Let $F= \Bbb{Q}(\zeta_9)$ with $\zeta_9 = e^{\frac{2i\pi}{9}}$.

a) What is the Galois group of $F$ over $\Bbb{Q}$?
b) Find all intermediate fields between $\Bbb{Q}$ and $F$. (Write each in the form $\Bbb{Q}(\alpha)$ for some specific $\alpha \in F$.)
c) For each intermediate field $E$ above, give the Galois group of $E$ over $\Bbb{Q}$.

a) Since $\Bbb{Q}(\zeta_9)$ is the splitting field of a cyclotomic polynomial, $Gal(F/\Bbb{Q})=\Bbb{Z}^{\times}_9$.

b) Since $Gal(F/\Bbb{Q})=\Bbb{Z}^{\times}_9$, we need to find the subgroups of $\Bbb{Z}^{\times}_9$. But $|\Bbb{Z}^{\times}_9|=6$ and there are only 2 groups of order 6 up to isomorphism: $\Bbb{Z}_6$ and $D_6$. Since $\Bbb{Z}^{\times}_9$ is abelian, we know that $\Bbb{Z}^{\times}_9 \cong \Bbb{Z}_6$. So we can use either of these diagrams, since they are both isomorphic

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Where $\langle 4 \rangle = \{1, 4, 7\}$ and $\langle 8 \rangle \{1, 8\}$.

So the corresponding diagram for the intermediate fields is enter image description here

We need to find $\alpha$ and $\beta$, where $[\Bbb{Q}(\alpha):\Bbb{Q}]=2$ and $[\Bbb{Q}(\beta):\Bbb{Q}]=3$.

We know that $(\zeta^3_9)^3 = 1 \implies (\zeta^3_9)^9-1=0$. So the minimal polynomial of $\zeta^3_9$ divides $x^3-1$. We know that $x^3-1=\Phi_1(x)\Phi_3(x)=(x-1)(x^2+x+1)$, where $\Phi_k(x)$ is the kth cyclotomic polynomial. Since $1 \in \Bbb{Q}$, we only need to check if $x^2+x+1$ is reducible over $\Bbb{Q}$. But since it is a cyclotomic polynomial, it must be. So the minimal polynomial of $\zeta_9^3$ over $\Bbb{Q}$ has degree $2 \implies \alpha=\zeta^3_9$.

Now we need to find $\beta$. Using Gerry Myerson's answer, we choose $\zeta_9 \in \Bbb{Q}(\zeta_9)$. Since we are looking at a the Galois group of degree $3$, we need to find a $\sigma$ such that $\sigma^3(\zeta_9 )=\zeta_9$. So we can choose $\sigma(\zeta_9)=\zeta_9^4$.

Now we need to find the minimal polynomial of $\zeta_9 + \zeta_9^4 + \zeta_9^7$. We see that $$(\zeta_9 + \zeta_9^4 + \zeta_9^7)^3 = 9\zeta_9^3 + 9\zeta_9^6 + 9$$

But $$\cos(6\pi/9) + i\sin(6\pi/9) + \cos(12\pi/9) + i\sin(12\pi/9)$$

$$= 2\cos(6\pi/9) = 2\cos(2\pi/3)=2(-1/2) = -1$$

since $2\pi6/9 - 2\pi = -6\pi/9$. It follows that $(\zeta_9 + \zeta_9^4 + \zeta_9^7)^3 = 9(-1+1)=0$. So we know that the min polynomial for $\zeta_9 + \zeta_9^4 + \zeta_9^7$ must divide $x^3$.

