Let R be a ring such that $a^2 = a$ , $\forall a$ $\in R$ . Prove that R is commutative.

Question

Im not sure what to do for this one, any help would be appreciated.

Answer 1

$$(a+b)^2 = (a+b) \Rightarrow a^2+ab+ba+b^2 = a+b$$ Since $a^2 = a$ and $b^2 = b$ we have that $ab = -ba$ by cancellation. Now $$(a+a) = (a+a)^2 = a^2+2a^2+a^2= a+a+a+a \Rightarrow a+a = 0 \text{ for all } a\in \mathbb{R}.$$ This says that each element is its own (additive) inverse and so $-ba = ba$ which finally gives $$ab = -ba = ba$$

Answer 2

$-1=-1^2=1$

Now notice $x+y=(x+y)^2=x^2+xy+yx+y^2=x+xy+yx+y\iff xy=-yx=yx$

Answer 3

First, we show that every element of $R$ is its own additive inverse. First, note that $2x = (2x)^2 = 4(x^2) = 2(x^2)$. This tells you that $4x = 2x$, and thus $2x = 0$, for all $x$ in $R$. Now, $x + y = (x+y)^2 = x^2 + xy + yx + y^2 = x + y + xy + yx$. From this we get $xy + yx = 0 \implies xy = -yx$, which coupled with our lemma and the fact that additive inverses are unique gives $xy = yx$.