# 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: , 2nd Edition, Springer, New York (1978).
Title direct product of partial algebras DirectProductOfPartialAlgebras 2013-03-22 18:43:40 2013-03-22 18:43:40 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 08A55 msc 08A62 msc 03E99 direct product