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I need to prove that $R$ is commutative, that is for each $x,y\in R$ $xy=yx$. I think for a ring to be commutative, it needs identity such that $1_R\cdot x=x=x\cdot 1_R$, $1_R\not=0_R$.

so, $xy=1_R(xy)=1_Rx\cdot 1_Ry=1_Ry\cdot 1_Rx=(yx)1_R=yx$ Dose it make sense? ----Ok does not make sense!

Pedro
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Wes
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    What is $R$? Also, unless your definition of a ring requires the existence of $1_R$, there's no need to have an identity in a commutative ring. Your proof also seems to assume that $1_R x$ and $1_R y$ commute, which is really what you're trying to prove. –  Oct 25 '13 at 02:51
  • What is $R$? If you purpose $R$ is ring, $R$ does not satisfy $xy=yx$ for each $x,y\in R$ in general. – Hanul Jeon Oct 25 '13 at 02:52
  • $R$ is a ring. And I am totally lost in the class. – Wes Oct 25 '13 at 02:56
  • So, what should I try? – Wes Oct 25 '13 at 02:56
  • @Wes Not all rings are commutative, and not all rings containing $1$ are commutative. Without a lot more information, the problem isn't true. –  Oct 25 '13 at 02:56
  • Ok, I think that's why i need to prove this case(?) is commutative somehow. – Wes Oct 25 '13 at 02:58
  • @Wes If $R$ is an arbitrary ring, then it's not guaranteed to be commutative. Please read the problem again. –  Oct 25 '13 at 03:00
  • @T.Bongers So, you mean I cannot be sure there is multiplicative inverse or identity either? Please give some guide line to start. – Wes Oct 25 '13 at 03:03
  • @Wes In an arbitrary ring, there need not be any multiplicative inverses or identity. As I've said, the claim is false. Please read the original question again. –  Oct 25 '13 at 03:04
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    Let $R$ be a ring such that for every $x\in R$, we have $x^2=x$. Prove that $R$ is commutative. In other words, for each $x,y\in R$, prove that $xy=yx$ – Wes Oct 25 '13 at 03:06
  • That condition $x^2=x$ is huge. – Karl Kroningfeld Oct 25 '13 at 03:06
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    @Wes That condition is extremely important and is completely missing from your original post. –  Oct 25 '13 at 03:07
  • @T.Bongers Oh...sorry.. I didn't realize that was important... What makes different? – Wes Oct 25 '13 at 03:11
  • @ZevChonoles Thanks. I looked through it but I don't understand the last part. Why $xy+yx=0$ becomes $yx=-yx=xy$ by characteristic 2? What is characteristic 2? – Wes Oct 25 '13 at 03:33

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Take $(x+x)^2=(2x)^2=4x^2=4x$. From the other hand, you have $(x+x)^2=x+x=2x$. So $4x=2x$, i.e. $2x=0$ for any $x$. In other words, $x=-x$ for any $x$.

Take $x+y=(x+y)^2$. Then you have $x+y=x^2+y^2+xy+yx=x+y+xy+yx$. From this you can conclude that $xy+yx=0$, i.e. $xy=-yx=yx$. The last equality follows from the first paragraph.

Sasha Patotski
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  • You do have to be a little bit careful on the first paragraph if the ring has characteristic 2. Then x+x = 0 so you get it anyway. – Tuo Jul 12 '15 at 02:18