebraic over a field if and only if its simple extension over that field is finite (An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite).

[L5]

Degrees multiply in a finite tower (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

[L6]

Every element of a finite extension is algebraic over the base field (Every finite field extension is algebraic).

[L7]

An algebraic closure of a field is an algebraic extension whose top field is algebraically closed (An algebraic closure of a field).

Proof

technique · direct

1.1L1

By [L1], the set $A$ is a subfield of $C$ containing $Q$ .

1.2L2

Let $f (x) = a_{n} x^{n} + \dots + a_{0} \in A [x]$ be nonconstant. Since $C$ is algebraically closed by [L2], the polynomial $f$ has a root $z \in C$ .

2.1L3L4L5L6step 1.2

The coefficients $a_{0}, \dots, a_{n}$ are algebraic over $Q$ , so $F : = Q (a_{0}, \dots, a_{n})$ is a finite extension of $Q$ by [L3]. The element $z$ is a root of a nonzero polynomial over $F$ , so it is algebraic over $F$ ; therefore [L4] makes $F (z) / F$ finite. By [L5], the extension $F (z) / Q$ is finite, and then [L6] makes $z$ algebraic over $Q$ . Hence $z \in A$ .

3.1L7step 2.1∎

Step 2.1 shows that every nonconstant polynomial in $A [x]$ has a root in $A$ , so $A$ is algebraically closed. Since every element of $A$ is algebraic over $Q$ by definition, [L7] makes $A$ an algebraic closure of $Q$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Fields and Galois Theory, v5.10, Corollary 5.7(b) (standard reference, not scraped)
P. L. Clark, Field Theory, Chapters 3 to 5 (standard reference, not scraped)

The algebraic numbers in CC form an algebraic closure of QQ