# Is my answer correct on this Galois Theory problem? Find the lattice of subfields of $\mathbb{Q}(\zeta_9)$

Problem: let $\zeta$ be a primitive $9$th root of unity, and $K = \mathbb{Q}(\zeta)$. Describe the lattice of subfields of $K$, give generators for each subfield and list its degree over $\mathbb{Q}$.

Okay, so $K/ \mathbb{Q}$ is Galois with group $G \cong U(\mathbb{Z}/9\mathbb{Z}) \cong \mathbb{Z}/6 \mathbb{Z}$, generated by the map $\phi$ given by $\phi(\zeta) = \zeta^2$. So $G = \{\phi, \phi^2, \phi^3, \phi^4 , \phi^5, 1\}$ and the nontrivial proper subgroups are $H_1 = \{\phi^2, \phi^4, 1\}$ and $H_2 = \{ \phi^3, 1\}$.

As for the fixed field $F_2$ of $H_2$, we have $\phi^3(\zeta^j) = \phi^3(\zeta)^j = \zeta^{8j}$ for any $j$. I take the basis $1, \zeta, ... , \zeta^5$ of $K/ \mathbb{Q}$ and look for all elements $\alpha = a_0 + a_1 \zeta + \cdots + a_5 \zeta^5$ which are equal to $a_0 + a_1 \zeta^8 + a_2 \zeta^{16} + a_3 \zeta^{24} + a_4 \zeta^{32} + a_5 \zeta^{40}$. The minimal polynomial of $\zeta$ is $X^6 + X^3 + 1$, so $\zeta^6 = - \zeta^3 - 1; \zeta^7 = - \zeta^4 - \zeta;$ and $\zeta^8 = \zeta^5 - \zeta^2$. Also $\zeta^{16} = \zeta^7, \zeta^{24} = \zeta^6$, $\zeta^{32} = \zeta^5$, and $\zeta^{40} = \zeta^4$.

Thus $$a_0 + a_1 \zeta^8 + a_2 \zeta^{16} + a_3 \zeta^{24} + a_4 \zeta^{32} + a_5 \zeta^{40}$$ $$= a_0 + a_1(\zeta^5 - \zeta^2) + a_2(-\zeta^4 - \zeta) + a_3(-\zeta^3 - 1) + a_4\zeta^5 + a_5 \zeta^4$$$$= (a_0 - a_3) - a_2 \zeta - a_1 \zeta^2 - a_3 \zeta^3 + (-a_2 + a_5) \zeta^4 + (a_1 +a_4) \zeta^5$$

So to have $\alpha \in F_2$, we need $a_0 = a_0 - a_3$ (which implies $a_3= 0$), $a_1 = -a_2, a_2 = -a_1$ (that's redundant but oh well), $a_3 = -a_3$ (i.e. $a_3 = 0$ again redundant). We also need $a_4 = -a_2 + a_5$ (or $a_4 = a_1 + a_5$) and $a_1 + a_4 = a_5$. Thus $a_4 = a_1 + a_1 + a_4 = 2a_1 + a_4$, or $a_1 = 0$ and $a_4 = a_5$. Thus also $a_2 = 0$.

Thus $F_2 = \{ a_0 + a_4 (\zeta^4 + \zeta^5) : a_0, a_4 \in \mathbb{Q} \}$

So $F_2 = \mathbb{Q}[\zeta^4 + \zeta^5]$. Also $[F_2 : \mathbb{Q}] = |Aut(F_2/ \mathbb{Q})| = |Aut(K/ \mathbb{Q})/ Aut(K/F_2)| = 6/2 = 3$.

As for $F_1$, the fixed field of $H_1 = \{\phi^2, \phi^4, 1\}$, it necessarily has degree $6/3 = 2$ over $\mathbb{Q}$. But I already know such a subfield. $\zeta^3$ is a primitive $3rd$ root of unity, and $\mathbb{Q}[\zeta^3]$ is a subfield of $K$ with degree $\phi(3) = 2$ over $\mathbb{Q}$, so it is the desired result.

