Consider a cluster algebra of finite type $\mathrm{A}$. The set of all (distinct) cluster variables is of finite cardinality, denote it by $k$, for such algebra. Is it true that, for an arbitrary choice of the ground ring, every such algebra is isomorphic to the polynomial ring (over the ground ring, I guess) in (suitably defined) $k$ variables moded out by the ideal generated by the exchange relations?
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2$\begingroup$ Since these are coordinate rings of Grassmannians of the form $\mathrm{Gr}(2,n)$, for which the defining equations are the 3-term Plucker relations, this should be true. $\endgroup$– Sam HopkinsCommented Jan 12, 2021 at 23:52
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$\begingroup$ Yes, when we don't invert the coefficients, this is true. I am still not sure if it holds if we do invert them (the coefficients). $\endgroup$– amator2357Commented Jan 13, 2021 at 9:55
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1$\begingroup$ (2,n) Grassmannians only provide a model of a type A cluster algebra with a particular choice of coefficients. The coordinate ring of its big Schubert cell is another, with different coefficients and different relations. It's not clear to me that using models in this way will give an answer to the question as asked. Nevertheless, I think that this should be true (being finite type A is very strong) but I don't know of a reference or how to prove it directly. $\endgroup$– Jan GrabowskiCommented Jan 13, 2021 at 12:33
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$\begingroup$ I see. Thank you, Jan. $\endgroup$– amator2357Commented Jan 13, 2021 at 14:44
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