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This is in connection with the question: Showing that the characteristic of a commutative ring R without zero divisors is 0 or prime.
I have found a solution without using commutativeness. I don't understand why commutativeness will be necessary to prove it.
Here is my solution:
If possible let char $R = pq$ where $p,q\in\mathbb N$ such that $p,q>1.$
There exists $a,b\in R$ such that $pa\ne0,qb\ne0.$
Then $(pa)(qb)\ne0,$ a contradiction since $(pa)(qb)=(pq)(ab)=0.$
Thus char $R = 0$ or prime.
Please tell me where did I go wrong.