= Solution
{c}
The required property is
$$
\boxed{P:\quad\text{being an integral domain}}.
$$
For a nonzero element $x$ of an <integral domain> $R$, this follows from
$$
\begin{aligned}
x\text{ is a prime element}
&\iff (x)\text{ is a prime ideal}\\
&\iff R/(x)\text{ is an integral domain}.
\end{aligned}
$$
Indeed, $\overline a\,\overline b=0$ in the <quotient ring> exactly when $x\mid ab$. The <prime element> condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of <zero divisors> in the <quotient ring>. The <nonunit> condition makes the <quotient ring> nonzero. Being an <integral domain> is preserved by every <ring isomorphism>.
With the usual definition of a <prime element>, $x\ne0$ is understood. The <prime ideal> formulation works for every <ideal> of any <commutative ring>, including the zero <ideal>; it avoids this convention about elements.
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