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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?

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  • $\begingroup$ 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 $\mathbb{F}_3\times \mathbb{F}_4$. $\endgroup$
    – orangeskid
    Nov 29, 2019 at 22:24

2 Answers 2

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

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  • $\begingroup$ 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. $\endgroup$ Nov 29, 2019 at 1:23
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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 $n \ge 1$.

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