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Let $R$ be the quotient ring ${\mathbb Z}_3[x]/\langle x^3 - x\rangle$. Show that for all $r \in R$, $r^3 = r$. Not even sure where to start or how to go about it. I noticed that it represents it is commutative but how does that help?

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Take any $f(x) = a_0 + a_1x + \ldots + a_nx^n \in \mathbb{Z}_3[x]$, then consider $g(x) = f^3(x) - f(x)$, then, since $$ a^3 = a \quad\forall a\in \mathbb{Z}_3 $$ we have $$ g(0) = g(1) = g(-1) = 0 $$ Hence, $x(x-1)(x+1) = x^3-x \mid g(x)$ in $\mathbb{Z}_3[x]$, whence $$ \overline{f(x)}^3 = \overline{f(x)} $$ in $\mathbb{Z}_3[x]/(x^3-x)$

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