Is this proof that $\mathbb Q(\sqrt[4]{-5})/\mathbb Q$ is not normal correct? I was asked to prove/disprove that $\mathbb Q(\sqrt[4]{-5})/\mathbb Q$ is normal. 
Here is my attempt. It feels a bit long winded, I appreciate any suggestions on how to make it more elegant.
Put $L:=\mathbb Q(\sqrt[4]{-5})$, and suppose $L/\mathbb Q$ is normal. Since $x^4+5$ is irreducible (by Eisenstein, say) and the root $\alpha:=\sqrt[4]{-5}\in L$, it follows that the other roots $\alpha i^k\in L$, $1\leq k \leq 3$. In particular, $i=\frac{\alpha i}\alpha\in L$, so $\mathbb Q(\sqrt[4]{-5},i)=L$. Now $[\mathbb Q(i):\mathbb Q]=2$ (since $\deg(m_{i,\mathbb Q})=\deg(x^2+1)=2$), so the tower law would suggest that $[L:\mathbb Q(i)]=2$ (since $[L:\mathbb Q]=4$).  Thus $\deg(m_{\alpha,\mathbb Q(i)})=2$, i.e., there are $p,q\in\mathbb Q(i)$ such that
            $$m_{\alpha,\mathbb Q(i)}(x)=x^2+px+q.$$
            Writing $p=a+bi$ and $q=c+di$ where $a,b,c,d\in\mathbb Q$, we have
            \begin{align*}
            0&=m_{\alpha,\mathbb Q(i)}(\sqrt[4]{-5})\\
            &= (a\sqrt[4]{-5}+c)+(\sqrt5+\sqrt[4]{-5}b+d)i,
        \end{align*}
            in particular, comparing real parts implies that $\sqrt[4]{-5}=-\frac ca\in\mathbb Q$, which is a contradiction. Thus $L/\mathbb Q$ is not normal.

I am not sure about the last part when you compare the real and imaginary part, for example, $a\sqrt[4]{-5}+c$ is not really real. However one can have a similar argument by noting that both $i,\sqrt{-5}=(\sqrt[4]{-5})^2\in L={\mathbb Q}(\alpha)$  with $\alpha=\sqrt[4]{-5}$ implies that $\sqrt{5}\in L$. Let $f(x)=x^2+px+q$ be the minimal polynomial for $\alpha$ over $\mathbb Q(\sqrt{5})$, where $$p=a+b\sqrt{5},q=c+d\sqrt{5},a,b,c,d\in \mathbb Q.$$ Then $f(\alpha)=0$ implies $$\sqrt{-5}+(a+b\sqrt{5})\sqrt[4]{-5}+(c+d\sqrt{5})=0$$ $$\Rightarrow (a+b\sqrt{5})\sqrt[4]{-5}=-[\sqrt{-5}+(c+d\sqrt{5})]$$ $$\Rightarrow (a+b\sqrt{5})^2\sqrt{-5}=-5+2(c+d\sqrt{5})\sqrt{-5}+(c+d\sqrt{5})^2$$ $$\Rightarrow [(a+b\sqrt{5})^2-2(c+d\sqrt{5})]\sqrt{-5}=-5+(c+d\sqrt{5})^2,$$ whence both sides should be real and equal $0$. This gives that $c+d\sqrt{5}=\pm \sqrt{5}$ and so $(a+b\sqrt{5})^2=2\sqrt{5}\Rightarrow (a^2+5b^2)+2ab\sqrt{5}=2\sqrt{5}$ for some $a,b\in \mathbb Q,$ which is not possible. Consequently $L$ is not normal over $\mathbb Q$.
