zero-divisors

[andy1224]

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?

[Tom Mattson]

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.

[Country Boy]

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.