Algebra, Problems For Fun (15)

**NonCommAlg** · June 7th 2009, 01:59 PM

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.

**AMI** · June 11th 2009, 12:09 AM

$\bullet$ $x:=a+b,y:=a-b$
$\Rightarrow ((a+b)^2-(a+b))(a-b)=(a-b)((a+b)^2-(a+b))$ $\Rightarrow (a^2+b^2+ab+ba-a-b)(a-b)=(a-b)(a^2+b^2+ab+ba-a-b)$ $\Rightarrow\dots\Rightarrow b^2a+ba^2-ba-a^2b-ab^2+ab=ab^2+a^2b-ab-ba^2-b^2a+ba$ (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 $ab=ba$ ( $\forall a,b\in R$ ).

**TheAbstractionist** · June 11th 2009, 01:58 AM

Originally Posted by AMI

$\bullet$ $x:=a+b,y:=a-b$
$\Rightarrow ((a+b)^2-(a+b))(a-b)=(a-b)((a+b)^2-(a+b))$ $\Rightarrow (a^2+b^2+ab+ba-a-b)(a-b)=(a-b)(a^2+b^2+ab+ba-a-b)$ $\Rightarrow\dots\Rightarrow b^2a+ba^2-ba-a^2b-ab^2+ab=ab^2+a^2b-ab-ba^2-b^2a+ba$ (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.

**AMI** · June 11th 2009, 04:17 AM

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.

**NonCommAlg** · June 11th 2009, 12:36 PM

good job! i just don't know why it took you guys so long to figure this out!

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