Intermediate fields of Gal($x^5-2$, $\mathbb{Q}$) I'm having trouble findind the fixed field of each subgroup of the Galois group of $\mathbb{Q(\alpha, \omega)}$, where $\alpha = 2^{\frac{1}{5}}$ and $\omega$ is a 5-th primitive root of unity. I list here the 20 automorphisms of the Galois group, $G$:
\begin{array}{|c|c|c|c|}
\hline
&I & \sigma & \sigma^2 & \sigma^3 & \sigma^4 & \tau & \tau^2 & \tau^3 & \sigma\circ\tau &\sigma\circ\tau^2\\ \hline
\alpha& \alpha & \alpha\omega &\alpha\omega^2&\alpha\omega^3&\alpha\omega^4&\alpha&\alpha&\alpha&\alpha\omega&\alpha\omega  \\ \hline
\omega& \omega  &\omega &\omega&\omega&\omega&\omega^2&\omega^4&\omega^3&\omega^2&\omega^4\\ \hline
\end{array}
\begin{array}{|c|c|c|c|}
\hline
 &\sigma\circ\tau^3 &\sigma^2\circ\tau &\sigma^2\circ\tau^2 &\sigma^2\circ\tau^3 &\sigma^3\circ\tau &\sigma^3\circ\tau^2 &\sigma^3\circ\tau^3 &\sigma^4\circ\tau &\sigma^4\circ\tau^2 &\sigma^4\tau^3 \\ \hline
\alpha& \alpha\omega &\alpha\omega^2 &\alpha\omega^2 &\alpha\omega^2 &\alpha\omega^3 &\alpha\omega^3 &\alpha\omega^3 &\alpha\omega^4 &\alpha\omega^4 &\alpha\omega^4\\ \hline
\omega &\omega^3 &\omega^2& \omega^4& \omega^3& \omega^2&\omega^4&\omega^3&\omega^2&\omega^4 &\omega^3\\ \hline
\end{array}
The subgroups are:

{I}, 
{I, $\sigma \circ \tau^2$}, {I,$\sigma^2 \circ \tau^2$ }, {I,$\sigma^3 \circ \tau^2$ }, {I,$\sigma^4 \circ \tau^2$ }, {I,$\tau^2$ }
{I, $\tau, \tau^2, \tau^3$}
{I, $\sigma, \sigma^2, \sigma^3, \sigma^4$}
{I,$\sigma, \sigma^2, \sigma^3, \sigma^4, \tau^2, \sigma \circ \tau^2, \sigma^2 \circ \tau^2, \sigma^3 \circ \tau^2, \sigma^4 \circ \tau^2 $}
$G$

{I} fixes $\mathbb{Q(\alpha, \omega)}$;
G fixes $\mathbb{Q}$;
{I, $\tau, \tau^2, \tau^3$} fixes $\mathbb{Q(\alpha)}$;
{I, $\sigma, \sigma^2, \sigma^3, \sigma^4$} fixes $\mathbb{Q(\omega)}$
However, I can't see what are the others fixed fields...
I believe {I,$\sigma, \sigma^2, \sigma^3, \sigma^4, \tau^2, \sigma \circ \tau^2, \sigma^2 \circ \tau^2, \sigma^3 \circ \tau^2, \sigma^4 \circ \tau^2 $} fixes $\mathbb{Q(\omega + \omega^4)}$, but then I'd have to show that $[\mathbb{Q(\omega + \omega^4)}:\mathbb{Q}]=2$, but I don't know what is the minimal polynomial of $\omega + \omega^4$.
Same for the subgroups of order 2. I'd have to find extensions of $\mathbb{Q}$ of order 10 that are fixed by the automorphisms in the subgroup. For example, I know that {I, $\sigma\circ\tau^2$} fixes $\mathbb{Q(\alpha\omega^3)}, \mathbb{Q(\alpha + \alpha\omega)}, \mathbb{Q(\alpha\omega^4+\alpha\omega^2)}, \mathbb{Q(\omega+\omega^4)}, \mathbb{Q(\omega^2+\omega^3)}$, so the fixed field must contain all of this subfields and have degree 10 over $\mathbb{Q}$, but how to find it?
