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Suppose $ p $ a prime integer and $ n $ a positive integer. Does anyone know off the top of their heads if the group ring $ \mathbb{F}_{p}[\mathbb{Z}/n] $ (perhaps regarding $ \mathbb{Z}/n $ as the $ n^{\mathrm{th}} $ roots of unity is helpful) is isomorphic to $ \mathbb{F}_{p^n} $ in general?

Peter Haine
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    This is not true. The splitting into summands depends on whether $p$ is a factor of $n$ and (more importantly) what is the smallest extension field of $\Bbb{F}_p$ that contains $n$th roots of unity (when $p$ and $n$ are coprime). Anyway, the block of the trivial representation will always be there, so the group ring cannot be a field. – Jyrki Lahtonen May 22 '15 at 05:18
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    See this answer for an explanation of what happens when $\gcd(n,p)=1$. – Jyrki Lahtonen May 22 '15 at 05:19

1 Answers1

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For any field $F$ and a cyclic group G of order $ n $ the group ring is the quotient of the polynomial ring in a varisble X by the ideal generated by $ X^ n-1$. As this polynomial is reducible this quotient ring will have zero divisors and hence cannot be field.

P Vanchinathan
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