Let R be a ring in which $x^2=x$ for all x ε R. (a) Prove that $x + x = 0$ for for all x ε R. (b) Prove that R is commutative.

Question

I am aware this question has been asked many times but I'm still struggling to understand a few things.

Let R be a ring in which $x^2=x$ for all x ε R. (a) Prove that $x + x = 0$ for for all x ε R. (b) Prove that R is commutative.

For (a):

I understand that $a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a$. But I don't understand how $a+a+a+a ⟹ a+a=0$.

For (b): For any $x, y$ in the ring $R$, $x+y=(x+y)^2=x^2+xy+yx+y^2=x+y+xy+yx,$ so $xy+yx=0$. But how do I get to $xy = yx$ ?

Answer 1

Both use the additive cancellation property: $a+b=a+c\implies b=c$, which is true in any ring, by subtracting $a$ from both sides.

For a), we have $a+a+a+a=a+a+0$.

For b), we have $xy+xy\overset{\text{from a)}}=0=xy+yx$.

Answer 2

But I don't understand how $a+a+a+a \Rightarrow a+a=0$.

You didn't prove $a+a+a+a$, this is actually meaningless. What you proved is that $$a+a=a+a+a+a$$

Now add $-a-a$ on both sides.

$xy+yx=0$. But how do I get to $xy=yx$ ?

Add $xy$ on both sides and use $xy+xy=0$.