# Thread: Rings

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

2. 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