I am doing Contemporary Abstract Algebra by Gallian and am stuck at the proof of the fact that integral domains have either $0$ or prime characteristic. The proof in the book goes as follows:
Let $R$ be an integral domain such that the additive order of $1$ is $n$. On the contrary, assume that $n$ is not prime and hence can be written as $n = s * t$ where, $1 \lt s, t \leq n $. Then $n \cdot 1 = (s* t) \cdot 1 = (s \cdot 1) (t \cdot 1) = 0$. Now since $R$ is an integral domain, one of these factors must be zero and hence we arrive at a contradiction.
But I then tried a different proof and can't seem to spot the mistake because my proof doesn't seem use the fact that $R$ is an integral domain and hence is giving the impression that any arbitrary ring with unity (at least) has either $0$ or prime characteristic. So, here goes my proof:
We have $n \cdot 1 = (s*t) \cdot 1 = \underbrace{(\underbrace{1+...+1}_{s\ times})+(\underbrace{1+...+1}_{s\ times})+\cdots+(\underbrace{1+...+1}_{s\ times})}_{t\ times}=0$.
Hence, we have $t \cdot (\underbrace{1+...+1}_{s\ times}) = t \cdot (s \cdot 1) = 0$. Now since $(s \cdot 1) \in R$ ($R$ is closed under addition), this is a contradiction to the fact that $n$ is the characteristic. Therefore, $(s \cdot 1) = 0$ which again is a contradiction to the fact that $n$ is the characteristic. Hence, $n$ must be a prime.
Now the thing is that nowhere in the proof did I use that $R$ is an integral domain and we could easily replace the integral domain condition with any arbitrary ring with unity. I feel that I am missing some fundamental difference between the definition of the characteristic of an integral domain and that of any arbitrary ring but still haven't been able to figure out.
Thanks in advance!
The problem is exactly the presence of zero divisors if the ring is not an integral domain. You might have $st \cdot 1 = 0$ with $s\cdot 1 \neq 0$ and $t\cdot 1 \neq 0$ (i.e. when $s\cdot 1$ and $t\cdot 1$ are zero divisors). Perhaps a simple counterexample can make it clearer. Consider $n\in \mathbb{N}$ a composite number (i.e. not prime), and look at the ring homomorphism $\varphi: \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}, m\mapsto 1 + \cdots + 1 \ ( \ m \mathrm{ \ times})$. It is clear that the kernel is $n\mathbb{Z}$ and hence the characteristic is $n$, a composite number.