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Nov 13, 2014 at 18:45 comment added Buschi Sergio Yes, I have defined $Simp(\mathcal{A}):=\mathbb{R}[\mathcal{A}]/I$ where $\mathbb{R}[\mathcal{A}]$ is the vector space with base the set of elements of $\mathcal{A}$, this is also a algebra: for distributivity its enough define the multiplication between elements of the base,and let $A\cdot B:= A \cap B$, and $I$ is the $\mathbb{R}$-ideal defined as above.
Nov 13, 2014 at 7:59 comment added Dominic van der Zypen OK I see. If I understand correctly, the natural map works this way: $A\in\mathcal{A} \mapsto [\chi_A]\in \mathbb{R}[\mathcal{A}]/I$. The natural map is injective if for $A\neq B\in\mathcal{A}$ we have $\chi_A - \chi_B \notin I$, right?
Nov 12, 2014 at 20:59 comment added Buschi Sergio My apologies for my mistakes on formalisms and for my bad English. Of course this true if $\mathcal{A}$ is a lattice of subset's of a set $X$, in this case we can prove that $Simp(\mathcal{A})=\mathbb{R}[\mathcal{A}]/I $ (where $I$ is the a ideal generated by $(*)$ or $(**)$, see above). I use this relation for define it for a general lattice $\mathcal{A}$, and then define $Simp(\mathcal{A}):=\mathbb{R}[\mathcal{A}]/I $ (where $I$ is defined with the some kind of relations) of course I asking about conjecture's about this generalization. THank you anyway.
Nov 12, 2014 at 10:28 history answered Dominic van der Zypen CC BY-SA 3.0