Finding the ring of integers of $\Bbb Q(\sqrt[4]{2})$

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 $\mathcal O_K = \mathbb Z[\sqrt[4]2]$.

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 $2^{-1}\mathfrak{p}^3$.