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Abstract Proof

  1. Apr 3, 2010 #1
    Let R be a ring with multiplicative identity 1R. Suppose that R is finite. The elemets xy1, xy2,...xyn are all different. So x y_i=1R for some i.

    A lemma that is not proven is given. If xyi=1R & yjx=1R, then yi=yj

    I need to show that yjx=1R.

    Right now I haven't got much. I took the contrapositive of the lemma, but I still get stuck as I'm not sure where I could go from there with the information that I'm given.

    The book gives a theorem which states Let R be a ring with identity and a, b of R. If a is a unit each of the equations ax=b & ya=b has a unique solution in R.

    Then it goes on to state that if the ring is not commutative, x may not be equal to y. But yeah, I'm still stuck.
    Last edited: Apr 4, 2010
  2. jcsd
  3. Apr 4, 2010 #2


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    Hi kuahji! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Hint: what's yjxyi ? :smile:
  4. Apr 4, 2010 #3


    The teacher also gave a proof for the lemma.


    Except of course we haven't shown yjx=1R yet.

    Still stuck, can't make the intuitive leap.
  5. Apr 4, 2010 #4


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    I assume you mean yi=(yjx)yi=yj(xyi)=yj1R=yj ?
    But that's the definition of yj

    yj is defined as the left inverse of x, and yi is defined as the right inverse …

    so you're home. :smile:
  6. Apr 4, 2010 #5
    Ok, sorry I don't see how it's the definition of yj.

    I mean with your hint we have

    But that is still different from yjx=1R
  7. Apr 4, 2010 #6


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    Because of …
    … yjx=1R is given. :wink:
  8. Apr 4, 2010 #7
    hehe but that lemma is what we're trying to prove. We can't use it to prove itself. But I got it now, had to work some algebraic magic. The "proof" of the lemma only showed that yi=yj. We were already given xyi=1R. So we still had to show the one part. Thanks for the help.
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