Prime ideal containing only zero divisors

Question

Let $R$ be a Noetherian ring and $p\subset R$ be a prime ideal containing only zero divisors. Does it follow that $p$ is an associated prime of $R$? In other words is there some $x\in R$ such that $p=Ann(x)$?

Answer 1

I think the answer is negative in general. (Update: with a little effort, you can see that the example given below is same as the example provided by user26857. And in the example, $(x)\cap (x,y,z)^2$ is abviously a primary decomposition of $0$, the user26857's answer points out another primary decomposition, we can see also that it is an example the uniqueness of primary decomposition of zero fails.)

Case 1: Let $A$ be a Noetherian ring.

Assume $\mathfrak{p}_1\subsetneq \mathfrak{p}_2\subsetneq \mathfrak{p}_3$ be three prime ideals of $A$ such that $\mathfrak{p}_3\in Ass(A)$.

Then by Krull's principal ideal theorem, there are infinitely many prime ideals $\mathfrak{p}$ with $\mathfrak{p}_1\subsetneq \mathfrak{p}\subsetneq \mathfrak{p}_3$. Since $\mathfrak{p}_3$ consists of zero divisors, those $\mathfrak{p}$ also consist of zero divisors. But there are only finitely many associated primes in a Noetherian ring. Thus there must be some $\mathfrak{p}$ which is not an associated prime.

For example, let $A=k[x,y,z]/(x)\cap (x,y,z)^2$ where $k[x,y,z]$ is the polynomial ring in variables $x,y,z$ over a field $k$. Then the associated primes are $(x)=(0:y)$ and $(x,y,z)=(0:x)$. But the prime ideal $(x,y)$ is not an associated prime ideal. As you can verify by hand there is no $f$ such that $(x,y)=(0:f)$.

Case 2 For the case a Noetherian ring $B$ whose all associated primes $\mathfrak{p}$ are of height $\leq 1$. Then a prime consisting of only zero-divisors is an associated prime.

The reason: since the set of zero-divisors is a union of the associated primes, by prime avoidance lemma, we can conclude that such a prime $\mathfrak{q}$ consisting of only zero-divisors is contained in some associated prime $\mathfrak{p}$, then either $\mathfrak{q}=\mathfrak{p}$ or $\mathfrak{q}$ is a minimal prime, in any case, $\mathfrak{q}$ is an associated prime.

So, combining all the above discussions, the Case 2 is the sufficient and necessary case for a noetherian ring such that any prime consisting of zero-divisors is an associated prime.

---

Update:

Let us say that we are given a Noetherian ring $A$ such that there exist prime ideals $\mathfrak{p}_1\subsetneq \mathfrak{p}_2\subsetneq \mathfrak{p}_3 $, we will show that (by Krull's principal ideal theorem) there are infinitely many prime ideals $\mathfrak{p}$ such that $\mathfrak{p}_1\subsetneq \mathfrak{p}\subsetneq \mathfrak{p}_3 $.

First, the Krull's principal ideal theorem here refers to Corollary 11.17, page 122, M.F. Atiyah and I.G. Macdonald, Introduction to commutative algebra, Addison-wesley series in mathematics, Addison-Wesley, Great Britain, 1969.

The statement is as following.

(Krull's principal ideal theorem) Let $A$ be a Noetherian ring and let $x$ be an element of $A$ which is neither a zero-divisor nor a unit. Then every minimal prime ideal $\mathfrak{p}$ of $(x)$ has height $1$.

You can also find various versions of this theorem in pages 233-238, David Eisenbud, Commutative algebra: With a view toward algebraic geometry, Springer, New York, 2004.

And see also chapter 7 (page 85-101) in A course in commutative algebra, GTM 256, Gregor Kemper. In fact, in this book, the exercise 7.5 (Can the spectrum by just one chain?) points out there must be infinitely many primes properly lies between two prime ideals in a Noetherian ring.

See also chapet 5 (dimension theory), theorem 13.5, Matsumura's Commutative ring theory, etc.

OK. Let us give a proof that there must be infinitely many primes properly lies between two prime ideals in a Noetherian ring.

We may assume $A$ is a Noetherian local domain with maximal ideal $\mathfrak{m}$ and of dimension $\geq 2$.

Suppose $A$ has only finitely non-zero, non-maximal prime ideals $\mathfrak{p}_i$, $i=1,\ldots,n$. Then $\mathfrak{m}\nsubseteq\cup_{i=1}^n \mathfrak{p}_i$ by prime avoidance lemma (Proposition 1.11 in M.F. Atiyah and I.G. Macdonald, Introduction to commutative algebra, Addison-wesley series in mathematics, Addison-Wesley, Great Britain, 1969). We can choose an $x\in \mathfrak{m}\setminus (\cup_i \mathfrak{p}_i)$. Now consider a minimal prime $\mathfrak{p}$ over $(x)$, then by Krull's principal ideal theorem, $\mathfrak{p}$ is of height $1$ and thus $\mathfrak{p}\neq \mathfrak{m}$. From the construction, prime ideal $\mathfrak{p}\neq \mathfrak{p}_i$ for $i=1,\ldots,n$. We win.

Answer 2

A simple (counter)example: $R=K[X,Y,Z]/(X^2,XY,XZ)$, $\mathfrak p=(x,y)$.

Let me give some details: the ideal $I=(X^2,XY,XZ)$ can be written as $(X)\cap(X^2,Y,Z)$ (see here). This is a primary decomposition of $I$ and thus the associated primes of $I$ are $(X)$ and $(X,Y,Z)$. Now consider the prime ideal $(X,Y)$ which is strictly contained between the two associated primes.