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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Nonsplit torsion-free abelian extensions preserve absence of group-ring zero divisors

Statement

Let k be a commutative field and let H⊴G be a normal subgroup with torsion-free abelian quotient G/H. If k[H] has no zero divisors, then k[G] has no zero divisors. The group extension need not split; neither finite generation of G/H nor an order on all of G is assumed.

Facts & Assumptions

Given: The field k, normal subgroup H⊴G, and torsion-free abelian quotient A=G/H.

[F1]

Group-ring elements have unique finite expansions, multiplication is k-bilinear with [g][h]=[gh], and each [g] is a unit (The group ring R[G] of finitely supported formal R-linear combinations of group elements, The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G]).

[F2]

Every finitely generated torsion-free abelian group is isomorphic to Zr for some finite r (Finite torsion-free abelian freeness by explicit rational coordinates without choice).

[A1]

The ring k[H] has no zero divisors.

Proof

Given: Two nonzero elements f,g∈k[G] and [A1].

1.1F1F2givenconstruct

Let B be the subgroup of A generated by the images of the finite supports of f and g. It is finitely generated, torsion-free and abelian, so [F2] identifies it with Zr. Transport the lexicographic order from Zr to B. This is a total order invariant under addition. No order on the whole quotient or on G is needed. Let X,Y⊆B be the finite quotient supports of f,g. Choose lifts tx∈G for the finitely many elements of X∪Y∪(X+Y), using the same lift for any repeated element. By grouping terms in each coset, [F1] gives unique expressions f=∑x∈Xrxtx and g=∑y∈Ysyty with 0≠rx,sy∈k[H]. Distinct cosets have disjoint group bases, so these grouped coefficients are nonzero.

2.1F1step 1.1algebra

For these lifts put αx(s)=txstx−1 on k[H] and hx,y=txtytx+y−1∈H. Normality makes αx a ring automorphism, with inverse conjugation by tx−1. Direct multiplication gives (rxtx)(syty)=rxαx(sy)hx,ytx+y. This is the crossed-product multiplication on the finite supports. The factor hx,y is retained and is a unit; it is not assumed to be 1. Its compatibility with iterated multiplication is inherited from associativity in G. Thus the calculation applies to a nonsplit extension.

3.1A1F1step 1.1step 2.1algebra∎

Let x0=max⁡X and y0=max⁡Y. Every other pair (x,y) satisfies x+y<x0+y0: if x<x0, then x+y≤x+y0<x0+y0, and if x=x0 but y<y0, addition invariance gives the same strict inequality. Hence the coefficient of tx0+y0 in fg is the single term rx0αx0(sy0)hx0,y0. It is nonzero by [A1], invertibility of hx0,y0 and injectivity of αx0. By disjointness of coset bases, fg≠0. Since f,g were arbitrary, k[G] has no zero divisors. Only finitely many lifts were chosen, so the proof introduces no choice axiom.

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