While doing my homework a question arose in my mind.
If I consider $p_1(x)=x^4+2$, I know its roots are $$\sqrt[4]{-2}\cdot e^{\frac{2\pi i k}{4}}$$ for $k=1,...,4$.
If I do the same for $p_2(x)=x^4-2$, I have $$\sqrt[4]{2}\cdot e^{\frac{2\pi i k}{4}}$$ for $k=1,...,4$.
From what I conclude that $\mathbb{Q}(i,\sqrt[4]{2})$ is the splitting field of both polynomials. Hence, $\mathbb{Q}\subseteq \mathbb{Q}(i,\sqrt[4]{2})$ is a normal extension.
I am asking this because I have seen that a similar reasoning for $\mathbb{Q}\subseteq \mathbb{Q}(i,\sqrt{2})$ with $p_3(x)=x^2+2$, $p_4(x)=x^2-2$ would not happen. We should chose $p_3(x)$ if we would like to say that $\mathbb{Q}(i,\sqrt{2})$ is the splitting field of a polynomial in $\mathbb{Q}[x]\setminus \mathbb{Q}$.
Is there any result related to this? Am I right in my reasoning?
The roots of $x^4+2$ are of the form $$ \sqrt[4]{2}\exp\Bigl(\frac{2k\pi i}{8}\Bigr),\qquad k=1,3,5,7 $$ Denote them by $r_1,r_2,r_3,r_4$. Then $$ \frac{r_2}{r_1}=\exp\Bigl(\frac{\pi i}{2}\Bigr)=i $$ so certainly $i$ belongs to the splitting field. Also $$ r_1+r_4=\dfrac{1+i}{\sqrt[4]{2}}+\dfrac{1-i}{\sqrt[4]{2}}=\dfrac{2}{\sqrt[4]{2}} $$ implies that also $\sqrt[4]{2}$ belongs to the splitting field. Now it's not difficult to express all roots of $x^4+2$ in terms of $i$ and $\sqrt[4]{2}$, so indeed the splitting field is $\mathbb{Q}(i,\sqrt[4]{2})$.
It's easier for $x^4-2$, because the roots are $\sqrt[4]{2},-\sqrt[4]{2},i\sqrt[4]{2},-i\sqrt[4]{2}$.
However, you don't need two polynomials in order to conclude that $\mathbb{Q}(i,\sqrt[4]{2})$ is a normal extension: one polynomial having this as the splitting field is sufficient.
For degree two polynomials in $\mathbb{Q}[x]$, it's always sufficient to add one root, in order to get the splitting field, because if $r$ is a root of $x^2+bx+c$, then the other root is $-b-r\in\mathbb{Q}(r)$.
Thus the splitting field of $x^2-2$ is $\mathbb{Q}(\sqrt{2})$, which is a subfield of $\mathbb{R}$. The splitting field of $x^2+2$ is $\mathbb{Q}(i\sqrt{2})$ which doesn't contain $\mathbb{Q}(\sqrt{2})$.
On the other hand, the splitting field of $x^4-4=(x^2-2)(x^2+2)$ must contain both $\sqrt{2}$ and $i\sqrt{2}$, so it also contains $i$. Therefore $\mathbb{Q}(i,\sqrt{2})$ is contained in the splitting field. Since every root of $x^4-4$ belongs to $\mathbb{Q}(i,\sqrt{2})$, we conclude that this is the splitting field, hence a normal extension.