PlanetMath: reduced direct product

Proof. Since

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

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

Since

is an ideal, $\operatorname{supp}(a,c)\in L$ , so $(a,c)\in \Theta_L$ , and $\Theta_L$ is an equivalence relation on

Next, let $\omega$ be an -ary operator on 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 -ary operators on . 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

$\displaystyle \operatorname{supp}(\omega(a_1,\ldots,a_n),\omega(b_1,\ldots,b_n))\subseteq \bigcup_{j=1}^n \operatorname{supp}(a_j,b_j).$

Since

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 $A=\prod \lbrace A_i\mid i\in I\rbrace$ ,

be a Boolean ideal of

and $\Theta_L$ be defined as above. The quotient algebra $A/\Theta_L$ is called the

-reduced direct product of

. The

-reduced direct product of

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

for short.

Example. Let $A=A_1\times \cdots \times A_n$ , and let

be the principal ideal generated by

. 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

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

-reduced direct product of

is isomorphic to

Remark. The definition of a reduced direct product in terms of a Boolean ideal can be equivalently stated in terms of a Boolean filter

. 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

. When

is prime, the

-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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