PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 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.

For finite subextensions in a common field, $[E E^{'} : F] \leq [E : F] [E^{'} : F]$

Statement

Let $E / F$ and $E^{'} / F$ be finite subextensions of a common field. Then their compositum is finite and

$[E E^{'} : F] \leq [E : F] [E^{'} : F] .$

Facts & Assumptions

Given: Finite subextensions $E / F$ and $E^{'} / F$ inside a field $Ω$ .

[L1]

The compositum $E E^{'}$ is the smallest subfield of $Ω$ containing $E \cup E^{'}$ (The composite of two subfields is the subfield generated by their union).

[L2]

Extension degree is the size of a finite basis (The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L3]

Assuming the Axiom of Choice, if $S \subseteq V$ spans $V$ then there is a basis $B$ of $V$ with $B \subseteq S$ (Every spanning subset of a vector space contains a basis).

[L4]

Every finite extension is algebraic (Every finite field extension is algebraic).

[L5]

A field generated by finitely many algebraic elements is finite over the base (An extension generated by finitely many algebraic elements is finite).

Proof

technique · direct

1.1

Choose $F$ -bases $(u_{1}, \dots, u_{m})$ of $E$ and $(v_{1}, \dots, v_{n})$ of $E^{'}$ . The $F$ -span $A$ of the $m n$ products $u_{i} v_{j}$ contains $E$ and $E^{'}$ and is closed under addition and multiplication, because products are reduced separately in the two bases.

givenL2algebra

2.1

By [L4], every $u_{i}$ and $v_{j}$ is algebraic over $F$ , so [L5] makes $E E^{'} = F (u_{1}, \dots, u_{m}, v_{1}, \dots, v_{n})$ finite over $F$ . Every $r \in A \subseteq E E^{'}$ is therefore algebraic over $F$ .

step 1.1L1L4L5

3.1

If $0 \neq r \in A$ , take a nonzero annihilating polynomial, factor out its largest power of $t$ , and cancel the corresponding nonzero power of $r$ in the ambient field. This gives $c_{0} + c_{1} r + \dots + c_{d} r^{d} = 0$ with $c_{0} \neq 0$ . Then $r^{- 1} = - c_{0}^{- 1} (c_{1} + c_{2} r + \dots + c_{d} r^{d - 1}) \in A$ , so $A$ is a field.

step 2.1algebra

4.1

Since $A$ is a field containing $E \cup E^{'}$ , [L1] gives $E E^{'} \subseteq A$ ; the reverse inclusion is clear from the product span, so $A = E E^{'}$ .

step 1.1step 3.1L1

5.1

The $m n$ products span $E E^{'}$ . By [L3] they contain a basis of at most $m n$ elements, so [L2] yields $[E E^{'} : F] \leq m n = [E : F] [E^{'} : F]$ .

step 4.1L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

For finite subextensions in a common field, [EE′:F]≤[E:F][E′:F][EE′:F]≤[E:F][E′:F]