Galois group of the polynomial $f(X)=(X^4+1)(X^4-2)$

Question

Here is my attempt:

Let $\alpha=e^{i\frac{\pi}{4}}$ and $\beta=\sqrt[4]{2}$ two roots of the polynomial. Then, the other ones are given by $\alpha\omega$, $\alpha\omega^2$, $\alpha\omega^3$ $\beta\omega$, $\beta\omega^2$ and $\beta\omega^3$, where $\omega=e^{i\frac{\pi}{2}}$. Hence, the splitting field of $f$ is $K_f=\mathbb{Q}(\alpha,\beta,\omega)$. The minimal polynomial of $\alpha$, $\beta$ and $\omega$ are $p_\alpha(X)=X^4+1$, $p_\beta(X)=X^4-2$ and $p_\omega(X)=X^2+1$, so the Galois group has order 32.

However, the solution given to this exercise says that the Galois group is $\mathcal{D}_4$. Can someone tell me where my mistake is?

Thanks in advance.

Answer 1

$\omega = \alpha^2$, so $K_f = \mathbb{Q}(\alpha, \beta)$. Also be careful:

$$\deg(\mathbb{Q}(\alpha, \beta, \omega) : \mathbb{Q}) \neq \deg(\mathbb{Q}(\alpha) : \mathbb{Q}) \deg(\mathbb{Q}(\beta) : \mathbb{Q}) \deg(\mathbb{Q}(\omega) : \mathbb{Q}).$$

You have to compute this adding one element at a time:

$$\deg(\mathbb{Q}(\alpha, \beta, \omega) : \mathbb{Q}) = \deg(\mathbb{Q}(\alpha,\beta,\omega) : \mathbb{Q}(\beta,\omega)) \deg(\mathbb{Q}(\beta,\omega) : \mathbb{Q}(\omega)) \deg(\mathbb{Q}(\omega) : \mathbb{Q}).$$

So for instance

$$\deg(\mathbb{Q}(\alpha,\beta,\omega) : \mathbb{Q}(\beta,\omega)) = 2$$ since $\alpha$ is a root of $X^2 - \beta \in \mathbb{Q}(\beta,\omega)[X]$.

Answer 2

The roots of $x^4-2$ are $\sqrt[4]{2}, i \sqrt[4]{2}, - \sqrt[4]{2}, -i \sqrt[4]{2}$. Thus, the splitting field of $x^4-2$ already contains $i$. Therefore, adding $\sqrt{i}$ makes $x^4+1$ split as well.