I'm solving problems from my Galois Theory course, and want to check if my solution to this one is correct. It says:
Find the set of automorphisms $Aut_{\mathbb Q}(L)$, where $L$ is splitting field of the polynomial $\text{Irr}_\alpha^{\mathbb Q}(X)$, being $\alpha=\sqrt{\sqrt2 - 1}$
The notation $\text{Irr}_\alpha^{\mathbb Q}(X)$ means the minimal polynomial of $\alpha$ over $\mathbb Q$. Here's my solution:
First, I found that $\text{Irr}_\alpha^{\mathbb Q}(X)=X^4+2X^2-1$, and prove it's irreducible (this part was easy). After that, I find that the roots of this polynomial are: $\pm\sqrt{\sqrt2 - 1}$, $\pm\sqrt{-\sqrt2 - 1}$, so I can conlcude that $$\mathbb{Q}\left(\sqrt{\sqrt2 - 1}, \sqrt{-\sqrt2 - 1}\right)$$ is splitting field of $\text{Irr}_\alpha^{\mathbb Q}(X)$ over $\mathbb{Q}$. I changed a bit the expressions to make it simpler to write, obtaining the splitting field $$\mathbb{Q}\left(\sqrt{\sqrt2 - 1}, i\sqrt 3 \right) = L.$$ Now I notice the extension $L/\mathbb Q$ has degree $8$, because $\mathbb Q(\sqrt{\sqrt2 - 1})/\mathbb Q$ has degree $4$ (due to the minimal polynomial) and the extension $L/\mathbb{Q}(\sqrt{\sqrt2 - 1})$ has degree $2$ (because it's either $1$ or $2$ for $\mathbb Q(i\sqrt3)/\mathbb Q$ having degree $2$, and it's not $1$ because $i\sqrt3\notin \mathbb Q(\sqrt{\sqrt2 - 1})$ since it's imaginary).
Now, if I'm not wrong, this $L$ is splitting field over $\mathbb Q$ not only of $X^4+2X^2-1$, but also of $(X^4+2X^2-1)(X^2+3)$, because $\text{Irr}_{i\sqrt3}^\mathbb{Q(\alpha)}(X)=\text{Irr}_{i\sqrt3}^\mathbb{Q}(X)=X^3+3$. Is this correct? If this was true, then I just have to check the different possibilites for the images under my possible automorphisms of $\sqrt{\sqrt2-1}$ and $i\sqrt3$, wich will be roots of $X^4+2X^2-1$ and$X^2+3$ respectively, giving 8 different possible asignations. Since the polynomial $(X^4+2X^2-1)(X^2+3)$ has no multiple roots, and the degree of $L/\mathbb{Q}$ is $8$, the it's followed that my possible automorphisms are: $$\sigma(\lambda)=\lambda, \quad \forall\lambda\in\mathbb Q$$ $$\sigma\left(\sqrt{\sqrt2-1}\right) \in \left\{ \pm\sqrt{\sqrt2-1}, \pm\sqrt{-\sqrt2-1} \right\} $$ $$\sigma\left(i\sqrt3\right)\in\left\{ \pm i \sqrt3 \right\}$$ And these different combinations make the 8 elements of $Aut_{\mathbb Q}(L)$.
Is my solution correct? If not, where did I go wrong? Is there anything I should explain better? Any help will be appreciated, thanks in advance.