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The ideal correspondence theorem identifies the ideals of $\mathbb{Z}/4\mathbb{Z}$ with the ideals of $\mathbb{Z}$ containing $4\mathbb{Z}$. The possible positive generators divide $4$, giving exactlyThe units are $\overline{1}$ and $\overline{3}$. The generators of the three displayed ideals are, respectively,Within the last pair, multiplication by the unit $\overline{3}$ exchanges the generators. The other two sets are singletons, whose members are associate elements to themselves by multiplication by $\overline{1}$. Thus equal principal ideals always have associate generators: this is an associator ring.

However, $\overline{2}\ne \overline{0}$ and ${\overline{2}}^2=\overline{0}$, so $\overline{2}$ is a nonzero zero divisor. Therefore

If $a=ub$ with $u$ a unit in a ring, then $a$ lies in the principal ideal generated by $b$. This gives $(a)\subseteq (b)$. Since $u$ has a multiplicative inverse, $b=u^{-1}a$, giving the reverse inclusion. ThereforeThus associate elements always generate the same principal ideal; the associator ring condition asks whether equality of these ideals also forces the generators to differ by a unit.

For two factors, the source's $R_1\oplus R_2$ is the direct product of rings with componentwise operations. Its principal ideals satisfyOne 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 $a_i=u_i{b}_i$. The pair $(u_1,u_2)$ is a unit of the direct product of rings, with inverse $(u_1^{-1},u_2^{-1})$, andHenceRepeating this argument proves the same assertion for every finite direct product of rings.

By the first isomorphism theorem for rings, the given surjection identifies $S$ with $R/\ker \varphi$. Since $R$ is a principal ideal domain, write $\ker \varphi =(a)$.

If $a=0$, the quotient ring is $R$, an integral domain, so part (a)(ii) applies. If $a$ is a unit, the quotient ring is the zero ring; its unique element is associate to itself by multiplication by its identity, so it too is an associator ring.

In the remaining case, unique factorization giveswhere $\epsilon$ is a unit, the $p_i$ are pairwise nonassociate prime elements, and $k_i\ge 1$. The principal ideals $(p_i^{k_i})$ are pairwise comaximal: different prime factors have greatest common divisor $1$, and Bézout's identity gives a linear combination equal to $1$. The Chinese remainder theorem therefore gives a ring isomorphismEvery factor is an associator ring by part (c), and their finite direct product of rings is an associator ring by part (b). A ring isomorphism preserves units and principal ideals, so the property transfers to $S$. Hence