Prove that if $a^2 = a$ for all elements of the ring $R$, then $R$ has characteristic 2. Is the converse of the statement true?
Attempt:
by the definition of characteristic of a ring, the first statement implies that characteristic of the ring is $\leq 2$. If the ring is assumed to be non trivial ring then, characteristic is 2.
About the last statement can we construct a field of say 8 elements and show that there exists a an element of order $\neq 2$?
Is there a simpler example ?
I don't think your attempt is correct ; you seem to be confused about what the characteristic of a ring is.
One way to do it is to notice that for all $a\in R$, $$2a=(2a)^2=4a^2=4a,$$hence $2a=4a-2a=0$.
As for finding a counterexample, you can indeed construct a field with $8$ elements, or more generally with $2^m$ elements for all $m\geq 1$; and since in a field, $a^2=a$ is equivalent to $a(a-1)=0$, any element $a$ other than $0$ or $1$ must be such that $a^2\neq a$.
There is, however, one example that might be simpler : the ring $\mathbb{Z}/2\mathbb{Z}[X]$ of polynomials with coefficients in $\mathbb{Z}/2\mathbb{Z}$ (or any other ring of characteristic $2$) also has characteristic $2$, but $X^2\neq X$.
