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A pair of congruences $\theta$ and $\theta^*$ are called factor congruences if

$\theta \vee \theta^*$ = full congruence. $\nabla$

$\theta \wedge \theta^*$ = trivial congruence. $\triangle$

I need to prove that every chain has only two factor congruences which are $\triangle$ and $\nabla$. But I do not seem to understand which property of chains can be used here to start with the proof.

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  • $\begingroup$ The property you want to use is that every element in a chain is comparable to every other element. $\endgroup$ – Eran Apr 14 '14 at 22:22
  • $\begingroup$ Yeah I thought of that property but how do I use it in congruences. I find them unrelated because when I visualize congruences I see them as a set of tuples. But how can I use each member being related (in orderly manner) to other in this $\endgroup$ – Alvis Apr 15 '14 at 3:47
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Let $\mathbf{C}=(C,\vee,\wedge)$ be a lattice that is a chain. Suppose $\mathbf{C}$ has two nontrivial factor congruences $\phi,\psi\in\mathrm{Con}\,\mathbf{C}$, i.e. $\phi\vee\psi=\nabla$, $\phi\wedge\psi=\Delta$, and $\phi,\psi\neq\nabla$.

Define $h:\mathbf{C}\to\mathbf{C}/\phi\times\mathbf{C}/\psi$ by $h(c)=(c/\phi,c/\psi)$. It is easy to check that this is a lattice isomorphism. Also, since $\phi,\psi\neq\nabla$, we have that $\mathbf{C}/\phi$ and $\mathbf{C}/\psi$ are not trivial.

Now you just need to convince yourself that the product of two nontrivial lattices cannot be a chain (Hint: find a pair of incomparable elements). Thus we have a contradiction! Therefore we do not have two nontrivial factor congruences.

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