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.

  • $\begingroup$ The ring of algebraic integers of $\mathbb{Q}(\sqrt{1+\sqrt{2}})$ is integrally closed by definition: it is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{1+\sqrt{2}})$. $\endgroup$ – Viktor Vaughn Dec 5 '16 at 22:04
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    $\begingroup$ Note $\sqrt {\sqrt 2+1}\cdot \sqrt {\sqrt 2-1}$ for example $\endgroup$ – Mark Bennet Dec 5 '16 at 22:10
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    $\begingroup$ It’s a unit in every algebraic integer ring that contains it. $\endgroup$ – Lubin Dec 8 '16 at 20:48
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    $\begingroup$ A couple of random musings: first, $1 + \sqrt 2$ is a unit in $\mathbb Z[\sqrt 2]$; secondly, did you try taking the reciprocal of $\sqrt{1 + \sqrt 2}$? $\endgroup$ – Mr. Brooks Dec 8 '16 at 22:19

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 $N_{K/\mathbb{Q}}(\alpha)$.)


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