Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ???

share|improve this question

1 Answer 1

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).

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.