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Idempotents in group rings

The von Neumann finiteness problem for k[G] is still open. Kaplansky proved it in characteristic zero. He used the nonvanishing of the trace: tr(e) = 0 implies e = 0 for any idempotent e ∊ k[G]. Assume now that char k = p > 0. Now tr can vanish on nonzero idempotents. Instead, we study the lifted trace ltr. For e = e² ∊ k[G], define ltr(e) by ê(1) where ê = Σ{x ∊ G}ê(x)x lifts e. Here ê is an infinite series with |ê(x)|<sub>p</sub>→0, where each ê(x) lives in the Witt vector ring of k. We prove that ltr(e) depends on e only, it is a p-adic integer and ltr(e) = ltr(f) if f is equivalent to e. Also ltr(e) ∊ Q and ltr(e) = 0 implies e = 0 if G is polycyclic-by-finite. We conjecture that -log<sub>p</sub>|ltr(e)|<sub>p</sub> < |supp(e)|. We prove this for e central and for e = e² ∊ k[G] with |G| ≤ 30. In the last section, we give the example of an idempotent e such ath supp(f) is infinite for all f ~ e. Finally we estimate |<supp(e)>| for central idempotents e. / Ph. D.

Identiferoai:union.ndltd.org:VTETD/oai:vtechworks.lib.vt.edu:10919/80271
Date January 1982
CreatorsMarciniak, Zbigniew
ContributorsMathematics, Farkas, Daniel R., Green, E.L., Dickman, R.F., Jr., Feustel, Charles D., Hannsgen, Kenneth B.
PublisherVirginia Polytechnic Institute and State University
Source SetsVirginia Tech Theses and Dissertation
Languageen_US
Detected LanguageEnglish
TypeDissertation, Text
Formatv, 85, [1] leaves, application/pdf, application/pdf
RightsIn Copyright, http://rightsstatements.org/vocab/InC/1.0/
RelationOCLC# 9008469

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