Simplifying the Splitting field of $x^n-a$

Question

Let $L/K$ be a field extension where $L$ is the splitting field of the polynomial $x^n-a\in K[x]$. Clearly $L=K(t,\zeta t,\ldots,\zeta^{n-1}t)$, where $t=\sqrt[n]{a}$ and $\zeta$ is the primitive $n^\text{th}$ root of unity. I can simplify $L$ to $L=K(t,\zeta)$, but can I simplify it further to $K(t,i)$ where $i^2=-1$?

Answer 1

No. For example, let $K=\mathbb{Q}$, $t=1$, and $n=3$. Then $L = K(t,\zeta) = \mathbb{Q}\left(\frac{-1+i\sqrt{3}}{2}\right) = \mathbb{Q}(i\sqrt{3})$, which is clearly unequal to $\mathbb{Q}(1,i)$.

In general, $[\mathbb{Q}[\zeta_n]:\mathbb{Q}] = \varphi(n)$ where $\varphi$ is the Euler totient function. Since $\zeta_4 = \pm i$, clearly in general $\mathbb{Q}[\zeta_n]\ne \mathbb{Q}[i]$.