The following question was given as homework by the prof. and I am not able to deduce it. So, I am asking for help here.
If F is normal over E and E is normal over K then prove that F need not be normal over K.
Prof. also gave hint to consider $\mathbb{Q} ((2)^{1/4})$ , $\mathbb{Q}(\sqrt(2))$ and $\mathbb{Q}$.
I proved that $\mathbb{Q} ((2)^{1/4})$ is normal over $\mathbb{Q}(\sqrt(2))$ and $\mathbb{Q}(\sqrt(2))$ is normal over $\mathbb{Q}$ but I am not able to prove that $\mathbb{Q} ((2)^{1/4})$ is nt normal over $\mathbb{Q}$.
My attempt for what I am not able to prove, If I assume $x^4-2 \in \mathbb{Q}$ then $(2)^{1/4}$ lies in $\mathbb{Q} ((2)^{1/4})$ and all the other roots of the equation $x^4-2=0$ also lie in $\mathbb{Q} ((2)^{1/4})$. So, It is normal in $\mathbb{Q}$.
which is wrong . SO, Can you please tellme what is wrong in my argument and what would be the correct argument.
$x^4-2=(x^2-\sqrt{2})(x^2+\sqrt{2})$. Thus there are two real roots $\pm\sqrt[4]{2}$ and two complex roots $\pm\sqrt[4]{2}i$.
Note that $\Bbb{Q}(\sqrt[4]{2})$ is the smallest field containing $\sqrt[4]{2}$ and $\Bbb{Q}$.
Therefore $\Bbb{Q}(\sqrt[4]{2})$ is not the splitting field of the polynomial $x^4-2$.
The extensions$\Bbb{ Q}(\sqrt[4]{2})/\Bbb{Q}( \sqrt{2})$ and $\Bbb{Q}( \sqrt{2})/\Bbb{Q }$are normal but the extension $\Bbb{Q}( \sqrt[4]{ 2})/\Bbb{Q}$is not a normal extension since the complex roots of the polynomial $x^4-2$ are not in $\Bbb{Q}(\sqrt{2}).$
It is not hard to see that $ \Bbb{Q}(\sqrt[4]{2},\sqrt[4]{2}i)$ is the splitting field of the said polynomial.