Determine the Galois group of $X^4 - 2X^2 +2$ over $\Bbb Q$
EDIT : I want to address the particular polynomial and not the full general case. In order to solve this problem, I want to have a self-contained proof/answer and am not willing to reproduce the general argument when even it might not be required. If you can provide help overcoming this simple obstacle then please help, instead of just lazily labeling the question as duplicate !
The result that is being linked\suggested, it boils down to the same problem whether $\sqrt 2$ is in the field $\Bbb Q(\alpha)$ or not, so it's pointless to refer the link just by merely looking at the form of the polynomial!
Mt attempt :
Obtained roots of $X^4 - 2X^2 +2$ to be : $\pm \frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} \pm i \sqrt{\sqrt 2 -1})$ . Then considered $\Bbb Q(\alpha)$ where $\alpha=\frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} + i \sqrt{\sqrt 2 -1})$ .
So I need to show whether $\frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} - i \sqrt{\sqrt 2 -1}) \in \Bbb Q(\alpha)$ or not (then the Galois group will be of order 4 or 8 respectively) .
I have observed that $\sqrt 2 \in \Bbb Q(\alpha) \implies $ $\frac{1}{\sqrt 2}(\sqrt{1+\sqrt 2} - i \sqrt{\sqrt 2 -1}) \in \Bbb Q(\alpha)$
So, (1) How to show (if it is true) $\sqrt 2 \in \Bbb Q(\alpha)$ ?
(2) How to obtain the Galois group ? If (1) is true I was tempting to use computational tricks like the ones exhibited in this answer
Thanks in Advance for help!
A question has been asked on this polynomial before, but the OP was confused himself , I could not follow his arguments and most importantly the question did not had any answer.
Let $K=\Bbb Q(\alpha)$ be the field generated by the one root $\alpha\in\Bbb C$ specified in the OP.
Here is a pedestrian proof for $\sqrt 2\not \in K$. Assume for this that we have a tower of fields $\Bbb Q\subset \Bbb Q[\sqrt 2]=k\subset \Bbb Q[\alpha]=K$. (So $K:k$ is an extension of degree two, it is Galois.) Then the polynomial $X^4-2X^2+2$, seen as a polynomial in $k[X]$ splits as $$ X^4-2X^2+2 =\Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)\cdot\Big(\dots\Big)\ , $$ with $a,b,c,d\in\Bbb Q$.
The factor with the root $\alpha$ (and its $(K:k)$-conjugate) was only shown. (The other factor does not have the root $\alpha$, so it is a different factor.) Consier the above relation as a product of polynomials in $k[X]$. Then we also have its conjugate polynomial as a factor. So from $$ \begin{aligned} X^4-2X^2+2 & =\Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)\cdot\Big(\dots\Big) \\ & =\Big(X^2+(a-b\sqrt 2)X+(c-d\sqrt 2)\Big)\cdot\overline{\Big(\dots\Big)} \end{aligned} $$ we get (second factor is different, unique factorization) $$ \begin{aligned} X^4-2X^2+2 =&\ \Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big) \\ \cdot&\ \Big(X^2+(a-b\sqrt 2)X+(c-d\sqrt 2)\Big)\ , \end{aligned} $$ then we multiply, look at the term in $X^3$, get $a=0$, then at the term in $X^1$, get $d=0$, so our factorization is $$ X^4-2X^2+2 = \Big(X^2+b\sqrt 2 X + c\Big) \Big(X^2-b\sqrt 2 X + c\Big)\ , $$ so $c^2=2$. Contradiction.
This negates (1) in the OP, so (2) is out of charge.
To have a quick structure description of the Galois closure $L$ of $K=\Bbb Q(\alpha)$, let us start a parallel construction. Let $u=\zeta_8$ be the (cyclotomic) unit with minimal polynomial $$ X^4+1 $$ over $\Bbb Q$, and to fix ideas, we will fix it with the complex image $\frac 1{\sqrt 2}(1+i)$ in the first quadrant. Its conjugates are $u, u^3, u^5, u^7$, with complex images $\frac 1{\sqrt 2}(\pm 1\pm i)$. It is clear that $$ M=\Bbb Q(u)=\Bbb Q(\zeta_8)=\Bbb Q(\sqrt{-1},\sqrt 2) $$ is (the cyclotomic field of order $8$, degree $4$ over $\Bbb Q$,) Galois over $\Bbb Q$ with Galois morphisms $\phi_j$ determined by $u\to\phi_j u=u^j$. So $\phi_5$ corresponds to $u\to u^5=-u$, and $\phi_7$ to the complex conjugation $u\to u^7=\frac 1u=\bar u$, they commute, the structure of $\operatorname {Gal}(M:\Bbb Q)$ is $\Bbb Z/2\times \Bbb Z/2$.
