Let $(R,+,*)$ be a ring so that $(x+y)^{2}=x^{2}+y^{2}$ $\forall\ x, y \in R$. Prove that

A)$xy=-yx$ $\forall\ x, y \in R$

B)$x^{2}+x^{2}=0$ $\forall\ x, y \in R$ and $x+x=0\ \forall\ x, y \in R$

I have already proved A) and $x^{2}+x^{2}=0$ $\forall\ x, y \in R$ and for the last part of B I had that:

$x^{2}+x^{2}=0 \Rightarrow\ (x+x)^{2}=0$ but I don´t know if this implies that $x+x=0$

So I would really appreciate your help for the last part of B

  • $\begingroup$ If your ring has a unit $1$, then you can use part (A), setting $y=1$. $\endgroup$ – ajd Mar 4 '14 at 5:16
  • 1
    $\begingroup$ If it doesn't have a unit, the result is false (try a zero ring on three elements). $\endgroup$ – Jack Schmidt Mar 4 '14 at 5:18
  • $\begingroup$ But how can I know that the ring has unit 1? Do I need to prove it? $\endgroup$ – user128422 Mar 4 '14 at 5:39
  • $\begingroup$ They bring up the question of $1$, because there are two different conventions about what the word "ring" means. Check your definition. $\endgroup$ – user14972 Mar 4 '14 at 6:02

Since $(x+y)^2=x^2+y^2+xy+yx$ , then by our assumption $x^2+y^2=(x+y)^2\Rightarrow xy+yx=0$ That is exactly what you wanted to show.note that taking $x=y$ immediately implies second part.it seems to me that ring $R$ ought to have multiplicative identity $1_R$.assuming that let $y=1_R$ then by first part $1x=-x1$ thus $x+x=0$

  • $\begingroup$ thanks a lot!!!!!! and one last thing: do I need to prove that the ring has unit 1? $\endgroup$ – user128422 Mar 4 '14 at 6:18
  • $\begingroup$ In some undergrad type textbooks they assume all their rings contain identity to begin with. im not sure if that statement still holds if we put the assumption of having identity aside.well i dont know any counterexample of top of my head. $\endgroup$ – BigM Mar 4 '14 at 6:24

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