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Edit: Badam Baplan gives a remarkable example below, which answers the question completely. I would still appreciate any other examples that others may come up with; in particular, is an example still possible if we ask for $R$ to be Noetherian?


Let $R$ be a (commutative, unital) ring and $I<R$ an ideal. Recall that we define $$\operatorname{ht}I=\min\{\operatorname{ht}P:P\in\operatorname{Spec}R,P\geqslant I\},$$ where $\operatorname{ht}P$ is defined in the usual way for a prime ideal $P$. Is there a nice example of an integral domain $R$ and an ideal $I$ such that (i) $I$ contains no non-zero prime ideals, and (ii) $\operatorname{ht}I>1$? Note: by Krull's height theorem, if $R$ is Noetherian then $I$ cannot be principal.


If we drop the condition that $R$ is an integral domain, then we can get examples where $\operatorname{ht}I$ is arbitrarily large. For instance, let $F$ be your favorite field, and define $A$ to be the polynomial ring $F[x_1,\dots,x_n]$. Choose any non-zero ideal $M\leqslant A$, with some $o\in M\setminus\{0\}$ of minimal degree. (For later, note that $o\notin x_1M+\dots+x_nM$; call this fact $(\star)$.) Now, consider the ring structure on the product $R:=A\times M$ defined by

  • $(a,m)+(b,n)=(a+b,m+n)$
  • $(a,m)\cdot(b,n)=(ab,an+bm)$

for all $a,b\in A$ and $m,n\in M$. Note that every element $(0,m)$ is nilpotent, so any prime ideal of $R$ must contain $0\times M$. Conversely, if $P$ is a prime ideal of $A$, then $P\times M$ is a prime ideal of $R$. With this in mind, consider the ideal $I<R$ generated by the elements $(x_i,0)$ for $1\leqslant i\leqslant n$. First note that $I$ contains no prime ideals; indeed, one can explicitly compute $I$ to be the product $$(x_1A+\dots+x_nA)\times(x_1M+\dots +x_nM),$$ where both direct factors are considered as ideals of $A$. In particular, by fact $(\star)$, we have $(0,o)\notin I$, and so (by the remark above, since $(0,o)$ is nilpotent), $I$ cannot contain any prime ideal.

On the other hand, consider the prime ideal $P:=\langle x_1,\dots,x_n\rangle\times M>I$. This is the unique minimal prime lying above $I$, since $P=I+(0\times M)$ and every prime must contain $0\times M$. However, we have $\operatorname{ht}P=n$, since $$0\times M<\langle x_1\rangle\times M<\langle x_1,x_2\rangle\times M<\dots<P$$ is a strictly ascending chain of primes below $P$. Thus, since $P$ is the unique minimal prime above $I$, we have $\operatorname{ht}I=n$.


One can extend this example in the natural way to find an example where $\operatorname{ht}I=\infty$. So this resolves the question completely if we do not require $R$ to be an integral domain. However, I'm struggling to find an example where $R$ is an integral domain. Can anyone think of a nice example?

I tried attempting to consider the example above as a quotient of an appropriate polynomial ring, but didn't make any progress. My thinking was that we could perhaps find a desirable prime lying below the kernel of the projection map, and then quotient by that to find a domain with some similar behavior. However, this doesn't seem to work out very nicely.

As a smaller follow-up question, what is the geometric significance (if any) of such a ring?

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One of my favorite examples of this kind of wonky behavior originates from Krull's (false) conjecture that every completely integrally closed domain is the intersection of its one-dimensional valuation overrings. Nakayama disproved this conjecture in On Krull's conjecture concerning completely integrally closed integrity domains, I and II. He did this by constructing an integral domain $D$ which possesses the following properties:

  • $D$ is completely integrally closed
  • $D$ is Bezout
  • Every nonzero prime of $D$ has infinite height

So for Nakayama's purposes every valuation overring of $D$ is a localization of $D$ at a prime ideal and thus has infinite dimension.

For the purposes of your question, every nonzero principal ideal of $D$ has infinite height, but a principal ideal of $D$ never contains a prime ideal.

Indeed, for any Archimedean domain (i.e. $\bigcap d^nD = 0$ for all $d \in D$) it is is impossible for a prime to be properly contained in a principal ideal. Consequently a principal prime in an Archimedean domain necessarily has height $1$. Hence in an Archimedean domain with no finite height nonzero primes, it is impossible for a nonzero prime to be contained in a principal ideal.

(Updated to mention low-dimensional Noetherian examples) In the 1970's there was focused interest in the possible poset structure of the prime ideals of a Noetherian ring. One of the motivating questions was a specialization of your own: do finite intersections of maximal ideals contain prime ideals?

Among the advances in this period was a construction of Heitmann, which takes in a number of compatible 'special' Noetherian domains $D_\alpha$ with prime posets $P_\alpha$ and spits out a Noetherian domain having (up to order isomorphism) as its prime poset $\coprod P_\alpha$, the disjoint union of the $P_\alpha$ glued together at $(0)$.

