Showing this field extension is not normal

Question

This question is part of a homework assignment. We are asked if the field extension $\mathbb{Q(\sqrt[4]{-7})}/\mathbb{Q}$ is normal. Here is what I have so far:

The obvious thing to do seems to be to look at the minimal polynomial of $\sqrt[4]{-7}$ over $\mathbb{Q}$, which is $f = x^{4} + 7$. The roots of this polynomial are $\sqrt[4]{-7}, -\sqrt[4]{-7}, i\sqrt[4]{-7}, -i\sqrt[4]{-7}$. So the question has come down to if these roots belong to $\mathbb{Q}(\sqrt[4]{-7})$. I am also certain that $i\sqrt[4]{-7}$ does not belong to this field, but have been unable to show this. Clearly this is true if and only if $i \in \mathbb{Q}(\sqrt[4]{-7})$, and since a $\mathbb{Q}$ basis for $\mathbb{Q(\sqrt[4]{-7})}/\mathbb{Q}$ is $1 , (-7)^{\frac{1}{4}}, (-7)^{\frac{1}{2}}, (-7)^{\frac{3}{4}}$, this is true if and only if:

$i = a + b(-7)^{\frac{1}{4}} + c(-7)^{\frac{1}{2}} + d(-7)^{\frac{3}{4}}$ where $a,b,c,d \in \mathbb{Q}$

It seems obvious but I am having trouble explicitly showing this equation has no solutions in $a, b, c, d$. I have tried taking the real and imaginary parts of the equation to get two equations but it gets a little bit messy, seems like there must be a simpler way of doing this?

EDIT: I have obtained the following solution. There was a lot of trial and error to get this and it is very long-winded but I believe it is correct. It would still be nice if someone could find a more elegant solution.

If $i\sqrt[4]{-7} \in \mathbb{Q(\sqrt[4]{-7})}$, then $i \in \mathbb{Q(\sqrt[4]{-7})}$ and thus $\frac{\sqrt{-7}}{i} = \sqrt{7} \in \mathbb{Q(\sqrt[4]{-7})}$. Hence $\mathbb{Q(\sqrt[4]{-7})} \supset \mathbb{Q(\sqrt{7})} \supset \mathbb{Q}$, from which we can deduce from the Tower Law that $[\mathbb{Q(\sqrt[4]{-7})}:\mathbb{Q(\sqrt{7})}]= 2$. It follows that $\exists a, b \in \mathbb{Q(\sqrt{7})}$ such that:

$(\sqrt[4]{-7})^{2} + a(\sqrt[4]{-7}) + b = 0$

Consider just the imaginary part of this equation, then we obtain:

$\sqrt{7} + a(\frac{\sqrt{2}}{2})\sqrt[4]{7} = 0$

Solving for $a$, and using the assumption $a \in \mathbb{Q}(\sqrt{7})$ we deduce $\sqrt{2} \in \mathbb{Q}(\sqrt{7})$. But since $1 , \sqrt{7}$ is a $\mathbb{Q}$ basis for $\mathbb{Q}\sqrt({7})$, $\exists p, q \in \mathbb{Q}(\sqrt{7})$ such that:

$\sqrt{2} = p + q\sqrt{7}$. Squaring both sides gives $2 = p^{2} + 7q^{2} + 2pq\sqrt{7}$. But the LHS is rational while the RHS is irrational, so we have obtained a contradiction.

Answer 1

Note that $[\mathbb{Q}(i):\mathbb{Q}]=2$ and the minimal polynomial of $\sqrt[4]{-7}$ over $\mathbb{Q}(i)$ is $f=x^{4}+7$ so $[\mathbb{Q}(\sqrt[4]{-7},i):\mathbb{Q}(i)]=4$.

This means that $[\mathbb{Q}(\sqrt[4]{-7},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{-7},i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]=8\neq 4=[\mathbb{Q}(\sqrt[4]{-7}):\mathbb{Q}]$.

Therefore $\mathbb{Q}(\sqrt[4]{-7},i)\neq \mathbb{Q}(\sqrt[4]{-7})$ so $i\notin \mathbb{Q}(\sqrt[4]{-7})$

Answer 2

The splitting field $K$ of $x^4+7$ is generated by $\sqrt[4]{-7}$ and $i\sqrt[4]{-7}$ subject to the relation between their squares: that is, it's $\mathbb{Q}[x,y]/(x^4=y^4=-7,x^2=-y^2)$. This has automorphisms $x\mapsto -x, y\mapsto y$ and $x\mapsto ix, y\mapsto iy$ of orders $2$ and $4$ respectively, which together generate the dihedral group $D_4$. Thus $K$ can't be 4-dimensional over $\mathbb{Q}$, in particular, it's not $\mathbb{Q}[\sqrt[4]{-7}]$.