Can someone give a help? Thanks!
 A: First of all, concerning your question about $[\mathbb{Q}(\omega+\omega^4):\mathbb{Q}]$. There is a nice fact that is often useful when dealing with roots of unity. Namely, if $\omega_n$ is an $n$th root of unity ($n$ at least 3) and if $\theta_n=\omega_n+\omega_n^{-1}$, we have $[\mathbb{Q}(\omega_n):\mathbb{Q}] = 2\cdot[\mathbb{Q}(\theta_n):\mathbb{Q}]$. Why is this? Well, $\theta_n\in\mathbb{R}$, whereas $\omega_n\in\mathbb{C}-\mathbb{R}$, so $[\mathbb{Q}(\omega_n):\mathbb{Q}(\theta_n)]\geq 2$.
Moreover, $\theta_n\omega_n=\omega_n^2+1$, so $\omega_n$ is the zero of $x^2-\theta_nx+1\in\mathbb{Q}(\theta_n)[x]$. This tells us that $[\mathbb{Q}(\omega_n):\mathbb{Q}(\theta_n)]\leq 2$. Combining this with the previous result, it has to equal $2$. In your particular case, this indeed gives you that $[\mathbb{Q}(\omega+\omega^4):\mathbb{Q}]=2$. To find an actual minimal polynomial of $\omega+\omega^4$, one can square this expression, and hope to find something containing $1+\omega+\omega^2+\omega^3+\omega^4=1$ there. Unless I'm much mistaken, one could obtain $x^2+x-1$ this way.
Now, for the subfields, you already gave four rather important ones and to find the rest is some sort of puzzling exercise. First of all, $\{1,\tau^2\}<\{1,\tau,\tau^2,\tau^3\}$, so the field fixed by this subgroup of order two has to be a field inbetween $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\alpha,\omega)$ and there's a perfectly suitable candidate: $\mathbb{Q}(\alpha,\omega+\omega^4)$. Moreover, you already managed to find that $\{1,\sigma,\sigma^2,\sigma^3,\sigma^4,\tau^2,\sigma\circ \tau^2,\sigma^2\circ \tau^2,\sigma^3\circ \tau^2,\sigma^4\circ \tau^2\}$ fixes $\mathbb{Q}(\omega+\omega^4)$ and are hopefully convinced of this by now.
The question that remains is to find four subfields of $\mathbb{Q}(\alpha,\omega)$, having $\mathbb{Q}(\omega+\omega^4)$ as a subfield and being of degree $5$ over this subfield. It looks an aweful lot like these should be $\mathbb{Q}(\alpha\omega,\omega+\omega^4)$, $\mathbb{Q}(\alpha\omega^2,\omega+\omega^4)$, $\mathbb{Q}(\alpha\omega^3,\omega+\omega^4)$ and $\mathbb{Q}(\alpha\omega^4,\omega+\omega^4)$. Can you fill in the gaps?
A: From $w^5=1$, after dividing with $w-1$, we get: $$w^4+w^3+w^2+w+1=0$$
As we know, $2 w^5=2$ and $w^8=w^3$, therefore previous equation is equal to:
 $$w^2+2 w^5+w^8+w+w^4-1=0,$$
which is equal to
$$(w+w^4)^2+(w+w^4)-1=0.$$
Therefore, the minimal polynomial for $w+w^4$ is $x^2+x-1$.
For other question, it is probably like this: {I, σ∘τ2} for fixed field has $Q(\alpha \omega^3, \omega^2+\omega^3)$, where the first element has order 5 and second has order 2, therefore $[Q(\alpha \omega^3, \omega^2+\omega^3):Q]=10$. It is easy to check: $w+w^4=(\omega^2+\omega^3)^2-2$, $\alpha+\alpha\omega=\alpha\omega^3(\omega^2+\omega^3)$ and $\alpha\omega^2+\alpha\omega^4=\alpha\omega^3(\omega+\omega^4)$.
