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.

Title	reduced direct product
Canonical name	ReducedDirectProduct
Date of creation	2013-03-22 17:10:11
Last modified on	2013-03-22 17:10:11
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	10
Author	CWoo (3771)
Entry type	Definition
Classification	msc 08B25
Synonym	ultraproduct
Defines	prime product