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Give an example of a ring in which there exist $x,y,z \in R\setminus\{1,0\}$ such that $x$ is a unit, $y$ is nilpotent, and $z$ is idempotent.

I have an idea of the form $R = \mathbb{Z}/ n\mathbb{Z}$. I just need to justify my answer. I have a feeling $n$ is a prime number. How can I justify this?

mdp
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Brandon
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    If $n$ is prime, then $R$ is a field and so there are no non-zero nilpotents, and no idempotents other than $0$ or $1$. So if there for an example of this form, $n$ is necessarily not prime. – mdp May 20 '14 at 14:48
  • thanks for the edit, any suggestions on how to find such an example? – Brandon May 20 '14 at 14:53
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    Do your rings have to be commutative? If not, then $2\times2$ matrices over a field will do. – mdp May 20 '14 at 14:59
  • yes we have been working with commutative rings only – Brandon May 20 '14 at 14:59
  • for it to be a unit such that there exists y $\in$ R that xy=yx=1, can x = y? For example, i think R = $Z$/12$Z$ where the unit may be [5][5] =1 . It works that [6] is nilpotent and [3] is idempotent. – Brandon May 20 '14 at 15:09
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    Brandon - I think you have the right idea, but $3^2=9$ is not idempotent according to the definition I know. – Mark Bennet May 20 '14 at 15:16
  • @Brandon Yes, $x$ can be equal to $y$. (The definition doesn't say anything to rule this out.) But Mark is right, $3$ is not idempotent in $\mathbb{Z}/12\mathbb{Z}$. – mdp May 20 '14 at 15:17

1 Answers1

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Hint $\ $ If $\,n\,$ is composite and not a prime power then $\,\Bbb Z/n\,$ has an idempotent $\not\equiv 1,0\,$ since then $\,n = ab,\,$ coprime $\,a,b>1$ so by CRT $\ \exists\,x\equiv (1,0)\ {\rm mod}\ (a,b)\,$ so $\,x^2\equiv (1,0)^2\equiv (1^2,0^2)\equiv x.\,$

Further if $\,n\,$ is not squarefree then a prime power $\,p^2\mid n\,$ so $\,{\rm mod}\ n\!:\ n/p\not\equiv 0,\,$ but $\, (n/p)^2\equiv 0$

Finally, $ $ if $\,n> 2\,$ then $\ {-1}\not\equiv 1,0,\ $ so $\ {-}1\,$ is a nontrivial unit.

Example $\ $ Simple cases arise for $\, n = p^2q,\,$ primes $\,p\ne q, \,$ e.g. $\ 2^2\cdot 3 = 12.\,$ Then, as above we have an idempotent $\,x\equiv (1,0)\ {\rm mod}\ (3,4)\!\!\overset{\rm CRT}\iff\! x\equiv 4\pmod{12},\,$ indeed $\,4^2\equiv 4\pmod{12}.$

Further we have the nonzero nilpotent $\,n/p = pq\,$ with $\,n^2\! = p^2q^2 \equiv 0\pmod {p^2q}.\,$ In our example $\, n/p = 12/2 = 6,\,$ and $\,6^2\equiv 0\pmod{12}.$

Remark $\,\ \color{#c00}{\rm Nontrivial\,}$ idempotents, i.e. elements satisfying $\,x^2 = x\,$ (and $\,\color{#c00}{x\neq 0,1})$ are intimately connected to coprime factorizations (of both elements and rings). As was explained above, by CRT, modulo any non-prime-power composite $\,n,\,$ there are nontrivial idempotents $\,(0,1),(1,0).$

Some integer factorization algorithms work by searching for nontrivial idempotents mod $\,n,\,$ which immediately yield a factorization of $\,n\,$ (generally one can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so any notrivial idempotent or nontrivial square-root will split $\,n,\,$ since it yields a quadratic with $\,\color{#c00}3\,$ roots).

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