Let a *fuzzy prime* be an ideal of a commutative unital ring whose radical is prime (I'm not sure if this kind of ideal already has a name). Is the intersection of all fuzzy primes $\{0\}$?

Note this is the same as those ideals $I$ for which $ab\in I$ implies $a\in\sqrt{I}$ or $b\in\sqrt{I}$: a fuzzy prime will have this property because $I\subseteq\sqrt{I}$, and if $I$ has this property and $ab\in\sqrt{I}$ then $a^nb^n\in I$ for some $n$ so $a^n\in\sqrt{I}\Rightarrow a\in\sqrt{I}$ or $b^n\in\sqrt{I}\Rightarrow b\in\sqrt{I}$, hence $\sqrt{I}$ is prime.

(Context: I was playing around with a modification of the $\text{Spec}$ construction which detects nilpotence topologically by taking fuzzy primes as points, but I'm not sure it does the job.)

If $A$ is reduced the answer is yes. If $A$ has prime nilradical the answer is again yes, because $(0)$ is a fuzzy prime. If $A$ is Noetherian then by Krull's intersection theorem the answer is yes since for any maximal $\mathfrak{m}$, $\mathfrak{m}^n$ is a fuzzy prime.

I tried modifying a proof that the intersection of the prime ideals is the nilradical (while avoiding the localization trick): if $a$ is in the intersection of all these ideals but is not $0$ then let $I$ be an ideal maximal among those not containing $a$ (by Zorn, because $a\notin(0)$). Supposing $I$ is not a fuzzy prime, by the above equivalent characterization there are $b,c$ with $bc\in I$ but neither is in $\sqrt{I}$ and thus they are not in $I$.

Now $(I:b)$ (the ideal of elements $r$ with $br\in I$) contains $c$ and everything in $I$ so by maximality of $I$ it contains $a$, i.e. $ab\in\sqrt{I}$. Similarly $(I:a)$ contains $b$ and $I$ so $a^2\in I$.

Not really sure what this tells us.