Is this proof that $\mathbb Q(\sqrt[4]{-5})/\mathbb Q$ is not normal correct?

Question

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.

Answer 1

You could have used a bit of Galois theory. Let $\alpha$ be a given root of $X^4+5=0$, for instance $\alpha=\zeta \sqrt [4] 5$, where $\zeta$ is a given square root of $i$, i.e. a primitive $8$-th root of unity. As you said, the extension $L=\mathbf Q(\alpha)$ has degree $4$ over $\mathbf Q$ (Eisenstein). If $L/\mathbf Q$ were normal, it would contain all the conjugates of $\alpha$, which are $\pm \alpha, \pm i\alpha$, i.e. $\pm \zeta \sqrt [4] 5, \pm \zeta^3\sqrt [4] 5$, thus actually $L$ would contain $\mathbf Q(\zeta^2, \sqrt [4] 5)$. But $\mathbf Q(\zeta^2)=\mathbf Q(i)$ is totally imaginary of degree $2$ and $\mathbf Q(\sqrt [4] 5)$ is totally real of degree $4$, so $L/\mathbf Q$ would have degree $\ge 8$ : contradiction.

A zest of Kummer theory (see any textbook) would even give a quicker answer. If $L/\mathbf Q$ were normal, it would obviously be $\mathbf Q(i,\alpha)=\mathbf Q(i)(\sqrt [4] {-5})$. Since $i$ is a $4$-th root of unity, Kummer's theorem asserts that $\mathbf Q(i)(\sqrt [4] {-5})/\mathbf Q(i)$ is cyclic of degree equal to the order of the class $-5$ mod $\mathbf Q{(i)^*}^4$ in $\mathbf Q(i)^*/\mathbf Q{(i)^*}^4$. Supposing that $-5 = z^4$ and taking norms from $\mathbf Q(i)$ to $\mathbf Q$, we get that $25$ is a $4$-th power in $\mathbf Q^*$ : impossible. So $[L:\mathbf Q]\ge 2.4=8$ : contradiction.

The two methods just exposed may seem overly complicated, but their advantage is their generality, which allows to treat more complicated cases than the one here.

Answer 2

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 $\mathbb Q$.