LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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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.

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

A finite extension with only finitely many intermediate fields is simple

Statement

Let $E / F$ be a finite extension. If it has only finitely many intermediate fields, then it is simple.

Facts & Assumptions

Given: A finite extension $E / F$ with finitely many intermediate fields.

[L1]

A finite extension of a finite field is simple (Every finite extension of a finite field is simple).

[L2]

A finite extension is a finite-dimensional vector space over its base (The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L3]

A finite-dimensional vector space over an infinite field is not a finite union of proper subspaces (A finite-dimensional vector space over an infinite field is not a finite union of proper subspaces).

[L4]

The field $F (α)$ is the smallest intermediate field containing $F$ and $α$ , and $E / F$ is simple when $E = F (α)$ for some $α$ (Field extensions, generated subrings $F [S]$ , generated subfields $F (S)$ , and simple extensions).

Proof

technique · direct

1.1L1

If $F$ is finite, [L1] supplies a primitive element.

1.2L4

Suppose $F$ is infinite. If every intermediate field $F (α)$ with $α \in E$ were proper, then the finitely many proper intermediate fields would cover $E$ , because every $α$ lies in its own $F (α)$ .

2.1step 1.2L2L3L4

Each proper intermediate field is a proper $F$ -linear subspace of the finite-dimensional space $E$ from [L2], so the cover in step 1.2 contradicts [L3]. Hence $E = F (α)$ for some $α$ , and the extension is simple.

3.1step 1.1step 2.1∎

Together with the finite-base case, this proves the assertion.

Depends on

Used by

A finite field extension is simple if and only if it has finitely many intermediate fields Theorem

Dependency tree · two levels

14 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, Chapter 5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Chapter 5 (standard reference, not scraped)