direct product of partial algebras

Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be two partial algebraic systems of type $\tau$ . The direct product of $\boldsymbol{A}$ and $\boldsymbol{B}$ , written $\boldsymbol{A}\times\boldsymbol{B}$ , is a partial algebra of type $\tau$ , defined as follows:

•

the underlying set of $\boldsymbol{A}\times\boldsymbol{B}$ is $A\times B$ ,
•

for each $n$ -ary function symbol $f\in\tau$ , the operation $f_{\boldsymbol{A}\times\boldsymbol{B}}$ is given by:

for $(a_{1},b_{1}),\ldots,(a_{n},b_{n})\in A\times B$ , $f_{\boldsymbol{A}\times\boldsymbol{B}}((a_{1},b_{1})\ldots,(a_{n},b_{n}))$ is defined iff both $f_{\boldsymbol{A}}(a_{1},\ldots,a_{n})$ and $f_{\boldsymbol{B}}(b_{1},\ldots,b_{n})$ are, and when this is the case,

$f_{\boldsymbol{A}\times\boldsymbol{B}}((a_{1},b_{1}),\ldots,(a_{n},b_{n})):=(f% _{\boldsymbol{A}}(a_{1},\ldots,a_{n}),f_{\boldsymbol{B}}(b_{1},\ldots,b_{n})).$

It is easy to see that the type of $\boldsymbol{A}\times\boldsymbol{B}$ is indeed $\tau$ : pick $a_{1},\ldots,a_{n}\in A$ and $b_{1},\ldots,b_{n}\in B$ such that $f_{\boldsymbol{A}}(a_{1},\ldots,a_{n})$ and $f_{\boldsymbol{B}}(b_{1},\ldots,b_{n}))$ are defined, then $f_{\boldsymbol{A}\times\boldsymbol{B}}((a_{1},b_{1}),\ldots,(a_{n},b_{n}))$ is defined, so that $f_{\boldsymbol{A}\times\boldsymbol{B}}$ is non-empty, where all operations are defined componentwise, and the two constants are $(0,0)$ and $(1,1)$ .

For example, suppose $k_{1}$ and $k_{2}$ are fields. They are both partial algebras of type $\langle 2,2,1,1,0,0\rangle$ , where the two $2$ ’s are the arity of addition and multiplication, the two $1$ ’s are the arity of additive and multiplicative inverses, and the two $0$ ’s are the constants $0$ and $1$ . Then $k_{1}\times k_{2}$ , while no longer a field, is still an algebra of the same type.

Let $\boldsymbol{A},\boldsymbol{B}$ be partial algebras of type $\tau$ . Can we embed $\boldsymbol{A}$ into $\boldsymbol{A}\times\boldsymbol{B}$ so that $\boldsymbol{A}$ is some type of a subalgebra of $\boldsymbol{A}\times\boldsymbol{B}$ ?

For example, if we fix an element $b\in B$ , then the injection $i_{b}:\boldsymbol{A}\to\boldsymbol{A}\times\boldsymbol{B}$ , given by $i_{b}(a)=(a,b)$ is in general not a homomorphism only unless $b$ is an idempotent with respect to every operation $f_{\boldsymbol{B}}$ on $B$ (that is, $f_{\boldsymbol{B}}(b,\ldots,b)=b)$ . In addition, $b$ would have to be the constant for every constant symbol in $\tau$ . Following from the example above, if we pick any $r\in k_{2}$ , then $r$ would have to be $0$ , since, $(s_{1}+s_{2},2r)=i_{r}(s_{1})+i_{r}(s_{2})=i_{r}(s_{1}+s_{2})=(s_{1}+s_{2},r)$ , so that $2r=r$ , or $r=0$ . But, on the other hand, $i_{r}(s^{-1})=(s,r)^{-1}=(s^{-1},r^{-1})$ , forcing $r$ to be invertible, a contradiction!

Now, suppose we have a homomorphism $\sigma:\boldsymbol{A}\to\boldsymbol{B}$ , then we may embed $\boldsymbol{A}$ into $\boldsymbol{A}\times\boldsymbol{B}$ , so that $\boldsymbol{A}$ is a subalgebra of $\boldsymbol{A}\times\boldsymbol{B}$ . The embedding is given by $\phi(a)=(a,\sigma(a))$ .

