$$ \newcommand{\abs}[1]{\left\vert #1 \right\vert} \newcommand\rme{\mathrm e} \newcommand\imu{\mathrm i} \newcommand\diff{\,\mathrm d} \DeclareMathOperator\sgn{sgn} \renewcommand \epsilon \varepsilon \newcommand\trans{^{\mathsf T}} \newcommand\F {\mathbb F} \newcommand\Z{\mathbb Z} \newcommand\R{\Bbb R} \newcommand \N {\Bbb N} \newcommand\Q{\Bbb Q} \newcommand\bbc{\Bbb C} $$ 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 $j,k \in \{1,2,4,5,7,8\}$.

  • $\begingroup$ Factor the cyclotomic polynomial $X^6+X^3+1$ of $z_8$ over $\Bbb Q(z_3)$. $\endgroup$
    – dan_fulea
    Commented Aug 14, 2019 at 16:44
  • $\begingroup$ Sage gives for instance for: K.<a> = QuadraticField(-3); R.<x> = K[]; factor(x^6 + x^3 +1) the result (x^3 - 1/2*a + 1/2) * (x^3 + 1/2*a + 1/2), where $a=\sqrt{-3}$. $\endgroup$
    – dan_fulea
    Commented Aug 14, 2019 at 16:46

1 Answer 1


$\mu=X^3-z_3 \in K[X] :=\mathbb{Q}(z_3)[X]$ is a polynomial that vanishes at $z_9$. Since $z_9$ has degree $3$ over $K$, $\mu$ is the minimal polynomial.

  • $\begingroup$ Oh right… what were I thinking… So simple. Thanks a lot! $\endgroup$
    – xbh
    Commented Aug 14, 2019 at 16:51
  • $\begingroup$ @xbh Alternatively; note that the roots of $X^6+X^3+1$ are precisely the cube roots of the roots of $X^2+X+1$, so the minimal polynomial of its root $z_9$ divides $X^3-z_3$. $\endgroup$
    – Servaes
    Commented Aug 14, 2019 at 17:29

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