Free distributive lattices on a finite set exist and are finite, while free modular lattices on a finite set exist but are not finite when the set has at least 4 elements.



> 
> **Question**: Is there a class (presumably, a variety in the sense of universal algebra) $C$ of lattices larger than the class of distributive lattices such that free $C$-lattices on a finite set exist (for all finite sets) and are finite?
> 
> 
> 


If yes, is there a "largest" such class $C$?



On free lattices
A variety is called *locally finite* if it has the property that its finitely generated algebras are finite. This is equivalent to the property that its finitely generated free algebras are finite. So, the questions may be rewritten as:
(1) Is there a locally finite variety of lattices larger than the variety of distributive lattices?   

(2) Is there a largest locally finite variety of lattices?
Every variety generated by a single finite algebra is locally finite, so there are lots of locally finite varieties of lattices. Any variety generated by a finite, nondistributive lattice will properly contain the variety of distributive lattices, hence will be an answer to Question (1).
[There do exist locally finite varieties of lattices that are not generated by a single finite lattice, like the variety generated by all lattices of height 2.]
If there were a largest locally finite variety of lattices, then by the answer to Question (1) it would have to contain every finite lattice. But the variety generated by all finite lattices is the variety of all lattices, and the variety of all lattices is not locally finite. Hence the answer to Question (2) is No. 
**A comment on the comments:**
The class of semidistributive lattices forms a quasivariety, and therefore
has free objects over any set. But this quasivariety is not locally finite. The free semidistributive lattice on 3 generators is infinite.










