1. ## Rings

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

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

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

2. Originally Posted by nmatthies1
Let $\displaystyle R$ be a ring in which $\displaystyle x^2 =x ~ \forall x \in R$. Show that $\displaystyle x+x = 0 ~ \forall x \in R$ and that $\displaystyle R$ is commutative.

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

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

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

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

Tonio