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Homedirect product of algebras

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# direct product of algebras

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

Let $\{A_{i}\mid i\in I\}$ 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).

## Mathematics Subject Classification

08A05*no label found*08A62

*no label found*

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