TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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An extension generated by finitely many algebraic elements is finite

Statement

If $a_{1}, \dots, a_{r}$ are algebraic over $F$ , then $F (a_{1}, \dots, a_{r}) / F$ is finite.

Facts & Assumptions

Given: A field extension containing elements $a_{1}, \dots, a_{r}$ algebraic over $F$ .

[L1]

The field $F (a_{1}, \dots, a_{r})$ is obtained by adjoining the finite list of generators (Finitely generated field extensions $F (a_{1}, \dots, a_{r})$ ).

[L2]

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

[L3]

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

[L4]

An element algebraic over $F$ satisfies a nonzero polynomial in $F [t]$ (Algebraic and transcendental elements and algebraic extensions).

Proof

technique · direct

1.1

Put $F_{0} = F$ and $F_{i} = F (a_{1}, \dots, a_{i})$ for $1 \leq i \leq r$ . By [L4], the nonzero polynomial over $F$ satisfied by $a_{i}$ also belongs to $F_{i - 1} [t]$ , so $a_{i}$ is algebraic over $F_{i - 1}$ . Thus [L2] makes $F_{i} / F_{i - 1}$ finite.

givenL1L2L4

2.1

Repeated application of [L3] makes $F_{r} / F$ finite, with degree equal to the product of the simple-step degrees.

step 1.1L3

3.1

By [L1], $F_{r} = F (a_{1}, \dots, a_{r})$ . If $r = 0$ , this is $F / F$ of degree one, so the boundary case also holds.

step 2.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 32 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. W. Knapp, Basic Algebra, 2nd ed., Chapter IX, Section 1 (standard reference, not scraped)