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I've came across this task in my book about abstract algebra. I know the definition of a cyclotomic polynomial, but I do not understand what $\Phi_{n,\mathbb{Z}_p}$ means: I only know what $\Phi_{n}$ is. Could you please explain to me what's going on here and how can we prove this equality?

$\deg(\Phi_{n,\mathbb{Z}_p})=ord_n(p)$

Jane Doe
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  • It is $\Phi_n$ over the finite field $\Bbb F_p$. Then it need not be irreducible anymore, but the irreducible factors have degree $ord_n(p)$, see this post and the links. – Dietrich Burde Aug 25 '24 at 11:29
  • @DietrichBurde, how is it defined (what's it's formula)? Do we use coeffs from $\mathbb{F}_p$? I mean, $\Phi_n$ is defined as product of $(x - a_i)$ where $a_i$ is a primitive root. So in case of $\Phi_n$ we do not use coeffs from a field, what about cyclotomic polynomial over the finite field $\mathbb{F}_p$? – Jane Doe Aug 25 '24 at 11:35
  • It is just defined modulo $p$, since the coefficients of $\Phi_n$ are integers. For example, $\Phi_7=x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible over $\Bbb Z$, but $\Phi_{7,\Bbb F_2}=(x^3+x+1)(x^3+x^2+1)$ is reducible. – Dietrich Burde Aug 25 '24 at 11:39
  • @DietrichBurde, well, I've found this step: suppose $\deg{\Phi_{n, \mathbb{Z}_p}} = k$, let $F = \mathbb{Z}_p(\xi)$, then $|F| = p^k$. my question is: what's $\mathbb{Z}_p(\xi)$ here and why order of $F$ is $p^k$? – Jane Doe Aug 25 '24 at 12:26
  • $F$ is a field extension of degree $k$, so that it is a vector space of dimension $k$ over $\Bbb F_p$. But this means $|F|=p^k$ by linear algebra (basis representation for each element) and counting. – Dietrich Burde Aug 25 '24 at 12:45

2 Answers2

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The equality $\Phi_{n,\Bbb F_p}=ord_n(p)$ is only true, if $\phi_n$ remains irreducible modulo $p$. Let $p\nmid n$. Then the cyclotomic polynomial $\Phi_n$ factors over $\Bbb F_p$ into $\phi(n)/r$ irreducible factors each of degree $r$, where $r=ord_n(p)$ is the multiplicative order of $p$ modulo $n$. To see an example, take $\Phi_7=x^6+x^5+x^4+x^3+x^2+x+1$ and $p=2$. Then the order of $2$ modulo $7$ is $3$, because $2^3\equiv 1\bmod 7$, but $2^2\not\equiv 1\bmod 7$. So the irreducible factors have degree $3$ over $\Bbb F_2$. Indeed, we have $$ x^6+x^5+x^4+x^3+x^2+x+1=(x^3+x+1)(x^3+x^2+1). $$

For a proof see this duplicate:

If a cyclotomic polynomial is reducible over a finite field, what does its factorisation look like?

Dietrich Burde
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I believe, without any provided references to guide thought otherwise that, $\Phi_{n,\mathbb{Z}_p}$ is the $n$-th cyclotomic polynomial modulo $p$. I'm assuming you are doing something regarding splitting fields of $P(x) = x^n-1$, which are specifically called cyclotomic fields. It is known that the $n$-th cyclotomic field over $\mathbb{Z}_p$ has degree $ord_n(p)$, if $\Phi_n$ is irreducible over the field. See Tengu's answer in here as well as a few other like the first comment on your post's link for some more info. Hope this helped!

Jasmine
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    Why has it degree $ord_n(p)$? In the example, with $n=7$ and $p=2$ this is only true for the irreducible factors. – Dietrich Burde Aug 25 '24 at 11:52
  • @DietrichBurde I didn't add that detail, thank you! I was going off of the definition of the irreducibilty, but didn't consider fields in which $Phi_n$ is reducible. Much appreciated – Jasmine Aug 25 '24 at 12:40