Let $\zeta$ be a primitive $2^k$th root of unity, and consider the ring $ \mathbb{Z}[\zeta, \frac{1}{2}] \subset \mathbb{C}$. Are there any elements of absolute value 1 other than the powers of $\zeta$?


Note that if $\frac{1}{2}$ were replaced by $\frac{1}{5}$, then the number $\frac{4}{5} + \frac{3}{5} i$ would be an example.


I can prove that the answer is 'no' for $k \leq 3$. In particular, for $k = 3$ we want to find integers $a, b, c, d$ and positive integer $n$ such that:


$$ (a + b \zeta + c \zeta^2 + d \zeta^3)(a + b \zeta^{-1} + c \zeta^{-2} + d \zeta^{-3}) = 4^n $$


Expanding this yields:


$$ a^2 + b^2 + c^2 + d^2 + (ab + bc + cd - da) \sqrt{2} = 4^n $$


By Jacobi's four-square theorem, either all but one of $a,b,c,d$ is zero (and we get a cyclotomic unit) or $a^2 = b^2 = c^2 = d^2 = 4^{n-1}$. In this case, $|ab| = |bc| = |cd| = |da|$ and an even number of them are negative. It follows that the coefficient of $\sqrt{2}$ is nonzero, whence we derive a contradiction.


Does this result hold for all $k$?



Elements of absolute value 1 in cyclotomic extension of dyadic rationals
The answer is yes. Let $K=\mathbb{Q}(\zeta)$ and suppose that there is an element $u\in\mathbb{Z}[\frac{1}{2},\zeta]$ such that for some embedding $\sigma\_0\colon K\hookrightarrow \mathbb{C}$ the absolute value of $\sigma\_0(u)$ is $1$. Then I claim that $u\in\mathbb{Z}[\frac{1}{2},\zeta]^\times$ and that for all embeddings $\sigma\colon K\hookrightarrow\mathbb{C}$ we have $\Vert\sigma(u)\Vert=1$. Indeed, $K$ is a $CM$ field, so there exists an automorphism $c\colon K\to K$ such that for all embeddings $\sigma$, the composition $\sigma\circ c$ equals $\overline{\sigma}$, or in other words $\sigma\bigl(c(x)\bigr)=\overline{\sigma(x)}$ for all $x\in K$. Thus the equation $\Vert\sigma\_0(u)\Vert=1$ can be written $\sigma\_0(u)\sigma\_0(cu)=1$ and as $\sigma\_0$ is injective this forces $cu=u^{-1}$. This proves my two claims at once, using that $c\bigl(\mathbb{Z}[\frac{1}{2},\zeta]\bigr)=\mathbb{Z}[\frac{1}{2},\zeta]$.


Now, set $\pi=(\zeta-1)$ which generates the *prime* ideal of $\mathbb{Z}[\zeta]$ above $2$ (in the sense that $2\mathbb{Z}[\zeta]$ is the principal ideal generated by $(\pi)^d$ with $d=\dim\_{\mathbb{Q}}(K)$). This $\pi$ verifies
$$
c\pi=\frac{1}{\zeta}-1=\frac{1-\zeta}{\zeta}=\xi\pi
$$
where $\xi=-\zeta^{-1}$ is again a root of unity. Our element $u$ can be written as $u=\alpha/\pi^n$ for some $\alpha\in\mathbb{Z}[\zeta]$ and $n\in\mathbb{N}$: we can also assume that $n$ is minimal, which amounts to saying that $\alpha\notin(\pi)$. Then
$$
1=u\cdot cu=\frac{\alpha}{\pi^n}\frac{c\alpha}{\pi^n\xi^n}=\frac{\alpha\cdot c\alpha\xi^{-1}}{\pi^{2n}}
$$
and this forces $n=0$ since the ideals $(\pi)$ and $(c\pi)$ coincide and hence $\alpha\cdot c\alpha\notin\pi$. But then $u\in\mathbb{Z}[\zeta]$, hence it is an *algebraic integer* all of whose conjugate lie on the unit circle. It is therefore a roots of unity, by [Kronecker's theorem.](https://mathoverflow.net/questions/33169/good-effective-version-of-kroneckers-theorem)


