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Math Help - Algebra, Problems For Fun (15)

  1. #1
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    Algebra, Problems For Fun (15)

    Let R be a ring (not necessarily with identity element). Suppose that (x^2-x)y=y(x^2-x), for all x,y \in R. Prove that R is commutative.
    Last edited by NonCommAlg; June 7th 2009 at 03:12 PM.
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  2. #2
    AMI
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    \bullet x:=a+b,y:=a-b
    \Rightarrow ((a+b)^2-(a+b))(a-b)=(a-b)((a+b)^2-(a+b)) \Rightarrow (a^2+b^2+ab+ba-a-b)(a-b)=(a-b)(a^2+b^2+ab+ba-a-b) \Rightarrow\dots\Rightarrow b^2a+ba^2-ba-a^2b-ab^2+ab=ab^2+a^2b-ab-ba^2-b^2a+ba (1)
    \bullet x:=a,y:=b
    \Rightarrow (a^2-a)b=b(a^2-a) \Rightarrow a^2b-ab=ba^2-ba
    Using this in (1), we obtain b^2a-ab^2=ab^2-b^2a \Rightarrow ab^2-b^2a=0 \Rightarrow ab^2=b^2a (2)
    \bullet x:=b,y:=a
    \Rightarrow (b^2-b)a=a(b^2-b) \Rightarrow b^2a-ba=ab^2-ab
    Using (2), we get ab=ba ( \forall a,b\in R).
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by AMI View Post
    \bullet x:=a+b,y:=a-b
    \Rightarrow ((a+b)^2-(a+b))(a-b)=(a-b)((a+b)^2-(a+b)) \Rightarrow (a^2+b^2+ab+ba-a-b)(a-b)=(a-b)(a^2+b^2+ab+ba-a-b) \Rightarrow\dots\Rightarrow b^2a+ba^2-ba-a^2b-ab^2+ab=ab^2+a^2b-ab-ba^2-b^2a+ba (1)
    \bullet x:=a,y:=b
    \Rightarrow (a^2-a)b=b(a^2-a) \Rightarrow a^2b-ab=ba^2-ba
    Using this in (1), we obtain b^2a-ab^2=ab^2-b^2a \Rightarrow ab^2-b^2a=0 \Rightarrow ab^2=b^2a (2)
    \bullet x:=b,y:=a
    \Rightarrow (b^2-b)a=a(b^2-b) \Rightarrow b^2a-ba=ab^2-ab
    Using (2), we get ab=ba ( \forall a,b\in R).
    Hi AMI.

    I think the proof can be simplified just a little.

    \bullet\,x=a+b,\,y=a
    ((a+b)^2-a-b)a\ =\ a((a+b)^2-a-b)
    \implies\hspace{30mm}\cdots

    \implies\ ba^2+b^2a-ba\ =\ a^2b+ab^2-ab\quad\ldots\fbox1


    \bullet\,x=a,\,y=a+b
    (a^2-a)(a+b)\ =\ (a+b)(a^2-a)
    \implies\hspace{30mm}\cdots

    \implies\ a^2b-ab\ =\ ba^2-ba\quad\ldots\fbox2


    \fbox1\,+\,\fbox2\ \implies\ b^2a\ =\ ab^2

    Putting x=b,\,y=a completes the proof.
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  4. #4
    AMI
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    Thanks, Abstractionist!
    Oh, and I think the reason why the problem said "not necessarily with identity element" is that if we had the identity 1, then we could take 1+x in the place of x, obtain (x^2+x)y=y(x^2+x), substract the initial identity, and we're done.
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  5. #5
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    good job! i just don't know why it took you guys so long to figure this out!
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