= Solution
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 exactly
$$
\boxed{\{\overline0\},\qquad
(\overline2)=\{\overline0,\overline2\},\qquad
\mathbb Z/4\mathbb Z.}
$$
The <units> are $\overline1$ and $\overline3$. The generators of the three displayed <ideals> are, respectively,
$$
\{\overline0\},\qquad\{\overline2\},\qquad
\{\overline1,\overline3\}.
$$
Within the last pair, multiplication by the <unit> $\overline3$ exchanges the generators. The other two sets are singletons, whose members are <associate elements> to themselves by multiplication by $\overline1$. Thus equal <principal ideals> always have associate generators: this is an <associator ring>.
However, $\overline2\ne\overline0$ and $\overline2^2=\overline0$, so $\overline2$ is a nonzero <zero divisor>. Therefore
$$
\boxed{\mathbb Z/4\mathbb Z\text{ is an associator ring but not an integral domain}.}
$$
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