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Thread: Algebra, Problems For Fun (15)

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

    Let $\displaystyle R$ be a ring (not necessarily with identity element). Suppose that $\displaystyle (x^2-x)y=y(x^2-x),$ for all $\displaystyle x,y \in R.$ Prove that $\displaystyle R$ is commutative.
    Last edited by NonCommAlg; Jun 7th 2009 at 02:12 PM.
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  2. #2
    AMI
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    $\displaystyle \bullet$ $\displaystyle x:=a+b,y:=a-b$
    $\displaystyle \Rightarrow ((a+b)^2-(a+b))(a-b)=(a-b)((a+b)^2-(a+b))$ $\displaystyle \Rightarrow (a^2+b^2+ab+ba-a-b)(a-b)=(a-b)(a^2+b^2+ab+ba-a-b)$$\displaystyle \Rightarrow\dots\Rightarrow b^2a+ba^2-ba-a^2b-ab^2+ab=ab^2+a^2b-ab-ba^2-b^2a+ba$ (1)
    $\displaystyle \bullet$ $\displaystyle x:=a,y:=b$
    $\displaystyle \Rightarrow (a^2-a)b=b(a^2-a)$ $\displaystyle \Rightarrow a^2b-ab=ba^2-ba$
    Using this in (1), we obtain $\displaystyle b^2a-ab^2=ab^2-b^2a$ $\displaystyle \Rightarrow ab^2-b^2a=0$ $\displaystyle \Rightarrow ab^2=b^2a$ (2)
    $\displaystyle \bullet$ $\displaystyle x:=b,y:=a$
    $\displaystyle \Rightarrow (b^2-b)a=a(b^2-b)$ $\displaystyle \Rightarrow b^2a-ba=ab^2-ab$
    Using (2), we get $\displaystyle ab=ba$ ($\displaystyle \forall a,b\in R$).
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by AMI View Post
    $\displaystyle \bullet$ $\displaystyle x:=a+b,y:=a-b$
    $\displaystyle \Rightarrow ((a+b)^2-(a+b))(a-b)=(a-b)((a+b)^2-(a+b))$ $\displaystyle \Rightarrow (a^2+b^2+ab+ba-a-b)(a-b)=(a-b)(a^2+b^2+ab+ba-a-b)$$\displaystyle \Rightarrow\dots\Rightarrow b^2a+ba^2-ba-a^2b-ab^2+ab=ab^2+a^2b-ab-ba^2-b^2a+ba$ (1)
    $\displaystyle \bullet$ $\displaystyle x:=a,y:=b$
    $\displaystyle \Rightarrow (a^2-a)b=b(a^2-a)$ $\displaystyle \Rightarrow a^2b-ab=ba^2-ba$
    Using this in (1), we obtain $\displaystyle b^2a-ab^2=ab^2-b^2a$ $\displaystyle \Rightarrow ab^2-b^2a=0$ $\displaystyle \Rightarrow ab^2=b^2a$ (2)
    $\displaystyle \bullet$ $\displaystyle x:=b,y:=a$
    $\displaystyle \Rightarrow (b^2-b)a=a(b^2-b)$ $\displaystyle \Rightarrow b^2a-ba=ab^2-ab$
    Using (2), we get $\displaystyle ab=ba$ ($\displaystyle \forall a,b\in R$).
    Hi AMI.

    I think the proof can be simplified just a little.

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

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


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

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


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

    Putting $\displaystyle x=b,\,y=a$ 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 $\displaystyle 1$, then we could take $\displaystyle 1+x$ in the place of $\displaystyle x$, obtain $\displaystyle (x^2+x)y=y(x^2+x)$, 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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