M3–N5 characterization of distributivity

Remarks

If $L$ is distributive, so is every sublattice, and $M_3$ and $N_5$ are not distributive. Conversely, if $L$ has no copy of $N_5$, then $L$ is modular by Dedekind's characterization. In a modular lattice in which $a \wedge (b \vee c) \ne (a \wedge b) \vee (a \wedge c)$, put $d = (a \wedge b) \vee (b \wedge c) \vee (c \wedge a)$ and $e = (a \vee b) \wedge (b \vee c) \wedge (c \vee a)$. Then $d < e$, and the elements $d$, $e$, $(a \wedge e) \vee d$, $(b \wedge e) \vee d$, $(c \wedge e) \vee d$ form a sublattice isomorphic to $M_3$.

Example: the lattice of subspaces of $\R^2$ is not distributive, because the zero subspace, three distinct lines and $\R^2$ form a copy of $M_3$. The lattice of divisors of $12$ under divisibility, with $\gcd$ as meet and $\lcm$ as join, is distributive, so it contains neither shape.

Going deeper

For a student

This theorem is the standard way to show that a lattice is not distributive: one exhibits a diamond or a pentagon. It refines Dedekind's characterization of modular lattices, which forbids only the pentagon. For a finite distributive lattice, the structure is then described completely by Birkhoff's representation theorem. The proof constructs the forbidden sublattice explicitly from any triple that violates distributivity.