Idempotents in group rings

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Virginia Polytechnic Institute and State University

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)|p→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 -logp|ltr(e)|p < |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.