5
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ I gather that by "ring" you mean ring without 1. $\endgroup$ Commented May 24 at 19:49
  • $\begingroup$ No it may have a unity $\endgroup$
    – Chaudhary
    Commented May 24 at 19:50
  • 11
    $\begingroup$ Well $1$ can never be a zero divisor. $\endgroup$ Commented May 24 at 19:53
  • $\begingroup$ " 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. $\endgroup$
    – Anixx
    Commented May 25 at 14:28

3 Answers 3

15
$\begingroup$

What about $\{0,3,6,9,12,15\}$ where addition and multiplication is $\pmod{18}$?

$\endgroup$
6
  • $\begingroup$ This should work. $\endgroup$
    – Chaudhary
    Commented May 24 at 20:06
  • 1
    $\begingroup$ or $\{0,2,4,6,8,10\}$ $\pmod{12}$? $\endgroup$ Commented May 24 at 21:47
  • $\begingroup$ 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. $\endgroup$
    – JAG131
    Commented May 28 at 21:10
  • 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
  • 1
    $\begingroup$ @JAG131: such as $n=k^2l$ with $\gcd(k,l)=1$ $\endgroup$ Commented May 28 at 21:48
6
$\begingroup$

I think this ring will work as well $R=\Bigg\lbrace \begin{pmatrix} 0& a \\ 0 & b \\ \end{pmatrix}|a,b \in \mathbb{Z} \Bigg\rbrace$.

$\endgroup$
1
$\begingroup$

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 $0$.

$\endgroup$
1
  • $\begingroup$ This is an interesting ring. $\endgroup$
    – Chaudhary
    Commented Jun 3 at 19:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .