Galois extension with $X^4-2aX^2+b\in\mathbb Q[X]$ 
Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result: 
  For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.

My thoughts so far:
The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:


*

*For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof.

*For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case?

 A: First assume that $\text{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$. Now the automorphisms are given by:
$$\text{Id}: \sqrt{a+\sqrt{a^2-b}} \to \sqrt{a+\sqrt{a^2-b}}$$
$$\sigma_1: \sqrt{a+\sqrt{a^2-b}} \to \sqrt{a-\sqrt{a^2-b}}$$
$$\sigma_2: \sqrt{a+\sqrt{a^2-b}} \to -\sqrt{a+\sqrt{a^2-b}}$$
$$\sigma_3: \sqrt{a+\sqrt{a^2-b}} \to -\sqrt{a-\sqrt{a^2-b}}$$
As the Galois group has no element of order $4$ we must have that all automorphisms are of order $2$. Now we have that: $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$. Apply $\sigma_1$ on both sides to get that:
$$\sigma_1(\sqrt{b}) = \sigma_1\left(\sqrt{a+\sqrt{a^{2}-b}}\right)\cdot\sigma_1\left(\sqrt{a-\sqrt{a^{2}-b}}\right)$$
$$=\sqrt{a-\sqrt{a^{2}-b}}\cdot\sqrt{a+\sqrt{a^{2}-b}} = \sqrt{b}$$
Similarly we have that $\sigma_3$ fixes $\sqrt{b}$. Hence we have that the subgroup of $\text{Gal}(L/\mathbb{Q})$ fixing $\sqrt{b}$ contains $\text{Id},\sigma_1$ and $\sigma_3$. So it must be the whole group. But then we get $\sqrt{b} \in \mathbb{Q} \subset{K}$.
Now consider the case when $\text{Gal}(L/\mathbb{Q}) \cong \mathbb{Z}_4$. Then it which has a unique quadratic field and so we must have that $\mathbb{Q}(\sqrt{b}) = \mathbb{Q}(\sqrt{a^2-b}) = K$. Hence $\sqrt{b} \in K$.
