PlanetMath: 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

of equivalence relations on

, the only remaining item to show is that it is a sublattice of

. In other words, we need to show that if $\mathcal{C}$ is a family of congruence relations on

, so is $\Theta:=\bigvee \mathcal{C}$ .

For convenience, let us introduce some notational devices:

$\mathbf{n}:=\lbrace 1,\ldots,n\rbrace$ ;
for any $a_k,b_k\in A$ and $\Theta_k\in\operatorname{Con}(A)$ , where $k\in \mathbf{n}$ , we mean
$\displaystyle (a_1,\ldots,a_n)\equiv (b_1,\ldots, b_n)\pmod{(\Theta_1,\ldots,\Theta_n)}$
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
$\displaystyle c_1\stackrel{\Theta_1}{\equiv} c_2\stackrel{\Theta_2}{\equiv} \cdots \stackrel{\Theta_{n-1}}{\equiv }c_n.$
Let us call this an equivalence chain (of length ).

Back to the proof. Let $\omega$ be an -ary operator on 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 equivalence chain

$\displaystyle a_i=c_1\stackrel{F_1}{\equiv} c_2\stackrel{F_2}{\equiv} \cdots \stackrel{F_{p-1}}{\equiv }c_p=b_i,$

where each is a congruence in $\mathcal{C}$ , and each $c_i\in A$ . Now the lengths of the sequences may vary by . 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 equivalence chains

$a_1=c_1\stackrel{F_1}{\equiv} c_2\stackrel{F_2}{\equiv} \cdots \stackrel{F_{p-1}}{\equiv }c_p=b_1$ , and
$a_2=d_1\stackrel{G_1}{\equiv} d_2\stackrel{G_2}{\equiv} \cdots \stackrel{G_{q-1}}{\equiv }d_q=b_2$ ,

where $c_i,d_j\in A$ and $F_i,G_j\in \mathcal{C}$ . If

, we may lengthen the second chain so it has the same length as the first one:

$\displaystyle a_2=d_1\stackrel{G_1}{\equiv} d_2 \cdots \stackrel{G_{q-1}}{\equi... ...equiv} d_{q+1}\stackrel{G_{q+1}}{\equiv} \cdots \stackrel{G_{p-1}}{\equiv }d_p,$

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 show that we can indeed make all the “expanded” equivalence chains the same length. As a result, without loss of generality, we assume that all the expanded chains have the same length, say .

Step 2 Complete the proof.

Instead of writing all

chains, let us use our notational device, and we have the following single equivalence chain (again, we may write it in this fashion because all chains are now assumed to have the same finite length of

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ (c_{11},\ldots,c_{1n})\ar@3{-}[r... ...3{-}[rr]^-{(F_{r1},\ldots,F_{rn})} && (c_{r+1,1},\ldots,c_{r+1,n}) } } \end{xy}$

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 equivalence chain of length as follows:

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ (c_{11},c_{12},\ldots,c_{1n})\ar... ...2}} & \cdots \ar@3{-}[r]^-{F_{1n}} & (c_{21},c_{22},\ldots,c_{2n}) } } \end{xy}$

The idea is to replace the coordinates one at a time, in sequence, from the first to the last, until all

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 -tuple to get a new equivalence chain

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

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

$\displaystyle \xymatrix{ \omega(a_1,\ldots,a_2)\ar@3{-}[r]^-{\Theta} & \omega(c... ...ar@3{-}[r]^-{\Theta} & \cdots \ar@3{-}[r]^-{\Theta} & \omega(b_1,\ldots,b_n) }.$

As $\Theta$ is an equivalence relation, $\omega(a_1,\ldots,a_2)\equiv \omega(b_1,\ldots,b_n)\pmod \Theta$ , completing the proof. $\qedsymbol$

Bibliography