This construction gives you a way to find $n$-dimensional Noetherian domain counterexamples to your question. For example, you can apply it to a couple of $n$-dimensional rings $D_1, D_2$. Pick a maximal ideal $M_i$ coming from the $P_i$ component. Clearly $M_1 \cap M_2$ is height $n$. But $M_1 \cap M_2$ cannot contain a nonzero prime, by disjointness of the $P_i$.

You can check out Heitmann's paper Prime ideal posets in Noetherian rings for details of the construction.

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About your example: preferring the notation $\mathfrak a$ to $M$, it is easier to think of your ring $A \times \mathfrak a$ as the Rees algebra of $\mathfrak a$ over $A$, namely $R = A[\mathfrak a t] = \bigoplus_{n \ge 0} \mathfrak a^n t^n$ (the variable $t$ is used to keep track of the degree of an element), modulo the ideal of all elements of $R$ of degree $\ge 2$. Under this notation, it is clear that if $\mathfrak p \in \mathrm{Spec}(A)$, then $(\mathfrak p, \mathfrak a t) \in \mathrm{Spec}(A)$ since $R/(\mathfrak p, \mathfrak a t) \simeq A/\mathfrak p$ is an integral domain. Now let $\mathfrak m = (x_1,\cdots,x_n)_A$ be the ideal of $A$ generated by the polynomial variables, and let us show that $\mathfrak b \overset{def}= (\mathfrak m)_R$ is a counter-example (as you did). Since one easily sees that $\mathfrak a t \subseteq \mathrm{Nil}(R)$, if $\mathfrak b$ contained a prime ideal $\mathfrak q$, then it would contain $\mathfrak a t \subseteq \mathrm{Nil}(R) \subseteq \mathfrak q$ as well, so we would expect $$ \mathfrak a t = \mathfrak b \cap \mathfrak at = \mathfrak m (\mathfrak a t) = (\mathfrak m \mathfrak a)t $$ because the only way to get an element of $t$-degree 1 by multiplying an element of $t$-degree $0$ in $\mathfrak m$ by something is if the element of $t$-degree $0$ in $\mathfrak m$ is multiplied by an element of $t$-degree $1$, and those all belong to $\mathfrak a t$. By Nakayama's Lemma (which we can use since $\mathfrak m$ is a maximal ideal of $\mathfrak a$), $\mathfrak a = \mathfrak m \mathfrak a$ implies $\mathfrak a = 0$, which is assumed false. So $\mathfrak b$ contains no prime ideal.

After that, as you mentioned, any minimal prime ideal above $\mathfrak b$ must contain $\mathfrak b \oplus \mathfrak a t$ since $\mathfrak a t \subseteq \mathrm{Nil}(R)$. If $\mathfrak q$ is a prime ideal containing $\mathfrak b$, then it contains $\mathfrak m$, and it contains $\mathfrak a t$ since those elements are nilpotent, so $$ \mathfrak q = \mathfrak m \oplus \mathfrak a t \supsetneq \mathfrak m \oplus \mathfrak m \mathfrak a t = \mathfrak b $$ because $\mathfrak q$ is maximal (and in fact $R/\mathfrak q \simeq A/\mathfrak m$). Therefore, $\mathfrak q$ is the unique prime ideal containing $\mathfrak b$, and thus $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak q)$. By construction, the ring extension $A \subseteq R$ is integral, and therefore $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak q) \ge \mathrm{ht}(\mathfrak m) > 1$.

I took care to not use anywhere the explicit expression of $\mathfrak m = (x_1,\cdots,x_n)_A$ that you picked in your example where $A = F[x_1,\cdots,x_n]$; nor did I exploit the fact that $\mathfrak a^2 = 0$, I only needed that $\mathfrak a t \subseteq \mathrm{Nil}(A)$. This means that for any non-zero ideal $\mathfrak a \trianglelefteq A$ that contains no non-zero prime ideals of $A$, the ideal $\mathfrak a t \subseteq A[\mathfrak a t]$ together with any maximal ideal $\mathfrak m$ of $A$ and the corresponding ideal $\mathfrak b = (\mathfrak m)_R$ provides an example of an ideal $\mathfrak b \trianglelefteq R$ containing no non-zero prime ideals and $\mathrm{ht}(\mathfrak b) \ge \mathrm{ht}(\mathfrak m)$, as long as $\mathfrak a t$ is nilpotent in $R$; furthermore, if $A$ is integrally closed, the integral extension $A \subseteq R$ satisfies the going-down theorem and $\mathrm{ht}(\mathfrak b) = \mathrm{ht}(\mathfrak m)$.

In particular, since you didn't explicitly use the fact that $A$ was an integral domain but only the fact that $F$ was a field to have your maximal ideal of height $n$ (and help you through the proof), it suffices to pick $A$ a ring with an ideal $0 \subsetneq \mathfrak a \subseteq \mathrm{Nil}(A)$ (you can take equality iff $\mathrm{Nil}(A)$ is not prime), since such an ideal cannot contain any prime ideals by the definition of $\mathrm{Nil}(A)$ and $\mathfrak at$ automatically becomes nilpotent. If you pick $A$ integrally closed and a maximal ideal $\mathfrak m$ of height $n$ (or infinity), you can also find an example $(R,\mathfrak b)$ where $\mathfrak b \trianglelefteq R$ and $\mathrm{ht}(\mathfrak b) = n$.

