LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Products of bases form a basis in a tower of finite extensions

Statement

Let $F \subseteq K \subseteq L$ be fields. If $(u_{1}, \dots, u_{m})$ is an $F$ -basis of $K$ and $(v_{1}, \dots, v_{n})$ is a $K$ -basis of $L$ , then

${u_{i} v_{j} : 1 \leq i \leq m, 1 \leq j \leq n}$

is an $F$ -basis of $L$ .

Facts & Assumptions

Given: A tower $F \subseteq K \subseteq L$ and the two finite bases in the Statement.

[L1]

A basis is a linearly independent spanning set (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L2]

Every vector has unique coordinates with respect to an ordered basis (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

Proof

technique · direct

1.1givenL1L2

For $x \in L$ , use the $K$ -basis to write $x = \sum_{j} b_{j} v_{j}$ with $b_{j} \in K$ , then use the $F$ -basis to write $b_{j} = \sum_{i} c_{i j} u_{i}$ . Thus $x = \sum_{i, j} c_{i j} u_{i} v_{j}$ , so the products span $L$ over $F$ .

1.2givenL1L2algebra

Suppose $\sum_{i, j} c_{i j} u_{i} v_{j} = 0$ with $c_{i j} \in F$ . Grouping by $v_{j}$ gives $\sum_{j} (\sum_{i} c_{i j} u_{i}) v_{j} = 0$ . Independence of the $v_{j}$ makes every inner coefficient zero, and independence of the $u_{i}$ then makes every $c_{i j} = 0$ .

2.1step 1.1step 1.2L1∎

The products span and are independent, hence form an $F$ -basis.

Depends on

Dependency tree · 2 levels

Sources

A. W. Knapp, Basic Algebra, 2nd ed., Chapter IX, Section 1