# 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

## 1 Answer

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.

The answer you're citing in the thread you linked to seems to be incorrect. If you go back to the thread that person linked to, you'll see that the answer they cite actually requires a stronger condition: there is a unique minimal prime containing all the zero-divisors (not the nilpotents).

• Yes, the linked thread is incorrect and you are right: +1 – Georges Elencwajg Feb 13 '17 at 0:39
• Great example, thanks – Prince M Feb 13 '17 at 6:12
• I have no idea what I was thinking. Looking at the timestamps, it looks like I wrote the two solutions two weeks apart, so perhaps I misremembered recent history in the second one (the one being cited here.) Having a prime nilradical obviously is only sufficient to make "half" the zero divisors nilpotent. – rschwieb Feb 13 '17 at 12:26
• @PrinceM You really should have read Manny's solution (which was parallel to the one of mine you read) which already included this example! His posts are very good. – rschwieb Feb 13 '17 at 14:49