Q:

Is $\mathbb{Q(\sqrt[4]{-7})/\mathbb{Q}}$ a normal extension?

I am trying to use the equivalence that the extension $\mathbb{Q(\sqrt[4]{-7})/\mathbb{Q}}$ is normal and separable $\iff$ $\mathbb{Q(\sqrt[4]{-7})}$ is the splitting field of a separable polynomial.
I'm considering $x^4+7$ as it seems appropriate. I've worked out the four roots of this to be $\zeta =\pm \frac{\sqrt[4]{7}\sqrt{2}}{2} \pm i\frac{\sqrt[4]{7}\sqrt2}{2}$ but I can't tell at all if all these roots lie in $\mathbb{Q(\sqrt[4]{-7})}$. I would probably guess not. Then since $x^4+7$ is irreducible over $\mathbb{Q}$ we must have that $\mathbb{Q(\sqrt[4]{-7})/\mathbb{Q}}$ is not normal.
But I don't know if $\mathbb{Q(\sqrt[4]{-7})}$ is indeed the splitting field.
I'm guessing there's another method. Any help would be appreciated.

A:

Note that the roots of $x^4 + 7$ are $\mu_k = (7^{1/4})e^{k \pi i /8}$, where $k$ is odd and less than 8. If $K = \Bbb{Q}((-7)^{1/4})$ is normal, it follows that $i = \mu_3/\mu_1 \in K$. Since $-\sqrt{7} = \mu_3 \mu_1$ is also in $K$, you get that $\sqrt{7} \in K$. Then $K = \Bbb{Q}(i, \sqrt{7})$ because the degrees are the same. Note that any $x \in K$ has imaginary part of the form $a + b \sqrt{7}$, with $a,b$ integers. If you write $(7)^{1/4}/\sqrt{2} = a + b\sqrt{7}$ and do some algebra, you see that $\sqrt{2}$ must be in $\Bbb{Q}((7)^{1/4})$. Here I find a contradiction by assuming $\sqrt{2} = r_1 + r_2(7)^{1/4} + r_2(7)^{1/2} + r_3(7)^{3/4}$ with all $r_i$ rational, squaring both sides and getting all the $r_i$ in terms of $r_2$ by solving the system of 4 equations.

