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I'm relatively new to ring theory so this is probably a simple question.

Its easy to see that a finite ring that is commutative and has no-zero divisors (i.e. an integral domain) must have multiplicative inverses.

I am wondering if we can rearrange these properties and always get implication or produce finite rings with only 2 of the above properties.

Explicitly my questions are:

1) Does a finite ring that is commutative with multiplicative inverses always have no-zero divisors? (EDIT: this one is pretty easy too, let xy=0 and assume x not 0. then (x^-1)xy=(x^-1)0. Hence, y=0 as desired.)

2) Is a finite ring that has multiplicative inverses and no zero divisors always commutative?

If these are actually simple exercises, I'd just like a hint to get started with the proof or any counterexamples.

Thanks :)

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You might want to look at: springer.com/engineering/electronics/book/978-1-4020-7039-6 –  Joseph Malkevitch Aug 31 '10 at 1:55

1 Answer 1

up vote 10 down vote accepted

HINT 1) If $\;\rm R\;$ is finite then $\;\rm x\to r\:x\;$ is onto iff 1-1, so $\;\rm R\;$ is a field iff $\;\rm R\;$ is a domain.

2) is Wedderburn's little theorem.

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Thanks! I see (1) now. Just say x,y are in R and then do xy=0. WLOG assume x not 0. then x has an inverse so (x^-1)xy=(x^-1)0, hence y=0. I think I need some more time with Weddernburn's little theorem, since I haven't really heard of a lot of the terms used in the link. –  WWright Aug 31 '10 at 2:39

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