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

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.

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
• $$b\vee (j\wedge c) = b\vee (j\wedge a\wedge c) = b\vee (j\wedge a\wedge b) = b$$,
• $$(b\vee j)\wedge c = (b\vee (b\wedge a)\vee j)\wedge c = (b\vee a)\wedge c = c$$,
• Is finiteness of $L$ used in any step or is it sufficient to provide that every down-directed set has a minimum? – Michał Zapała Dec 9 at 18:35
• @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.) – Keith Kearnes Dec 9 at 19:07