Integral basis of $\mathbb{Q}(\sqrt{2+\sqrt{2}})$?

Question

So far, I know the Galois group is $C_4$, the conjugates are $\sqrt{2+\sqrt{2}}$, $-\sqrt{2+\sqrt{2}}$, $\sqrt{2-\sqrt{2}}$, $-\sqrt{2-\sqrt{2}}$. At first I thought these would form the integral basis, but then I realised it's not that simple, and I'm not sure how to proceed.

Answer 1

Let $a = \sqrt{2+\sqrt2}$, so $(a^2 - 2)^2 - 2 = 0$, i.e. $a^4 - 4a^2 + 2 = 0$. Also let $K = \Bbb Q(a)$.

This polynomial is Eisenstein at $2$, so the ring of integers of $K_2$ is $\Bbb Z_2[a]$, i.e. $\{1, a, a^2, a^3\}$ is a basis. I'm not sure if this helps.

The discriminant of $\{1,a,a^2,a^3\}$ is $N(f'(a))$ $=$ $N(4a^3-8a)$ $=$ $2^8 N(a) N(a^2 - 2) = 2^8 \times 2 \times N(\sqrt2)$ $=$ $2^8 \times 2 \times 2^2 = 2^{11}$. This means $[\mathcal O_K : \Bbb Z[a]]$ is a power of $2$, so for every $x \in \mathcal O_K$ there is $n$ such that $2^n x \in \Bbb Z[a]$.

However, we know that $\mathcal O_{K_2} = \Bbb Z_2[a]$, so if $x \in \mathcal O_K$, then $x \in \Bbb Z_2[a]$, so $2^n x \in 2^n \Bbb Z_2[a] \cap \Bbb Z[a] = 2^n \Bbb Z[a]$, so $x \in \Bbb Z[a]$. Therefore, $\mathcal O_K = \Bbb Z[a]$.

PS: The equality $2^n \Bbb Z_2[a] \cap \Bbb Z[a] = 2^n \Bbb Z[a]$ is because: $$\begin{array}{cl} & 2^n \Bbb Z_2[a] \cap \Bbb Z[a] \\ =& 2^n \Bbb Z[a]_{(a)} \cap \bigcap_{\mathfrak p \ne (a)} \Bbb Z[a]_\mathfrak p \\ =& 2^n \Bbb Z[a]_{(a)} \cap \bigcap_{\mathfrak p \ne (a)} 2^n \Bbb Z[a]_\mathfrak p \\ =& 2^n \bigcap_\mathfrak p \Bbb Z[a]_\mathfrak p \\ =& 2^n \Bbb Z[a] \end{array}$$