= Solution
For two factors, the source's $R_1\oplus R_2$ is the <direct product of rings> with componentwise operations. Its <principal ideals> satisfy
$$
\bigl((a_1,a_2)\bigr)=(a_1)\times(a_2),
$$
One inclusion follows by projection to the factors. For the reverse inclusion, multiplication by $(1,0)$ and $(0,1)$ puts $(a_1,0)$ and $(0,a_2)$ in the generated <ideal>; taking <ideal> combinations separately produces every pair in $(a_1)\times(a_2)$.
Suppose $(a_1,a_2)$ and $(b_1,b_2)$ generate the same <principal ideal>. Projecting to the factors gives $(a_i)=(b_i)$ for $i=1,2$. Since each factor is an <associator ring>, there are <units> $u_i\in R_i$ with $a_i=u_ib_i$. The pair $(u_1,u_2)$ is a <unit> of the <direct product of rings>, with inverse $(u_1^{-1},u_2^{-1})$, and
$$
(a_1,a_2)=(u_1,u_2)(b_1,b_2).
$$
Hence
$$
\boxed{R_1\oplus R_2\text{ is an associator ring}.}
$$
Repeating this argument proves the same assertion for every finite <direct product of rings>.
Back to article page
