# Integral domain without unity has prime characteristic?

By an integral domain, we mean here, a ring (not necessarily with unity) in which $ab=0$ implies $a=0$ or $b=0$.

Question: If an integral domain without unity has positive characteristic, is it necessarily prime?

An integral domain $D$ is said to be of finite characteristic if there exists a positive integer $m$ such that $ma=0$ for all $a\in D$. [cf. Topics in Algebra- I. N. Herstein, 2nd Ed., p. 129]

My question is a slight modification of Problem 6 in [Topics in Algebra- I. N. Herstein, 2nd Ed., p. 130]

Topics in Algebra- Herstein, 2nd Ed.

• Related, maybe duplicate – Daniel Fischer Dec 22 '14 at 10:46
• I think to be safe you should add a definition of characteristic of a non-unital ring. The one used in the link suspiciously does not coincide with the characteristic of a unital ring. – Pavel Čoupek Dec 22 '14 at 10:50
• @Daniel: My definition of characteristic is according to Herstein, which do not match with that in the Duplicate's Link. – Groups Dec 22 '14 at 10:58
• In part, because the definition of characteristic used there is unusual, I said "maybe duplicate" and not "duplicate". The argument of the answers there also applies with the more common definition, however. But $D = \{0\}$ is a special case, its characteristic is not prime (it's $1$). – Daniel Fischer Dec 22 '14 at 11:02

I will assume the characteristic of a non-unital ring is simply the exponent of the additive group, i.e. the smallest $n$ such that $na=0$ for every $a \in A$.
So assume that $n=kl, \;\;\; 0<k , l <n$ is a composite number. Choose $a \in A$ such that $ra \neq 0$ for every $r = 1, 2, \dots, n-1$ (if there is no such element, the characteristic is ncessarily smaller than $n$). Thus, $ka \neq 0, \; la \neq 0$ but $(ka) \cdot (la)=kl a^2=na^2=0,$ contradicting the fact that $A$ was an integral domain. Thus, if the characteristic is positive, it is necessarily prime.