Subdirect products

Question

My question actually regards to slightly degenerate case of a subdirect product:

Is every algebra a subdirect product of itself alone I.e. of the product only involving itself once?

Answer 1

From Burris and Sankappanavar, Ch. II.

Definition 8.1. An algebra $\mathbf A$ is a subdirect product of an indexed family $(\mathbf A_i)_{i \in I}$ of algebras if (i) $\mathbf A \leq \prod_{i \in I} \mathbf A_i$
(ii) $\pi_i(\mathbf A) = \mathbf A_i$, for each $i \in I$.

(Here, $\pi_i$ is the i-th projection, $\pi_j: \prod_{i\in I} \mathbf A_i \to \mathbf A_j$ defined by $\pi_j(a) = a(j)$.)

Now if $\mathbf A = \prod_{i \in I} \mathbf A_i$, then $\pi_i(\mathbf A) = \mathbf A_i$, by definition of the projections.
So a direct product is always a subdirect product.
In particular, you can take $I = \{1\}$ and $\mathbf A_1 = \mathbf A$, and conclude that $\mathbf A$ is a subdirect product of itself.