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Math Help - Prove that an equation with a zero divisor can't have a unique solution

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    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.
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    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?
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    Quote Originally Posted by emakarov View Post
    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.
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    Quote Originally Posted by uberbandgeek6 View Post
    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.
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