Units of $\mathbb{Z}[\sqrt[4]2]$

Question

How would one compute the units in $\mathbb{Z}[\sqrt[4]2]$? According to one source, it can be shown that the fundamental units are $1 + \sqrt[4]2$ and $1 + \sqrt{2}$, but it does not specify the proof of this.

Answer 1

Here is an argument to at least prove that $1+\sqrt{2}$ is not the power of any unit in $\mathbb Z[2^{1/4}].$

Suppose that there were a unit $u \in \mathbb Z[2^{1/4}]$ for which $u^2 = 1+\sqrt{2}$. Since $1+\sqrt{2}$ is a fundamental unit of $\mathbb Z[\sqrt{2}],$ we see that $u \not\in \mathbb Q(\sqrt{2})$, and so $\mathbb Q(2^{1/4}) = \mathbb Q(\sqrt{2})(u^{1/2}).$

But a local calculation at $2$ will show that the power of $2$ dividing the discriminant of $\mathbb Z_2[2^{1/4}]$ is greater than that dividing the discriminant of $\mathbb Z_2[\sqrt{2},u^{1/2}]$ for any unit $u \in \mathbb Z_2[\sqrt{2}]^{\times}$.

Thus the claimed equality can't hold, and so $u$ must not exist.