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.)




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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.



What is the intersection of all ideals whose radicals are prime?
**Is the intersection of all fuzzy primes $\{0\}$?**
Not always. Let me describe a commutative ring where the intersection of the fuzzy primes is nonzero.
**Plan.** The idea will be to construct a commutative ring $R$ that has a nonzero element $c$ such that $c^2=0$ and $c$ is contained in every nonzero ideal of $R$. If the intersection of all fuzzy primes of $R$ was $(0)$, then some fuzzy prime ideal would not contain $c$, and that ideal could only be $(0)$ itself. However, if $(0)$ was a fuzzy prime ideal, then the radical of $R$ would have to be prime. To make sure that the radical of $R$ is not prime, we make sure that $R$ contains elements $a$ and $b$ such that $ab=0$, but neither $a$ nor $b$ is contained in the radical.
**Execution.** This suggests a presentation of a ring $S$ by generators and relations:
$$S = \langle a, b, c, d\_n, e\_n\;(n=1,2,\ldots)\;| a\cdot b=0,\; c^2=0,\; a^n\cdot d\_n = c,\; b^n\cdot e\_n = c\rangle.$$


Assume for now that in $S$ we have $c\neq 0$. Let $I\lhd S$ be an ideal maximal for $c\notin I$. Let $R=S/I$. In this quotient, $\overline{c}$ will be nonzero and will be contained in every nonzero ideal of $R$. The other relations from the presentation will hold in $R$, and they imply that $\overline{a}\cdot \overline{b}=0$
and that neither $\overline{a}$ not $\overline{b}$ is contained in the radical (since, e.g., $\overline{a}^n\cdot \overline{d}\_n=\overline{c}\neq 0$ for every $n$).
**Resolving the issues.**
We are done if we can show that $c\neq 0$ in the ring $S$ presented by
$\langle a, b, c, d\_n, e\_n\;(n=1,2,\ldots)\;| a\cdot b=0,\; c^2=0,\; a^n\cdot d\_n = c,\; b^n\cdot e\_n = c\rangle$. In other words, we have to show that there is a model for the set of sentences expressing 'I am a commutative ring' and 'I have elements $a, b, c, d\_n, e\_n$ satisfying 'the desired relations' $a\cdot b=0,\; c^2=0,\; a^n\cdot d\_n = c,\; b^n\cdot e\_n = c$ AND $c\neq 0$'.   
It suffices to show that any finite subset of these sentences has a model, so it suffices to show that, for every $n$, there is a commutative ring $R\_n$ with elements $a, b, c, d, e$ such that
$$\tag{1} ab=0, c^2=0, a^nd=c, b^ne = c,\;\textrm{AND}\; c\neq 0.$$ (Comment: If there is a single $d=d\_n$ such that $a^nd=c$, then for $k<n$ we also have $a^k(a^{n-k}d) = c$, so we can let $d\_k = a^{n-k}d$ and get $a^id\_i=c$ for $i=1,\ldots,n$. Thus, $R\_n$ satisfies 'the desired relations' truncated at $n$.)


There is a commutative monoid $M$ with elements $\{1, a, b, a^2, b^2, \ldots, a^{n-1}, b^{n-1}, a^n=b^n, 0\_M\}$ where all listed elements except $a^n$ and $b^n$ are distinct, and the products are the obvious ones (e.g. $a^ia^j=a^{i+j}, b^ib^j=b^{i+j}$) along with $ab=0\_M$, $a^{n+1}=b^{n+1}=0\_M$, and $10\_M=a0\_M=b0\_M=0\_M$. Form the monoid ring
$T:=\mathbb Z[M]$ and define $c:=a^n=b^n$ and $d=e:=1$. This ring is additively free over $M$ and the additive summand $\mathbb Z\cdot 0\_M$ is an ideal of $T$ not containing $c$.
In the quotient $R=T/(0\_M)$ the element $0\_M$ is identified with $0\_T$, while $c$ is not identified with $0\_T$. All relations from (1) are satisfied in $R$.












