Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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A full-rank unimodular sublattice has index one

Statement

Let n≥1, let L be a free abelian group of rank n with an integer-valued symmetric bilinear form whose Gram determinant is 1 or −1, and let f:L→Zn be an isometric homomorphism, where Zn has its standard dot product. Then f(L)=Zn.

Facts & Assumptions

Given: A rank-n free abelian group L, a unimodular integral symmetric form on L, and an isometric homomorphism f to the standard dot-product lattice Zn.

[A1]

The stated Gram determinant has absolute value one; the argument is finite integral linear algebra and uses no choice axiom.

[F1]

If A∈Mn(Z) has nonzero determinant, then the subgroup AZn has index ∣det⁡A∣ in Zn. The index of a full-rank subgroup of Zn is the absolute determinant of a generating matrix

Proof

1.1givenA1chooseconstruct

Choose an integral basis of L and let B be the integer n×n matrix whose columns are the images of that basis under f. Since f is isometric, the Gram matrix of L is BTB.

2.1step 1.1algebra

Taking determinants gives 1=∣det⁡(BTB)∣=∣det⁡B∣2, hence det⁡B=1 or −1.

3.1F1step 2.1algebra∎

By [F1] and step 2.1, the image subgroup BZn=f(L) has index ∣det⁡B∣=1 in Zn. The quotient therefore has one element, which means BZn=Zn and f(L)=Zn.

Depends on

Used by

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