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.


2 Answers 2


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 $R$.

  • $\begingroup$ Very nice. I think I can give a direct construction of your ring $S$: Let $A = k[x,y]/(xy)$, which we consider as a graded ring $\bigoplus A_i$. Let $M$ be the graded dual: $\bigoplus \text{Hom}(A_i, k)$, so $M$ is an $A$-module. Your ring $S$ is $A \oplus M$ where $A$ multiplies $M$ by the $A$-module structure, and $M^2 =0$. $\endgroup$ May 29, 2023 at 2:59
  • $\begingroup$ Very nice. Maybe properly speaking you're taking the quotient of the monoid ring $\mathbf Z[M]$ by the obvious relation $[0] = 0$ (where $[m]$ denotes the class of $m \in M$ in $\mathbf Z[M]$). I suppose this is the "monoid with absorbing element"-ring, left adjoint to $R \mapsto (R,\times,0)$. $\endgroup$ May 29, 2023 at 3:14
  • $\begingroup$ @R.vanDobbendeBruyn: I agree that I should identify the zero element of $M$ and the zero element of the monoid ring. I have edited to reflect this. $\endgroup$ May 29, 2023 at 16:21
  • $\begingroup$ @DavidESpeyer I think this is a construction of $R$, not $S$: You've added extra relations, like $d_ne_n=0$, but in your ring every nonzero ideal contains $c$ (it's not hard to check $c$ is a multiple of every nonzero element by some case analysis). $\endgroup$
    – Will Sawin
    May 30, 2023 at 15:00
  • $\begingroup$ @WillSawin Thanks, Will, you are right. $\endgroup$ May 30, 2023 at 15:19

This is not true, and we can produce explicit examples by applying standard modifications (quotients, fibre products) to higher rank valuation rings.

Example. Let $M$ be the submonoid of $\mathbf Z^2$ generated by $(0,1)$ and $(1,a)$ for all $a \in \mathbf Z$. By the usual correspondence between cancellative commutative monoids and partially ordered abelian groups, it corresponds to $\mathbf Z^2$ with the lexicographic order given by $(a,b) \geq (c,d)$ if and only if either $a > c$, or $a = c$ and $b \geq d$.

Let $A$ be a valuation ring with value group $M$ that contains its residue field $k$, and let $x$ and $y$ be elements of valuation $(1,0)$ and $(0,1)$ respectively. For instance, endow $k[x^{\pm 1},y^{\pm 1}]$ with the valuation $v \colon k[x^{\pm 1},y^{\pm 1}] \to \mathbf Z^2$ taking $\sum_{i,j} a_{i,j}x^iy^j$ to the lexicographically smallest index $(i,j)$ with $a_{i,j} \neq 0$, and then take $A = \{f \in k(x,y)\ |\ v(f) \in M\}$. Note that $A$ is a $2$-dimensional domain with prime ideals $(0) \subsetneq (xy^{\mathbf Z}) \subsetneq (y)$. Since $A$ is a valuation ring, its ideals are totally ordered by inclusion.

Now set $B = A/(xy)$ with quotient map $\pi \colon A \twoheadrightarrow B$. Then (the image of) $x$ is contained in any nonzero ideal of $B$. Indeed, if $I \subseteq B$ is a nonzero ideal, then $\pi^{-1}(I) \subseteq A$ contains some element of valuation $(i,j) < (1,1)$, hence it contains $x^iy^j$ since any two elements of the same valuation differ by a unit. But then $(i,j) \leq (1,0)$ (separate the cases $i=0$ or $i=1$ and $j < 1$), so $I$ contains $x$.

Next, glue two copies of $B$ along itself at the origin: consider the fibre product $C = B \times_k B$, where $B \twoheadrightarrow k$ is the quotient by the maximal ideal (see [Tag 0D2G] for the spectrum of a fibre product). Then any nonzero ideal of $C$ contains either $(x,0)$ or $(0,x)$. Note that $J = \{(ax,-ax)\ |\ a \in k\}$ is an ideal in $C$: multiplication by $(b_1,b_2) \in C$ on $J$ only depends on the constant terms of $b_1$ and $b_2$, which agree by definition of $C$. So finally set $R = C/J$, and note that any nonzero ideal of $R$ contains $\overline{(x,0)} = \overline{(0,x)}$. We are done since the nilradical is generated by $(xy^i,xy^j)$ for $i,j \in \mathbf Z$, which is not prime as $(x,y)(y,x) = 0$ but neither $(x,y)$ nor $(y,x)$ is nilpotent. $\square$

This answer is very similar to Keith Kearnes's (nearly simultaneous) answer, but my answer focuses a bit more on producing an explicit model. Both answers need to get rid of unwanted ideals (namely the ones not containing $c$); I do this by passing to valuation rings, whereas Kearnes uses Zorn's lemma. Ultimately, the underlying monoid constructions are nearly identical. I can take $d_n = (0,xy^{-n})$ and $e_n = (xy^{-n},0)$ in the notation of Kearnes's answer.

  • $\begingroup$ I'm confused. It seems to me that the radical of $(0)$ is prime in $R$: For any $n$, we have $(xy^{-n})^2 = x^2 y^{-2n} = (xy)(x y^{-2n-1}) = 0$ in $R$. So every element in $x y^{\mathbb{Z}}$ is nilpotent, and the quotient by $\sqrt{0}$ is a dvr. $\endgroup$ May 29, 2023 at 2:37
  • $\begingroup$ @DavidESpeyer right; I had just realised the same. But this is easily fixed by glueing two copies of $B$ (known as $R$ in the initial answer) at the origin. $\endgroup$ May 29, 2023 at 4:41

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