$$ \newcommand{\abs}[1]{\left\vert #1 \right\vert} \newcommand\rme{\mathrm e} \newcommand\imu{\mathrm i} \newcommand\diff{\,\mathrm d} \DeclareMathOperator\sgn{sgn} \renewcommand \epsilon \varepsilon \newcommand\trans{^{\mathsf T}} \newcommand\F {\mathbb F} \newcommand\Z{\mathbb Z} \newcommand\R{\Bbb R} \newcommand \N {\Bbb N} \newcommand\Q{\Bbb Q} \newcommand\bbc{\Bbb C} $$ This is a question from Michael Artin's Algebra [2nd ed.].
Let $z_n = \exp (2\pi\imu /n)$. How do we find the irreducible polynomial $f$ of $z_9$ over the field $\Q(z_3)$?
I made some attempts to reduce the question.
Easy to see $[\Q(z_9): \Q] = 6$, since the irreducible polynomial of $z_9$ over $\Q$ is $x^6 + x^3 + 1$. For $z_3$, the rational irreducible polynomial is $x^2 + x + 1$, then $[\Q (z_3): \Q] = 2$. Also $z_3 \in \Q(z_9)$, so $\Q(z_3)$ is a subfield of $\Q(z_9)$ and $\Q(z_9, z_3) = \Q(z_9)$. By the Multiplicative Property, $$6 = [\Q(z_9):\Q] = [\Q(z_9):\Q(z_3)] \cdot [\Q(z_3):\Q ], $$ hence $[\Q(z_9):\Q(z_3)] = 3$.
For convenience we consider the polynomial over $\bbc$ where we have $(x-z_9)\mid f(x)$, thus $$ f(x) = (x-z_9) (x-z_9^j)(x-z_9^k)\quad [j,k > 1]. $$ The remained part is: how do we determine $j,k$ without ordinary "trial and error"? i.e. plug different $(j,k)$, check $f(x)$, accept or move on ?
Thanks in advance.
Update: the range of $j,k$ could also be determined. Since $\Q \subset \Q(z_3) \subset \Q(z_9)$, $f$ shall divide the irreducible polynomial of $z_9$ over $\Q$, namely, $f(x)\mid (x^6 + x^3 + 1)$. Easy to see that $x^6 + x^3 + 1$ has complex roots $z_9, z_9^2, z_9^4, z_9^5, z_9^7, z_9^8$. Therefore $j,k \in \{1,2,4,5,7,8\}$.
$\mu=X^3-z_3 \in K[X] :=\mathbb{Q}(z_3)[X]$ is a polynomial that vanishes at $z_9$. Since $z_9$ has degree $3$ over $K$, $\mu$ is the minimal polynomial.