CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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.

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

For an element $a$ of an extension $K / F$ ,

$a is algebraic over F ⟺ F (a) / F is finite .$

If $a$ is algebraic with minimal polynomial of degree $n$ , then $[F (a) : F] = n$ .

Given: A field extension $K / F$ and an element $a \in K$ .

[L1]

If $a$ is algebraic with minimal polynomial of degree $n$ , then $1, a, \dots, a^{n - 1}$ is a basis of $F (a) / F$ and its degree is $n$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L2]

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

technique · direct

1.1givenL1

If $a$ is algebraic, [L1] gives a finite power basis of $F (a)$ and the stated degree.

1.2givenL2

Conversely, if $F (a) / F$ is finite, [L2] says every element of $F (a)$ , in particular $a$ , is algebraic over $F$ .

2.1step 1.1step 1.2∎

These are the two implications of the equivalence.

Depends on