Galois Group of the 11th roots of Unity

Question

It is clear that the Galois group of $\mathbb{Q}(\omega):\mathbb{Q}$ where $\omega$ is a primitive root of unity is a cyclic group of order 10. Let $\sigma$ be a primitive element then $\sigma^2$ has order 5.

I want to find an $\alpha$ such that $Fix(<\sigma^2>)=\mathbb{Q}(\alpha)$.

Using the Fundamental Theorem of Galois Theory I know that any such field $\mathbb{Q}(\alpha)$ must satisfy $|\mathbb{Q}(\alpha):\mathbb{Q}|=2$

Let the primitive element $\sigma : \omega \mapsto \omega^2$ be the primitive element meaning $\sigma^2: \omega \mapsto \omega^4$. But I don't know where to go from here. I would assume that you find a basis for $\mathbb{Q}(\omega)$ and work out the $Fix$ after applying $\sigma^2$ to the basis but $\sigma^2$ just permutes the $\omega^i$.

Answer 1

Add up the five $11^{th}$ roots in an orbit of your group element $\sigma^2$ of order $5$, and that is the $\alpha$ you seek.

In other words $\omega +\omega^4 +\omega^5 +\omega^9 +\omega^3$.

You can then see that this has minimal polynomial $x^2+x+3$.

Answer 2

Consider the orbit of $\omega$ under $<\sigma>$, which is $\{\omega^i|1\leq i\leq10\}$. In cycle notation, we can write $\sigma^2$ as $(\omega^\text{ } \text{ } \omega^4\text{ } \omega^5\text{ } \omega^9\text{ } \omega^3)(\omega^2 \text{ } \omega^8\text{ } \omega^{10}\text{ } \omega^7\text{ } \omega^6)$. From this, we can see that $\omega + \omega^4+ \omega^5+ \omega^9+ \omega^3$ and $\omega^2 + \omega^8+ \omega^{10}+ \omega^7+ \omega^6$ are fixed by $\sigma^2$. Now, suppose, that $\alpha$ is fixed by $\sigma^2$. Then $\alpha$ can be written uniquely as $a_0+a_1\omega+a_2\omega^2+\cdots+a_{10}\omega^{10}$. But, since it is fixed, we get that $a_1=a_4=a_5=a_9=a_3$ and $a_2=a_8=a_{10}=a_7=a_6$. So 1, $\omega + \omega^4+ \omega^5+ \omega^9+ \omega^3$ and $\omega^2 + \omega^8+ \omega^{10}+ \omega^7+ \omega^6$ must be a basis for $Fix(<\sigma^2>)$ (which by the way is often denoted as $\Bbb{Q}(\omega)^{<\sigma^2>}$).