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

Question

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?

Answer 1

$\mathbb{Q}(\sqrt[4]2)$ 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 $\sqrt3 \mapsto \pm\sqrt3$.

Answer 2

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 $i\sqrt[4]2\notin\Bbb Q(\sqrt[4]2)$.

Answer 3

It seems to me the easiest solution is likely to be to show that $x^2-3$ has no solutions in $\Bbb Q [\sqrt[4]{2}]$.

Answer 4

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