As for the geometric significance of this construction, the Rees algebra $A[\mathfrak a t]$ corresponds to the blow-up of the affine scheme $\mathrm{Spec}(A)$ at the sheaf of ideals generated by $\mathfrak a$ (in algebraic geometry terms); if we restrict our attention to non-zero nilpotent ideals $\mathfrak a$ not equal to $\mathrm{Nil}(A)$, I can't really give you a geometric picture that's comfortable since the only comfortable blow-ups I know are those at a geometric point like a concrete point in $n$-dimensional affine space over a field or blowing up at some subvariety. Blowing up at a nilpotent ideal is a bit strange because it satisfies the two following properties:

  • Nilpotent elements $a \in \mathfrak a$ now satisfy $(at)^2 = 0$ in $A[\mathfrak at]$
  • The map $\mathfrak p \mapsto \mathfrak p \oplus \mathfrak at$ from $\mathrm{Spec}(A)$ to $\mathrm{Spec}(A[\mathfrak at])$ is a homeomorphism and comes from a closed immersion induced by the quotient map $A[\mathfrak at] \subseteq A[t] \to A$ sending $t$ to $0$. (In fact, seeing it like this, one notices that the property that $A$ is integrally closed is not really necessary in the above arguments; the homeomorphism property follows from the fact that $\mathfrak a$ is nilpotent.)

Let's look at a specific example to develop some intuition. Let $F$ be a field and consider $A = F[x_1,\cdots,x_n, x]/(x^n)$. We have $\mathrm{Nil}(A) = (x)/(x^n)$ which contains no non-zero prime ideals, so we can actually take $\mathfrak a = (x^k)/(x^n) \subseteq \mathrm{Nil}(A)$ for some $1 < k \le n$ (we can't take $k=1$ since in this case $(x^k)/(x^n)$ is prime). The Rees algebra $R$ of $\mathfrak a$ in $A$ is equal to $F[x_1,\cdots,x_n,x,x^kt]/(x^n) \subseteq F[x_1,\cdots,x_n, x,t]/(x^n)$. Clearly $x$ and $t$ are algebraically independent, so $R \simeq F[x_1,\cdots,x_n,x,u]/(x^n, u^{\lceil n/k \rceil})$ (just set $u = x^kt$). So the ring $F[x_1,\cdots,x_n,x]/(x^n)$ is like the point $0 \in \mathbb A^n_F$ that has been "fattened", and blowing-up at $(x^k)/(x^n)$ for $1 < k \le n$ is like making the point $0 \in \mathbb A^n_F$ "even fatter, but in a different direction" because you're blowing it up in this new direction $t$ while still preserving the fatness of that point you already had. But since we just fattened the already fat point, we keep the same points as before blowing up.

Our examples of ideals $\mathfrak b$ are sort of ideals representing "different degrees of fatness" of this point. They cannot contain any prime ideal $\mathfrak q$, because the inclusion $\mathfrak q \subseteq \mathfrak b$ is equivalent to the existence of a closed immersion $A/\mathfrak q \to A/\mathfrak b$, which would mean our "fat point" would contain a proper subvariety; $\mathfrak b$ is a fat point, so the only thing this ideal contains is "fatter points", and going fatter and fatter, you eventually obtain the zero ideal (the fattest point). But none of these fatter points are prime since they are all "fattened by nilpotents". However, the height of this fat point is the same as that of the corresponding point in the non-fattened variety.

Replace "geometric point in $\mathbb A^n_F$" by "some closed point of $\mathrm{Spec}(A)$" and you go back to the arbitrary setting of algebraic geometry and affine schemes. As a note, when $\mathfrak a \trianglelefteq A$ is an ideal, the associated graded ring $\mathrm{gr}_A(\mathfrak a) \overset{def}= (A/\mathfrak a)[(\mathfrak a/\mathfrak a^2)t]$ has also been extensively studied; in our context, $R/\mathfrak b = \mathrm{gr}_A(\mathfrak a) / \mathfrak m\mathrm{gr}_A(\mathfrak a)$.

Hope that helps,

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  • $\begingroup$ this is an absolutely beautiful answer, thank you!! I will need some time to digest all of the general details in full, but the example you give is extremely helpful. thank you for the time and energy you have put in $\endgroup$ – Atticus Stonestrom Feb 10 at 6:24
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    $\begingroup$ @AtticusStonestrom My pleasure! I'm reviewing my blowups and a study of graded rings and projective morphisms these days, so it was a nice coincidence to bump on this question. To be fair, you were half-way there! Usually when I see proofs like this that have a very strong geometric intuition behind them (like the fact that your example was precisely a nilpotent blowup), I try to go coordinate-free and remove any clutter whatsoever in the proof. Seeing a proof in full generality is harder but tends to reveal the core of the argument. $\endgroup$ – Patrick Da Silva Feb 10 at 12:50

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