CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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An algebraic extension need not be finite

Statement refuted

Every algebraic field extension is finite.

Facts & Assumptions

Given: For $m \geq 0$ , let $a_{m} = 2^{1 / 2^{m}}$ be the positive real root and $K_{m} = Q (a_{m})$ , and put $K = ⋃_{m \geq 0} K_{m}$ .

[L1]

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

[L3]

The degree of an intermediate field divides the total finite degree (The degree of an intermediate field divides the degree of a finite extension).

[L4]

Positive real $n$ -th roots exist (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[L5]

Eisenstein's criterion proves $t^{2^{m}} - 2$ irreducible at $2$ (Eisenstein criterion over the integers).

[L6]

A simple extension has degree equal to its minimal-polynomial degree (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

Counterexample

technique · contradiction

1.1givenL4

By [L4], the elements $a_{m}$ exist, and $a_{m + 1}^{2} = a_{m}$ , so $K_{m} \subseteq K_{m + 1}$ and the union $K$ is a field.

1.2L5L6

By [L5] and [L6], $[K_{m} : Q] = 2^{m}$ for every $m$ .

2.1step 1.1L1L5L6

Every element of $K$ lies in some $K_{m}$ . By [L5] and [L6], the extension $K_{m} / Q$ has finite degree $2^{m}$ , so [L1] makes each of its elements algebraic over $Q$ . Thus $K / Q$ is algebraic.

2.2step 1.2L3assume-contrachoose

Suppose, for contradiction, that $K / Q$ has finite degree $N$ . Then [L3] makes $2^{m} = [K_{m} : Q]$ divide $N$ for every $m$ . Choosing $m$ with $2^{m} > N$ is impossible.

3.1step 2.1step 2.2discharge-contradiction∎

Thus $K / Q$ is algebraic by step 2.1 but not finite, refuting the statement.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 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

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