= Solution
{c}
\b[True.] An <irreducible ideal> $I$ is a proper <ideal> such that $I=J\cap K$ forces $I=J$ or $I=K$. Passing to the <quotient ring> $A=R/I$ turns this into the assertion that zero cannot be the intersection of two nonzero <ideals>. Also $A$ is <Noetherian>: ascending chains of <ideals> lift to ascending chains of <ideals> containing $I$ in $R$.
Suppose $ab=0$ in $A$, and suppose $b$ is not a <nilpotent element>. The <ascending chain condition> makes the chain of <annihilators>
$$
\operatorname{Ann}(b)\subseteq\operatorname{Ann}(b^2)\subseteq\cdots
$$
stabilize. Choose $n\geq1$ with $\operatorname{Ann}(b^n)=\operatorname{Ann}(b^{n+1})$. We claim
$$
(a)\cap(b^n)=0.
$$
If $z=ua=vb^n$ lies in this intersection, then $bz=uab=0$, so $vb^{n+1}=0$. The equality of <annihilators> implies $vb^n=0$, hence $z=0$. Since $b$ is not nilpotent, $(b^n)\ne0$. Irreducibility of zero therefore forces $(a)=0$, or $a=0$.
We have proved that $ab=0$ and $a\ne0$ force $b$ to be a <nilpotent element>. This says precisely that zero is a <primary ideal> in $A$, and therefore that $I$ is a <primary ideal> in $R$.
If its <radical of an ideal> is to be denoted by a <prime ideal>, this requires no extra hypothesis: for any <primary ideal> $I$, $uv\in\sqrt I$ and $u\notin\sqrt I$ imply $u^nv^n\in I$ with $u^n\notin I$, so some power of $v$ lies in $I$. Thus $v\in\sqrt I$.
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