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An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

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

Suppose $R$ is an integral domain and $(a)=(b)$. If $a=0$, then $(b)=0$, so $b=0$ and $a=1\cdot b$.

Otherwise membership in the equal principal ideals supplies $r,s\in R$ with $a=rb$ and $b=sa$. Substituting gives $a=rsa$, or $(1-rs)a=0$. There are no nonzero zero divisors in an integral domain, and $a\ne 0$, so $rs=1$. Hence $r$ is a unit with inverse $s$, and $a=rb$ makes $a$ and $b$ associate elements. We have proved

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.

Put $A=R/(p^k)$. A principal ideal domain is a unique factorization domain, so every nonzero element of this quotient ring can be writtenIndeed, remove the largest power of the prime element $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have greatest common divisor $1$ up to a unit, Bézout's identity in the principal ideal domain gives $xs+y{p}^k=1$. Thus $\overline{s}$ is a unit in $A$.

Consequently every nonzero principal ideal in $A$ is generated by one of $1,\overline{p},\dots ,{\overline{p}}^{k-1}$. Include the zero ideal by assigning it exponent $e=k$. These ideals are distinct: their inverse images in $R$ are $(1),(p),\dots ,(p^k)$, and equality of two different powers would, by cancellation in the integral domain, make the nonunit $p$ a unit.

Now suppose $(\overline{a})=(\overline{b})$. If this is the zero ideal, both generators are zero and are associate elements. Otherwise their exponents agree, so writewith $\overline{s},\overline{t}$ both units. Thenand the multiplier is a unit. ThusThe exponent detects the principal ideal, while its remaining factor is invertible; this is why zero divisors cause no obstruction here.

Take the integral domain $R=\mathbb{Z}[\sqrt{-5}]$, embedded in $\mathbb{C}$. It exhibits failure of unique factorization in the ring of integers generated by square root of minus five:Here is why these are genuinely distinct factorizations into irreducible elements. The multiplicative algebraic norm $N(a+b\sqrt{-5})=a^2+5b^2$ is a nonnegative integer and equals one only for $\pm 1$, which are therefore the only units. Neither two nor three is a value of $N$: if $b\ne 0$ the value is at least five, and if $b=0$ it is a square. A factorization of $2$ into nonunits would require norms two and two; one of $3$ would require norms three and three; one of $1\pm \sqrt{-5}$ would require norms two and three. All are impossible. Thus all four factors are irreducible. The factors of norms four and nine cannot be associate elements of those of norm six, and $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are not related by a sign. Therefore $R$ is an integral domain which is not a unique factorization domain.

Since the irreducible element $x$ is not a unit, the principal ideal $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a principal ideal domain, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an irreducible element says that $d$ or $r$ is a unit.

If $d$ is a unit, then $J=R$. If $r$ is a unit, then $d$ and $x$ are associate elements, so $J=(x)$. Thus $(x)$ is a maximal ideal. By the permitted criterion that a quotient ring by a maximal ideal is a field,