Cyclic quartic field from a nested quadratic radical ($\mathbb{Q}(\sqrt{p+\sqrt{p}}),p=n^2+1\text{ prime}$)

For a prime $p=n^2+1$, this real number field is a Galois extension of degree four with cyclic Galois group. If $\theta =\sqrt{p+\sqrt{p}}$, its minimal polynomial of an algebraic element is $X^4-2p{X}^2+p^2-p$, irreducible by Eisenstein's criterion at $p$. The other positive root $\beta =\sqrt{p-\sqrt{p}}$ belongs to the same field because $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$. Thus all four roots lie in the field, proving normality. The Galois automorphism $\theta \mapsto \beta$ sends $\sqrt{p}$ to its negative and consequently sends $\beta$ to $-\theta$. It has order four, so the group is $C_4$, rather than a product of two order-two groups.

The prime condition is used for irreducibility, while the square condition on $p-1$ supplies the missing conjugate inside the field. The example $p=2$ yields $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, the real subfield of the sixteenth cyclotomic field.