direct product of partial algebras

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

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

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

  for $(a_1,b_1),\mathrm{\dots },(a_n,b_n)\in A\times B$, $f_{𝑨\times 𝑩}((a_1,b_1)\mathrm{\dots },(a_n,b_n))$ is defined iff both $f_{𝑨}(a_1,\mathrm{\dots },a_n)$ and $f_{𝑩}(b_1,\mathrm{\dots },b_n)$ are, and when this is the case,

  $$f_{𝑨\times 𝑩}((a_1,b_1),\mathrm{\dots },(a_n,b_n)):=(f_{𝑨}(a_1,\mathrm{\dots },a_n),f_{𝑩}(b_1,\mathrm{\dots },b_n)).$$

It is easy to see that the type of $𝑨\times 𝑩$ is indeed $\tau$: pick $a_1,\mathrm{\dots },a_n\in A$ and $b_1,\mathrm{\dots },b_n\in B$ such that $f_{𝑨}(a_1,\mathrm{\dots },a_n)$ and $f_{𝑩}(b_1,\mathrm{\dots },b_n))$ are defined, then $f_{𝑨\times 𝑩}((a_1,b_1),\mathrm{\dots },(a_n,b_n))$ is defined, so that $f_{𝑨\times 𝑩}$ 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 $⟨2,2,1,1,0,0⟩$, 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 $𝑨,𝑩$ be partial algebras of type $\tau$. Can we embed $𝑨$ into $𝑨\times 𝑩$ so that $𝑨$ is some type of a subalgebra of $𝑨\times 𝑩$?

For example, if we fix an element $b\in B$, then the injection $i_b:𝑨\to 𝑨\times 𝑩$, given by $i_b(a)=(a,b)$ is in general not a homomorphism only unless $b$ is an idempotent with respect to every operation $f_{𝑩}$ on $B$ (that is, $f_{𝑩}(b,\mathrm{\dots },b)=b)$. 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 $\sigma :𝑨\to 𝑩$, then we may embed $𝑨$ into $𝑨\times 𝑩$, so that $𝑨$ is a subalgebra of $𝑨\times 𝑩$. The embedding is given by $\varphi (a)=(a,\sigma (a))$.

Proof.

Suppose $f_{𝑨}(a_1,\mathrm{\dots },a_n)$ is defined. Since $\sigma$ is a homomorphism, $f_{𝑩}(\sigma (a_1),\mathrm{\dots },\sigma (a_n))$ is defined, which means $f_{𝑨\times 𝑩}((a_1,\sigma (a_1)),\mathrm{\dots },(a_n,\sigma (a_n)))=f_{𝑨\times 𝑩}(\varphi (a_1),\mathrm{\dots },\varphi (a_n))$ is defined. Furthermore, we have that

$f_{𝑨\times 𝑩}((a_1,\sigma (a_1)),\mathrm{\dots },(a_n,\sigma (a_n)))$$=$$(f_{𝑨}(a_1,\mathrm{\dots },a_n),f_{𝑩}(\sigma (a_1),\mathrm{\dots },\sigma (a_n)))$

$=$$(f_{𝑨}(a_1,\mathrm{\dots },a_n),\sigma (f_{𝑨}(a_1,\mathrm{\dots },a_n)))$

$=$$\varphi (f_{𝑨}(a_1,\mathrm{\dots },a_n)),$

showing that $\varphi$ is a homomorphism. In addition, if $f_{𝑨\times 𝑩}(\varphi (a_1),\mathrm{\dots },\varphi (a_n))$ is defined, then it is clear that $f_{𝑨}(a_1,\mathrm{\dots },a_n)$ is defined, so that $\varphi$ is a strong homomorphism. So $\varphi (𝑨)$ is a subalgebra of $𝑨\times 𝑩$. Clearly, $\varphi$ is one-to-one, and therefore an embedding, so that $𝑨$ is isomorphic to $\varphi (𝑨)$, and we may view $𝑨$ as a subalgebra of $𝑨\times 𝑩$.∎

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

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

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

  $$f_{𝑨}(a)(i):=f_{{𝑨}_{𝒊}}(a(i)).$$

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