Understanding an observation between the 3rd and 15th cyclotomic polynomial over $\mathbb{F}_7$

Question

We consider the $3$rd and $15$-th cyclotomic polynomial over $\mathbb{Z}$ first:

$$\Phi_3 = x^2 + x + 1, \quad \Phi_{15} = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1.$$

If we reduce them modulo $7$, we obtain the following facorization in irreducible factors over $\mathbb{F}_7[x]$:

\begin{eqnarray*} \Phi_3&=&(x-2)(x-4),\\ \Phi_{15}&=&(x^4+2x^3+4x^2+x+2)(x^4+4x^3+2x^2+x+4). \end{eqnarray*}

Let us denote by $f$ and $g$ the first and second factor of $\Phi_{15}$, respectively. Let us also choose $\alpha \in \mathbb{F}_{7^4}$ with minimal polynomial $\min_{\mathbb{F}_7}(\alpha) = f$. Then my teacher computed

\begin{eqnarray*} f&=&(x-\alpha^1)(x-\alpha^{7})(x-\alpha^{4})(x-\alpha^{13}),\\ g&=&(x-\alpha^2)(x-\alpha^{14})(x-\alpha^8)(x-\alpha^{11}). \end{eqnarray*}

Here, she concluded immediately that $\alpha^5 = 4$. Also, if I understood her correctly, she mentioned that if we would have chosen the minimal polynomial of $\alpha$ to be $g$ instead of $f$, we could immediately say $\alpha^5 = 2$.

The thing which bugs me right now is that these two possible results of $\alpha^5$ ($4$ and $2$, depending of the choice of the minimal polynomial) are exactly the roots of $\Phi_3$ in $\mathbb{F}_7[x]$. Also, they lie in the diffrent cosets $\{1,7,4,13\}$ and $\{2,14,8,11\}$ of $(\mathbb{Z}/15 \mathbb{Z})^\times$ where the equivalence relation $a \sim b \: :\Leftrightarrow \: a = b \cdot 7^k$ for some $k$.

My question: Is the last observation just a coincidence or does it go back to some general result?

Answer 1

As a root of $\Phi_{15}$, $\alpha$ is a primitive $15$th root of unity (in general, the roots of the $n$th cyclotomic polynomial $\Phi_n$ are precisely the primitive $n$th roots of unity). Therefore, $k=15$ is the smallest positive integer with $\alpha^k = 1$. From $1 = \alpha^{15} = (\alpha^5)^3$ we find that $\alpha^5$ is a primitive $3$rd root of unity and thus a root of $\Phi_3$.

Answer 2

Because $\alpha$ is a root of $\Phi_{15}$ we have $\alpha^{15}=1$, and $k=15$ is the smallest positive integer satisfying $\alpha^k=1$. Then $k=3$ is the smallest positive integer satisfying $(\alpha^5)^k=1$, so $\alpha^5$ is a root of $\Phi_3$. So this is no coincidence.

This also shows that the roots of $\Phi_{15}$ are the (nontrivial) fifth roots of the roots of $\Phi_3$. Indeed $$f(2X)=(2X)^4+2(2X)^3+4(2X)^2+(2X)+2=2^4(X^4+X^3+X^2+X+1)=2\Phi_5,$$ which shows that the roots of $f$ are of the form $(\tfrac{\zeta_5}{2})^k=(4\zeta_5)^k$ where $\zeta_5\in\Bbb{F}_{7^4}$ is a primitive fifth root of unity. The same argument shows that the roots of $g$ are of the form $(2\zeta_5)^k$.

The fact that $2$ and $4$ lie in different cosets of $(\Bbb{Z}/15\Bbb{Z})^{\times}$ is a coincidence though; we could repeat the same argument over $\Bbb{F}_{13}$ to find that \begin{eqnarray*} \Phi_3&=&(X-3)(X-9),\\ \Phi_{15}&=&(X^4+3X^3+9X^2+X+3)(X^4+9X^3+3X^2+X+9), \end{eqnarray*} and if $\alpha\in\Bbb{F}_{13}$ is a root of the first factor $f$ of $\Phi_{15}$ we again find that \begin{eqnarray*} X^4+3X^3+9X^2+X+3&=&(X-\alpha^1)(X-\alpha^4)(X-\alpha^7)(X-\alpha^{13})\\ X^4+9X^3+3X^2+X+9&=&(X-\alpha^2)(X-\alpha^8)(X-\alpha^{11})(X-\alpha^{14}), \end{eqnarray*} but now $3$ and $9$ don't even occur in the cosets because they aren't coprime to $15$. So your second observation is a coincidence. The situation is even 'worse' over $\Bbb{F}_{37}$; here we have $$\Phi_3=(X-10)(X-26),$$ where $10$ isn't coprime to $15$ and $26\equiv11\pmod{15}$. This also illustrates why we shouldn't expect a connection; the roots of $\Phi_3$ live in $\Bbb{F}_7$ while the exponents live in $(\Bbb{Z}/15\Bbb{Z})^{\times}$. Whatever connection you might see should be invariant under adding multiples of $7$, and under adding multiples of $15$ (or perhaps $\varphi(15)=8$ in some way). Either way it should be invariant under addition of two coprime integers, and hence under addition of any integer. That wouldn't be much of a pattern.