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Let $p$ be prime and $\mathbb{F}_p$ the field of characteristic $p$. Let $\mathbb{Z}/n$ be the cyclic group of order $n$ where $p\not|n$. I know $\mathbb{F}_p[C_n]\cong\mathbb{F}_p[x]/(x^n-1)$ and I know $(x^n-1)=\prod_{d|n}\Phi_d(x)$, where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. I get the feeling that $\mathbb{F}_p[C_n]\cong\prod_{d|n}\mathbb{F}_p$, but unsure why that is. Can anyone help?

Moreover, does this then mean $\mathbb{F}_p[C_5]\cong\mathbb{F}_p\times\mathbb{F}_p$, because this seems surprising. I would have expected five copies of $\mathbb{F}_p$ (again $p$ does not divide 5).

Sam Jefferies
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    A basic problem here is that the cyclotomic polynomials don't remain irreducible over $\Bbb{F}p$. Assuming $\gcd(p,n)=1$ the Wedderburn decomposition will be a product of several extension fields $\Bbb{F}{p^{n_i}}$, where the exponents $n_i$ range over the degrees of irreducible factors of $\Phi_d(x)$. – Jyrki Lahtonen Nov 07 '17 at 13:54
  • See this answer by yours truly for the details. I think the question is a duplicate of that, but given that I answered the linker version may be I should refrain from voting. – Jyrki Lahtonen Nov 07 '17 at 13:56
  • @JyrkiLahtonen So let $n=q$ some other prime. So we definitely have $gcd(p,q)=1$. Then would $x^q-1$ not decompose into $(x-1)(x-\zeta)\cdots(x-\zeta^{q-1})$. From this would we not then get $q$ copies of $\mathbb{F}_p$? – Sam Jefferies Nov 07 '17 at 13:57
  • Oh thanks you, I'll take a look now. – Sam Jefferies Nov 07 '17 at 13:58
  • For the specific cases of $C_5$ it goes as follows. If $p\equiv1\pmod 5$, then $\Bbb{F}p[C_5]$ is, indeed, a product of five copies of $\Bbb{F}_p$. If $p\equiv -1\pmod 5$, then $p^2\equiv1\pmod5$, and $\Bbb{F}_p[C_5]\simeq \Bbb{F}_p\oplus\Bbb{F}{p^2}^2$. If $p\equiv\pm2\pmod5$, then $\Bbb{F}p[C_5]\simeq \Bbb{F}_p\oplus\Bbb{F}{p^4}$. – Jyrki Lahtonen Nov 07 '17 at 13:59
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    Sam, the problem is that those roots $\zeta^i$ may belong to extension fields of varying degrees. – Jyrki Lahtonen Nov 07 '17 at 14:00
  • Ah yes, I think I see. Is there anywhere you could guide me to for a general proof of of the case $Fp[Cq]$ where $p\equiv 1(modq)$? So I can see precisely why this is happening – Sam Jefferies Nov 07 '17 at 14:19
  • @JyrkiLahtonen Sorry to pester you, but am I understanding this right. The problem basically comes down to one of whether the cyclotomic polynomials are irreducible over $\mathbb{F}_p$. If they are, then $\mathbb{F_p}[C_q]$ will be $q$ lots of $\mathbb{F}_p$. If not, then you have other things to check. Is this right? And if so, does it then mean that we need $p\equiv 1 (mod q)$, then the cyclotomic polynomials are irreducible? – Sam Jefferies Nov 07 '17 at 14:27
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    Sam, no.$\Bbb{F}p[C_q]$ is a direct product of $q$ copies of $\Bbb{F}_p$ if and only if $x^n-1$ splits into a product linear factors modulo $p$ if and only if $p\equiv1\pmod q$. So the cyclotomic polynomials are the opposite of irreducible in that case. The degrees of the irreducible factors of $x^n-1$ give the dimensions $n_i$ of the irreducible represesentations of $\Bbb{F}[C^n]$. And also the extension degrees of the Wedderburn components $\Bbb{F}_{p^{n_i}}$. – Jyrki Lahtonen Nov 07 '17 at 15:32
  • @JyrkiLahtonen Right okay, I think I see. Why does $x^n-1$ split in this case? I guess once it does split, then the answer follows from the Chinese Remainder Theorem. – Sam Jefferies Nov 07 '17 at 15:49
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    Sam, remember that $\Bbb{F}_p^*$ is cyclic of order $p-1$. So if $q\mid p-1$ then all those $\zeta$s are elements of $\Bbb{F}_p$. So $x^q-1$ splits completely in $\Bbb{F}_p[x]$ because it has $q$ distinct zeros is $\Bbb{F}_p$. – Jyrki Lahtonen Nov 07 '17 at 20:48
  • @JyrkiLahtonen Ah yes, of course. Thank you very much – Sam Jefferies Nov 09 '17 at 10:03

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