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Show given bi-unary algebra is subdirectly irreducible.

Let $A$ be the the set of functions from $\omega$ to {$0,1$} and define the bi-unary algebra $\langle A, f, g \rangle $ as follows:

$f(a)(i) = a(i +1) \\ g(a)(i) = a(0)$

A definition states that an algebra $A$ is subdirectly irreducible if for ever subdirect embedding

$\alpha : A \to \prod_{i \in I}A_i$ there is an $i \in I$ such that $\pi_i \circ \alpha : A \to A_i $ is an isomorphism.

So if $A$ embeds into $\prod_{i \in I}A_i$ then the ith projection to $A_i$ is isomorphic to $A$ then we are subdirectly irreducible.

Additionally, there is a theorem that states an algebra $A$ is subdirectly irreducible iff there exists a congruence which is a minimum in $Con(A) - \{\Delta\}$ or $A$ is trivial.

In this case $A$ is not trivial. I don't think using congruences are the right idea here.

I just am not sure how to make this mapping work so that we get an isomorphism. I need to construct an $A_i$ such that its projection after embedding is isomorphic to $A$ but prior to that to construct it so that we have $A \le \prod_{i \in I} A_i$.

oliverjones
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1 Answers1

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Let us call $\mu_{\mathbf{A}}$, the monolith of $\mathbf{A}$, to that minimal congruence with respect to being different from $\Delta$.

As you point out, $\mathbf{A}$ is subdirectly irreducible iff there is such a congruence.

Define $a_0 : \omega \to \{0,1\}$ and $a_1 : \omega \to \{0,1\}$ by making $$a_0 : n \mapsto 0, \qquad a_1 : n \mapsto 1,$$ that is, they're both constants. Now we'll see that $\Theta(a_0,a_1)$, the least congruence relation having the pair $\langle a_0,a_1 \rangle$, is $\mu_{\mathbf{A}}$.

So let $\theta \in \mathbf{Con} \mathbf{A}$ be such that $\theta \neq \Delta$ and let $b,c : \omega \to \{0,1\}$ be such that $b \neq c$, and $\langle b, c \rangle \in \theta$.
Fix $i_0 \in \omega$ such that $b(i_0) \neq c(i_0)$ and suppose, without loss of generality, that $b(i_0)=0$ and $c(i_0)=1$. Then $$b \,\theta\, c \Longrightarrow f(b) \,\theta\, f(c) \Longrightarrow f^2(b) \,\theta\, f^2(c) \Longrightarrow \cdots \Longrightarrow f^{i_0}(b) \,\theta\, f^{i_0}(c).$$ But $f^{i_0}(b)(0) = b(i_0) = 0$ and $f^{i_0}(c)(0) = c(i_0) = 1$.
Thus, $a_0 = g(f^{i_0}(b)) \;\theta\; g(f^{i_0}(c)) = a_1$, that is $\Theta(a_0,a_1) \subseteq \theta$, whence $\mu_{\mathbf{A}} = \Theta(a_0,a_1)$.

amrsa
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