Proof.

Suppose $f_{\boldsymbol{A}}(a_{1},\ldots,a_{n})$ is defined. Since $\sigma$ is a homomorphism, $f_{\boldsymbol{B}}(\sigma(a_{1}),\ldots,\sigma(a_{n}))$ is defined, which means $f_{\boldsymbol{A}\times\boldsymbol{B}}((a_{1},\sigma(a_{1})),\ldots,(a_{n},% \sigma(a_{n})))=f_{\boldsymbol{A}\times\boldsymbol{B}}(\phi(a_{1}),\ldots,\phi% (a_{n}))$ is defined. Furthermore, we have that

$\displaystyle f_{\boldsymbol{A}\times\boldsymbol{B}}((a_{1},\sigma(a_{1})),% \ldots,(a_{n},\sigma(a_{n})))$	$\displaystyle=$	$\displaystyle(f_{\boldsymbol{A}}(a_{1},\ldots,a_{n}),f_{\boldsymbol{B}}(\sigma% (a_{1}),\ldots,\sigma(a_{n})))$
	$\displaystyle=$	$\displaystyle(f_{\boldsymbol{A}}(a_{1},\ldots,a_{n}),\sigma(f_{\boldsymbol{A}}% (a_{1},\ldots,a_{n})))$
	$\displaystyle=$	$\displaystyle\phi(f_{\boldsymbol{A}}(a_{1},\ldots,a_{n})),$

showing that $\phi$ is a homomorphism. In addition, if $f_{\boldsymbol{A}\times\boldsymbol{B}}(\phi(a_{1}),\ldots,\phi(a_{n}))$ is defined, then it is clear that $f_{\boldsymbol{A}}(a_{1},\ldots,a_{n})$ is defined, so that $\phi$ is a strong homomorphism. So $\phi(\boldsymbol{A})$ is a subalgebra of $\boldsymbol{A}\times\boldsymbol{B}$ . Clearly, $\phi$ is one-to-one, and therefore an embedding, so that $\boldsymbol{A}$ is isomorphic to $\phi(\boldsymbol{A})$ , and we may view $\boldsymbol{A}$ as a subalgebra of $\boldsymbol{A}\times\boldsymbol{B}$ . ∎

Remark. Moving to the general case, let $\{\boldsymbol{A_{i}}\mid i\in I\}$ be a set of partial algebras of type $\tau$ , indexed by set $I$ . The direct product of these algebras is a partial algebra $\boldsymbol{A}$ of type $\tau$ , defined as follows:

•

the underlying set of $\boldsymbol{A}$ is $A:=\prod\{A_{i}\mid i\in I\}$ ,
•

for each $n$ -ary function symbol $f\in\tau$ , the operation $f_{\boldsymbol{A}}$ is given by: for $a\in A$ , $f_{\boldsymbol{A}}(a)$ is defined iff $f_{\boldsymbol{A_{i}}}(a(i))$ is defined for each $i\in I$ , and when this is the case,

$f_{\boldsymbol{A}}(a)(i):=f_{\boldsymbol{A_{i}}}(a(i)).$

Again, it is easy to verify that $\boldsymbol{A}$ is indeed a $\tau$ -algebra: for each symbol $f\in\tau$ , the domain of definition $\operatorname{dom}(f_{\boldsymbol{A_{i}}})$ is non-empty for each $i\in I$ , and therefore the domain of definition $\operatorname{dom}(f_{\boldsymbol{A}})$ , being $\prod\{\operatorname{dom}(f_{\boldsymbol{A_{i}}})\mid i\in I\}$ , is non-empty as well, by the axiom of choice.

References

1 G. Grätzer: Universal Algebra, 2nd Edition, Springer, New York (1978).

Title	direct product of partial algebras
Canonical name	DirectProductOfPartialAlgebras
Date of creation	2013-03-22 18:43:40
Last modified on	2013-03-22 18:43:40
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	7
Author	CWoo (3771)
Entry type	Definition
Classification	msc 08A55
Classification	msc 08A62
Classification	msc 03E99
Defines	direct product