Elements of absolute value 1 in cyclotomic extension of dyadic rationals 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 $k$?
 A: 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.
A: I believe that there are only the trivial solutions. Set $m=2^{k-2}$, let $\zeta$ be a primitive $4m$-th root of unity, and $u=\sum_{\nu=0}^{2m-1}a_\nu\zeta^\nu$ be an arbitrary element in $\mathbb Z[\zeta]$. The following seems to hold: If $u\bar u$ is a rational integer which in addition is divisible by $4$, then all $a_\nu$ are even. If that holds, then the claim follows.
The assertion about the $a_\nu$ being even seems to follow in two steps: Computing modulo $2$ seems to imply that the system of quadratic equations for the $a_\nu$ implies that $a_\nu-a_{m+\nu}$ is even for all $\nu$. Using this information, all the coefficients of the new system are even. Dividing by $2$, and then again computing modulo $2$, seems to apply that $a_\nu$ is even for $0\le\nu\le m-1$, and the claim follows.
I tried and verified that using Groebner bases for $k\le 5$. So I'm pretty sure that a clever manipulation of the equations will confirm this observation in general.
Note: For the first step, we need to know the following: Let $r$ be a power of $2$, and let $a_0,a_1,\dots,a_{r-1}$ be elements from $\mathbb F_2$ such that $\sum_{i=0}^{r-1}a_ia_{i+j}=0$ for all $j=0,1,\dots,r-1$ (indices taken modulo $r$). Then $a_i=a_{i+r/2}$ for all $i$. Strangely, this holds for all even $r\le26$ except for $r=14$. (I'm sure that this kind of circulant equation should have been studied somewhere.) 
