Let R be a ring such that $x^2=x$ for each x in R. Which of the following must be true?
I‐ x= ‐x for all x in R.
II‐ R is commutative
III‐ xy+yx=0 for all x in R
I know that for R to be a ring there are three conditions that must be met. They are:
1)[A,+] is an abelian group (i.e. its commutative)
2)multiplication is associative in A, and
3)multiplication is left and right- distributed over addition.
Therefore I know by definition that II is correct, but need help figuring out if I and III are correct.
If $x\in R$, then $$2x=(2x)^2 = (x+x)^2 = (x+x)(x+x) = x^2 + x^2 + x^2 + x^2 = 4x^2 = 4x$$ hence $4x=2x$, so $2x=0$.
Thus, $2x=0$, for all $x\in R$.
Equivalently, $x=-x$, for all $x\in R$.
If $x,y\in R$, then $$x+y = (x+y)^2 = (x+y)(x+y) = x^2 + xy + yx + y^2 = (x + y) + (xy + yx)$$ hence $xy+yx=0$.
But also, since $yx=-yx$, we have $$xy-yx=xy + (-yx) = xy+yx =0$$ so $R$ is commutative.