0 divisors and solutions in ax=b

[Glass]

Homework Statement

Prove that if a is a 0 divisor of some ring, then for any b of the ring, ax=b cannot have one x that satisfies the equation (i.e. cannot have one solution)

Homework Equations

The Attempt at a Solution

It seems to me this isn't even right. Say a is a right 0 divisor but not a left. Then ax=0 has only one solution (namely x=0) so unless all right 0 divisors are left 0 divisors (which I didn't think is true) isnt' this a contradiction of the theorem I'm trying to prove?

 

[morphism]

It's telling you that a is a 0 divisor, i.e. both a right and left 0 divisor.

 

[matt grime]

Firstly, morphism is correct, but I wanted to add this: you say

"Say a is a right 0 divisor but not a left. Then ax=0 has only one solution (namely x=0)"

This means that a right 0 divisor means a *is not* a left 0 divisor. Be careful of saying things like that. The status of a as a right 0 divisor means nothing about it as a left 0 divisor (unless the ring is abelian under *).