Do there exist elements in cyclotomic fields with certain absolute value? Consider the $m$-th cyclotomic field $K$ over $\mathbb Q$. For an integer $n\ge2$, we know it is possible that there does not exist integer $\alpha\in\mathcal O_K$ such that $\alpha\bar\alpha=n$. What if we let $\alpha\in K$? Does it hold for arbitrary $n$ and $m$?
I suppose this is not a hard problem, but I can't find references. Thank you in advance.
 A: Of course you cannot factor an arbitrary integer $n$ this way.
Write $(\alpha) = {\mathfrak a}/{\mathfrak b}$ for coprime integral ideals ${\mathfrak a}$ and ${\mathfrak b}$. Then $(\overline{\alpha}) = \overline{\mathfrak a}/\overline{\mathfrak b}$. Since $\alpha \overline{\alpha} = n $ is an integer,   ${\mathfrak b}$ must divide $\overline{\mathfrak a}$, say $\overline{\mathfrak a} = {\mathfrak b} {\mathfrak c}$, and then $(\alpha)(\overline{\alpha}) = {\mathfrak c} \overline{\mathfrak c}$. If $K$ has class number $1$, then ${\mathfrak c}$ is principal, hence there do not exist any integers $n$ that can be written as $\alpha \overline{\alpha}$ for proper fractional elements $\alpha$. In any case $n$ must split into two ideal factors in the maximal real subfield.
A more challenging question is whether the equation  $\alpha\overline{\alpha} = n$ is possible for $\alpha \in K$ when it is impossible in the ring of integers. We know that ${\mathfrak a} \sim {\mathfrak b}$, hence ${\mathfrak c} = \overline{\mathfrak a}/{\mathfrak b} \sim \overline{\mathfrak a}/{\mathfrak a} = {\mathfrak a}^{\sigma-1}$, where $\sigma$ denotes complex conjugation. This implies that ${\mathfrak c}^{1+\sigma} \sim 1$. Thus the ideal class of ${\mathfrak c}$ must be in the kernel of the relative norm to the maximal real subfield. This is a necessary condition; I do not know whether it is sufficient.
A: If $\overline{\alpha}$ means the complex conjugate then $\alpha\overline{\alpha}=n$ gives that
$$N_{\Bbb{Q}(\zeta_m)/\Bbb{Q}}(\alpha) =N_{\Bbb{Q}(\zeta_m+\zeta_m^{-1})/\Bbb{Q}}(n)=n^{\phi(m)/2}$$
If $n=p$ is an inert prime number (ie. it stays prime in $\Bbb{Q}(\zeta_m)$ which happens when $p$ has order $\phi(m)$ modulo $m$) then this is impossible,
Because any $\beta\in \Bbb{Q}(\zeta_m)^*$ is of the form $p^r u$ with $r\in \Bbb{Z}$ and $u$ a unit of the DVR $\Bbb{Z}[\zeta_n]_{(p)}$, the conjugates of $u$ are units too, so $$N_{\Bbb{Q}(\zeta_m)/\Bbb{Q}}(\beta)=p^{\phi(m) r} N_{\Bbb{Q}(\zeta_m)/Q}(u)\in p^{\phi(m) r} \Bbb{Z}_{(p)}$$
A: We can prove that if $(p)$ is prime in $\mathcal{O}_K$, there isn't an $\alpha \in K$ such that $|\alpha|^2=p$.
Fix $m$, and choose a prime $p$ such that the ideal $(p)$ is inert in $\mathbb{Z}[\zeta_m]. \ \ $ Thus $(p)$ is prime in $\mathbb{Z}[\zeta_m]. \ $ If there were $\alpha \in \mathbb{Q}(\zeta_m)$ such that $ \alpha \overline{\alpha}=p$, then we may choose $k \in \mathbb{N}$ minimal, such that $k\alpha \in \mathbb{Z}[\zeta_m]$. Set $\beta = k \alpha$, then $\beta \overline{\beta}=k^2p \in (p).$ But $\beta,\overline{\beta} \notin (p).$ Thus the ideal $(p)$ is not prime. This is a contradiction.
