Abstract
AbstractWe investigate the problem of determining when IA(Fn(AmA)) is finitely generated for all n and m, with n ≥ 2 and m ≠ 1. If m is a nonsquare free integer then IA(Fn(AmA)) is not finitely generated for all n and if m is a square free integer then IA(Fn(AmA)) is finitely generated for all n, with n ≠ 3, and IA(F3(AmA)) is not finitely generated. In case m is square free, Bachmuth and Mochizuki claimed in ([7], Problem 4) that TR(AmA) is 1 or 4. We correct their assertion by proving that TR(AmA) = ∞.
Publisher
Canadian Mathematical Society
Cited by
3 articles.
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