Let $L$ be the splitting field of $x^4-2x^2-\sqrt2 \in \mathbb{Q(\sqrt2)}[x]$.
Definition:
A $K$-automorphism $\sigma:L \rightarrow L$ is a morphism such that $\sigma_{|K}=\text{id}_K$, where $K \leq L$ are fields.
How do I determine the automorphism group of $L$ fixing $\mathbb{Q}(\sqrt2)$?
My attempt:
The solutions to the equations are:
$\alpha_1=\sqrt{1-\sqrt{1+\sqrt{2}}}$
$\alpha_2=-\sqrt{1-\sqrt{1+\sqrt{2}}}$
$\alpha_3=-i\sqrt{-1+\sqrt{1+\sqrt{2}}}$
$\alpha_4=i\sqrt{-1+\sqrt{1+\sqrt{2}}}$
As far as I did understand, for a map $\sigma:L \rightarrow L$ to be a $\mathbb{Q}(\sqrt2)$-Automorphisms we need that $\sigma_{|\mathbb{Q}(\sqrt2)}$=id
This means that the only mappings possible are permutations of $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$
This means that there exist $4!=24$ Automorphisms.
Is my attempt correct?
Best Answer
Although there are $24$ permutations of the roots, not all permutations induce an automorphism fixing $\mathbb{Q}(\sqrt{2})$. If so, then the automorphism group of any separable splitting field would simply be $S_n$.
For example, consider the cyclic permutation $(\alpha,-\alpha,\beta)$. This cannot induce an automorphism $\sigma$ fixing $\mathbb{Q}(\sqrt{2})$. If that were to be the case, then the following relation $$ \alpha+(-\alpha)=0 $$ would become $$ \begin{align*} \sigma(\alpha)+\sigma(-\alpha) &=\sigma(0) \\ -\alpha+\beta &= 0 \end{align*} $$ which is false. This happens when the roots have some "unexpected" symmetry among one another (in this case if $\alpha$ is a root, then $-\alpha$ is a root).
First, your $\alpha_1,\alpha_2$ are calculated incorrectly.$\newcommand{\rat}{\mathbb{Q}}$ The four roots should be $$ \begin{align*} \alpha &= \sqrt{1+\sqrt{1+\sqrt{2}}} \\ \beta &= i\sqrt{-1+\sqrt{1+\sqrt{2}}} \end{align*} $$ and $-\alpha, -\beta$. Then the splitting field is $L=\rat(\alpha,\beta)$. Note that $\alpha\beta=i2^{1/4}$, so $L=\rat(\alpha,i2^{1/4})$.
Now, $\rat(\alpha)\subset\mathbb{R}$, so $\rat(\alpha)\ne L$ and $[L:\rat(\alpha)]\ge 2$. Also, the quadratic polynomial $x^2+\sqrt{2}\in\rat(\alpha)[x]$ vanishes $i{2^{1/4}}$, so $[L:\rat(\alpha)]=2$. This shows that $[L:\rat(\sqrt{2})]=2\cdot 4=8$.$\newcommand{\g}{\gamma}\newcommand{\a}{\alpha}$
Let $\g=i2^{1/4}$. Let $\sigma\in\text{Aut}(L/\rat(\sqrt{2}))$. Then $\sigma$ is completely determined by its images of $\a$ and $\g$. The automorphism $\sigma$ must map each element to its conjugate over $\rat(\sqrt{2})$. The $\rat(\sqrt{2})$-conjugates of $\a$ are $\a, -\a, \g/\a, -\g/\a$, and $\rat(\sqrt{2})$-conjugates of $\g$ are $\g,-\g$. The $8$ automorphisms induced by them account for all $8$ elements in $\text{Aut}(L/\rat(\sqrt{2}))$.$\newcommand{\u}{\mu}\newcommand{\t}{\tau}$
Let $\u$ be the automorphism induced by $\a\mapsto \g/\a$ and $\g\mapsto -\g$. Let $\t$ be the automorphism induced by $\a\mapsto\a$ and $\g\mapsto -\g$. We have the following multiplication table: $$ \begin{array}{c|cccccccc} & \text{id} & \u & \u^2 & \u^3 & \t & \t\u & \t\u^2 & \t\u^3\\ \hline \a & \a & \g/\a & -\a & -\g/\a & \a & -\g/\a & -\a & \g/\a \\ \g & \g & -\g & \g & -\g & -\g & \g & -\g & \g \end{array} $$ We see that $\u$ has order $4$ and $\t$ has order $2$, and $\u\t=\t\u^{-1}$. Thus, $\text{Aut}(L/\rat(\sqrt{2}))\cong D_8$, the dihedral group of order $8$.