Let $H$ be a Heyting Algebra based on the $2 \times 3$ lattice. Find the subvarieties of $V(H)$.
$V(H)$ is definitely a locally finite variety since it is based on a finite algebra. In previous problems looking at finitely generated subdirectly subalgebras since they will generate the variety. Heyting algebras have distributive lattice reducts. And for distributive lattices, the two-element chain is the only subdirectly irreducible. Also, every Heyting Algebra with a unique coatom is SI. So that gives the the 2 and 3 element chain is SI.
I suspect that he subvarieties are the trivial one, $V(2), V(3)$.
Does locally finite need to be used here or is the SI subalgebras enough to look at?
Since $\mathbf 2$ is a subalgebra of ${\mathbf 3}$, $V({\mathbf 2}\times {\mathbf 3})$ is contained in $V({\bf 3}\times {\mathbf 3})=V({\mathbf 3})$. The reverse containment follows from the fact that ${\mathbf 3}$ is a quotient of ${\mathbf 2}\times {\mathbf 3}$. Thus $V({\mathbf 2}\times {\mathbf 3}) = V(\mathbf 3)$. Now convince yourself that the only SI's in $V(\mathbf 3)$ are ${\mathbf 2}$ and ${\mathbf 3}$, hence the only subvarieties are trivial, $V(\mathbf 2)$, and $V(\mathbf 3)$.
Does locally finite need to be used here or is the SI subalgebras enough to look at?
You might use Jonsson's Theorem to prove that the SI's in $V({\mathbf 3})$ are just ${\mathbf 2}$ and ${\mathbf 3}$. The fact that $V({\mathbf 3})$ is finitely generated enters in here.