November 14th, 2013, 09:56 AM  #1 
Newbie Joined: Sep 2013 Posts: 17 Thanks: 0  Division Ring
Let R be a ring with 1 . Prove that R is a division ring if and only if the only left ideals of R are 0 and R.

November 14th, 2013, 02:47 PM  #2  
Senior Member Joined: Aug 2012 Posts: 2,259 Thanks: 686  Re: Division Ring Quote:
(edit) No response for a while, here's more of a hint. A left ideal is characterized by the property that it's closed under arbitrary (left) multiplication by any element of the ring R. I think of ideals as "absorbing" multiplication. So suppose a left ideal X contains 1. For any element r of the ring R, r1 = r is in X. So X contains everything in R, in other words X = R. So if X contains 1 it's the entire ring. Ok that's the background. To refresh our memories and provide clarity, here is the original problem. Quote:
(=>) Let R be a division ring and show that its only left ideals are R and 0. Ok, suppose we have an ideal X. If it consists of only 0 we're done. So there's some x in X that's not zero. Since R is a division ring, if x is not zero then 1/x exists in R. But we can multiply the elements of X by any elements of R, including 1/x. But now 1/x * x = 1, so 1 is in X. And what do we know happens the moment we know that 1 is in an ideal? We know that the ideal is the whole ring. Because if 1 is in X then so is r1 for any r in R. So that's the (=>) direction. Next, the other direction. (<=) Suppose that R is a ring such that its only ideals are R and 0. We must prove that R is a division ring. Do you want to take a shot at carrying this forward?  

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