Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers 2

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.

share|improve this answer
    
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
add comment

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,

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.