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I'd like to prove (exercise 9.5 in Roman's Lattices and Ordered Sets, p.203) that the lattice $M_3$

enter image description here

is simple, meaning that the only congruences on $M_3$ are the trivial ones (the 'equality' congruence, i.e. $\{(x,x); x\!\in\!M_3\}$, and the 'everything' congruence, i.e. $\{(x,y); x,y\!\in\!M_3\}$).

Attempt of proof: By the symmetry of $M_3$, it suffices to prove that for any congruence $\theta$ on $M_3$: (i) if $0 \theta a$, then $\theta\!=\!M_3\!\times\!M_3$; (ii) if $a \theta b$, then $\theta\!=\!M_3\!\times\!M_3$.

(i): If $0 \theta a$, then by the definition of a congruence, $(0\!\vee\!b)\theta(a\!\vee\!b)$, i.e. $b\theta1$. Then $(c\!\wedge\!b)\theta(c\!\wedge\!1)$, i.e. $0\theta c$. Then $(a\!\vee\!0)\theta(a\!\vee\!c)$, i.e. $a\theta 1$. Then $(b\!\wedge\!a)\theta(b\!\wedge\!1)$, i.e. $0\theta b$. Then $(c\!\vee\!0)\theta(c\!\vee\!b)$, i.e. $c\theta 1$. Thus $\theta\!=\!M_3\!\times\!M_3$.

(ii): If $a \theta b$, then ???

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up vote 3 down vote accepted

If $a \theta b$, then $(a\land b)\theta(b\land b)$, i.e. $0\theta b$, and you're back to case (i).

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Ah, of course, it didn't occur to me to take $b$ twice; I was afraid I'd have to use some 'methods'. Thank you. – Leon May 25 '12 at 5:28

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