A fundamental result in universal algebra is the Subdirect Representation Theorem, which tells us how to decompose an algebra \(A\) into its "basic parts". Formally, we say that \(A\) is a subdirect product of \(A_1\), \(A_2\), ..., \(A_n\) when \(A\) is a subalgebra of the product
\[
A_1\times A_2\times\cdots\times A_n
\]
and for each index \(1\le i\le n\) we have for the projection \(\pi_i\) that \(\pi_i(A)=A_i\). In other words, a subdirect product "uses each component completely", but may be smaller than the full product.
A trivial circumstance is that \(\pi_i:A\to A_i\) is an isomorphism for some \(i\). The remaining components would then be superfluous. If an algebra \(A\) has the property than any way of representing it as a subdirect product is trivial in this sense, we say that \(A\) is "subdirectly irreducible".
Subdirectly irreducible algebras generalize simple algebras. Subdirectly irreducible groups include all simple groups, as well as the cyclic \(p\)-groups \(\mathbb{Z}_{p^n}\) and the Prüfer groups \(\mathbb{Z}_{p^\infty}\).
In the case of lattices, there is no known classification of the finite subdirectly irreducible (or simple) lattices. This page (https://math.chapman.edu/~jipsen/posets/si_lattices92.html) by Peter Jipsen has diagrams showing the 92 different nontrivial subdirectly irreducible lattices of order at most 8. See any patterns?
We know that every finite subdirectly irreducible lattice can be extended to a simple lattice by adding at most two new elements (Lemma 2.3 from Grätzer's "The Congruences of a Finite Lattice", https://arxiv.org/pdf/2104.06539), so there must be oodles of finite simple lattices out there.
)
