## Equivalent characterizations of subdirect irreducibility

Let $S$ be a nontrivial semigroup. Show that the following are equivalent.

1. $S$ is subdirectly irreducible.
2. The set of proper congruences on $S$ is closed under arbitrary intersections.
3. $S$ has a $\subseteq$-least proper congruence.

We will follow the strategy $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. Note that the condition that $S$ is nontrivial is required to have the set $\mathsf{Q}(S)$ of nonidentity congruences be nonempty.

$(1) \Rightarrow (2)$: Suppose $S$ is a subdirectly irreducible semigroup. Let $Q \subseteq \mathsf{Q}(S)$ be a nonempty set of proper congruences on $S$, and consider the product $\prod_{\sigma \in Q} S/\sigma$.

Suppose $\bigcap Q = \Delta$. (That is, suppose the intersection over $Q$ is the identity congruence.) By II.1.3, the family of congruences in $Q$ separate $S$. By II.1.4, $S$ is a subdirect product of $\{S/\sigma\}_Q$ via the map $s \mapsto ([s]_\sigma)$. Since $S$ is subdirectly irreducible, then for some congruence $\sigma$, the map $s \mapsto [s]_\sigma$ is an isomorphism. Clearly then $\sigma = \Delta$, a contradiction. So $\bigcap Q \supsetneq \Delta$.

$(2) \Rightarrow (3)$: If $\mathsf{Q}(S)$ is the set of all proper (i.e. nontrivial) congruences on $S$, then (by our hypothesis) $\bigcap \mathsf{Q}(S) = \tau \in \mathsf{Q}(S)$ is a nontrivial congruence. Certainly if $\sigma \in \mathsf{Q}(S)$ then $\tau \subseteq \sigma$, so that $\tau$ is $\subseteq$-least among the elements of $\mathsf{Q}(S)$.

$(3) \Rightarrow (1)$: Suppose $S$ has a $\subseteq$-least proper congruence $\tau$. Suppose further that $\varphi : S \rightarrow \prod_B S_\beta$ is a subdirect product; that is, $\varphi$ is injective and each $\pi_\beta \circ \varphi$ is surjective. By II.1.4, the family of congruences $\mathsf{ker}\ \pi_\beta \circ \varphi$ separate the elements of $S$. By II.1.3, we have $\bigcap \mathsf{ker}\ \pi_\beta \circ \varphi = \Delta$. Since $S$ has a $\subseteq$-least proper congruence, one of the $\mathsf{ker}\ \pi_\beta \circ \varphi$ must be $\Delta$. So $\pi_\beta \circ \varphi : S \rightarrow S_\beta$ is an isomorphism.