Let $K=\mathbb{Q}(\sqrt[11]{7},i)$. Is $K$ the splitting field of some polynomial over $\mathbb{Q}$?
My attempt: My first intuition would be no. Since if $K$ is the splitting field of some polynomial then $K/\mathbb{Q}$ must be a finite, normal extension. Considering the irreducible polynomial $x^{11}-7$ over $\mathbb{Q}$, it is clear that one of its roots, namely $\sqrt[11]{7}$ lies in $K$.
It suffices to show that some other roots of $x^{11}-7$ does not lie in $K$. But I ended up with expressions in terms of $cos\left(\frac{n\pi}{11}\right)$ and$sin\left(\frac{n\pi}{11}\right)$ whose values are nearly impossible to compute.
Is there an easier way?
We need a primitive 11th root of unity $\zeta_{11}$ in $K$ (because all roots of $X^{11}-7$ are of the form $\zeta_{11}^k\sqrt[11]7$). Now $$22 = [K:\Bbb Q]=[K:\Bbb Q(\zeta_{11})]\cdot [\Bbb Q(\zeta_{11}):\Bbb Q]=10\cdot [K:\Bbb Q(\zeta_{11})],$$ contradiction.