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

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?

• Note that the indentity $y^8=y$ implies $(-1)^8=-1$, so $1=-1$, $2=0$. An example of a ring with $y^7=y$ is $\mathbb{F}_3\times \mathbb{F}_4$. Nov 29, 2019 at 22:24

The field $$\mathbb{F}_8$$ is a counterexample. $$\mathbb{F}_8$$ is the splitting field of $$T^8-T$$ over $$\mathbb{F}_2$$.
• A counterexample of such a ring without identity I can think of is $\bigoplus_{k=1}^\infty \Bbb F_8$. It would be interesting to know if there is a counterexample of a finite ring without identity. Nov 29, 2019 at 1:23
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 $$n \ge 1$$.