Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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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.

Finite elementary abelian p-groups have bases, basis extension, and a well-defined dimension

Statement

Every finite elementary abelian p-group has a basis; every independent subset extends to a basis, every spanning subset contains a basis, and all bases have the same finite size.

Facts & Assumptions

Given: A finite elementary abelian p-group E, an independent subset IE, and a spanning subset SE.

[F1]

A basis of an elementary abelian p-group is an independent spanning subset for its canonical Fp-linear structure (Fp-spanning sets, independence, and bases in an elementary abelian p-group).

[L1]

The cardinality of a finite Cartesian product is the product of the cardinalities of its factors (The product rule: A×B=AB, and i<mAi=i<mAi).

[L3]

Every nonempty subset of N has a least element (The well-ordering principle).

Proof

technique · direct
1.1

The set of cardinalities of spanning subsets of E is nonempty because E spans itself, so [L3] gives its least member; choose a spanning subset B of that size. It is inclusion-minimal, and if a nontrivial linear relation existed in B, one member with nonzero coefficient could be solved for using its inverse scalar, contradicting minimality. Thus B is a basis by [F1].

givenF1L3algebra
2.1

Starting from I, adjoin an element outside its span while one exists; adjoining such an element preserves independence, and finiteness makes the process terminate at a spanning independent set. This extends I to a basis. Applying the deletion argument of step 1.1 inside S extracts a basis from every spanning set.

step 1.1F1givenalgebra
3.1

If B is a basis, uniqueness of coordinates gives a bijection FpBE, so [L1] gives E=pB. For two bases B,C, the equality pB=pC and uniqueness of the exponent of the prime p in [L2] give B=C.

step 2.1F1L1L2algebra

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