# Thread: Prove that an equation with a zero divisor can't have a unique solution

1. ## Prove that an equation with a zero divisor can't have a unique solution

Let R be a commutative ring, and suppose b is a zero divisor in R. Let c be any element in R. Prove that the equation bx = c can never have a unique solution in R. (If it has any solution, there is more than one solution.)

I know that if b is a zero divisor, there must be some c != 0 in R where bc = 0, and that you can't cancel a zero divisor from a product. I guess that if there is a solution to bx = c, there must also be a y such that by = c, where x != y. I could set the two equal to each other: bx = by, and then simplify to b(x-y) = 0. I'm not sure where to go from here, or even if this was the right way to start the proof.

2. I guess that if there is a solution to bx = c, there must also be a y such that by = c, where x != y. I could set the two equal to each other: bx = by, and then simplify to b(x-y) = 0.
Note that assuming that there is another solution y besides x would not work in the final proof because the existence of this y is precisely what you need to prove. In the initial attempt, however, it is fine to assume that such y exists in order to see what properties it must have. So you concluded that b(x-y) = 0 must hold. Can it in fact hold since b is a zero divisor?

3. Originally Posted by emakarov
So you concluded that b(x-y) = 0 must hold. Can it in fact hold since b is a zero divisor?
x-y would also have to be a zero divisor, right? Still, I really have no idea where to start with this proof.

4. Originally Posted by uberbandgeek6
x-y would also have to be a zero divisor, right? Still, I really have no idea where to start with this proof.
So, consider b and x fixed but y a variable.

bx = by iff b(x-y) = 0

because b is a zero divisor there exists y != x such that b(x-y) = 0. (From the definition of zero divisor, there exists a != 0 such that ba = 0; so let y = x-a.) Thus there exists y != x such that bx = by.

Putting the pieces together, if bx = c has a solution, then there exists y != x such that bx = by = c.