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...
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Format: | Others |
Language: | en_US |
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Virginia Polytechnic Institute and State University
2017
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Online Access: | http://hdl.handle.net/10919/80271 |