Jtdylktuy

I'm trying to compute the Galois group of the polynomial $P(X) = (X^4 - 2)(X^2 + 2)$, but I think that I made a mistake. This is what I thought:

$P(X)$ has roots $pm i sqrt{2}, pm i sqrt[4]{2}, pm sqrt[4]{2}$, then the splitting field of $P(X)$ is $K := mathbb{Q}[sqrt[4]{2}, i sqrt[4]{2}, i sqrt{2}]$, but $sqrt{2} = left( sqrt[4]{2} right)^2 in mathbb{Q}[sqrt[4]{2}, i]$, $i sqrt{2} in mathbb{Q}[sqrt[4]{2}, i]$, $i sqrt[4]{2} in mathbb{Q}[sqrt[4]{2}, i]$ and $i = (i sqrt[4]{2})^3 (sqrt[4]{2}) left( frac{-1}{2} right) in K$, then $K = mathbb{Q}[sqrt[4]{2}, i]$.

Observing that $X^4 - 2$ is monic, has $sqrt[4]{2}$ as a root and is irreducible on $mathbb{Q} [X]$ by Eisenstein's criterion applied for $p = 2$, we have that $X^4 - 2$ is a minimal polynomial over $mathbb{Q}$ which has a root $sqrt[4]{2}$, then $left[ mathbb{Q} [sqrt[4]{2}] : mathbb{Q} right] = 4$. Furthermore, $X^2 + 1$ is monic, which has a root $i$ and is irreducible over $mathbb{Q}[sqrt[4]{2}] [X]$ since $X^2 + 1$ is a polynomial of degree $2$ and $i notin mathbb{Q}[sqrt[4]{2}]$, then $X^2 + 1$ is a minimal polynomial over $mathbb{Q}[sqrt[4]{2}] [X]$, then $left[ mathbb{Q} [sqrt[4]{2},i] : mathbb{Q} [sqrt[4]{2}] right] = 2$.

By the tower law,

$$left[ mathbb{Q} [sqrt[4]{2},i] : mathbb{Q} right] = left[ mathbb{Q} [sqrt[4]{2},i] : mathbb{Q} [sqrt[4]{2}] right] left[ mathbb{Q} [sqrt[4]{2}] : mathbb{Q} right] = 2 cdot 4 = 8,$$

but there are only $4$ automorphisms of $mathbb{Q} [sqrt[4]{2},i]$ which fixes $mathbb{Q}$ and this contradicts the fact that the cardinality of the Galois group of $K$ over $mathbb{Q})$ is equal to $left[ mathbb{Q} [sqrt[4]{2},i] : mathbb{Q} right]$

What am I doing wrong?

You are wrong that there are only $4$ automorphisms of $mathbb{K}:=mathbb{Q}(sqrt[4]{2},text{i})$ that fix $mathbb{Q}$. The automorphism group $G:=text{Aut}_mathbb{Q}(mathbb{K})$ is isomorphic to the unique nonabelian semidirect product $C_4rtimes C_2cong D_4$, where $C_k$ is the cyclic group of order $k$ and $D_k$ is the dihedral group of order $2k$. The group $G$ is generated by $tau$ and $sigma$ where $$tau(sqrt[4]{2})=text{i}sqrt[4]{2},,,,tau(text{i})=text{i},,,,sigma(sqrt[4]{2})=sqrt[4]{2},,text{ and }sigma(text{i})=-text{i},.$$
Note that $langle tauranglecaplangle sigmarangle=text{Id}$, with $langle tauranglecong C_4$ and $langle sigmaranglecong C_2$. In addition,
$$(taucircsigmacirctau)(sqrt[4]{2})=(taucirc sigma)(text{i}sqrt[4]{2})=tau(-text{i}sqrt[4]{2})=sqrt[4]{2}=sigma(sqrt[4]{2})$$
and
$$(taucircsigmacirctau)(text{i})=(taucirc sigma)(text{i})=tau(-text{i})=-text{i}=sigma(text{i}),.$$
Hence, $taucircsigmacirctau=sigma$, or $taucircsigma=sigmacirctau^{-1}$. That is,
$$G=biglangle tau,sigma,big|,tau^{circ 4}=text{id},,,,sigma^{circ 2}=text{id},,text{ and }taucircsigma=sigmacirctau^{-1}ranglecong D_4,.$$