Characteristic of a Non-unital Integral Ring

If $R$ is a unital integral ring, then its characteristic is either $0$ or prime. If $R$ is a ring without unit, then the char of $R$ is defined to be the smallest positive integer $p$ s.t. $ pa = 0 $ for some nonzero element $a \in R$. I am not sure how to prove that the characteristic of an integral domain without a unit is still either $0$ or a prime $p$. I know that if $p$ is the char of $R$, then $px = 0 $ for all $x \in R$. If we assume $ p \neq 0 $ and $R$ has nonzero char, and $p$ factors into $nm$, then $ (nm) a = 0 $ , which means $ n (ma) = 0 $. Well $ma \neq 0$, because this would contradict the minimality of $p$ on $a$. But I don't know where to go from this point w/o invoking a unit.

Edit: I had left out the assumption that $R$ is assumed to be a integral domain. This has been corrected.

Solution 1:

Suppose $p$ is the characteristic of $R$ and not prime, so that $p=mn$ for some positive integers $m$,~$n>1$. In particular, $p>n$ and $p>m$. According to the definition you are using, $p$ is the least positive number such that there exists a non-zero $a\in R$ with $pa=0$: it follows that $na\neq0$, and then that moreover $m(na)\neq0$. This is absurd, of course, because $m(na)=(mn)a=pa$ because the addition in $R$ is associative.