Showing $[\mathbb{Q}(\sqrt[4]{2},\sqrt{3}):\mathbb{Q}]=8$. I have a question about calculating the degree of a finite field extension over $\mathbb{Q}$.
This is problem 18 in chapter 1 of Patrick Morandi's Field and Galois Theory.
The problem asked to show that $$ \left[ \mathbb{Q}{ \left( \sqrt[4]{2},\sqrt{3} \right):\mathbb{Q} }\right] =8 $$
Here's my work :
First, we have $$  \left[ \mathbb{Q} ( \sqrt[4]{2},\sqrt{3}):\mathbb{Q}\right] = \left[ \mathbb{Q}(\sqrt[4]{2},\sqrt{3}) : \mathbb{Q}({\sqrt[4]{2}}) \right] . \left[ \mathbb{Q}{(\sqrt[4]{2})}: \mathbb{Q} \right]$$
It's easy to show that $\left[ \mathbb{Q}{(\sqrt[4]{2})}: \mathbb{Q} \right] =4 $ because the minimal polynomial of $\sqrt[4]{2}$ over $\mathbb{Q}$ is $x^4 -2$. Next, I'm going to show that 
$$ \left[ \mathbb{Q}(\sqrt[4]{2},\sqrt{3}) : \mathbb{Q}({\sqrt[4]{2}}) \right] =2$$
Consider the polynomial $f(x) = x^2 -3$. It's monic and $\sqrt{3}$ is a root of $f$, but I'm stuck in proving $\sqrt{3} \notin \mathbb{Q}(\sqrt[4]{2})$ (hence $f$ is irreducible over $\mathbb{Q}(\sqrt[4]{2}))$.
My question is: Is there any proof to show that $\sqrt{3} \notin \mathbb{Q}(\sqrt[4]{2})$ or another way to caculate the degree of this field extension?
Thank you for your help.
 A: I think one can use reduction modulo $7$ to show that $\sqrt 3$ is not in $\mathbb{Q}[\sqrt[4]{2}]$, since in $\mathbb{F}_7$ the equation $X^4-2$ has a solution (i.e.\ $x=5$) but $X^2-3$ has no solution.
Edit: Easier proof: Assume that $\sqrt{3}\in \mathbb{Q}(\sqrt[4]{2})$. Then $\mathbb{Q}(\sqrt{2},\sqrt{3})\subseteq \mathbb{Q}(\sqrt[4]{2})$ and from the equality of degrees over $\mathbb{Q}$ those are equal. But this implies that $\mathbb{Q}(\sqrt[4]{2})$ is Galois which is contradiction, since $e^{2\pi i/4} \sqrt[4]{2}\not\in \mathbb{Q}(\sqrt[4]{2})$.
A: Here is an alternative method to show that $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt[4]{2})$ results in a contradiction, without using any knowledge of Galois extensions (since the problem is asked in the textbook before this topic is covered).
A basis for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ is $\{ 1,\sqrt{2},\sqrt{3},\sqrt{6} \}$. By assumption, $\sqrt[4]{2} \in \mathbb{Q}(\sqrt{2},\sqrt{3})$, so there exist $a,b,c,d \in \mathbb{Q}$ such that
$$
\begin{align}
& & \sqrt[4]{2} &= a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}\\
&\implies &\sqrt{2} &= (a^2 + 2b^2 + 3c^2 + 6d^2) + 2(ab + 3cd)\sqrt{2} + 2(ac + 2bd)\sqrt{3} + 2(ad+bc)\sqrt{6}\\
&\implies &0 &= \color{blue}{(a^2 + 2b^2 + 3c^2 + 6d^2)} + 2(ab + 3cd - 1)\sqrt{2} + 2(ac + 2bd)\sqrt{3} + 2(ad+bc)\sqrt{6}
\end{align}
$$
Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is a $\mathbb{Q}$-linearly independent set, each of the coefficients must be zero. In particular,
$$
a^2 + 2b^2 + 3c^2 + 6d^2 = 0 \implies a=b=c=d=0 \implies \sqrt[4]{2} = 0,
$$
which is a contradiction.
