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Math Help - invertible?

  1. #1
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    invertible?

    Hi
    Let A be a ring, a,b \in A such that ab=1. Suppose that X=\{x \in A;\ ax=0\} is finite.
    Prove that ba=1


    We can see that 1-ba \in X, so if X=\{0\}, it's over.
    Any idea?
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  2. #2
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    Quote Originally Posted by clic-clac View Post
    Hi
    Let A be a ring, a,b \in A such that ab=1. Suppose that X=\{x \in A;\ ax=0\} is finite.
    Prove that ba=1


    We can see that 1-ba \in X, so if X=\{0\}, it's over.
    Any idea?
    we have: a^{k-1} - ba^k \in X, \ \forall k \in \mathbb{N}. hence, since X is finite, there exist n, m \in \mathbb{N} such that a^n-ba^{n+1}=a^m -ba^{m+1}, \ n > m, which gives us: (1-ba)a^n=a^m-ba^{m+1}. \ \ \ \ (1)

    now multiply both sides of (1) from the right by b^n and note that \forall k \in \mathbb{N}: \ a^kb^k=1. then you'll get: 1-ba=(a^m-ba^{m+1})b^n=a^mb^n - ba^{m+1}b^n=b^{n-m}-b^{n-m}=0. \ \ \ \ \ \ \ \ \ \ \Box
    Last edited by NonCommAlg; November 16th 2008 at 12:02 PM.
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  3. #3
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    Thanks for your answer.

    I don't see why there exists n such that a^n-ba^{n+1}=a^{n-1}-ba^n.
    The conclusion is still the same with a^n-ba^{n+1}=a^{k}-ba^{k+1} for a k<n , but there is something I'm missing...
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  4. #4
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    Quote Originally Posted by clic-clac View Post
    Thanks for your answer.

    I don't see why there exists n such that a^n-ba^{n+1}=a^{n-1}-ba^n.
    The conclusion is still the same with a^n-ba^{n+1}=a^{k}-ba^{k+1} for a k<n , but there is something I'm missing...
    you're right, thanks! it was just one of those really weird mistakes that i make sometimes! it's fixed now!
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  5. #5
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    You're welcomed.
    But for me the problem is that doing this, we assume that every element of X is equal to a a^{q}-ba^{q+1}, q\in \mathbb{N}.
    Why can we say that?
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  6. #6
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    Quote Originally Posted by clic-clac View Post

    we assume that every element of X is equal to a a^{q}-ba^{q+1}, q\in \mathbb{N}.
    Why can we say that?
    i didn't say that! what we have is this: A=\{a^{k-1} - ba^k: \ k \in \mathbb{N} \} \subseteq X, and we're given that X is finite. so A must also be finite and thus we can find two elements of A which are equal.
    Last edited by NonCommAlg; November 16th 2008 at 12:51 PM.
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  7. #7
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    Great NonCommAlg, I finally got it! Thank you.
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