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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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The elements of an extension algebraic over the base field form a subfield

Statement

For a field extension K/F, the set

A={a∈K:a is algebraic over F}

is a subfield of K containing F.

Facts & Assumptions

Given: A field extension K/F and algebraic elements a,b∈K.

[L1]

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

[L2]

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

[L3]

A subset containing 1 and closed under subtraction, multiplication, and inverses of nonzero elements is a subfield (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

[L4]

Algebraic means annihilated by a nonzero polynomial over the base field (Algebraic and transcendental elements and algebraic extensions).

Proof

technique · direct
1.1L4

Every c∈F is algebraic over F, since it is a root of t−c, so F⊆A and in particular 0,1∈A.

1.2givenL1L2

By [L1], F(a,b)/F is finite. Thus [L2] makes all of its elements algebraic over F, including a−b and ab.

1.3givenL1L2

If a≠0, then a−1∈F(a)⊆F(a,b) and is algebraic by the same argument.

2.1step 1.1step 1.2step 1.3L3∎

Therefore A satisfies the subfield criterion [L3] and is a subfield containing F.

Depends on

Used by

Dependency tree · two levels

19 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