How to prove these algebras are subdirectly irreducible, using the claim about their congruence relations? Am I proceeding correctly?

Solution 1:

I will try and answer my own question, with huge help from @amrsa and his/her comments.

$P^l_5$

Congruence relations on $P^l_5$ $\Delta$ = $|0|1|2|3|4|$. (By $\Delta$, I denote the diagonal congruence relation).

Other congruence relations:

For any $a \in P^l_5$ and any congruence relation $\theta_1$: If $0 \theta a$, then $1 = f(0) \theta_1 f(a) = a+1$ and so on, which gets us to $3 \theta_1 4 => 4 \theta_1 4$ in the end.

Hence, $\theta_1 = |01234|$. (The congruence relation relating all elements is denoted $\bigtriangledown$ in Burris & Sankappanavar, however, I am not sure, how to call it and will use $\theta_i$).

This observation shows that any congruence relation, apart from $\Delta$, has to either have $0$ in a separate class, or all elements are in one class with zero. This will help me generate other congruence relations on $P^l_5$.

If I start from $1$ this time and assume another congruence relation, $\theta_2$, then $1 \theta_2 2 => 2 = f(1) \theta_2 f(2) = 3 => 3 \theta_2 4 => 4 \theta_2 4$. (Assuming that zero is in separate congruence class).

Hence, $\theta_2 = |0|1234|$.

Now I will proceed by assuming $1$ is also in separate class and generating another congruence class, starting from $2 \theta_3 3$. This gives me $\theta_3 = |0|1|234|$.

By similar procedure, $\theta_4 = |0|1|2|34|$.

In conclusion, the non-diagonal congruence relations on $P^l_5$ are: $|01234|$, $|0|1|2|34|$, $|0|1|234|$, $|0|1234|$.

Finding the monolith

The monolith here is the relation $|0|1|2|34|$ (also possible to be written as $\{3,4\}^2 \cup \bigcup_{x \in A}\{x\}^2$). It is because for aby $a,b \in P^l_5$, whenever $a \neq b$ and $a \theta b$ by any congruence relation $\theta$, then $3 \theta 4$ too.

To put it another way, from the relation $3 \theta 4$, no other relation can be derived.

Hence the $|0|1|2|34|$ is the monolith because it's contained in each of the other congruence relation (apart from the diagonal one). If there are any two different elements which are related, then $3$ and $4$ must be related too. Thus it's the least congruence among the non-trivial ones.

Whenever $a \neq b$ and $a \alpha b$ by some congruence relation $\alpha$, it follows that $a + 1 \alpha b + 1 => a + 2 \alpha b + 2$ ... $4 \alpha 4$. And the step before $4 \alpha 4$ hass to be $3 \alpha 4$ if $a \neq b$, so $3 \alpha 4$ always follows.

The pair $(3,4)$ is therefore the "witness" of subdirect irreducibility of the $P^l_5$.

$P^l_\infty$

Congruence relations on $P^l_\infty$ $\Delta$ = $|0|1|2|3|4|...$.

Other congruence relations:

I will use similar thought process as with $P^l_5$. However, this time, $0$ will be in the role of $4$, because it is mapped to itself. So now, any element will be either in separate class, or together in a class with $0$.

This means that naturally, one congruence relation is immediately $\theta_1 = |01234...|$.

Then, I will proceed by putting more and more elements into the class together with zero and this way generate the rest.

For $\theta_2$, suppose $|01|$ is one congruence class (which is possible, because $0 = f(0) \theta_2 f(1) = 0$ holds). Then, if $2 \theta_2 a$ for any $a \neq 0,1,2$, then $1 \theta_2 a$ and $0 \theta_2 a$ ass well, which goes against the assumption that $0, 1$ are in a separate class. Similar thinking about other elements shoes that there is always one congruence class containing zero, and the other elements are either in this class with zero, or they are in one-element class.

Hence, the (non-diagonal) congruence relations on $P^l_\infty$ look like this:

$|01|2|3|4...|$, $|012|3|4...|$, $|0123|4...|$, ...

Indeed, the $P^l_\infty$ is an infinite algebra and there is an infinite number of congruence relations on $P^l_\infty$.

Finding the monolith

Here, the monolith is $|01|2|3|4...|$, because for any $a,b \in P^l_\infty$, $a \neq b$, if they are related by any congruence $\theta$, then $0 \theta 1$ too.

Whenever $a \neq b$ and $a \alpha b$ by some congruence relation $\alpha$, it follows that $a - 1 \alpha b - 1 \implies a - 2 \alpha b - 2 \implies$ ... $0 \alpha 0$. And the "step" before $0 \alpha 0$ has be $0 \alpha 1$ if $a \neq b$, so $0 \alpha 1$ always follows.

Hence, this algebra is again subdirectly irreducible and the pair $(0,1)$ is the "witness" of subdirect irreducibility.