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?