6. Intersections of congruence relations

Suppose $\theta _ 1, \theta _ 2 \in$   Con$({\cal A})$. Let $\pi _ 1: {\cal A} \rightarrow {\cal A}/\theta _ 1$ and $\pi _ 2: {\cal A} \rightarrow {\cal A}/\theta _ 2$ be the natural homomorphisms. Combining these, we get a a homomorphism $\pi _ 1 \times \pi _ 2: {\cal A} \rightarrow {\cal A}/\theta _ 1 \times {\cal A}/\theta _ 2$ (not necessarily onto). What is its kernel? By considering when $a,a' \in A$ have equal images, we see that the kernel is $\theta _ 1 \cap \theta _ 2$. From this and the first isomorphism theorem we get this fact:

6.1 Theorem (subdirect embedding theorem). For an algebra ${\cal A}$ and $\theta _ 1, \theta _ 2 \in$   Con$({\cal A})$, there is a natural embedding of ${\cal A}/(\theta _ 1 \cap \theta _ 2) \hookrightarrow {\cal A}/\theta _ 1 \times {\cal A}/\theta _ 2$.

(``Subdirect'' means that the image of the embedding inside the product is large enough to map onto each factor. This will be important later.)

6.2 Corollary. If ${\cal A}$ has congruence relations $\theta _ 1, \theta _ 2$ with $\theta _ 1 \cap \theta _ 2 = 0$ (the equality relation), then ${\cal A} \hookrightarrow {\cal A}/\theta _ 1 \times {\cal A}/\theta _ 2$ (an embedding).