# Irreducible Spectrum implies Zero-Divisors are Nilpotent

Let $A$ be a commutative unital ring, and let $\mathrm{Spec}(A)$ be it's set of prime ideals endowed with the Zariski Topology.

In this topological space, my text had me prove a couple facts, specifically:

1. $\{X_a \colon a \in A\}$ forms a basis for the topology, where $X_a = \{ x \in \mathrm{Spec}(A) \colon a \notin x\}$
2. $X_a \cap X_b = X_{ab}$
3. $X_a = \emptyset \iff a$ is nilpotent in $A$

I have a conjecture that $\mathrm{Spec}(A)$ irreducible implies that every zero-divisor is nilpotent. The definition I have of irreducible is

$\mathrm{Spec}(A)$ is irreducible if every two non-empty open sets have a non-empty intersection.

I thought I successfully proved my claim, but after review I think I actually proved a weaker claim (technically a stronger claim, depending on how you look at it). Here is my attempted proof.

Proof: Suppose $z \in A$, $z \neq 0$, and $zy = 0$, for some $y \neq 0$. Proceeding by way of contrapositive suppose $z$ is not nilpotent. Then by fact 3 above we have $X_z \neq \emptyset$, but $X_z \cap X_y = X_{zy} = X_0 = \emptyset$. Thus $\mathrm{Spec}(A)$ is reducible.

It appears to be an obvious flaw that I do not make an assertion about whether or not $y$ is nilpotent. Therefore the claim I actually proved is

If $\mathrm{Spec}(A)$ is irreducible and $xy = 0$, for non-zero $x$ and $y$, then one of $x$ or $y$ is nilpotent.

Upon inspection, this actually made sense because the text had me prove the characterization that $\mathrm{Spec}(A)$ is irreducible if and only if the nilradical is a prime ideal. With this characterization, if the nilradical is prime and $xy = 0$, since $0$ is in the nilradical then $xy$ is in the nilradical, and if the nilradical is prime then $x$ is in or $y$ is in. Which is consistent with what I proved, that atleast $\textbf{one}$ of $x$ or $y$ is nilpotent.

Obviously it is no such luck that one could prove if $xy = 0$ and $x$ is not nilpotent, then $y$ is not nilpotent, since we are not in an integral domain (otherwise my original claim would be vacuously true).

$\textbf{HOWEVER:}$ I still believe my conjecture to be true. I think either 1) I am overlooking something very obvious, a quick fix, or 2) In order to prove my original claim in full generality I might need more machinery than I have at my disposal right now. Some thoughts I had to try and get my claim proven were to 1. Consider when and why the nilradical is a prime ideal, 2. Consider the fact that the set of zero divisors is a union of prime ideals, 3. quit wasting time thinking about this and move forward in my text and maybe when I finish the book I will either realize why this claim is not even important and thus proving it is not a good use of time, or the proof will be obvious.

Any help, thoughts, opinions appreciated.

Edit: I should also mention that the reason I still believe my conjecture is true is not because I have some intuition (although I did check some specific examples), it is because I googled "when are zero divisors nilpotent" and found an answer to this question Under what conditions is a zero divisor element $a$ in commutative ring $R$ nilpotent? that says the nilradical being prime as a sufficient condition for zero divisors being nilpotent. As mentioned the nilradical being prime is iff the spectrum is irreducible, thus my claim should hold.

Also can someone give me a counter example for why the converse does not hold? That is an example where the zero divisors and nilpotents coincide but the nilradical is not prime? There is probably a simple $\mathbb{Z}_n$ case that is apparent using elementary number theory but I don't have anything.

• To help avert any future confusion: I went ahead and removed the spurious claim about $Nil(R)$ being prime from my solution (the one referred to in this post.) – rschwieb Feb 13 '17 at 14:48

Counterexample: Consider $A=k[x,y]/(x^2,xy)$. Then $Nil(A)=(x)$ is prime, so $\textrm{Spec}(A)$ is irreducible. But $y$ is a zero-divisor which is not nilpotent.
The key to this example is that $A$ has an embedded prime $(x,y)$. The nilradical of any ring is the intersection of the minimal primes, so we have $\textrm{Spec}(A)$ is irreducible iff $A$ has only a single minimal prime. But $A$ still may have embedded primes, and then any element in an embedded prime but not in the minimal prime will be a zero-divisor but not nilpotent.