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The following problem is taken from Burris and Sankappanavar's A Course in Universal Algebra (11, pg 42).

Suppose $L$ is a distributive lattice and $a,b,c,d\in L$. Then $\langle a, b\rangle\in\Theta(c,d)$ if and only if $c\wedge d\wedge a = c\wedge d \wedge b$ and $c\vee d\vee a = c\vee d \vee b$.

Here, $\Theta(c,d)$ denotes the smallest congruence on $L$ that contains $(c,d)$, i.e., the principal congruence on $L$ generated by $(c,d)$.

Proving sufficiency is straightforward. However, I am not sure about where to get started on the necessary part. Any help is appreciated.

Mashiath Khan
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1 Answers1

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A possible way to tackle this problem (I did it that way years ago when I was studying that book, but there might be a more straightforward one...) is to consider $$\Phi = \{ \langle a,b\rangle \in L^2 : a\wedge c\wedge d = b\wedge c\wedge d \;\text{and}\; a\vee c\vee d = b\vee c\vee d\}.$$ Prove that $\Phi$ is a congruence (it's immediate that it's an equivalence relation).
Next, notice that clearly $\langle c, d \rangle \in \Phi$, whence $\Theta(c,d) \subseteq \Phi$, and then prove the reverse equality.

To prove this last equality, what I did was to show that if $\theta$ is a congruence such that $c\equiv_{\theta}d$ then from $a\equiv_{\Phi}b$ we can conclude that $a\equiv_{\theta}b$.

If you find something unclear in this reasoning, or if you still can't complete the proof, I can elaborate on this later on.

amrsa
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  • Thanks, that was helpful. The other inclusion is immediate from the sufficiency part of the proof. The important thing is the observation that $\Phi$ is indeed a congruence, and the rest falls into place. – Mashiath Khan Dec 26 '22 at 17:11
  • I have a question : Where is the distributive lattice hypothesis used? – UDAC Oct 11 '24 at 05:12
  • @UDAC In proving that $\Phi$ is a congruence, at some point you'll have to use it. Exactly which point may depend on your strategy, but perhaps you pick $(a_1,b_1),(a_2,b_2)\in\Phi$, and you need to prove that $$(a_1\wedge a_2,b_1\wedge b_2)=(a_1,b_1)\wedge(a_2,b_2)\in\Phi,$$ and likewise for the join. In doing so, you have two equalities to prove for each of the operations. One of these only uses meets or joins; in the other, they get entangled, and the way to prove the equalities is to use distributivity. – amrsa Oct 11 '24 at 08:42
  • @UDAC If you're having difficulties in proving $\Phi$ is a congruence relation, just ask another question, perhaps with a link to this question to make it clear that, although related, it's not a duplicate. Or it isn't clear to you that $\Phi$ is not a congruence relation if the lattice is not distributive; that's another possibility. – amrsa Oct 11 '24 at 08:45