Was my field $F_1$ correct? Is there a faster way to find it?

• An error: $\zeta^8=-\zeta^5-\zeta^2$. You forgot one of the minus signs. This leaves you a 3-dimensional subfield fixed by $\phi^3$. – Jyrki Lahtonen Jun 11 '14 at 5:29

$F_1$ is correct, and your method seems direct enough. However, you might better recognize $F_2$ as the maximal real subfield of $K$, i.e. $\mathbb{Q}(\zeta+\zeta^{-1})$. Your expression for $F_2$ seems incorrect. I hesitate to say where you went wrong.

• It's been fixed, thank you! – Billy Madison Jun 11 '14 at 3:20
• If I didn't know to look for the maximal real subfield, is there a faster way to figure out what $F_2$ is? – Billy Madison Jun 11 '14 at 3:35
• Actually it has not been fixed. There is still a mistake somewhere because the field cannot be spanned by two elements. – Billy Madison Jun 11 '14 at 3:49

Here's a trick that sometimes helps finding elements fixed by an automorphism. If $\sigma$ is an automorphism of order two on a field $F$, and $z\in F$ is arbitrary, then $u=z+\sigma(z)$ is fixed under $\sigma$. This is because $$\sigma(u)=\sigma(z+\sigma(z))=\sigma(z)+\sigma^2(z)=\sigma(z)+z=u.$$

Similarly, if $\tau$ is another automorphism of order three, then $v=z+\tau(z)+\tau^2(z)$ is fixed under $\tau$, because $$\tau(v)=\tau(z+\tau(z)+\tau^2(z))=\tau(z)+\tau^2(z)+\tau^3(z)=\tau(z)+\tau^2(z)+z=v.$$

The same trick gives elements under automorphisms of a higher order, or even under a non-cyclic subgroup. Instead of a sum you can use other elementary symmetric polynomials. Symmetry is the key here.

This is not guaranteed to produce anything useful, but it is a shortcut. Let's take a look at your case. Here $\phi^3$ is of order two, so let's check. The following elements are fixed under $\phi^3:\zeta\mapsto\zeta^8$: \begin{aligned} \zeta+\phi^3(\zeta)&=\zeta+\zeta^8=\zeta+\zeta^{-1},\\ \zeta^2+\phi^3(\zeta^2)&=\zeta^2+\zeta^7=\zeta^2+\zeta^{-2},\\ \zeta^3+\phi^3(\zeta^3)&=\zeta^3+\zeta^6=-1,\quad\leftarrow\text{not useful},\\ \zeta^4+\phi^3(\zeta^4)&=\zeta^4+\zeta^5. \end{aligned} Leaving it to you write the first two entries in terms of your basis, and spot the linear dependency among these fixed points, and thus finding a vector space basis for the fixed field.

You $\phi^2:\zeta\mapsto\zeta^4$ is of order three, so the following elements are fixed under $H_1$:

\begin{aligned} \zeta+\phi^2(\zeta)+\phi^4(\zeta)&=\zeta+\zeta^4+\zeta^7=0,\quad\leftarrow\text{not useful}\\ \zeta^2+\phi^2(\zeta^2)+\phi^4(\zeta^2)&=\zeta^2+\zeta^8+\zeta^5=0,\quad\leftarrow\text{not useful}\\ \zeta^3+\phi^2(\zeta^3)+\phi^4(\zeta^3)&=3\zeta^3,\quad\leftarrow\text{Duh. Good job, Sherlock}. \end{aligned}

When applying this method you need to know the dimension of the fixed field in advance, so that you know when to stop. Of course, if you find one fixed element, then its powers give you others. The pro is that you don't need to set up those systems of equations as you did in your attempt. The con is that sometimes the choice of $z$ that gives you something non-trivial is a bit tricky, and you will waste a bit of time there. This is just a nice tool to have in your bag, not a universal solution.