Determine the irreducible polynomial


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 ?

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 $$j,k \in \{1,2,4,5,7,8\}$$.
• Factor the cyclotomic polynomial $X^6+X^3+1$ of $z_8$ over $\Bbb Q(z_3)$. – dan_fulea Aug 14 at 16:44
• 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 $a=\sqrt{-3}$. – dan_fulea Aug 14 at 16:46
$$\mu=X^3-z_3 \in K[X] :=\mathbb{Q}(z_3)[X]$$ is a polynomial that vanishes at $$z_9$$. Since $$z_9$$ has degree $$3$$ over $$K$$, $$\mu$$ is the minimal polynomial.
• @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 $X^3-z_3$. – Inactive - avoiding CoC Aug 14 at 17:29