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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.

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marked as duplicate by rschwieb ring-theory Aug 9 at 13:11

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    $\begingroup$ The upvoted answer in your linked post doesn't use commutativity either... $\endgroup$ – Arthur Aug 9 at 11:26
  • $\begingroup$ The other existing answer in the question covers what you wrote in this post. It's good that you found that post... but it's strange that you didn't seem to read the answers... $\endgroup$ – rschwieb Aug 9 at 13:08
  • $\begingroup$ Also, in case this is the point of confusion, as it often seems to be, just because a question contains an assumption doesn't always mean the assumption is necessary, or that something weaker can't replace it. Quite often stronger assumptions are included to make the problem more accessible to people learning the subject for the first time. $\endgroup$ – rschwieb Aug 9 at 13:11

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