$\omega$ is the $11$th primitive root of unity. Find minimal polynomial of $\beta=\omega+\omega^3+\omega^4+\omega^5+\omega^9$

Question

I am studying for my Galois theory final tomorrow, I am stuck on this question.

Let $\omega$ is the $11$th primitive root of unity.

1. Find $|\mathbb{Q}(\beta):\mathbb{Q}|$ [Hint: Consider $\beta^{\tau}$ where $\omega^{\tau}=\omega^3$ etc.]

2. Find minimal polynomial of $\beta=\omega+\omega^3+\omega^4+\omega^5+\omega^9$.

To be honest, I am not sure where to start, I haven't really seen similar questions.

I know that $G=\text{Gal}(\mathbb{Q}(\omega):\mathbb{Q}) \cong \mathbb{U}_{11}\cong C_{10}$. Let $\sigma_3 : \omega \to \omega^3 \in G$. Then $\sigma_3(\beta)=\beta$. And that subgroups of $C_{10}$ are of order $2$ or $5$.

Say that the subgroup corresponding to $\mathbb{Q}(\beta)$ is $H$. So $\text{Fix}H=\mathbb{Q}(\beta)$. As above I see that $1,\sigma_3, \sigma_9 \in H$ so we must have $|H| = 5$, so $|\mathbb{Q}(\beta):\mathbb{Q}|=2$.

Is this correct and if so how to find the minimal polynomial?

Answer 1

Since $2^2\not\equiv 1\pmod{11}$ and $2^5\not\equiv 1\pmod{11}$, the Galois group $\operatorname{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})$ is generated by $\sigma_2\colon\omega\mapsto\omega^2$.

Note that $\sigma_2^2(\beta)=\beta$, $\sigma_2(\beta)\neq\beta$, so $$ \beta'=\sigma_2(\beta)=\omega^2+\omega^6+\omega^7+\omega^8+\omega^{10} $$ is the conjugate of $\beta$ in the quadratic extension $\mathbb{Q}(\beta)/\mathbb{Q}$.

We have $\beta+\beta'=-1$, and $\beta\beta'=3$ by easy computation. Thus $\beta$ has minimal polynomial $x^2-x+3$.