characteristic prime or zero Let $R$ be a ring with $1$ and without zero-divisors. I have to show that the characteristic of $R$ is a prime or zero. This is my attempt:
This is equivalent to finding the kernel of the homomorphism $f\colon \mathbb{Z}\rightarrow R$ which has the form $n\mathbb{Z}$. There are two cases. Suppose $f$ is injective, then only $f(0)=0$. Suppose now that $f$ is not injective, thus there exist $m,n\in \mathbb{Z}$ such that $f(m)=f(n)$. This implies $f(m-n)=0$ thus $m-n$ is in the kernel and so is every multiple of $m-n$ thus $(m-n)\mathbb{Z}$ is the kernel (I think this is not 100% correct). Suppose now that $m-n$ is not a prime. Then there exist $a,b\in \mathbb{Z}$ such that $m-n=ab$. $f(a)\neq 0\neq f(b)$ because then $a\mathbb{Z}$ or $b\mathbb{Z}$ would be the kernel because $a<m-n$ and also $b<m-n$. Then $0=f(ab)=f(a)f(b)$. This contradicts the fact that there are no zero-divisors.
Is this proof correct? If not what is wrong? Thanks.
 A: Though the machinery of homomorphisms may provide a convenient language to discuss this issue, it is not really necessary to address the essential question here.  But before proceeding, I should say that the OP's proof, once adapted according to Daniel Fischer's comments, seems OK to me.  You just needs to make sure you get your hands on what Daniel Fischer calls $k$, the smallest positive integer in $\ker f$.  Of course, to be technically complete, you need to establish that the map $f:Z \to R$ defined by $f(1_Z) = 1_R$ is in fact a (group and/or ring) homomorphism, but that's a pretty straightforward exercise from the definitions.  (Here I have used the notations $1_Z$, $1_R$ to denote the "$1$" elements of $Z$, $R$, respectively.)
Having said these things, I'd like to return for a moment to the problem as stated in the first two sentences of the OP's question:
"Let R be a ring with 1 and without zero-divisors. I have to show that the characteristic of R is a prime or zero."
I think it's a little simpler to dispense altogether with the homomorphism $f:Z \to R$ in this case, and just argue straight from $R$ in the usual manner, viz. suppose $1_R$, when added to itself some finite number $n$ of times, yields the zero element of $R$; that is, we have
$1_R + 1_R + . . . + 1_R= 0$,
where there are $n$ occurrences of $1_R$ on the left-hand side of this equation.  Granting the existence of such $n \in Z_+$, where $Z_+$ denotes the set of positive integers, it makes sense to assert the existence of a least such $n$, call it $n_0$.  We must have $n_0 > 1$, lest $1_R = 0$, and the whole ring $R$ collapses to $\{0\}$ via $a = 1_Ra = 0a = 0$ for all $a \in R$.  Ruling out this trivial or perhaps, better, excluded by definition of ring $R$ with unit $1_R \ne 0$, case, we observe that if $n_0 > 1$ is prime, we are done.  If $n_0$ is composite, say, $n_0 = m_1m_2$ with $1 < m_1, m_2 < n_0$, then a simple re-arrangement of the $n_0$ terms $1_R$ in our sum yields zero divisors $0 \ne m_11_R, m_21_R \in R$.  Of course, if no such $n \in Z_+$ exists, we can only obtain an equation such as
$1_R + 1_R + . . . + 1_R= 0$
by "adding together" zero $1_R$'s on the left (forgive my slight abuse of notation/language), so $char R = 0$ in this case.
In all honesty, I should add that this is the standard argument I and about a gazillion other math undergrads witnessed in our first abstract algebra courses.
Well, there's two cents worth if there ever was.  Not a completely valueless two cents, however.  Hope this clarifies.  Cheers.
A: The claim being proved here is not true; it fails to hold for the zero ring. It doesn't have nontrivial zero divisors (because it doesn't have nontrivial elements at all), yet its characteristic $1$ is neither zero nor prime.
The flaw in the proof is that it assumes that just because $m-n$ is positive and nonprime, it is the product of integers strictly between $0$ and $m-n$. That is not true for $m-n=1$.
