I am studying field theory and I am trying to find the normal closure of $\mathbb{Q}(\sqrt{2 + \sqrt{2}})/\mathbb{Q}$.
I know that normal extension (i.e. an algebraic extension in which every irreducble polynomial having a root in the extension has all its roots in the extension) is the same thing as a splitting field of some set of polynomial over the base field.
Let $y_0 = \sqrt{2 + \sqrt{2}}$. Then $y_0^2 - 2 = \sqrt{2}$, and thus $y_0$ is a root of $f(y) := y^4 - 4y^2 + 4 - 2 y^4 - 4y^2 + 2 = 0$. Let $z := y^2$. Then by quadratic formula, $z = 2 \pm \sqrt{2}$. Since $z = y^2$, $\pm \sqrt{2 \pm \sqrt{2}}$ are all the roots of $f(y)$ (here, the signs of the two $\pm$ are unrelated). $f(y)$ is irreducible over $\mathbb{Z}[y]$ over Eisenstein, and thus is irreducible over $\mathbb{Q}[y]$ by Gauss's lemma. However, I am not sure whether, for example, $\sqrt{2 - \sqrt{2}} \in \mathbb{Q}(\sqrt{2 + \sqrt{2}})$ or not.