# Identify the fixed field of each of these subgroups.

Let $\zeta_0=exp \left(\frac{2\pi i}{5} \right)$. Note that $\zeta_0, \zeta_0^2, \zeta_0^3, \zeta_0^4$ are the four roots of the irreducible polynomial $P(x)=1+x+x^2+x^3+x^4 \in \mathbb{Q}[x]$.

1. Find all automorphisms of the field extension $\mathbb{Q}(\zeta_0):\mathbb{Q}$.
2. Identify the group $\Gamma (\mathbb{Q}(\zeta_0):\mathbb{Q})$. (In other words find (with explanation) a well known group with which this group is isomorphic.
3. Draw the lattice diagram of the subgroups of $\Gamma (\mathbb{Q}(\zeta_0):\mathbb{Q})$. The identify (with proof) the fixed field of each of these subgroups in as concise a manner as possible. Draw the corresponding lattice diagram for these intermediate fields.

$\textbf{Part i: Find all automorphisms of the field extension$\mathbb{Q}(\zeta_0):\mathbb{Q}$.}$

Let $K=\mathbb{Q}(\zeta_0)$, where $\zeta_0=exp \left(\frac{2\pi i}{5} \right) \in \mathbb{C}$. Now $\zeta_0^5=1$ and $\mathbb{Q}(\zeta_0)$ consists of all elements $$p+q\zeta_0+r\zeta_0^2+s\zeta_0^3+t\zeta_0^4,$$ where $p,q,r,s,t \in \mathbb{Q}$. So all the automorphisms of the field extension are \begin{equation*} \begin{aligned} \alpha_1 & : & p+q\zeta_0+r\zeta_0^2+s\zeta_0^3+t\zeta_0^4 & \mapsto & p+q\zeta_0+r\zeta_0^2+s\zeta_0^3+t\zeta_0^4 \\ \alpha_2 & : & & \mapsto & p+r\zeta_0+s\zeta_0^2+t\zeta_0^3+q\zeta_0^4 \\ \alpha_3 & : & & \mapsto & p+s\zeta_0+t\zeta_0^2+q\zeta_0^3+r\zeta_0^4 \\ \alpha_3 & : & & \mapsto & p+t\zeta_0+q\zeta_0^2+r\zeta_0^3+s\zeta_0^4 \\ \end{aligned} \end{equation*}

It is easy, but tedious, to check that all of these are $\mathbb{Q}$-automorphisms. Hence we are going to take that fact for granted.

$\textbf{Part ii: Identify the group$\Gamma (\mathbb{Q}(\zeta_0):\mathbb{Q})$.}$ For convenience, we are going to use permutation notation to describe $\alpha_i$, for $i=1,\dots, 4$. So $$\alpha_1 = \begin{pmatrix} p & q & r & s & t \\ p & q & r & s & t \end{pmatrix}$$ $$\alpha_2 = \begin{pmatrix} p & q & r & s & t \\ p & r & s & t & q \end{pmatrix}$$
$$\alpha_3 = \begin{pmatrix} p & q & r & s & t \\ p & s & t & q & r \end{pmatrix}$$ $$\alpha_4 = \begin{pmatrix} p & q & r & s & t \\ p & t & q & r & s \end{pmatrix}$$

Through computation of the elements of this field extension, we can construct the following Cayley Table.

\begin{tabular}{|c|c|c|c|c|} \hline $\cdot$ & $\alpha_1$ & $\alpha_2$ & $\alpha_3$ & $\alpha_4$ \\ \hline $\alpha_1$ & $\alpha_1$ & $\alpha_2$ & $\alpha_3$ & $\alpha_4$ \\ \hline $\alpha_2$ & $\alpha_2$ & $\alpha_3$ & $\alpha_4$ & $\alpha_1$ \\ \hline $\alpha_3$ & $\alpha_3$ & $\alpha_4$ & $\alpha_1$ & $\alpha_2$ \\ \hline $\alpha_4$ & $\alpha_4$ & $\alpha_1$ & $\alpha_2$ & $\alpha_3$ \\ \hline \end{tabular}

Since we have four elements in the field extension, the only groups with four elements are $\mathbb{Z}_4$ or Klein $4$-group, where each element has order 2. To be able to determine which group is isomorphic to the table above we need to see if the order of each of the elements are equal to $2$. Looking along the the diagonal we don't see that this is so. Hence $\Gamma(\mathbb{Q}(\zeta_0):\mathbb{Q})\cong \mathbb{Z}_4$.

$\textbf{Part iii: Draw the lattice diagram of the subgroups of$\Gamma (\mathbb{Q}(\zeta_0):\mathbb{Q})$.}$

\begin{center} \begin{tikzpicture}[scale=1.5,auto] \node (i) at (1,1) {${i }$}; \node (Z2) at (1,2) {$\mathbb{Z}_2$}; \node (Z4) at (1,3) {$\mathbb{Z}_4$};

\path[-] (i) edge (Z2) (Z2) edge (Z4) ; \end{tikzpicture} \end{center}

How would I find the fixed fields of $\Gamma(\mathbb{Q}(\zeta_0):\mathbb{Q})$, with a proof?

You’re making too much of a megillah of this. Your $\zeta$ (I’ll leave off the subscript) is a primitive fifth root of unity, and any element $\sigma$ of the Galois group must send $\zeta$ to a $\zeta^n$, where $n$ is well-defined only modulo $5$, and where $n$ must be incongruent to $0$. I hope you see that what $\sigma$ does to $\zeta$ determines what it does to any $\zeta^j$. Our $\sigma=\sigma_n$ is determined by the value of $n$, and you should write down what $\sigma_m\circ\sigma_n$ is, that is what it does to $\zeta$. That will give you the structure and nature of the Galois group very quickly.
As to an identification of what the intermediate field is, let me tell you that experience has shown that if you have a subgroup $S$ of the Galois group, very often the trace with respect to $S$ of your generator $\zeta$ will give a generator of the fixed field. (Not guaranteed). At any rate, your subgroup will have order two, $S=\{e,\tau\}$, and you should try $e(\zeta)+\tau(\zeta)=\zeta+\tau(\zeta)$ for $\tau$ the nonidentity element of $S$.
• According to what I said above, the top Galois group is the numbers $\{1,2,3,4\}$, representing $\zeta\mapsto\zeta^n$, where $n$ is one of those numbers. You can call the corresponding automorphism $\sigma_n$. You see that the way these compose is that $\sigma_n\circ\sigma_m=\sigma_{nm}$, always reading the numbers $n$ and $m$ modulo $5$. Then you see that $\sigma_2$ generates all (check this!), so that the group is cyclic of order four. Thus there’s only one intermediate field, and I told you how to find an element of this, too. – Lubin Nov 16 '15 at 21:23