Is it true that $y^7=y$ in a commutative ring with $r^8=r$?

Question

Let $R$ be a commutative ring with identity such that $\forall r \in R[r^8=r]$. Let $y \in R$. Is it true that $y^7=y$?

It is obviously true if $R$ is the trivial ring (i.e. $0=1$ ) so suppose $R$ is not the trivial ring.

I see that, in the arithmetic of $R$, $(1+1)^8=1+1$, so $254=0$. I see also that $(y+y)^8=256y^8=2y^8$, yet I don't see how this helps.

Bonus: If it is true that $y^7=y$, would this still be true even if $R$ is a commutative ring without identity?

Answer 1

The field $\mathbb{F}_8$ is a counterexample. $\mathbb{F}_8$ is the splitting field of $T^8-T$ over $\mathbb{F}_2$.

Answer 2

If $y^7 = y$, then you must have $y = y^8 = y \cdot y^7 = y^2$. Conversely, if $y = y^2$ then $y = y^n$ for all $n \ge 1$.