I know that the collection of all nilpotent elements in a commutative ring form such a ring where all elements are zero divisors. For example, if we take $S=\lbrace 0,2,4,6 \rbrace$ as a subring of $\mathbb{Z_8}$, then all elements of $S$ are zero divisors and nilpotent. If we look at $\mathbb{Z_{p^n}}$, then we arrive at another example of this. But I'm struggling to find a well-known example where not every element is nilpotent. More specifically, I want to find such a ring that at least one of the element of the ring is a zero divisor but not nilpotent. How should I think of such an example?
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2$\begingroup$ I gather that by "ring" you mean ring without 1. $\endgroup$– Sassatelli GiulioCommented May 24 at 19:49
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$\begingroup$ No it may have a unity $\endgroup$– ChaudharyCommented May 24 at 19:50
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11$\begingroup$ Well $1$ can never be a zero divisor. $\endgroup$– Cameron WilliamsCommented May 24 at 19:53
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$\begingroup$ " I want to find such a ring that atleast one of the element of the ring is a zero divisor but not nilpotent." - split-complex numbers. $\endgroup$– AnixxCommented May 25 at 14:28
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3 Answers
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What about $\{0,3,6,9,12,15\}$ where addition and multiplication is $\pmod{18}$?
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1$\begingroup$ or $\{0,2,4,6,8,10\}$ $\pmod{12}$? $\endgroup$ Commented May 24 at 21:47
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$\begingroup$ More generally, $k\mathbb{Z}/n\mathbb{Z}\subset\mathbb{Z}/n\mathbb{Z}$ (where "$k$" divides "$n$") will always give an example of a finite ring in which every element is a zero divisor, as well -- in most cases -- where no elements are nilpotent. $\endgroup$– JAG131Commented May 28 at 21:10
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1$\begingroup$ @JAG131: if $n=k^2$, all the elements of $k\mathbb Z/n\mathbb Z$ are nilpotent; if $n=kl$, where $k $ and $l$ are distinct primes, then $k\mathbb Z/n\mathbb Z$ does not have nonzero zero divisors $\endgroup$ Commented May 28 at 21:31
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1$\begingroup$ @JAG131: such as $n=k^2l$ with $\gcd(k,l)=1$ $\endgroup$ Commented May 28 at 21:48
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I think this ring will work as well $R=\Bigg\lbrace \begin{pmatrix} 0& a \\ 0 & b \\ \end{pmatrix}|a,b \in \mathbb{Z} \Bigg\rbrace$.
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The residues $0,6,10,15\bmod 30$ form a multiplicative ring in which every element is idempotent while the product of any two distinct elements is $0$.
