Embedding a free algebra in a product.

Question

Let $A$ be an algebra; set $V=HSP(A)$. Then for any $X\neq\emptyset$, there is an embedding of the free algebra $F_{V}(X)$ into the algebra $A^{A^{X}}$, $\Psi\colon F_V(X)\rightarrow A^{A^X}$ given by $t\mapsto \epsilon_t$, where $\epsilon_t\colon A^X\rightarrow A$ is given by $f\mapsto \bar{f}(t)$ (here, $\bar{f}$ is the homomorphism from $F_V(X)$ to $A$ obtained as an extension of $f\in A^X$).

Here is my question: suppose $A$ has no proper subalgebra. The claim is that the embedding $\Psi$ is also subdirect. Can anyone help me see how this is the case? Thanks in advance!

Answer 1

I'll use the notation of the book I think you took this from (Burris and Sankappanavar).

If $K$ is a family of algebras of the same type and $X$ is a set of variables, then $$\Phi_K(X) = \{ \phi \in \mathrm{Con}\,\mathbf T(X): \mathbf{T}(X)/\phi \in IS(K) \}.$$

I'll also use the result of another (previous) exercise:

Let $\mathbf A$ be an algebra and $(\theta_i)_{i \in I}$ a family of congruences of $\mathbf A$. Let $\theta = \bigcap_{i \in I}\theta_i$. Then $\mathbf A/\theta$ is subdirectly embeddable into $\prod_{i \in I}\mathbf A/\theta_i$.

Now in this case, we have $K = \{\mathbf A\}$, and so if $\theta \in \Phi_K(X)$, then $\mathbf T(X)/\theta$ is embeddable into $\mathbf A$ (by the definition of $\Phi_K(X)$).
By the exercise mentioned above, $\mathbf F_V(X)$ is subdirectly embeddable into $\prod_{\theta \in \Phi_K(X)}\mathbf T(X)/\theta$.
Now, since $\mathbf T(X)/\theta$ is embeddable into $\mathbf A$, then this embedding is an isomorphism in the case in which $\mathbf A$ has no proper subalgebras.
Hence $\prod_{\theta \in \Phi_K(X)}\mathbf T(X)/\theta$ is isomorphic to $\prod_{\theta \in \Phi_K(X)}\mathbf A$ which is $\mathbf A^{|\Phi_K(X)|}$, and so $\mathbf F_V(X)$ is subdirectly embeddable into $\mathbf A^{|\Phi_K(X)|}$.
Now you're left with the task of showing that $|\Phi_K(X)| = |A|^{|X|}$.