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Thread: Need help with invertibility proof

  1. #1
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    Need help with invertibility proof

    Hello.

    I would use some help with following task:
    Prove that when in ring R products xy and yx are invertible then elements x and y are also invertible.

    Thanks in advance.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Need help with invertibility proof

    What does it mean by definition that $\displaystyle xy$ is invertible?

    If you have the definition then you can use the fact that the multiplication is an associative operation.
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    Re: Need help with invertibility proof

    It means that there exists (xy)^-1 so that (xy)^-1 * xy = 1
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    Re: Need help with invertibility proof

    Quote Originally Posted by rain1 View Post
    It means that there exists (xy)^-1 so that (xy)^-1 * xy = 1
    Yes, $\displaystyle xy$ is invertible is there exists an unique $\displaystyle z \in R$ such that $\displaystyle (xy)z=1$.
    Now, use the associtativity.
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  5. #5
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    Re: Need help with invertibility proof

    You mean like this?$\displaystyle (xy)^{-1}\cdot (xy) = x\cdot (y^{-1}\cdot y)\cdot x^{-1} = x\cdot 1x^{-1} = 1 \cdot 1 =1$? Thats all?
    Last edited by rain1; Mar 24th 2013 at 12:38 PM.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Need help with invertibility proof

    Quote Originally Posted by rain1 View Post
    You mean like this?$\displaystyle (xy)^{-1}\cdot (xy) = x\cdot (y^{-1}\cdot y)\cdot x^{-1} = x\cdot 1x^{-1} = 1 \cdot 1 =1$? Thats all?
    In fact, what I mean is the following.
    Suppose $\displaystyle xy$ is invertible then there exists an unique $\displaystyle z \in R$ such that $\displaystyle (xy)z=1$. Since $\displaystyle 1=(xy)z=x(yz)$ (here I use the associativity), we find that $\displaystyle x$ is invertible with $\displaystyle yz$ as the inverse element.
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