Magnus embeddings and residual nilpotence
Let G be a group, \({\mathbb{Z}}G\) its integral group ring, and \(\Delta\) (G) the augmentation ideal. Put \(\Delta^{\omega}(G)=\cap^{\infty}_{m=1}\Delta^ m(G)\). If \(G=F/R\), where F is a non cyclic free group, then \(\Delta^{\omega}(G)=0\) if and only if \(\cap^{\infty}_{m=1}({\mathfrak f}^ m+{\mathfrak r})={\mathfrak r}\), where \({\mathfrak f}=\Delta (F)\) and \({\mathfrak r}=\Delta (R){\mathbb{Z}}F\). For \(q\in \{0,1\}\), write \({\mathfrak r}^ q={\mathfrak r}\) if \(q=1\) and \({\mathfrak r}^ q={\mathfrak f}\) if \(q=0\). Then a theorem of \textit{C. K. Gupta} and \textit{N. D. Gupta} [Houston J. Math. 7, 43-52 (1981; Zbl 0467.20010)] states that \[ \cap^{\infty}_{m=1}({\mathfrak f}^ m+{\mathfrak r}^{q_ 1}...{\mathfrak r}^{q_ n})={\mathfrak r}^{q_ 1}...{\mathfrak r}^{q_ n}\text{ if and only if }\cap^{\infty}_{m=1} ({\mathfrak f}^ m+{\mathfrak r})={\mathfrak r}, \] that is, the augmentation ideal of G is residually nilpotent. Here \((q_ 1,...,q_ n)\) is an n-tuple different from (0,0,...,0) with \(q_ i\in \{0,1\}\). Write \(M(q_ 1,...,q_ n)=F\cap (1+{\mathfrak r}^{q_ 1}...{\mathfrak r}^{q_ n})\). In the same paper it is further claimed that \(\Delta^{\omega}(G)=0\) if and only if \(\Delta^{\omega}(F/M(q_ 1,...,q_ n))=0\). In the paper under review it is pointed out that the proof is incorrect, and the authors are able to go some distance towards re-proving this assertion. Firstly, if \(\Delta^{\omega}(F/R)=0\), then \(\Delta^{\omega}(F/M(q_ 1,...,q_ n))=0\). This is proved by using a faithful representation of \(F/M(q_ 1,...,q_ n)\), due to Gupta and Gupta, by triangular matrices whose entries are polynomials with coefficients in the group ring of a direct power of \(G=F/R\). In the converse direction the important case is when \(M(q_ 1,...,q_ n)\) is minimally expressed, that is, it differs from \(M(q_ 1,...,q_{n-1})\) and \(M(q_ 2,...,q_ n)\). In that case, if G is periodic or torsion free, \((q_ 1,q_ n)\neq (0,0)\), and \(F/M(q_ 1,...,q_ n)\) is residually nilpotent, then \(\Delta^{\omega}(G)=0\).
scientific article; zbMATH DE number 4008633