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So I have this practice problem. I see that $x^3 - x - 1$ is irreducible over $Z/3$ and so I guess the number of elements in $R$ would just be the same as a degree $3$ field extension over $F_3$ so $27$ elements I think? Then since its irreducible it must also be a maximal ideal in $ \space F[x]$ which is a PID and so the quotient ring is a field which must be isomorphic to $F_{27}$ I think? So I guess the splitting field would be $x^{3^3-1}-x$, but I don't know how to see which roots of unity which has. I'm not even sure if what I have so far is correct

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  • $\begingroup$ Have you proved that the polynomial in question is irreducible ? If you have then what you said is correct. I don't know what you mean by "the splitting field would be $x^{27-1} -x$". As for the roots of unity, have you tried Lagrange's theorem ? $\endgroup$ – Max Apr 26 '18 at 10:36
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You're on the right track.

Since $R=\mathbb F_{27}$ is a field, the multiplicative group $R^\times$ is cyclic of order $26$. Therefore, it has elements of all orders dividing $26$. These are the primitive roots of unity in $R$.

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