Galois group of weird polynomial

Question

Find $\text{Gal}(f)$ of $f=x^{2018}+1\in\mathbb{Q}(X)$.

The roots are: $x=\pm i\zeta_{4036}^k$ with $k=0, 1, ..., 1009$ so at least $\Phi_{4036}\vert x^{2018}+1$ where $\deg({\Phi_{4036}})=2016$ by Euler's totient function.

Splitting field: $\Omega_{\mathbb{Q}}^{f}=\mathbb{Q}(i\zeta_{4036})$. Now, we know that $f=x^{2018}+1=(x^2+1)\Phi_{4036}$ for both $i$ and $\zeta_{4036}$ should be roots.

Can we now state that $\text{Gal}(f)=\text{Gal}(\Phi_{4036})=\big(\mathbb{Z}/4036\mathbb{Z})^{*}$?

Answer 1

Yes. $f(x)$ is a factor of $x^{4036}-1$ so $f$ splits over $\Bbb{Q}(\zeta_{4036})$. Also, as you observed, $\Phi_{4036}(x)\mid f(x)$, so the splitting field $\Phi_{4036}$ is contained in the splitting field of $f$.

Hence $Gal(f)=Gal(\Phi_{4036})$.