PlanetMath: reduced direct product

Let $\lbrace A_i\mid i\in I\rbrace$ 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 $$\operatorname{supp}(a,b):=\lbrace k\in I\mid a(k)\ne b(k)\rbrace.$$ Consider a Boolean ideal $L$ of the Boolean algebra $P(I)$ of $I$ . Define a binary relation $\Theta_L$ on $A$ as follows: $$(a,b)\in \Theta_L\quad \mbox{ iff }\quad \operatorname{supp}(a,b)\in L.$$

Proof. Since $L$ is an ideal $\varnothing\in L$ . Therefore, $(a,a)\in \Theta_L$ , since $\lbrace k\in I\mid a(k)\ne a(k)\rbrace=\varnothing$ . Clearly, $\Theta_L$ is symmetric. For transitivity, suppose $(a,b,(b,c)\in \Theta_L$ . 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 $$\operatorname{supp}(a,c) \subseteq \operatorname{supp}(a,b) \cup \operatorname{supp}(b,c).$$ Since $L$ is an ideal, $\operatorname{supp}(a,c)\in L$ , so $(a,c)\in \Theta_L$ , and $\Theta_L$ is an equivalence relation on $A$ .

Next, let $\omega$ be an $n$ -ary operator on $A$ and $a_j\equiv b_j\pmod {\Theta_L}$ , where $j=1,\ldots,n$ . We want to show that $\omega(a_1,\ldots,a_n)\equiv \omega(b_1,\ldots,b_n)\pmod {\Theta_L}$ . Let $\omega_i$ be the associated $n$ -ary operators on $A_i$ . If $\omega(a_1,\ldots,a_n)(k)\ne \omega(b_1,\ldots,b_n)(k)$ , then $\omega_k(a_1(k),\ldots,a_n(k))\ne \omega_k(b_1(k),\ldots,b_n(k))$ , which implies that $a_j(k)\ne b_j(k)$ for some $j=1,\ldots,n$ . This implies that $$\operatorname{supp}(\omega(a_1,\ldots,a_n),\omega(b_1,\ldots,b_n))\subseteq \bigcup_{j=1}^n \operatorname{supp}(a_j,b_j).$$ Since $L$ is an ideal, and each $\operatorname{supp}(a_j,b_j)\in L$ , we have that $\operatorname{supp}(\omega(a_1,\ldots,a_n),\omega(b_1,\ldots,b_n))\in L$ as well, this means that $\omega(a_1,\ldots,a_n)\equiv \omega(b_1,\ldots,b_n)\pmod {\Theta_L}$ . $\qedsymbol$

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

Example. Let $A=A_1\times \cdots \times A_n$ , and let $L$ be the principal ideal generated by $1$ . Then $L=\lbrace \varnothing, \lbrace 1\rbrace\rbrace$ . The congruence $\Theta_L$ is given by $(a_1,\ldots,a_n)\equiv (b_1,\ldots, b_n)\pmod {\Theta_L}$ iff $\lbrace i\mid a_i\ne b_i\rbrace =\varnothing$ or $\lbrace 1\rbrace$ . This implies that $a_i=b_i$ for all $i=2,\ldots,n$ . In other words, $\Theta_L$ is isomorphic to the direct product of $A_2\times\cdots\times A_n$ . 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 $$\prod {}_{P(J)} \lbrace A_i\mid i\in I\rbrace \cong \prod \lbrace A_i\mid i\in I-J\rbrace.$$ For $a\in A=\prod \lbrace A_i\mid i\in I\rbrace$ , write $a=(a_i)_{i\in I}$ . It is not hard to see that the map $f:\prod_{P(J)} \lbrace A_i\mid i\in I\rbrace \to \prod \lbrace A_i\mid i\in I-J\rbrace$ 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 $\operatorname{supp}(a,b)$ by its complement: $\operatorname{supp}(a,b)^c:=\lbrace k\in I\mid a(k)=b(k)\rbrace$ . The congruence relation is now $\Theta_{F'}$ , where $F'=\lbrace I-J \mid J\in F\rbrace$ is the ideal complement of $F$ . When $F$ is prime, the $F'$ -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.

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