Past exam of the algebra 2 course of the University of Oxford / 2015 / 1 / a

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.

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

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