# Let $F= \Bbb{Q}(\zeta_9)$ with $\zeta_9 = e^{\frac{2i\pi}{9}}$. Find the Galois group and the intermediate fields.

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

or

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

So the corresponding diagram for the intermediate fields is

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:

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?

• Is it $\zeta_9=\exp(2\mathrm i\pi/9)$?
– Did
Aug 16, 2013 at 21:44
• Won't you correct your post?
– Did
Aug 16, 2013 at 22:41
• By the way, could it be that (c) asks for $\operatorname{Aut}(E/\mathbb{Q})$ whereas you give $\operatorname{Gal}(F/E)$? Aug 17, 2013 at 0:47
• @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 $\operatorname{Gal}(F/E)$. Aug 17, 2013 at 9:28
• (With $\beta=\zeta_9+\zeta_9^{-1}$, as in my answer.) Aug 17, 2013 at 9:38

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 $\mathbb{Q}$.