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Math Help - Rings

  1. #1
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    Rings

    Let R be a ring in which x^2 =x ~ \forall x \in R. Show that x+x = 0 ~ \forall x \in R and that R is commutative.

    Using closure under addition I can show that x+x = 0. However, I'm not too sure about this. Do I use the fact that x + (-x) = 0 so that 4x=2x \iff 2x=0 \iff x+x=0 ?

    I can't think of a way to show that \forall x, y \in R, ~ xy=yx.
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  2. #2
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    Quote Originally Posted by nmatthies1 View Post
    Let R be a ring in which x^2 =x ~ \forall x \in R. Show that x+x = 0 ~ \forall x \in R and that R is commutative.

    Using closure under addition I can show that x+x = 0. However, I'm not too sure about this. Do I use the fact that x + (-x) = 0 so that 4x=2x \iff 2x=0 \iff x+x=0 ?

    I can't think of a way to show that \forall x, y \in R, ~ xy=yx.

    If you already reached the equality 4x=2x then of course that you can deduce 2x=x+x=0 since x+(-x)=0 is true in any abelian group and thus in rings, too.

    Now, use this with x+y=(x+y)^2=x^2+xy+yx+y^2... to show that xy=yx

    Tonio
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