Subdirectly irreducible algebra

Remarks

Equivalently, $\mathbf{A}$ is subdirectly irreducible when, for every subdirect embedding $\alpha \colon \mathbf{A} \to \prod_{i \in I} \mathbf{A}_i$, some composite $\pi_i \circ \alpha$ is an isomorphism. This is the form that gives the name, through subdirect products. In terms of the congruence lattice: the intersection of all congruences other than $\Delta_A$ is not $\Delta_A$.

Example: the group $\Z/4\Z$ is subdirectly irreducible. Its congruences come from the subgroups $\set{0} \subset \set{0, 2} \subset \Z/4\Z$, which form a chain, so the congruence of $\set{0, 2}$ is the monolith.

Non-example: the group $\Z/6\Z$ is not subdirectly irreducible. The congruences modulo $2$ and modulo $3$ are both different from $\Delta$, and their intersection is $\Delta$, so there is no least one.

The ring $\Z$ is not subdirectly irreducible either, since the congruences modulo the primes intersect in $\Delta_\Z$.