= Cyclic quartic field from a nested quadratic radical
{title2=$\mathbb Q(\sqrt{p+\sqrt p}),\quad p=n^2+1\text{ prime}$}
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 $X^4-2pX^2+p^2-p$, 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 $\mathbb Q(\sqrt{2+\sqrt2})$, the real subfield of the sixteenth <cyclotomic field>.
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