reduced direct product

Let $\{A_i\mid i\in I\}$ be a set of algebraic systems of the same type, indexed by $I$. Let $A$ be the direct product of the $A_i$’s. For any $a,b\in A$, set

$$\mathrm{supp}(a,b):=\{k\in I\mid a(k)\ne b(k)\}.$$

Consider a Boolean ideal $L$ of the Boolean algebra $P(I)$ of $I$. Define a binary relation ${\mathrm{\Theta }}_L$ on $A$ as follows:

$$(a,b)\in {\mathrm{\Theta }}_L\mathit{\hspace{1em}}\text{ iff }\mathit{\hspace{1em}}\mathrm{supp}(a,b)\in L.$$

Lemma 1.

${\mathrm{\Theta }}_L$ defined above is a congruence relation on $A$.

Proof.

Since $L$ is an ideal $\mathrm{\varnothing }\in L$. Therefore, $(a,a)\in {\mathrm{\Theta }}_L$, since $\{k\in I\mid a(k)\ne a(k)\}=\mathrm{\varnothing }$. Clearly, ${\mathrm{\Theta }}_L$ is symmetric. For transitivity, suppose $(a,b,(b,c)\in {\mathrm{\Theta }}_L$. If $a(k)\ne c(k)$ for some $k\in I$, then either $a(k)\ne b(k)$ or $b(k)\ne c(k)$ (a contrapositive argument). So

$$\mathrm{supp}(a,c)\subseteq \mathrm{supp}(a,b)\cup \mathrm{supp}(b,c).$$

Since $L$ is an ideal, $\mathrm{supp}(a,c)\in L$, so $(a,c)\in {\mathrm{\Theta }}_L$, and ${\mathrm{\Theta }}_L$ is an equivalence relation on $A$.

Next, let $\omega$ be an $n$-ary operator on $A$ and $a_j\equiv b_j(mod{\mathrm{\Theta }}_L)$, where $j=1,\mathrm{\dots },n$. We want to show that $\omega (a_1,\mathrm{\dots },a_n)\equiv \omega (b_1,\mathrm{\dots },b_n)(mod{\mathrm{\Theta }}_L)$. Let ${\omega }_i$ be the associated $n$-ary operators on $A_i$. If $\omega (a_1,\mathrm{\dots },a_n)(k)\ne \omega (b_1,\mathrm{\dots },b_n)(k)$, then ${\omega }_k(a_1(k),\mathrm{\dots },a_n(k))\ne {\omega }_k(b_1(k),\mathrm{\dots },b_n(k))$, which implies that $a_j(k)\ne b_j(k)$ for some $j=1,\mathrm{\dots },n$. This implies that

$$\mathrm{supp}(\omega (a_1,\mathrm{\dots },a_n),\omega (b_1,\mathrm{\dots },b_n))\subseteq \bigcup _{j=1}^n\mathrm{supp}(a_j,b_j).$$

Since $L$ is an ideal, and each $\mathrm{supp}(a_j,b_j)\in L$, we have that $\mathrm{supp}(\omega (a_1,\mathrm{\dots },a_n),\omega (b_1,\mathrm{\dots },b_n))\in L$ as well, this means that $\omega (a_1,\mathrm{\dots },a_n)\equiv \omega (b_1,\mathrm{\dots },b_n)(mod{\mathrm{\Theta }}_L)$.∎

Definition. Let $A=\prod \{A_i\mid i\in I\}$, $L$ be a Boolean ideal of $P(I)$ and ${\mathrm{\Theta }}_L$ be defined as above. The quotient algebra $A/{\mathrm{\Theta }}_L$ is called the $L$-reduced direct product of $A_i$. The $L$-reduced direct product of $A_i$ is denoted by ${\prod }_L\{A_i\mid i\in I\}$. Given any element $a\in A$, its image in the reduced direct product ${\prod }_L\{A_i\mid i\in I\}$ is given by $[a]{\mathrm{\Theta }}_L$, or $[a]$ for short.

Example. Let $A=A_1\times \mathrm{\cdots }\times A_n$, and let $L$ be the principal ideal generated by $1$. Then $L=\{\mathrm{\varnothing },\{1\}\}$. The congruence ${\mathrm{\Theta }}_L$ is given by $(a_1,\mathrm{\dots },a_n)\equiv (b_1,\mathrm{\dots },b_n)(mod{\mathrm{\Theta }}_L)$ iff $\{i\mid a_i\ne b_i\}=\mathrm{\varnothing }$ or $\{1\}$. This implies that $a_i=b_i$ for all $i=2,\mathrm{\dots },n$. In other words, ${\mathrm{\Theta }}_L$ is isomorphic to the direct product of $A_2\times \mathrm{\cdots }\times A_n$. Therefore, the $L$-reduced direct product of $A_i$ is isomorphic to $A_1$.

The example above can be generalized: if $J\subseteq I$, then

$$\prod {}_{P(J)}\{A_i\mid i\in I\}\cong \prod \{A_i\mid i\in I-J\}.$$

For $a\in A=\prod \{A_i\mid i\in I\}$, write $a={(a_i)}_{i\in I}$. It is not hard to see that the map $f:{\prod }_{P(J)}\{A_i\mid i\in I\}\to \prod \{A_i\mid i\in I-J\}$ given by $f([a])={(a_i)}_{i\in I-J}$ is the required isomorphism.

Remark. The definition of a reduced direct product in terms of a Boolean ideal can be equivalently stated in terms of a Boolean filter $F$. All there is to do is to replace $\mathrm{supp}(a,b)$ by its complement: $\mathrm{supp}{(a,b)}^c:=\{k\in I\mid a(k)=b(k)\}$. The congruence relation is now ${\mathrm{\Theta }}_{F^{\prime }}$, where $F^{\prime }=\{I-J\mid J\in F\}$ is the ideal complement of $F$. When $F$ is prime, the $F^{\prime }$-reduced direct product is called a prime product, or an ultraproduct, since any prime filter is also called an ultrafilter. Ultraproducts can be more generally defined over arbitrary structures.