A question about Birkhoff's Theorem and subdirectly irreducible algebras

Question

By Birkhoff's Theorem, we know that every algebra is a subdirect product of subdirectly irreducible algebras. So every finite algebra is a subdirect product of finitely many subdirectly irreducible algebrass. Is it known that for which infinite algebra $A$, we can conclude that $A$ is a subdirect product of finitely many subdirectly irreducible algebras? Thanks in advance

Answer 1

As far as I know, the answer to your question is 'It is not known'. But in Birkhoff's original paper he proves a result related to your question:

One way to define 'subdirectly irreducible' is to say that $A$ is subdirectly irreducible if any intersection of nonzero congruences of $A$ is nonzero. Birkoff defines $A$ to be 'weakly (subdirectly) irreducible' if any finite intersection of nonzero congruences of $A$ is nonzero. Theorem 1 of Birkhoff's paper Subdirect unions in universal algebra (Bulletin of the AMS 1944) asserts

Theorem 1. (paraphrased) If the congruence lattice of $A$ satisfies the ascending chain condition, then $A$ is a subdirect product of finitely many weakly subdirectly irreducible algebras.

(I) this is only a sufficient condition, and (II) it involves weakly SI factors instead of truly SI factors.

Theorem 1 was proved earlier by Emmy Noether for commutative rings. Birkhoff's contributions were to note that (i) Noether's Theorem holds for any type of algebraic structure, and (ii) it is possible to remove the finiteness conditions in Noether's Theorem to obtain his general subdirect representation theorem.

Let me add that if the congruence lattice of $A$ satisfies the descending chain condition, then any weakly subdirectly irreducible factor is actually subdirectly irreducible. Thus, from Theorem 1 one can derive that if the congruence lattice of $A$ satisfies both the ascending and the descending chain conditions, then $A$ is a finite subdirect product of subdirectly irreducible algebras.