# congruence lattice

###### Proof.

It is not hard to see that $\operatorname{Con}(A)$ is a topped intersection structure. As a result, $\operatorname{Con}(A)$ is a complete lattice. Since it is also a subset of the lattice $E(A)$ of equivalence relations  on $A$, the only remaining item to show is that it is a sublattice of $E(A)$. In other words, we need to show that if $\mathcal{C}$ is a family of congruence relations on $A$, so is $\Theta:=\bigvee\mathcal{C}$.

For convenience, let us introduce some notational devices:

• $\mathbf{n}:=\{1,\ldots,n\}$;

• for any $a_{k},b_{k}\in A$ and $\Theta_{k}\in\operatorname{Con}(A)$, where $k\in\mathbf{n}$, we mean

 $(a_{1},\ldots,a_{n})\equiv(b_{1},\ldots,b_{n})\pmod{(\Theta_{1},\ldots,\Theta_% {n})}$

by $a_{k}\equiv b_{k}\pmod{\Theta_{k}}$, for each $k\in\mathbf{n}$;

• $(a_{1},\ldots,a_{n})\equiv(b_{1},\ldots,b_{n})\pmod{\Theta}$ means $a_{k}\equiv b_{k}\pmod{\Theta}$, for each $k\in\mathbf{n}$;

• Finally, when $c_{k}\equiv c_{k+1}\pmod{\Theta_{k}}$, where $k\in\mathbf{n-1}$, we write

 $c_{1}\lx@stackrel{{\scriptstyle\Theta_{1}}}{{\equiv}}c_{2}\lx@stackrel{{% \scriptstyle\Theta_{2}}}{{\equiv}}\cdots\lx@stackrel{{\scriptstyle\Theta_{n-1}% }}{{\equiv}}c_{n}.$

Let us call this an (of $n$).

Back to the proof. Let $\omega$ be an $n$-ary operator on $A$ and $(a_{1},\ldots,a_{n})\equiv(b_{1},\ldots,b_{n})\pmod{\Theta}$. We want to show that $\omega(a_{1},\ldots,a_{n})\equiv\omega(b_{1},\ldots,b_{n})\pmod{\Theta}$.

The proof now breaks up into two main steps:

###### Step 1.

For each $i\in\mathbf{n}$, $a_{i}\equiv b_{i}\pmod{\Theta}$ means we have a finite

 $a_{i}=c_{1}\lx@stackrel{{F_{1}}}{{\equiv}}c_{2}\lx@stackrel{{% F_{2}}}{{\equiv}}\cdots\lx@stackrel{{F_{p-1}}}{{% \equiv}}c_{p}=b_{i},$

where each $F_{i}$ is a congruence in $\mathcal{C}$, and each $c_{i}\in A$. Now the lengths of the sequences  may vary by $i$. The idea is to show that we can in fact pick the sequences so that they all have the same length.

To see this, take the first two pairs of congruent elements $a_{1}\equiv b_{1}\pmod{\Theta}$ and $a_{2}\equiv b_{2}\pmod{\Theta}$, and expand them into two finite

1. 1.

$a_{1}=c_{1}\lx@stackrel{{F_{1}}}{{\equiv}}c_{2}\lx@stackrel{{% F_{2}}}{{\equiv}}\cdots\lx@stackrel{{F_{p-1}}}{{% \equiv}}c_{p}=b_{1}$, and

2. 2.

$a_{2}=d_{1}\lx@stackrel{{G_{1}}}{{\equiv}}d_{2}\lx@stackrel{{% G_{2}}}{{\equiv}}\cdots\lx@stackrel{{G_{q-1}}}{{% \equiv}}d_{q}=b_{2}$,

where $c_{i},d_{j}\in A$ and $F_{i},G_{j}\in\mathcal{C}$. If $q, we may lengthen the second chain so it has the same length as the first one:

 $a_{2}=d_{1}\lx@stackrel{{G_{1}}}{{\equiv}}d_{2}\cdots\lx@stackrel% {{G_{q-1}}}{{\equiv}}d_{q}\lx@stackrel{{G_{q}}}{{% \equiv}}d_{q+1}\lx@stackrel{{G_{q+1}}}{{\equiv}}\cdots% \lx@stackrel{{G_{p-1}}}{{\equiv}}d_{p},$

where $G_{q-1}=G_{q}=\cdots=G_{p-1}$ and $d_{q}=\cdots=d_{p}=b$.

