In this entry, let $O$ be a fixed operator set. All algebraic systems have the same type (they are all $O$ -algebras).

Let $\lbrace A_i\mid i\in I\rbrace$ be a set of algebraic systems of the same type ($O$ ) indexed by $I$ . Let us form the Cartesian product of the underlying sets and call it $A$ : $$A:=\prod_{i\in I} A_i.$$ Recall that element $a$ of $A$ is a function from $I$ to $\bigcup A_i$ such that for each $i\in I$ , $a(i)\in A_i$ .

For each $\omega\in O$ with arity $n$ , let $\omega_{A_i}$ be the corresponding $n$ -ary operator on $A_i$ . Define $\omega_A: A^n\to A$ by $$\omega_A(a_1,\ldots,a_n)(i)=\omega_{A_i}(a_1(i),\ldots,a_n(i))\quad\mbox{ for all }i\in I.$$ One readily checks that $\omega_A$ is a well-defined $n$ -ary operator on $A$ . $A$ equipped with all $\omega_A$ on $A$ is an $O$ -algebra, and is called the direct product of $A_i$ . Each $A_i$ is called a direct factor of $A$ .

If each $A_i=B$ , where $B$ is an $O$ -algebra, then we call $A$ the direct power of $B$ and we write $A$ as $B^I$ (keep in mind the isomorphic identifications).

If $A$ is the direct product of $A_i$ , then for each $i\in I$ we can associate a homomorphism $\pi_i:A\to A_i$ called a projection given by $\pi_i(a)=a(i)$ . It is a homomorphism because $\pi_i(\omega_A(a_1,\ldots, a_n))=\omega_A(a_1,\ldots, a_n)(i)=\omega_{A_i}(a_1(i),\ldots,a_n(i))=\omega_{A_i}(\pi_i(a_1),\ldots, \pi_i(a_n))$ .

Remark. The direct product of a single algebraic system is the algebraic system itself. An empty direct product is defined to be a trivial algebraic system (one-element algebra).