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Let $A$ a domain, i.e. $ab\in A\implies a=0$ or $b=0$. It's written that all domains are commutative. Is it by definition, or can we prove that domains are commutative? I mean, do we only consider domaind for commutative rings, or is a ring that is a domain then commutative?

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It is "by definition" though, as the other answer indicates, some might disagree with the terminology.

It is perfectly possible to have a non-commutative ring that has no zero-divisors (except for $0$). Examples would be, for example, non-commutative division rings, also called skew fields. (These are rings where every non-zero element has a multiplicative inverse.)

As an aside, for division rings there is an interesting result that all finite division rings in fact are commutative. I mention this as an example where commutativity does follow from at first glance unrelated properties.

Maybe even more to the point, and as mentioned on the page lined in the other answer, as in a finite ring (with identity) without non-trivial zero-divisors every non-zero element has an inverse for finite rings indeed commutativity follows from the property you mentioned.

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That's a mistake: a domain doesn't have to be commutative. When it is, we call it an integral domain.

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Some authors do allow "domain" to refer to a noncommutative ring. But nearly universally "integral domain" will imply commutativity.

Here are nine examples of noncommutative rings with no nonzero zero divisors.

I'll call out at least two of them, the Hurtwitz and Lipschitz quaternions.

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