I'm studying for a qualifying exam and wanted to know if my reasoning on this problem was correct:
Let $L$ be the splitting field of $f(x) = 1 + x^3 + x^6 + x^9$ over $\mathbb{Q}$. Find the Galois group of $L$ over $\mathbb{Q}$.
My solution: Let $\xi$ be a primitive $12$th root of unity. We have $$f(x) = 1 + x^3 + (x^3)^2 + (x^3)^3 = \frac{(x^3)^4 - 1}{x^3-1} = \frac{x^{12}-1}{x^3-1}$$ so the roots of $f$ are all the twelfth roots of unity which aren't third roots of unity. Therefore the set of roots of $f$ is $S = \{1, \xi, \xi^2, ... , \xi^{11}\} \setminus \{1, \xi^4, \xi^8\}$. But from the remaining roots we still have $\xi$, so clearly $L = \mathbb{Q}(S)$ is equal to $\mathbb{Q}(\xi)$. Therefore $L$ is cyclotomic of order $12$ over $\mathbb{Q}$, so its Galois group is isomorphic to $$U(\mathbb{Z}/12\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$