So we need to check if $(\zeta_9 + \zeta_9^4 + \zeta_9^7)^2=0$. We see that $(\zeta_9 + \zeta_9^4 + \zeta_9^7)^2 = 3\zeta_2^7 + 3\zeta_9^5 + 3\zeta_9^8$. But we can see from the unit circle, that the sum of the three vectors ends up $50^{\circ}$ above the horizontal axis; so it is not zero $\implies$ the minimal polynomial for $\zeta_9 + \zeta_9^4 + \zeta_9^7$ is of degree $3$ $\implies$ $\zeta_9 + \zeta_9^4 + \zeta_9^7$ is a possible choice for $\beta$.

c) The Galois group of $\Bbb{Q}(\alpha)$ is isomorphic to $\Bbb{Z}_3$, and that of $\Bbb{Q}(\beta)$ is isomorphic to $\Bbb{Z}_2$.

I have two questions:

  1. Is my answer correct?

  2. For part b), we know that $\Bbb{Z}^{\times}_9 \cong \Bbb{Z}_6$, so finding the lattice of the subgroups is easy. But what if it was not isomorphic to a cyclic group? Would it be ok to find only $1$ subgroup for each possible order, or do we need to find all subgroups of a certain order? For example, we know that $\Bbb{Z}^{\times}_9$ has subgroups of orders $2$ and $3$. So after finding $\langle 4 \rangle$ of order $3$, and $\langle 8 \rangle$ of order $2$, can I just stop? Or do I need to search for more groups of orders $2$ and $3$? The reason why I'm asking this is because we know that for any cyclic group, there is at most 1 subgroup of a given order. So we can just stop when we find $\Bbb{Z}_2$ and $\Bbb{Z}_3$ in $\Bbb{Z}_6$. But we can't guarantee that with other groups, right?

Thanks in advance

  • 1
    $\begingroup$ Is it $\zeta_9=\exp(2\mathrm i\pi/9)$? $\endgroup$
    – Did
    Aug 16, 2013 at 21:44
  • 1
    $\begingroup$ Won't you correct your post? $\endgroup$
    – Did
    Aug 16, 2013 at 22:41
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    $\begingroup$ By the way, could it be that (c) asks for $\operatorname{Aut}(E/\mathbb{Q})$ whereas you give $\operatorname{Gal}(F/E)$? $\endgroup$
    – ccorn
    Aug 17, 2013 at 0:47
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    $\begingroup$ @Artus: Here, $\operatorname{Aut}(E/\mathbb{Q})=\operatorname{Gal}(E/\mathbb{Q})$ because $E/\mathbb{Q}$ is Galois. Look at $E=\mathbb{Q}(\beta)$ over $\mathbb{Q}$: The extension has degree $3$, and there are $3$ automorphisms of $\mathbb{Q}(\beta)$ that fix $\mathbb{Q}$, namely those generated by the $\sigma$ taking $\beta$ to $\beta^2-2$ (the one that would take $\zeta_9$ to $\zeta_9^{-2}$ in $F$). So $\operatorname{Gal}(E/\mathbb{Q})$ is cyclic of order $3$. You give order $2$ which applies to $\operatorname{Gal}(F/E)$. $\endgroup$
    – ccorn
    Aug 17, 2013 at 9:28
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    $\begingroup$ (With $\beta=\zeta_9+\zeta_9^{-1}$, as in my answer.) $\endgroup$
    – ccorn
    Aug 17, 2013 at 9:38

1 Answer 1


As to question (1): Alas, $\zeta_9+\zeta_9^4+\zeta_9^7=\zeta_9(1+\zeta_3+\zeta_3^2)=0$. That is obviously no useful $\beta$.

You need a $\beta$ of degree $3$ over $\mathbb{Q}$, so in order to apply Gerry Myerson's trick, the $\sigma$ you look for should have order $6:3=2$. Pick $\sigma(\zeta_9)=\zeta_9^8=\zeta_9^{-1}$. Then $\beta=\zeta_9+\zeta_9^{-1}$ and you will find $\beta^3=3\beta-1$ which yields a monic, irreducible, hence minimal, polynomial for $\beta$ over $\mathbb{Q}$.

As to question (2): By the fundamental theorem of Galois theory, different subgroups correspond to different intermediate fields, even if orders, resp. extension degrees, are the same. Wikipedia has an example.


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