4. The internal point of view

4.1 Observation. If ${\cal A}$ has two congruence relations $\theta _ 1$ and $\theta _ 2$ with $\theta _ 1 \cap \theta _ 2 = 0$, then ${\cal A}$ has a subdirect representation ${\cal A} \hookrightarrow {\cal A}/\theta _ 1 \times {\cal A}/\theta _ 2$.

The reason is that the two natural homomorphisms of ${\cal A}$ onto ${\cal A}/\theta _ i$ ($i=1,2$) give a homomorphism of ${\cal A}$ into the direct product with kernel $\theta _ 1 \cap \theta _ 2 = 0$, so the homomorphism is an embedding. Composing with the projections gives back the natural homomorphisms, so this is a subdirect product.

More generally, if ${\cal A}$ has congruence relations $\theta _ \gamma, \gamma \in \Gamma$ with $\cap _ \gamma \theta _ \gamma = 0$, then ${\cal A}/\cap _ {\gamma \in \Gamma} \theta _ \gamma \hookrightarrow \prod _ {\gamma \in \Gamma} A/\theta _ \gamma$.

4.2 Observation. Up to isomorphism, any subdirect representation of ${\cal A}$ is the same as an appropriate subdirect representation of the form given in Observation .

The reason: Given a subdirect representation $\phi: {\cal A} \hookrightarrow \prod _ {\gamma \in \Gamma} {\cal B} _ \gamma$, let ${\cal A}' = \phi({\cal A})$, the image of $\phi$. Then for each $\gamma \in \Gamma$, the coordinate projection $\pi _ \gamma$ takes ${\cal A}'$ onto ${\cal B} _ \gamma$ with some kernel $\theta _ \gamma$. The intersection of these kernels is the 0 congruence relation, since in any product two elements are equal when their projections on all factors are the same. Moreover, by the first isomorphism theorem, ${\cal B} _ \gamma \cong A'/\theta _ \gamma$. The mappings

${\cal A} \hookrightarrow \prod _ {\gamma \in \Gamma} {\cal B} _ \gamma \stackrel{\pi _ \gamma}{\to} {\cal B} _ \gamma$

become

${\cal A}' \hookrightarrow \prod _ {\gamma \in \Gamma} {\cal A}'/\theta _ \gamma \stackrel{\pi _ \gamma}{\to} {\cal A}'/\theta _ \gamma$, up to isomorphism.

4.3 Proposition. The following conditions are equivalent:

(1) ${\cal A}$ is subdirectly irreducible;

(2) $\cap _ {\gamma \in \Gamma} \theta _ \gamma = 0$ implies $\theta _ \gamma = 0$ for some $\gamma \in \Gamma$;

(3) $0 \in$   Con$({\cal A})$ is completely meet irreducible;

(4) Con$({\cal A})$ has a least element $> 0$ (the monolith of ${\cal A}$).

This gives an internal description of subdirect irreducibility.