showing that $20^{1/4} \omega^3$ is not inside $\mathbb{Q}(20^{1/4} \omega)$, $\omega=e^{i \pi /4}$ Define $\omega=e^{i \pi /4}$. Is there an elegant way of showing that $20^{1/4} \omega^3$ is not inside $\mathbb{Q}(20^{1/4} \omega)$?
The way i am doing it is by observing that $20^{1/4} \omega$ is algebraic over $\mathbb{Q}$ with minimal polynomial $x^4+20$ and i assume that $20^{1/4} \omega^3$ is in the span over $\mathbb{Q}$ of $1, \alpha, \alpha^2,\alpha^3$, with $\alpha=20^{1/4} \omega$. But it is very messy to arrive at a contradiction.
Thanks.
 A: Here's one way that I can think of. We want to show that $\omega^2\not\in \mathbb{Q}(20^{1/4}\omega)$. It follows that it's sufficient to show that $x^4+20$ is irreducible over $\mathbb{Q}(i)=\mathbb{Q}(\omega^2)$.
Any factorization in $\mathbb{Q}(i)$ will actually be a factorization in the Gaussian integers $\mathbb{Z}[i]$ by Gauss lemma. Assume that $x^4+20$ is not irreducible over $\mathbb{Z}[i]$. Since no root of $x^4+20$ is contained in $\mathbb{Z}[i]$ the polynomial needs to factorize as a product of quadratic polynomials. But the only quadratic polynomials dividing it in $\mathbb{C}[x]$ hence in $\mathbb{Z}[i][x]$ are $x^2\pm i\sqrt{20}$ which are not in $\mathbb{Z}[i][x]$, so it must be irreducible.
If $\omega^2=i\in \mathbb{Q}(20^{1/4}\omega)$, then it follows by the irreducibility of $x^4+20$ that $\mathbb{Q}(20^{1/4}\omega)/\mathbb{Q}(i)$ is of degree $4$, hence $\mathbb{Q}(20^{1/4}\omega)/\mathbb{Q}$ is of degree $8$, which is a contradiction.
A: Let $K= \mathbb{Q}(20^{1/4}\omega) $,  a field of dimension $4$ over $\mathbb Q$ .
 Clearly  $\omega^2\in K \iff K \;$is the splitting field of $f(X)=X^4+20$  over  $\mathbb Q$ .
 It suffices to show that this is not the case.
The  resolvent of $f(X)$ is $r(Y)=Y^3-80Y $ .   This implies   that the Galois group of $X^4+20$ is the dihedral group $D_4$ of order $8$ , (cf. for example   Morandi's  Field and Galois Theory,  Theorem 13.4, page 126).
 Thus the splitting field of $f(X)$ over $\mathbb Q$ has dimension $8$ 
   and   is thus not equal to $K$.
