Let $u=\sqrt{5 + \sqrt2} \in \mathbb{C}$.
Hoping I have done well, are there errors in my considerations? A help will be well appreciated, thank you.
Let $f=\min_{\mathbb{Q}}(u) = x^4-10x^2+23$, the splitting field $\mathbb{F}$ for $f$ on $\mathbb{Q}$ is $\mathbb{F} = \mathbb{Q}[u,v]$, where $v = \sqrt{5 - \sqrt2}\,\,$. After proving that $\sqrt 2$, $\sqrt{23} \in \mathbb{F}$, which is quite easy, consider $\mathbb{K} = \mathbb{Q}[\sqrt 2]\,\,$ and $\,\,g =\min_{\mathbb{K}}(v) = x^2-5+\sqrt 2$.
How to prove that $u \notin \mathbb{K}[v]$?
In addition, $[\mathbb{F} : \mathbb{Q}] = [\mathbb{F} : \mathbb{Q}[u]][\mathbb{Q}[u] : \mathbb{Q}] = 8$, which is a Galois extension because it's finite, normal and separable; hence $[\mathbb{F} : \mathbb{Q}] = 8 = |G| :=|\mathrm{Gal}(\mathbb{F} | \mathbb{Q})|$. Now, $G$ acts transitively on the set $R_f=\{u, -u, v, -v\}\,$, therefore $G \simeq K \leqslant \mathrm{Sym}(R_f) \simeq S_4$ and the only subgroup of $S_4$ of order $8$ is $D_8$, so $G \simeq D_8$.
Finally, there are two sub-fields $\mathbb{L}$ of $\mathbb{F}$ of grade $4$ on $\mathbb{Q}$ s.t. $\mathbb{L} | \mathbb{Q}$ is normal, which are $\mathbb{K}[u]$ and $\mathbb{K}[v]$.
