TITLE: Is $\sqrt{1 + \sqrt{2}}$ a unit in some ring of algebraic integers?
QUESTION [7 upvotes]: Since $\sqrt{1 + \sqrt{2}}$ has a minimal polynomial $x^4 - 2x^2 - 1$, it seems to me like this number should be a unit in some ring of algebraic integers.
My first thought was that maybe it's a unit in the ring of algebraic integers of $\mathbb{Q}(\root 4 \of 2)$, but $$\frac{\sqrt{1 + \sqrt{2}}}{1 - \root 4 \of 2} = -\sqrt{17 + 12 \sqrt{2} + 2 \sqrt{140 + 99 \sqrt{2}}},$$ an algebraic integer of degree $8$. So much for that.
Next it occurs to me that maybe what I'm looking for is the ring of algebraic integers of $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$, which may be integrally closed but I can't say for sure.

REPLY [9 votes]: Letting $\alpha = \sqrt{1 + \sqrt{2}}$ and $K = \mathbb{Q}(\sqrt{1 + \sqrt{2}})$, then $\alpha$ is indeed a unit in the ring of integers $O_K$, and even in the ring $\mathbb{Z}[\alpha]$ (which may or may not be the full ring of integers).  One can see this from the minimal polynomial that you found:
$$
\alpha^4 - 2 \alpha^2 - 1 = 0 \implies 1 = (\alpha^3 - 2 \alpha)\alpha \implies \frac{1}{\alpha} = \alpha^3 - 2 \alpha \, .
$$
In fact, an element is a unit in the ring of algebraic integers iff the constant term of its minimal polynomial is a unit in $\mathbb{Z}$, i.e., $\pm 1$.  (Note that the constant term of the minimal polynomial is $\pm$ the field norm $N_{K/\mathbb{Q}}(\alpha)$.)