Jump to content

zero-divisors


Recommended Posts

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

  • 1 month later...

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

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.