LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 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.

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A simple finite extension has only finitely many intermediate fields

Statement

If $E / F$ is a finite simple extension, then there are only finitely many intermediate fields $F \subseteq M \subseteq E$ .

Facts & Assumptions

Given: A finite simple extension $E = F (α)$ .

[L1]

The notation $F (α)$ denotes the smallest subfield containing $F$ and $α$ (Field extensions, generated subrings $F [S]$ , generated subfields $F (S)$ , and simple extensions).

[L2]

An algebraic element has a unique monic irreducible minimal polynomial over its base field (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[L3]

A polynomial ring over a field is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

[L4]

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

Proof

technique · direct

1.1L1L2

Let $f$ be the minimal polynomial of $α$ over $F$ . For an intermediate field $M$ , let $g_{M}$ be the minimal polynomial of $α$ over $M$ and let $M_{0}$ be the subfield of $M$ generated over $F$ by the coefficients of $g_{M}$ .

2.1step 1.1L2

The polynomial $g_{M}$ divides $f$ in $M [x]$ and therefore in $E [x]$ . It is irreducible over $M_{0}$ , since a factorisation over $M_{0}$ would be one over $M$ , so it is also the minimal polynomial of $α$ over $M_{0}$ .

3.1step 2.1L4

Thus $[E : M_{0}] = \deg g_{M} = [E : M]$ ; the tower law [L4] in $M_{0} \subseteq M \subseteq E$ gives $[M : M_{0}] = 1$ , so $M = M_{0}$ . Hence the coefficients of $g_{M}$ determine $M$ .

4.1step 3.1L3∎

By unique factorisation [L3], the fixed polynomial $f$ has only finitely many monic divisors in $E [x]$ . The injective assignment $M \mapsto g_{M}$ therefore proves that there are only finitely many intermediate fields.

Depends on

Used by

Dependency tree · two levels

17 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, Chapters 3 to 5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Chapter 5 (standard reference, not scraped)