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.

-
It is false that the characteristic of a unital ring is either $0$ or a prime. $\mathbb{Z}/n\mathbb{Z}$ has a natural structure of a unital ring for any $n\gt 1$, and its characteristic is $n$, which of course need not be a prime. The characteristic of an integral domain is either $0$ or a prime. Your definition of "characeristic" for nonunital rings is also, in my opinion, rather off; it should be "for all $a$", not "for some a"... –  Arturo Magidin Feb 21 '12 at 0:28

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.

-
This is just Patrick's answer reworded in a way I understand it :D –  Mariano Suárez-Alvarez Feb 21 '12 at 0:37
BTW, that $R$ be an integral domain has nothing to add to this. Your definition of characteristic, though, is a bit strange... –  Mariano Suárez-Alvarez Feb 21 '12 at 0:39

You don't need to invoke units. As your proof stated, if we assume $(nm)a = 0$ for some $a \in R$ non-zero, then $n(ma) = 0$, and since $nm$ is the least integer with the property that $m(na) = 0 = n(ma)$, then $na \neq 0 \neq ma$. Since $$0 = 0a = ((nm)a)a = (nm)a^2 = (na)(ma) \neq 0,$$ we have a contradiction (the last part is because $na \neq 0 \neq ma$ and $R$ is an integral domain).

Hope that helps,

-
This is just the generalization of the trick in a unital ring where we write $(nm)1 = (n1)(m1)$, but in fact this is the same proof as in the unital ring case, except that in unital rings we have $1^2 = 1$ which makes the proof a little simpler in the sense that we can "look" at integers in $R$ more explicitly. –  Patrick Da Silva Feb 20 '12 at 21:37
The question defines the characteristic as the smallest positive $p$ such that $pa=0$ for some nonzero $a$, not all of them. –  Mariano Suárez-Alvarez Feb 20 '12 at 21:52
When I first saw the question that's what was written there... I edited my proof to suit this definition. My comment still applies. –  Patrick Da Silva Feb 20 '12 at 21:59