# finiteness of a non-commutative ring

• Apr 24th 2010, 07:44 AM
xixi
finiteness 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 AM
nimon
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 PM
NonCommAlg
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 AM
xixi
Quote:

Originally Posted by NonCommAlg
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)?

for my question , I meant there are non-units other than zero and the set of these elements is finite , now by this assumption ,can't we still say that \$\displaystyle R\$ is finite?