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In T.Y. Lam's book Lectures on Modules and Rings, a ring $R$ is said to satisfy the strong rank condition if, for every natural number $n$, there is no right $R$-module monomorphism $R^{n+1}\to R^n$. Is there any known example of a ring that fails to satisfy this condition such that the smallest $n$ witnessing the failure of the condition is larger than $1$?

In other words, can anyone provide an example of ring $R$ such that there is no right $R$-module monomorphism $R^2\to R$ but there is a right $R$-module monomorphism $R^{n+1}\to R^n$ for some $n>1$?

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  • $\begingroup$ 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). $\endgroup$
    – YCor
    Commented Dec 11, 2018 at 16:52
  • $\begingroup$ 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 I am not sure who originally made the conjecture, but it may have been Matt Brin. $\endgroup$
    – Karl Lorensen
    Commented Dec 11, 2018 at 17:19
  • $\begingroup$ 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). $\endgroup$
    – YCor
    Commented Dec 11, 2018 at 17:38
  • $\begingroup$ 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 $R^2\to R$. $\endgroup$
    – Karl Lorensen
    Commented Dec 11, 2018 at 17:47
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Dec 11, 2018 at 17:54

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