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

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 \begin{eqnarray*} f_{\boldsymbol{A}\times \boldsymbol{B}}((a_1,\sigma(a_1)), \ldots, (a_n,\sigma(a_n))) &=& (f_{\boldsymbol{A}}(a_1,\ldots, a_n),f_{\boldsymbol{B}}(\sigma(a_1),\ldots, \sigma(a_n))) \\ &=& (f_{\boldsymbol{A}}(a_1,\ldots, a_n),\sigma(f_{\boldsymbol{A}}(a_1, \ldots, a_n))) \\ &=& \phi(f_{\boldsymbol{A}}(a_1,\ldots, a_n)), \end{eqnarray*}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 $\lbrace \boldsymbol{A_i} \mid i\in I\rbrace$ 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 $A:= \prod \lbrace A_i\mid i\in I\rbrace$ ,
• 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, $$f_{\boldsymbol{A}}(a)(i):= f_{\boldsymbol{A_i}}(a(i)).$$

Bibliography

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