($\sqrt2$ is in $M$: $\zeta_8^2=\zeta_4=i$, so $\Bbb Q[i]$ is a subfield, but then from $\zeta_8=(1+i)/\sqrt 2$ also $\Bbb Q[\sqrt 2]$ is a subfield, and by dimension reasons we have $M= \Bbb Q[\zeta_8]=\Bbb Q[\sqrt{-1},\sqrt 2]$. Then $\phi_5$ maps $i=u^2\to (-u)^2=u^2=i$, and thus $\sqrt 2=(1+i)/\zeta_8\to -\sqrt 2$. And $\phi_7$, the complex conjugation, does the other job, $i\to -i$, $\sqrt 2\to \sqrt 2$.)
We have now the "hint", that may be useful below, $$ X^4-2X^2+2 = (X^2-1)^2-(\sqrt{-1})^2=(X^2-(1+u^2))(X^2-(1-u^2))\ . $$ So we consider $N=M(a)=\Bbb Q(u,a)$, the quadratic extension of $M$ that introduces $a=\alpha=\sqrt{1+u^2}=\sqrt{1+\sqrt{-1}}$ as a root of the first factor. All roots of $X^4-2X^2+2$ are then $$ a,\ -a,\ u^3a,\ -u^3a\ . $$ We have a Galois substitution $\Psi:N\to N$ (over $M$, thus also over $\Bbb Q$), determined algebraically by $\Psi:(u\to u,\ a\to-a)$, and we lift $\phi_j$ as Galois morphisms $\Phi_j$ of $N$ as follows: $$ \begin{aligned} \Phi_5:&\ u\to\phi_5u=u^5=-u\ ,& (X^2-(1+u^2))&\text{ stays, so we can take } &\ a&\to a\ ,\\ \Phi_7:&\ u\to\phi_7u=u^7=\frac 1u=\bar u\ ,& (X^2-(1+u^2))&\text{ changes, so chose } & a&\to -u^3a\ .\\ &\qquad\text{Recall also:} \\ \Psi:&\ u\to u\ , && & a&\to -a\ . \end{aligned} $$ ($\Phi_7$ is well defined, in the relation $a^2-(1+u^2)=0$ the LHS is moved to $(-u^3a)^2-(1+u^6)=u^6a^2-1-u^6=u^6(a^2-1) -1 = (-i)(1+i-1)-1=0$. (Use $X$ instead of $a$ for a pedant factorization.)
In particular $$ \boxed a \overset{\Phi_7}{\longrightarrow} \boxed {-u^3a} \overset{\Phi_7}{\longrightarrow} - u^{21}\; u^3a= \boxed{-a} \overset{\Phi_7}{\longrightarrow} \boxed {u^3a} \ , $$ so $\Phi_7$ permutes as a rotation the vertices of the square with the vertices marked as $a,-u^3a,-a,u^3 a$ in this cyclic order. Then $\Psi$ is the reflection w.r.t. the "virtual center" of the square, and $\Phi_5$ acts as the transposition of $\pm u^3 a$. (Symmetry w.r.t. the "virtula line" through $a,-a$, which are invariated.)
So $\Phi_7$ generates a Galois subgroup $\cong\Bbb Z/4$ and $\Phi_5$ implements / acts as a "twist" of it, the relation $$ \Phi_5\Phi_7\Phi_5=\Phi_7^{-1} $$ can be tested either geometrically (in the virtual picture) or algebraically on the generators, e.g. $$ \begin{aligned} \left( \frac1u\leftarrow \frac1{-u}\leftarrow -u\leftarrow u\right) &= \left( \frac1u\leftarrow u\right) \ , \\ (u^3 a\leftarrow -u^3a \leftarrow a \leftarrow a) &= (u^3 a\leftarrow a)\ . \end{aligned} $$
It is a good idea to investigate the situation using computer power. Here sage. The following code gives information on the given situation.