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.

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 $m-n=1$.