The above argument and an induction  step on $n$ show that we can indeed make all the “expanded” the same length. As a result, without loss of generality, we assume that all the expanded chains have the same length, say $r+1$.

Instead of writing all $n$ chains, let us use our notational device, and we have the following single (again, we may write it in this fashion because all chains are now assumed to have the same finite length of $r+1$):

 $\xymatrix{(c_{11},\ldots,c_{1n})\ar@3{-}[rr]^{-}{(F_{11},\ldots,F_{1n})}&&(c_{% 21},\ldots,c_{2n})\ar@3{-}[rr]^{-}{(F_{21},\ldots,F_{2n})}&&\cdots\ar@3{-}[rr]% ^{-}{(F_{r1},\ldots,F_{rn})}&&(c_{r+1,1},\ldots,c_{r+1,n})}$

where each $c_{ij}\in A$, each $F_{ij}\in\mathcal{C}$, with $(i,j)\in(\mathbf{r+1})\times\mathbf{n}$, and that $(a_{1},\ldots,a_{n})=(c_{11},\ldots,c_{1n})$ and $(c_{r+1,1},\ldots,c_{r+1,n})=(b_{1},\ldots,b_{n})$.

Look at the first congruence pair $\xymatrix{(c_{11},\ldots,c_{1n})\ar@3{-}[rr]^{-}{(F_{11},\ldots,F_{1n})}&&(c_{% 21},\ldots,c_{2n})}$. This can be expanded into an of length $n$ as follows:

 $\xymatrix{(c_{11},c_{12},\ldots,c_{1n})\ar@3{-}[r]^{-}{F_{11}}&(c_{21},c_{12},% \ldots,c_{1n})\ar@3{-}[r]^{-}{F_{12}}&\cdots\ar@3{-}[r]^{-}{F_{1n}}&(c_{21},c_% {22},\ldots,c_{2n})}$

The idea is to replace the coordinates   one at a time, in sequence, from the first to the last, until all $n$ coordinates are completely replaced from $(c_{11},\ldots,c_{1n})$ to $(c_{21},\ldots,c_{2n})$.

Now, since each $F_{1j}$ is a congruence, apply $\omega$ to each $n$-tuple to get a new

 $\xymatrix{\omega(c_{11},c_{12},\ldots,c_{1n})\ar@3{-}[r]^{-}{F_{11}}&\omega(c_% {21},c_{12},\ldots,c_{1n})\ar@3{-}[r]^{-}{F_{12}}&\cdots\ar@3{-}[r]^{-}{F_{1n}% }&\omega(c_{21},c_{22},\ldots,c_{2n})}.$

But this chain implies that $\omega(a_{1},\ldots,a_{n})=\omega(c_{11},\ldots,c_{in})\equiv\omega(c_{21},% \ldots,c_{2n})\pmod{\Theta}$. So what we have shown is that the images of the first congruence pair are congruent modulo $\Theta$. But this process can be easily applied to subsequent congruence pairs, so that we end up with

 $\xymatrix{\omega(a_{1},\ldots,a_{2})\ar@3{-}[r]^{-}{\Theta}&\omega(c_{21},% \ldots,c_{2n})\ar@3{-}[r]^{-}{\Theta}&\cdots\ar@3{-}[r]^{-}{\Theta}&\omega(b_{% 1},\ldots,b_{n})}.$

As $\Theta$ is an equivalence relation, $\omega(a_{1},\ldots,a_{2})\equiv\omega(b_{1},\ldots,b_{n})\pmod{\Theta}$, completing the proof. ∎

Remarks.

## References

• 1 G. Grätzer: , 2nd Edition, Springer, New York (1978).
• 2 F. Wehrung: http://hal.archives-ouvertes.fr/docs/00/11/93/14/PDF/CLP.pdfA Solution to Dilworth’s Congruence Lattice Problem, (2005).
Title congruence lattice CongruenceLattice 2013-03-22 17:06:31 2013-03-22 17:06:31 CWoo (3771) CWoo (3771) 8 CWoo (3771) Definition msc 08A30 lattice of congruences