We consider the 15-th cyclotomic polynomial over $\mathbb{Z}$ first:
$$\Phi_{15} = x^8 – x^7 + x^5 – x^4 + x^3 – x + 1.$$
If we reduce it modulo $7$, we obtain two irreducible factors of $\Phi_{15}$ over $\mathbb{F}_7[x]$:
$$\Phi_{15} = (x^4+2x^3+4x^2+x+2)(x^4+4x^3+2x^2+x+4).$$
Let us name the first factor $f$ and the second factor $g$. Let us also choose $\alpha \in \mathbb{F}_{7^4}$ with minimal polynomial $\min_{\mathbb{F}_7}(\alpha) = f$. Then my teacher immediately concluded
$$f = (x-\alpha^1)(x-\alpha^{7})(x-\alpha^{4})(x-\alpha^{13})$$
and
$$g = (x-\alpha^2)(x-\alpha^{14})(x-\alpha^8)(x-\alpha^{11}).$$
I noticed that the powers ($\{ 1,7,4,13\}$ and $\{2,14,8,11\}$) of the different factors are different cosets in $(\mathbb{Z}/15 \mathbb{Z})^\times$ where the equivalence relation is defined by $a \sim b \: :\Leftrightarrow \: a = b \cdot 7^k$ for some $k$.
My question: Does this go back to a general result regarding cyclotomic polynomials? Or is this merely a coincidence?
Best Answer
If $p$ is a prime and $p\nmid n$ then the polynomial $X^n-1$ has distinct roots $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ in some extension $k=\Bbb F_{p^t}$ of $\Bbb F_p$. The roots of the cyclotomic polynomial $\Phi_n$ are the $\alpha^j$ where $\gcd(j,n)=1$. The Galois group of $k$ over $\Bbb F_p$ is generated by the Frobenius map $F:x\mapsto x^p$. Its orbits on the roots of $\Phi_n$ have the form $$\{\alpha^j,\alpha^{pj},\alpha^{p^2j},\ldots,\alpha^{p^{t-1}j}\}$$ which corresponds to the irreducible factor $$(X-\alpha^j)(X-\alpha^{pj})(X-\alpha^{p^2j})\cdots(X-\alpha^{p^{t-1}j})$$ of $\Phi_n$ which is irreducible over $\Bbb F_p$. This shows that $t$ is the least positive integer with $p^t\equiv1\pmod n$. The exponents in this factor are the integers $j$, $pj$, $p^2j,\ldots,p^{t-1}j$ considered modulo $p$; this is an equivalence class in $\Bbb Z/n\Bbb Z$ under $a\sim b\iff a\equiv p^sb\pmod n$ for some $s$.