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Is there a finite lattice such that every join-irreducible element is left-modular and that is not semimodular?

I've been able to prove there are no such bounded atomistic lattices (not necessarily finite), using the fact that such a lattice is semimodular iff it is geometric.

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I argue the contrapositive, that if $L$ is a finite lattice that is not semimodular, then $L$ has a join irreducible that is not left modular.

Since $L$ is not semimodular, it contains elements $a, b$ such that $a\wedge b\prec a$ but $b\not\prec a\vee b$. Choose any $c$ such that $b < c < a\vee b$. Now choose any $j\leq a$ minimal for the property that $j\not\leq b$; $j$ is join irreducible. Moreover, $j$ is not left modular, since $(j,c)$ is not a modular pair. To verify this last claim, compute that

  • $b\vee (j\wedge c) = b\vee (j\wedge a\wedge c) = b\vee (j\wedge a\wedge b) = b$,
  • $(b\vee j)\wedge c = (b\vee (b\wedge a)\vee j)\wedge c = (b\vee a)\wedge c = c$,
    and these are different.

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    • $\begingroup$ Is finiteness of $L$ used in any step or is it sufficient to provide that every down-directed set has a minimum? $\endgroup$ – Michał Zapała Dec 9 at 18:35
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      $\begingroup$ @MichałZapała: The only place I use finiteness is where I assume that the difference of intervals $(a] - (a\wedge b]\;\; (= \{x\leq a|x\not\leq b\})$ has a minimal element. (This is not a down-closed set.) $\endgroup$ – Keith Kearnes Dec 9 at 19:07

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