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Say I got a lattice $L$, a bounded lattice $K$ with top-element $1$ and a homomorphism $\varphi : L \to K$, then $\varphi^{-1}(1)$ is a filter in L.

I wondered whether I can represent every filter $F \subset L$ in that way, i.e. I find some other bounded-above lattice such that $F$ is some preimage of the top element. This should be some kind of contraction of $F$ into a new, single element.

My idea would be to define an equivalence relation on $L$

$$a \sim b :\Leftrightarrow a = b \text{ or } a, b \in F$$

and show that it's in fact a congruence. Then $L/\sim$ should be the lattice I look for. Is this proposition in fact true and is there an easier proof? Thanks

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2 Answers 2

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This lattice

enter image description here

(known as "the diamond"), has no proper congruence and three proper filters.

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  • $\begingroup$ Nice, thank you. Has the fact that filters don't correspond to congruences have something to do with non-distributivity of the diamond (at least, in distributive lattices we had ultrafilters)? $\endgroup$
    – Dario
    Aug 4, 2013 at 20:52
  • $\begingroup$ @Dario Yes. In fact, the filters (or dually, ideals) that are (co)kernels of a congruence can be characterized in a very nice way math.bme.hu/~schmidt/papers/9.pdf . Allow me to note that, in general lattice theory, the prevailing convention is to use ideals (not filters) when formulating a theorem. You should bear this fact in mind when googling for an answer. $\endgroup$ Aug 4, 2013 at 22:15
  • $\begingroup$ Oh, eccelent, this characterisation was what I was looking for. Thanks! $\endgroup$
    – Dario
    Aug 4, 2013 at 23:06
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Not every ideal is a kernel of a congruence [Steven Roman, Lattices and Ordered Sets, Springer, 2008, p.78, Example 3.40]. Dually, not every filter is a preimage of the top element.

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  • $\begingroup$ Thanks. Could you elaborate a bit please, though? $\endgroup$
    – Dario
    Aug 4, 2013 at 11:28
  • $\begingroup$ To elaborate what? Duality? $\endgroup$ Aug 4, 2013 at 11:38
  • $\begingroup$ I can't access the book right now as can't people who just stumble upon this question/answer. It would be great if you could give a brief summary how to find the above ideal and in which lattices. $\endgroup$
    – Dario
    Aug 4, 2013 at 13:02
  • $\begingroup$ It is easier for me to send to you the book, if you will write me by e-mail. OK? $\endgroup$ Aug 4, 2013 at 13:13

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