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I've looked at earlier similar questions and, as far as I could see, the examples of zero divisors that are not nilpotent are idempotents. I tried to prove that those are the only examples, at least in some cases, but could not. So:

Let $k$ be a field and let $R$ be a finitely generated and reduced $k$-algebra such that the only idempotents in $R$ are $0$ and $1$. Is it the case that the only zerodivisor of $R$ is $0$?

Boogie
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    Consider the $k$-algebra $k[x,y]/(xy)$. – Geoffrey Trang Dec 31 '21 at 21:24
  • Is it obvious that it is reduced and does not have idempotents? – Boogie Dec 31 '21 at 21:33
  • @Boogie Not absolutely obvious, but easy to check. – Mark Dec 31 '21 at 22:06
  • I appreciate it was a bit of a dumb question. Still I think there is something I need to understand. Let ${A = k[x, y]/(xy)}$ be the $k$-algebra you proposed. The elements ${(0,1) , (1,0) \in k^2}$ determine maximal ideals of $A$, say, ${I, J \subseteq A}$. (I would not know how to write them popertly but they are there.) Is there a primary ideal ${P \subseteq I\cap J \subseteq A}$? Happy new year. – Boogie Jan 01 '22 at 19:40

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In general, it is not said that if a zero-divisor is not nilpotent it is idempotent. The ring $\mathbb Z/(15)$ is Noetherian, where $3$ and $5$ are zero-divisors; however $3^2=9$ and $5^2=10$, so they are not idempotent.

The example you're looking for could be the algebra $A:=k[X,Y]/(XY)$. It is reduced (because $(XY)$ is radical, being the intersection of the two minimal primes in which it is contained, $(X)$ and $(Y)$), but $x$ and $y$ (the images of $X$ and $Y$ in $A$) are two zero-divisors clearly not idempotent.

Dr. Scotti
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    Actually, $\mathbb{Z}/(15)$ does not work, because $6$ and $10$ are two nontrivial idempotents. – Geoffrey Trang Jan 16 '22 at 02:46
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    Of course you're right, it was just to show that if a zero-divisor is not nilpotent, it is not necessarily idempotent (that was the original doubt of the OP). – Dr. Scotti Jan 16 '22 at 07:00
  • @GeoffreyTrang I still have the feeling that the abscence of idempotents should play some role, and that I am asking the wrong question. Anyway, how do you prove that ${k[x,y]/(xy)}$ lacks idempotents? – Boogie Jan 21 '22 at 15:25
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    @Boogie first observe that any non-zero polynomial in $k[x,y]$ of the form ($$) $\sum_{n\in \mathbb N}(a_nx^n+b_ny^n)$, with all the $a_n,b_n\in k$, is not divisible by $xy$. This means that we can choose, to represent the equivalence class of an element in $k[x,y]/(x y)$, an unique polynomial $f\in k[x,y]$ of the form ($$). Now it's easy to see that $f^2-f$ is not divisible by $xy$, unless $f$ was the zero polynomial. – Dr. Scotti Jan 21 '22 at 16:43