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Exterior powers of modules for group rings of polycyclic groups. (English) Zbl 0911.20006
The authors prove that every finitely presented abelian-by-polycyclic group has a metanilpotent normal subgroup of finite index (Theorem A), so answering in the negative the question (1973) by G. Baumslag, whether every finitely generated (f.g.) abelian-by-polycyclic group can be embedded into a finitely presented (f.p.) group in the same class. This question of G. Baumslag [Proc. 2nd internat. Conf. Theory of Groups, Canberra 1973, Lect. Notes Math. 372, 65-74 (1974; Zbl 0304.20015)] has been the primary motivation for this paper.
Theorem A is deduced from the following fact (Theorem B): Let $$H$$ be a polycyclic group and $$M$$ a f.g. $$\mathbb{Z} H$$-module. If the exterior power $$\bigwedge^r M$$ (for some $$r\geq 2$$) is f.g., regarded as a $$\mathbb{Z} H$$-module with $$H$$ acting diagonally, then the factor-group of $$H$$ by the unique largest normal subgroup $$\eta_H(M)$$ of $$H$$ which acts nilpotently on $$M$$, is virtually nilpotent. To prove Theorem B the authors suppose it to be false and take $$H$$ to be a polycyclic group of the least possible Hirsch number for which the assertion of Theorem B fails. This last holds then also for all subgroups of finite index in $$H$$, so allowing to further suppose that $$H$$ is an orbitally sound, torsion-free polycyclic group, whose centre coincides with its FC-centre. It follows that $$C_H(M)=1$$ and $$M$$ is a prime $$\mathbb{Z} A$$-module for each abelian normal subgroup $$A$$ in $$H$$. Using J. Roseblade’s control theorems [Proc. Lond. Math. Soc., III. Ser. 36, 385-447 (1978; Zbl 0391.16008)] for prime ideals in group algebras of polycyclic groups, this argument is further reduced to consideration of polycyclic groups $$H$$ with $$A$$ a self-centralizing plinth of $$H$$ whose rank is greater than 1, with $$Q=H/A$$ being free abelian and with $$\eta_H(M)=1$$. Here, according to J. Roseblade [(*) J. Pure Appl. Algebra 3, 307-328 (1973; Zbl 0285.20008)], $$A$$ is a plinth of $$H$$ if $$A\otimes_{\mathbb{Z}}\mathbb{Q}$$ when treated as a $$\mathbb{Q} H_0$$-module is irreducible for every subgroup $$H_0$$ of finite index in $$H$$. It follows that $$Q$$ embeds into the unit group of some number field and the Dirichlet unit theorem gives $$r_0(Q)<r_0(A)$$ for ranks. Then, the following result (Theorem C) is used: Let $$G$$ be a split extension of a non-trivial f.g. free abelian group $$A$$ by a f.g. free abelian group $$Q$$ and let $$\mathbf k$$ be a locally finite field. Further, let $$S$$ be a $$Q$$-invariant subring of $${\mathbf k}A$$ and suppose that there exists a module for $$SQ$$ – this last being the subring generated by $$S$$ and $$Q$$ – which is f.g. for $${\mathbf k}Q$$ and not torsion for $$S$$. If $${\mathbf k}A$$ is integral over $$S$$, then $$r_0(Q)\geq r_0(A)$$. Supposing that Theorem C holds, together with the supposition that $$r\geq 2$$, $${\mathbf k}$$ is a finite field, $$M$$ is a f.g. $${\mathbf k}H$$-module which is torsion-free as a $${\mathbf k}A$$-module of rank $$\geq r$$ and $$\bigwedge^r M$$ is f.g. as a $${\mathbf k}H$$-module, the authors prove $$r_0(Q)\geq r_0(A)$$. This contradicts $$r_0(Q)<r_0(A)$$ above and so giving Theorem C $$\Rightarrow$$ Theorem B.
At last, it is proved Theorem C using J. Roseblade’s theorem C* [in (*)] and $$\dots$$ the theory of Hilbert-Serre dimension. For interesting details and wealth of ideas used in this truely remarkable paper on group rings the reader should consult the paper itself.

##### MSC:
 20C07 Group rings of infinite groups and their modules (group-theoretic aspects) 16S34 Group rings 20C12 Integral representations of infinite groups 20F19 Generalizations of solvable and nilpotent groups 20F05 Generators, relations, and presentations of groups 20E07 Subgroup theorems; subgroup growth 20F16 Solvable groups, supersolvable groups
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