Principal congruences defined by terms

Question

I'm having trouble understanding an exercise of the book "A course in Universal Algebra".

Show that for any algebra $A$ and $a, b \in A$, $Θ(a,b) = t^*(s(\{(p(a,\bar{c}), p(b, \bar{c})): p(x, y_1, ...,y_n) \ is \ a \ term, c_1,...,c_n \in A\}))\cup \Delta_A$, where $t^∗( \ \ )$ is the transitive closure operator, i.e., for $Y \subseteq A × A$, $t^∗(Y)$ is the smallest subset of $A × A$ containing $Y$ and closed under $t$.

Here $Θ(a,b)$ is the principal congruence generated by $(a,b)$, i.e. the smallest congruence containing $(a,b)$.

In precedence $Θ$ was shown to be the algebraic closure operator generating the subuniverses of an algebra with universe $A × A$ and with the following foundamental operations:

$s((a,b))=(b, a)$

$t((a, b), (c, d))= \begin{cases} (a, d) \ \ if \ \ b=c \\ (a,b) \ \ otherwise \end{cases}$

For every $f$ in the type of $A$:

$f^{A × A}((a_1, b_1),...,(a_n, b_n))=(f^A(a_1,...,a_n), f^A(b_1,...,b_n))$

I suppose in the above notation $p(a, \bar{c})$ stands for $p(a, c_1, ..., c_n)$.

I'm looking for a sketch of the demonstration.

Answer 1

This seems to be rather simple:

For the direct inclusion (i.e., $\Theta(a,b) \subseteq t^*(...) \cup \Delta_A$), just use the fact that since a congruence is reflexive (closed under $\Delta$), symmetric (closed under $s$), transitive (closed under $t$) and compatible, just use the fact that for each $n$-ary basic operation $f$, we have that $f(x_1,\ldots,x_n)$ is a term; so plug it in the expression.

For the reverse inclusion (to show that that composition of operations applied to $(a,b)$ is contained in $\Theta(a,b)$), use (from the same book)

Theorem 10.3(a) Let $p$ be an $n$-ary term, $\theta$ a congruence of $\mathbf A$, and suppose $(a_i,b_i) \in \theta$. Then $$p(a_1,\ldots,a_n)\theta p(b_1,\ldots,b_n).$$

In this case, you would have for some $i$'s, $a_i=a$ and $b_i=b$, and for the others $a_i=b_i$ (the vector $\overline{c}$ on both sides).

You might have to polish the explanation a bit, but since you asked for a sketch...