= Solution
Use the standard definition that a <prime ideal> is proper, in addition to the product condition. The quotient operations are
$$
(r+I)+(s+I)=(r+s)+I,\qquad(r+I)(s+I)=rs+I.
$$
Its zero and identity are $I$ and $1+I$, which differ because $I$ is proper. If $(r+I)(s+I)=I$, then $rs\in I$, so primality gives $r\in I$ or $s\in I$. Thus one factor is zero. The commutative <quotient ring> has no nonzero <zero divisors> and is therefore an <integral domain>.
The printed definition omits properness. If that omission were read literally, $I=R$ would satisfy its product condition, but $R/I$ would be the zero <ring>, which is not an <integral domain> under the standard convention $1\ne0$. Properness is the necessary qualification, retained explicitly in this proof.
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