ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Assuming Choice, real algebraic numbers embed properly in an algebraic closure of $Q$

Example

Assume the Axiom of Choice. Let $A = {r \in R : r is algebraic over Q}$ . There is an algebraic closure $\overline{Q}$ containing a $Q$ -isomorphic copy of $A$ , and that copy is a proper subfield which is not algebraically closed.

Facts & Assumptions

Given: The Axiom of Choice, the rational subfield of the ordered field $R$ , and the set $A$ displayed above.

[L1]

Assuming Choice, every field has an algebraic closure (Assuming Choice, every field has an algebraic closure).

[L2]

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

[L3]

Assuming Choice, a base embedding extends across every algebraic extension into an algebraically closed field (Assuming Choice, a base-field embedding extends across every algebraic extension).

[L4]

The real numbers form an ordered field (The reals form a totally ordered field).

Verification

technique · direct

1.1L1L2choose

By [L2], $A$ is a subfield of $R$ containing $Q$ , and $A / Q$ is algebraic. Choose an algebraic closure $Ω / Q$ by [L1].

1.2L4algebra

The polynomial $x^{2} + 1$ has no root in the ordered field $A \subseteq R$ , since every square is nonnegative and $- 1 < 0$ by [L4].

2.1step 1.1L3

The identity on $Q$ extends by [L3] to an embedding $A \to Ω$ ; denote its image by $A^{'}$ .

3.1step 2.1step 1.2algebra∎

If $A^{'}$ contained a root of $x^{2} + 1$ , its preimage under the isomorphism $A \to A^{'}$ would be a root in $A$ , contrary to step 1.2. The algebraically closed field $Ω$ does contain such a root, so $A^{'}$ is proper in $Ω$ and is not algebraically closed. No use of $C$ is required.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

P. L. Clark, Field Theory, Corollary 3.4 (standard reference, not scraped)

Assuming Choice, real algebraic numbers embed properly in an algebraic closure of QQ