andy1224 Posted May 16, 2007 Share Posted May 16, 2007 I'm not really sure how to approach the following: Let A be a commutative ring for which aA=A for all a not equal to 0. prove A has no zero divisors. any ideas? Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 18, 2007 Share Posted May 18, 2007 That can't be the complete problem statement. Where do A and a come from? Do they both belong to the commutative ring? And are you sure that aA=A for all a not equal to zero? Seems to me it would make a lot more sense if it were for all A not equal to zero. The way you've worded it we get the following. aA=A aAA-1=AA-1 a1=1 a=1 Mind you, that's a=1 for all a not equal to zero. That doesn't make much sense to me. Link to comment Share on other sites More sharing options...
Country Boy Posted July 6, 2007 Share Posted July 6, 2007 Tom, I would interpret this to mean that A is the ring and a some member of A. aA= {ab| b contained in A}. If it is true that aA= A for all a, then x-> ax is a one-to-one mapping from A onto aA. If a were a zero-divisor, then there exist b such that ab= 0= a0, so that the mapping is not one-to-one. Link to comment Share on other sites More sharing options...
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