$\sqrt[7]{11}$ is not contained in the splitting field of $x^7-12$ over $\mathbb{Q}$ I want to prove: 

$\sqrt[7]{11}$ is not contained in the splitting field of $x^7-12$ over $\mathbb{Q}$. Is there any direct way to prove?

I have computed that the splitting field of $x^7-12$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[7]{12},\zeta)$ where $\zeta^7=1$ and $\zeta\neq 1$. I just use a basis $\{(\sqrt[7]{12})^i\zeta^j\mid i=0,1,\dots,6; j=1,2,\dots,6\}$ of the field extension $\mathbb{Q}(\sqrt[7]{12},\zeta):\mathbb{Q}$ to show it is impossible that $\sqrt[7]{11}$ be expressed as a linear combination of these basis with coefficients in $\mathbb{Q}$. 

In general, what can we obtain for the minimal polynomial $g(x)$ of an element  in the splitting field of a polynomial $f(x)\in \mathbb{Q}[x]$? Can $g$ be coprime to $f$?

 A: Yes, there is a simple argument from Galois theory (or rather, “Kummer” theory).
It works with any parameters in place of $11$ and $12$.
Let $K={\mathbb Q}(\zeta)$ and $L=K(\sqrt[7]{11})$. We know that
$G={\sf Gal}(L/K)$ is canonically isomorphic to $(\frac{\mathbb Z}{7\mathbb Z})^*$, 
generated by the unique $K$-automorphism
$\sigma$ of $L$ satisfying $\sigma(\sqrt[7]{11})=\zeta\sqrt[7]{11}$.
Suppose, by contradiction, that $\sqrt[7]{12}\in L$.  Since the polynomial
$X^7-12$ is irreducible over $K$, $\sigma(\sqrt[7]{12})$ must be a conjugate
of $\sqrt[7]{12}$ : we have an index $k\in (\frac{\mathbb Z}{7\mathbb Z})^*$
such that $\sigma(\sqrt[7]{12})=\zeta^k\sqrt[7]{12}$. We then have
$$
\frac{\sigma(\sqrt[7]{12})}{\sqrt[7]{12}}=
\zeta^k=\bigg(\frac{\sigma(\sqrt[7]{11})}{\sqrt[7]{11}}\bigg)^k
$$
So
$$
\sigma(\sqrt[7]{12^6 \times 11^k})=\sqrt[7]{12^6 \times 11^k}
$$
So $N=\sqrt[7]{12^6 \times 11^k}$ is a rational algebraic number ; we deduce that
it is an integer. This is impossible because of the prime factorization of
$11$ and $12$.
