Timeline for Rings that fail to satisfy the strong rank condition
Current License: CC BY-SA 4.0
10 events
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| Dec 11, 2018 at 18:08 | comment | added | Karl Lorensen | Although there are no known examples for group rings, I thought that there might be some other sorts of rings in the literature for which the smallest $n$ such that $R^{n+1}$ embeds in $R^n$ is larger than 1. | |
| Dec 11, 2018 at 17:55 | comment | added | Karl Lorensen | Also, this is equivalent to $R$ being a right Ore domain. | |
| Dec 11, 2018 at 17:54 | comment | added | YCor | OK thanks for clarifying. There exists torsion-free non-amenable groups with arbitrary large Tarski number. If, as Kaplansky's conjecture predicts, they have no zero divisor in their group algebra, then it follows that $(KG)^2$ embds into $KG$. So my approach is probably worthless. | |
| Dec 11, 2018 at 17:47 | comment | added | Karl Lorensen | Yes, it is an elementary fact that, if $R$ is a domain, then the strong rank condition is equivalent to there not being any right $R$-module monomorphism $R^2\to R$. | |
| Dec 11, 2018 at 17:38 | comment | added | YCor | Oh, but then reading more carefully Kielak's appendix to Bartholdi's paper seems to rather suggest that my expectation is not correct, and that non-amenability is equivalent to the existence of an embedding of $(KG)^2$ into $KG$ as $KG$-module (at least, it is when $KG$ has no zero divisor, which is conjecturally true when $G$ is torsion-free and proved in a number of cases). | |
| Dec 11, 2018 at 17:19 | comment | added | Karl Lorensen | YCor, Thanks for your interesting comments! That the smallest such $n$ may depend on the Tarski number certainly seems very plausible. Moreover, if true, it would disprove Conjecture 4.1 in Bartholdi's earlier paper "Gardens of Eden and amenability on cellular automata," J. Eur. Math Soc. 12 (2010), 241-248. I am not sure who originally made the conjecture, but it may have been Matt Brin. | |
| Dec 11, 2018 at 16:52 | comment | added | YCor | A large source of example is due to a construction of Bartholdi (arxiv.org/abs/1605.09133) namely if $G$ is a nonamenable group and $K$ any field, then there exists $n$ such that $(KG)^{n+1}$ has an injective right $KG$-module homomorphism into $(KG)^n$ (I think $n$ can be chosen regardless of $K$). This actually characterizes non-amenable groups. It's likely that $n$ cannot be chosen equal to $2$ in general, and even that the smallest $n$ depends on the size of a paradoxical decomposition of $G$ (Tarski number). | |
| Dec 11, 2018 at 16:48 | history | edited | YCor |
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| Dec 11, 2018 at 16:20 | review | First posts | |||
| Dec 11, 2018 at 16:29 | |||||
| Dec 11, 2018 at 16:17 | history | asked | Karl Lorensen | CC BY-SA 4.0 |