Let be a nontrivial semigroup. Show that the following are equivalent.
- is subdirectly irreducible.
- The set of proper congruences on is closed under arbitrary intersections.
- has a -least proper congruence.
We will follow the strategy . Note that the condition that is nontrivial is required to have the set of nonidentity congruences be nonempty.
: Suppose is a subdirectly irreducible semigroup. Let be a nonempty set of proper congruences on , and consider the product .
Suppose . (That is, suppose the intersection over is the identity congruence.) By II.1.3, the family of congruences in separate . By II.1.4, is a subdirect product of via the map . Since is subdirectly irreducible, then for some congruence , the map is an isomorphism. Clearly then , a contradiction. So .
: If is the set of all proper (i.e. nontrivial) congruences on , then (by our hypothesis) is a nontrivial congruence. Certainly if then , so that is -least among the elements of .
: Suppose has a -least proper congruence . Suppose further that is a subdirect product; that is, is injective and each is surjective. By II.1.4, the family of congruences separate the elements of . By II.1.3, we have . Since has a -least proper congruence, one of the must be . So is an isomorphism.