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April 7th, 2010, 10:05 PM  #1 
Senior Member Joined: Dec 2009 Posts: 150 Thanks: 0  Ring with unity and two right ideals
I got the following wrong on a practice GRE math subject test and I don't know how to do it: Let R be a ring with a multiplicative identity, and U an additive subgroup such that u in U and r in R implies u*r is in U. (Compare with a normal ideal which requires both u*r, and r*u to be in U) . U is called a right ideal of R. R has exactly two right ideals, which of the following is true: I) R is commutative II) R is a division ring (all elements except additive identity have multiplicative inverses) III) R is infinite The answer is (II) only. While (I)is clearly not true because there are two such ideals, it's that (II) is necessary that I don't know how to show. Also how can we be sure the ring is NOT infinite... Square matrices with nonzero determinants come to my mind, because they lack commutativity and have multiplicative inverses, but this is a finite ring. Perhaps the quaternions have this property? Everything except the part (Compare with ... to be in U) is straight from the practice exam word for word. Thank you for your time in advance. 
April 8th, 2010, 08:49 AM  #2 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Ring with unity and two right ideals
It's possible that the problem is a bit of a trick problem or at least, unclear wording. It says it has exactly two right ideals. Does it say anywhere that these need to be proper and/or nontrivial? Also, does "which of these hold?" mean "which must hold?" or "exactly one of the following holds. Which?" ? Anyway, if my suspicion about the lack of clarity is correct, we need to show that a ring with no proper (and nonzero) right ideals is necessarily a division ring. Suppose R is not a division ring. Choose a nonzero element a with no inverse in R. define aR = {a*r  r in R }. You can show that aR is a right ideal of R (show it's a subgroup using distributivity, and show it consumes elements on the right using associativity of multiplication.) And clearly , so aR is a proper, nonzero right ideal in R. So we've shown that any ring without inverses will have proper nonzero right ideals. Of course, if my suspicion is wrong (which is possible, because normally ideals are explicitly defined to be proper), I'm not sure how to proceed. 

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