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I am studying Universal Algebra using the book of Clifford Bergman and I encounter trouble in one of the exercises.

Task (2.12 - 3.) Let $L$ be a copy of $N_5$ with elements $\{a,b,c,u,v\}$ and $u<a<b<v$. Let $\theta$ be a congruence on $L$. Prove that $\theta\neq O_L$ implies $(a,b)\in\theta$.

Notation Here $N_5$ is the pentagon-shaped five element lattice, where $v$ serves as 1, $u$ as 0 and we have two linear sublattices $u<a<b<v$ and $u<c<v$. Other then this, elements are uncomparable.

By $0_L$ we mean trivial congruence where each block of equivalence is a singleton.

My approach So one can understand congruences just as partitioning elements of lattice into subsets (classes of equivalence) such that if I take two elements from one class of elements then both their meet and their join still lie in a given class of equivalence.

But if I take partition $\theta$ as $\{a\}$ and $\{b,c,u,v\}$ clearly $(a,b)\notin \theta$ but it seems to me that this is indeed a congruence.

I tried taking couples of elements, I tried thinking about it but it seems to me that if I take $x,y\in \{b,c,u,v\}$ then both their meet and join cannot be $a$

So I am clearly missing something here. Can anyone point out my mistake? Why $\theta$ is not congruence?

Shaun
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dmk
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2 Answers2

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Your definition is not quite correct. You state that an equivalence relation on a lattice is a congruence if each of its equivalence classes is a sublattice (i.e. closed under joins and meets). You need the equivalence classes to interact together as well:

if $x\mathrel{\theta}y$ and $z\mathrel{\theta}w$, then $x\wedge z\mathrel{\theta}y\wedge w$ and $x\vee z\mathrel{\theta}y\vee w$. (In particular, $x$ and $z$ do not need to belong to the same equivalence class.)

In other words, the equivalence classes need to form a lattice with well-defined operations. Can you find an example where your $\theta$ fails this condition?

Eran
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Consider the interval $[u,b]$ in $L$. Writing $x\equiv y$ for $\langle x,y\rangle\in\theta$, we have

$$u=u\land a\equiv b\land a=a\,,$$

since $u\equiv b$ and $a\equiv a$. But then $u\equiv a$, contradicting the definition of $\theta$. (The same argument shows that in general lattice congruence classes are convex.)

Brian M. Scott
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