# What is the intersection of all ideals whose radicals are prime?

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.

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 $$k we also have $$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 $$R$$.

• 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 $M^2 =0$. May 29, 2023 at 2:59
• 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 $R \mapsto (R,\times,0)$. May 29, 2023 at 3:14
• @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. May 29, 2023 at 16:21
• @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). May 30, 2023 at 15:00
• @WillSawin Thanks, Will, you are right. 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.

• 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. May 29, 2023 at 2:37
• @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. May 29, 2023 at 4:41