Rings with a finite number of non-units

Posted: September 17, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,

Problem. Let R be an infinite ring with 1 and let U(R) be the set of units of R. Prove that if R \setminus U(R) is finite, then R is a division ring.

Solution. Suppose, to the contrary, that there exists some 0 \neq x \in R \setminus U(R). First note that if I \neq R is a left or right ideal of R, then I is finite because otherwise I \cap U(R) \neq \emptyset and so I=R. Therefore Rx and xR cannot both be infinite. Suppose that Rx is finite and let I=\{r \in R: \ rx = 0 \}. Then I is a left ideal of R and I \neq R because x \neq 0. Hence I is finite. Now the R-module isomorphism R/I \cong Rx implies that R is finite. Contradiction!  

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s