Arithmeticity of stabilizers of ideals in group rings
A conjecture of Bachmuth, Baumslag, Dyer and Mochizuki states that if G is a two-generator metabelian group then the image of Aut(G) in \(Aut(G_{ab})\) is finitely presented. This can be reduced to the question of whether the stabilizer in GL(2,\({\mathbb{Z}})\) of an ideal in \({\mathbb{Z}}[x,x^{-1},y,y^{-1}]\) is finitely presented. This question is answered in the affirmative. This is done by first proving the main theorem: if A is the free Abelian group of finite rank n\(>1\) and I an ideal of either of the group rings \({\mathbb{Q}}[A]\) or \({\mathbb{Z}}[A]\) such that the stabilizer Stab(I) in GL(n,\({\mathbb{Z}})\) and all its subgroups of finite index are irreducible subgroups of GL(n,\({\mathbb{Q}})\) (i.e. Stab(I) and all its subgroups of finite index act rationally irreducibly on A) then Stab(I) is commensurable with an arithmetic subgroup of GL(n,\({\mathbb{Z}})\). (Two subgroups are commensurable if their intersection has finite index in each.) Given this theorem it is easy to deduce that Stab(I) is finitely presented if I is an ideal of \({\mathbb{Z}}[x,x^{-1},y,y^{-1}]\). For, if Stab(I) satisfies the hypothesis of the main theorem then it is arithmetic and hence finitely presented by a result of Borel and Harish- Chandra [see \textit{A. Borel}, Proc. Internat. Congr. Math. Stockholm 1962, 10-22 (1963; Zbl 0134.165)] and if not then Stab(I) has a subgroup H of finite index which leaves a nontrivial subspace of \({\mathbb{Q}}^ 2\) invariant, H is solvable and hence polycyclic by a theorem of \textit{A. I. Mal'tsev} [Mat. Sb., Nov. Ser. 28(70), 567-588 (1951; Zbl 0043.023); transl. in Am. Math. Soc. Transl., II. Ser. 2, 1-21 (1956)] so again Stab(I) is finitely presented. The starting point for the proof of the main theorem is a result of \textit{G. M. Bergman} [Trans. Am. Math. Soc. 157, 459-470 (1971; Zbl 0197.17102)] which states that if I is an ideal of \({\mathbb{Z}}[A]\) satisfying the hypotheses of the main theorem then either \(I=0\) or \(\dim_ kk[A]/I<\infty\). Arithmetic groups enter the picture via a linearization technique introduced by \textit{L. Auslander} and \textit{G. Baumslag} [Bull. Am. Math. Soc. 73, 716-717 (1967; Zbl 0149.269)]. For prime ideals the proofs are more straightforward because of \textit{J. E. Roseblade's} generalization of Bergman's Theorem [Proc. Lond. Math. Soc., III. Ser. 36, 385-447 (1978; Zbl 0391.16008)]. Here the authors prove that if J is a field or \({\mathbb{Z}}\) and I a semiprime ideal of J[A] then Stab(I) is commensurable with an arithmetic group.
scientific article; zbMATH DE number 3843939
scientific article