= Solution
{c}
If the <quotient ring> $R/I$ is an <integral domain>, then $I\ne R$, because an <integral domain> has $1\ne0$. For $ab\in I$,
$$
(a+I)(b+I)=0+I.
$$
There are no nonzero <zero divisors> in an <integral domain>, so $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$, proving \b[$I$ is a prime ideal].
If $R/I$ is a <field>, again $I$ is proper. Let $J$ be an <ideal> properly containing $I$, and choose $a\in J\setminus I$. The nonzero element $a+I$ has a <multiplicative inverse> $b+I$, so $ab-1\in I\subset J$. Since $ab\in J$, subtraction gives $1\in J$, hence $J=R$. Therefore \b[$I$ is a maximal ideal]. These arguments establish the relevant <prime ideal quotient criterion> and <maximal ideal quotient criterion>.
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