Let $\displaystyle R$ be a non-commutative ring . Suppose that the number of non-units of $\displaystyle R$ is finite . Can we say that $\displaystyle R$ is a finite ring?

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- Apr 24th 2010, 07:44 AMxixifiniteness of a non-commutative ring
Let $\displaystyle R$ be a non-commutative ring . Suppose that the number of non-units of $\displaystyle R$ is finite . Can we say that $\displaystyle R$ is a finite ring?

- Apr 24th 2010, 08:52 AMnimon
I think that we can say $\displaystyle R$ is a finite ring. The only way it could be infinite is if the group of units, call it $\displaystyle U$, were infinite. But then $\displaystyle rU$ for any $\displaystyle r\notin U$ with $\displaystyle r\neq 0$ would be an infinite number of non-unital elements, which is a contradiction.

So I guess we must assume that there is at least one non-zero, non-unital element, otherwise an infinite ring of nothing but units with an additive identity thrown in (such as $\displaystyle \mathbb{R}$) would destroy the argument.

In short: yes, unless $\displaystyle R$ is a field! - Apr 24th 2010, 06:48 PMNonCommAlg
**nimon**'s proof works only for commutative*domains*because $\displaystyle rU$ is not necessarily infinite even if $\displaystyle U$ is infinite. the answer to**xixi**'s question is also negative for non-commutative rings.

for example $\displaystyle \mathbb{H},$ the ring of quaternions over $\displaystyle \mathbb{R},$ has this property because every non-zero element of $\displaystyle \mathbb{H}$ is a unit. in general, every (infinite) division ring has the property because every

non-zero element of a division ring is a unit. here's a less trivial version of**xixi**'s problem:

let $\displaystyle R$ be an infinite commutative (resp. non-commutative) ring. suppose that the number of non-unit elements of $\displaystyle R$ is finite. is $\displaystyle R$ necessarily a field (resp. division ring)? - Apr 25th 2010, 06:58 AMxixi