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 ${a}_{j}\equiv {b}_{j}\phantom{\rule{veryverythickmathspace}{0ex}}(mod{\mathrm{\Theta}}_{L})$, where $j=1,\mathrm{\dots},n$. We want to show that $\omega ({a}_{1},\mathrm{\dots},{a}_{n})\equiv \omega ({b}_{1},\mathrm{\dots},{b}_{n})\phantom{\rule{veryverythickmathspace}{0ex}}(mod{\mathrm{\Theta}}_{L})$. 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 $\omega ({a}_{1},\mathrm{\dots},{a}_{n})\equiv \omega ({b}_{1},\mathrm{\dots},{b}_{n})\phantom{\rule{veryverythickmathspace}{0ex}}(mod{\mathrm{\Theta}}_{L})$. ∎
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 $({a}_{1},\mathrm{\dots},{a}_{n})\equiv ({b}_{1},\mathrm{\dots},{b}_{n})\phantom{\rule{veryverythickmathspace}{0ex}}(mod{\mathrm{\Theta}}_{L})$ 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^{}.
References
- 1 G. Grätzer: Universal Algebra^{}, 2nd Edition, Springer, New York (1978).
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 |