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.
Identifer | oai:union.ndltd.org:VTETD/oai:vtechworks.lib.vt.edu:10919/80271 |
Date | January 1982 |
Creators | Marciniak, Zbigniew |
Contributors | Mathematics, Farkas, Daniel R., Green, E.L., Dickman, R.F., Jr., Feustel, Charles D., Hannsgen, Kenneth B. |
Publisher | Virginia Polytechnic Institute and State University |
Source Sets | Virginia Tech Theses and Dissertation |
Language | en_US |
Detected Language | English |
Type | Dissertation, Text |
Format | v, 85, [1] leaves, application/pdf, application/pdf |
Rights | In Copyright, http://rightsstatements.org/vocab/InC/1.0/ |
Relation | OCLC# 9008469 |
Page generated in 0.002 seconds