Define $\operatorname{char}(R)$ as the least positive integer $\bar n$ for which: $\bar n\cdot x=\underbrace{x+x+\ldots+x}_{\bar n\text{ times}}=0$ for all $x\in R$. We say $\bar n=0$ when no positive integer will suffice. I choose this definition, because it is also applicable when $R$ is not unital.

Throughout this post I will use the "bar notation" (e.g. $\bar n$) if and only if, something represents an integer. This way there will be no ambiguity when I write $\bar n x$.

If we have that $\bar nx=0$, for some postive integer $\bar n$, but $\bar mx\neq 0$ for every positive integer $\bar m<\bar n$, then we say $\operatorname{ord}(x)=\bar n $.

Consider a (possibly nonunital) ring $R$, with $\operatorname{char}(R)=\bar n$. I would like to study the cases where $\bar m x=0$ for a nonzero $x$ and a positive $\bar m<\bar n$. I have already shown that without loss of generality we may assume that $\bar m|\bar n$. This means ${\bar n}/{\bar m}$ is a positive integer. I want to show that this can only happen in a "trivial" manner. Let me make that more precise:

If $\operatorname{ord}(x)=\bar m<\operatorname{char}(R)=\bar n$, then there exists a nonzero $y\in R$, such that $x=\dfrac{\bar n}{\bar m} y$.

This holds in all rings without zero divisors, but I don't know how to show this for the general case. Could anyone help me towards a proof, or towards a counterexample?

  • $\begingroup$ @Peter Which is not a counterexample if that's what you're suggesting. $2+2=0\implies \exists\, y\neq 0$, such that $2={y+y}$. And there is, namely $y=1$. $\endgroup$
    – gebruiker
    Dec 22, 2015 at 17:24
  • $\begingroup$ Ok, now I understand what you mean. $\endgroup$
    – Peter
    Dec 22, 2015 at 17:26
  • $\begingroup$ Can we assume $m|n$, or do we need to prove it ? $\endgroup$
    – Peter
    Dec 22, 2015 at 17:29
  • $\begingroup$ You may assume that $\bar m|\bar n$. I already proved that this can be done without loss of generality. $\endgroup$
    – gebruiker
    Dec 22, 2015 at 17:30
  • $\begingroup$ What about $\mathbb{Z}/2 \oplus \mathbb{Z}/4$? $\endgroup$
    – Ofir
    Dec 28, 2015 at 21:35


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