Q:

$\mathbb{Q}(\sqrt2, \sqrt3)$ is not isomorphic to $\mathbb{Q}(\sqrt[4]2)$

I wish to show that $\mathbb{Q}(\sqrt2, \sqrt3)$ is not isomorphic to $\mathbb{Q}(\sqrt[4]2)$. 

What I did so far was to show that $\mathbb{Q}(\sqrt2,\sqrt3) =\text{span}\{1,\sqrt2,\sqrt3,\sqrt6\}$.
Assuming they are isomorphic, denoting $f$ the isomorphism, then $f(1) = 1$, so $x^2-3$ has a root in $\mathbb{Q}(\sqrt2,\sqrt3)$ so $\sqrt 3\in\mathbb{Q}(\sqrt[4]2)$. ($\sqrt2 \in \mathbb{Q}(\sqrt[4]2)$, $\sqrt[4]2\cdot\sqrt[4]2=\sqrt2)$. 
Thus under the assumption, $\mathbb{Q}(\sqrt2, \sqrt3)\subseteq \mathbb{Q}(\sqrt[4]2)$. As a vector spaces of the same dimension over $\mathbb{Q}$, it means that $\mathbb{Q}(\sqrt2, \sqrt3)= \mathbb{Q}(\sqrt[4]2)$.
Now all is left to do it to show a contradiction: $\sqrt[4]2\in\mathbb{Q}(\sqrt2,\sqrt3)$, then $\sqrt[4]2 = a+b\sqrt2+c\sqrt3+d\sqrt6$ so $2=(a+b\sqrt2+c\sqrt3+d\sqrt6)^4$, I believe that opening the expression, I would find out that it can't be an element of $\mathbb{Z}$. However this is a really inelegant way (if it really works), and I wish for a better way to do so.
An other approach: $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2 + \sqrt3)$ by the irreducibility of $x^4-10x^2+1$ over $\mathbb{Q}$ (an some dimension theorem usage). Also, $x^4-10x^2+1$ splits in $\mathbb{Q(\sqrt2, \sqrt3)}$. 
Is there a way to show that this polynomial doesn't split over $\mathbb{Q}(\sqrt[2]4)$? with different method then assuming that $\sqrt3\in\mathbb{Q}(\sqrt[4]2)$ and getting a contradiction? Maybe the fact that $\sqrt[4]2$ has complex roots might help?

A:

Another approach: $\Bbb Q(\sqrt2,\sqrt3)$ is normal over $\Bbb Q$, as it's
the splitting field of $(X^2-2)(X^2-3)$.
But $\Bbb Q(\sqrt[4]2)$ is not normal over $\Bbb Q$. The number $\sqrt[4]2$
has minimal polynomial $X^4-2$ which also has the zero
$i\sqrt[4]2\notin\Bbb Q(\sqrt[4]2)$.

A:

Compute the discriminants of the fields $\Bbb Q(\sqrt{2},\sqrt{3})=\Bbb Q(\sqrt{2}+\sqrt{3})$ and $\Bbb Q(\sqrt[4]{2})$. They are $2304=2^{8}\cdot 3^2$ and $-2048=-2^{11}$ respectively. Since they are different, the fields are not isomorphic.
References:
Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
Discriminant of Number Fields

A:

$\mathbb{Q}(\sqrt[4]2)$ has exactly two automorphisms: those induced by $\sqrt[4]2 \mapsto \pm\sqrt[4]2$.
$\mathbb{Q}(\sqrt2, \sqrt3)$ has at least four automorphisms: those induced by $\sqrt2 \mapsto \pm\sqrt2$ and $\sqrt3 \mapsto \pm\sqrt3$.

