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I know$^{(1)}$ that the ring of integers of $K=\Bbb Q(\sqrt[4]{2})$ is $\Bbb Z[\sqrt[4]{2}]$ and I would like to prove it.

A related question is this one, but it doesn't answer mine. I computed quickly the discriminant $\text{disc}(1,\sqrt[4]{2},\sqrt[4]{4},\sqrt[4]{8})=-2^{11}$. According to this answer, this means that $\mathcal{O}_K \subset \frac{1}{m}\Bbb Z\left[\sqrt[4]{2}\right]$ where $m$ is an integer whose square divides $2^{11}$, so $m=1,2,2^2,\dots,2^5$ are possible. But how could I rule out the values $m>1$?

I am aware that it can be a tricky problem. Any reference providing a description of $\mathcal{O}_{\Bbb Q(\sqrt[4]{2})}$ would be satisfactory. I will be grateful for any help!

$^{(1)}$ I tested with SAGE the following code

K.<a> = NumberField([x^4-2]);
K.integral_basis()

and I got the expected answer, namely $[1,a,a^2,a^3]$.

Community
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Watson
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  • Possibly related: http://math.stackexchange.com/questions/99913 – Watson Jul 23 '16 at 14:51
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    T Funakura completely determined integral basis of $\mathbb{Q}(\sqrt[4]{m})$. Here is the link: http://ousar.lib.okayama-u.ac.jp/en/33322 – Sungjin Kim Jul 23 '16 at 23:35
  • Note: I learned from https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf that it is not true that the ring of integers of $\Bbb Q(\sqrt[n]{2})$ is equal to $\Bbb Z[\sqrt[n]{2}]$ in general! This is true for $n=2, 3, 4$ and in fact for all $n \leq 1000$, but not for $n=1093$... ! – Watson Jun 04 '21 at 09:48

2 Answers2

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Following the approach of Keith Conrad, suppose that $$\alpha = a + b \sqrt[4]2+c\sqrt[4]4+d\sqrt[4]8,\quad a,b,c,d\in\mathbb Q$$ is an element of $\mathcal O_K$. We will show that $\alpha\in\mathbb Z[\sqrt[4]2]$. Calculating traces, $$ \mathrm{Tr}_{K/\mathbb Q}(\alpha) = 4a\\ \mathrm{Tr}_{K/\mathbb Q}(\sqrt[4]2\alpha) = 8d\\ \mathrm{Tr}_{K/\mathbb Q}(\sqrt[4]4\alpha) = 8c\\ \mathrm{Tr}_{K/\mathbb Q}(\sqrt[4]8\alpha) = 8b $$ are all integers, and therefore, the denominators of $a,b,c$ and $d$ can only involve powers of $2$.

This enables us to solve our problem $2$-adically - indeed, it suffices to show that $\mathcal O_{\mathbb Q_2(\sqrt[4]2)} = \mathbb Z_2[\sqrt[4]2]$, since if $\alpha=\frac{1}{2^k}\alpha'$, where $\alpha'\in\mathbb Z[\sqrt[4]2]$, then $\alpha$ can only be an element of $\mathbb Z_2[\sqrt[4]2]$ if $k\le 0$.

But $\mathbb Q_2(\sqrt[4]2)$ is totally ramified with uniformiser $\sqrt[4]2$ (by observation, or since $X^4-2$ is Eisenstein at $2$), so it follows by Lemma $1$ in Conrad's notes that $\mathcal O_{\mathbb Q_2(\sqrt[4]2)} = \mathbb Z_2[\sqrt[4]2]$. Hence $\mathcal O_K = \mathbb Z[\sqrt[4]2]$.

Mathmo123
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  • For the prime $2$ being totally ramified in $\mathbb{Q}(\sqrt[4]{2})$, we can use $(\sqrt[4]{2}) ^4 = (2)$. So, I think we can avoid using Eisenstein. – Sungjin Kim Jul 24 '16 at 02:32
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    @i707107 you're right. I guess I'm not really using the lemma in Conrad's notes. Rather I'm using the fact that if $\alpha$ is a root of a polynomial that is Eisenstein at $p$, then $\alpha$ generates the $p$-adic ring of integers. This is the difficult part. I've edited my answer to be more consistent. – Mathmo123 Jul 24 '16 at 09:28
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There is a quick argument available in this case. Write $\alpha=\sqrt[4]{2}$ and $K=\mathbb{Q}(\alpha)$.

A prime number $p$ divides the index $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ if and only if there is a non-invertible prime $\mathfrak{p}$ of $\mathbb{Z}[\alpha]$ lying over $p$. In this case we have $$ -2^{11}=\Delta(\mathbb{Z}[\alpha])=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\cdot\Delta(\mathcal{O}_K), $$ so to prove that $[\mathcal{O}_K:\mathbb{Z}[\alpha]]=1$, it suffices to show that every prime $\mathfrak{p}$ of $\mathbb{Z}[\alpha]$ lying over $2$ is invertible.

If $\mathfrak{p}$ is such a prime, we have $\alpha\in\mathfrak{p}$ because $\alpha^4=2\in\mathfrak{p}$. As $(\alpha)$ is already a prime ideal (note that $\mathbb{Z}[\alpha]/(\alpha)\cong\mathbb{F}_2$), we have $(\alpha)=\mathfrak{p}$. But then the identity $(2)=\mathfrak{p}^4$ proves that $\mathfrak{p}$ is invertible; its inverse is the fractional ideal $2^{-1}\mathfrak{p}^3$.

Tim.ev
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