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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.

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Well if $20^{1/4}\omega^3$ is in that field then the quotient $20^{1/4}\omega^3/(20^{1/4}\omega)=\omega^2$ is too. I would guess it's much easier to derive a contradiction that $\omega^2\not\in\mathbb{Q}(20^{1/4}\omega)$. –  pki Dec 4 '11 at 2:11

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up vote 4 down vote accepted

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.

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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 $K$.

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