Rings

**nmatthies1** · March 7th 2010, 02:44 PM

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 $\forall x, y \in R, ~ xy=yx$ .

**tonio** · March 7th 2010, 06:09 PM

Originally Posted by nmatthies1

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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