3 Thank D.Speyer for link to math.se/39856 with alternative proof via Kronecker instead of Dirichlet

A.Quas already noted in his comment that ${\bf Z}[\omega_l]$ might contain a $2l$-th root of unity (he gave $l=3$ but even $l=1$ works...). But that's the only possibility: we show that the only algebraic integers $z$ in ${\bf Q}(\omega_l)$ satisfying $|z| = 1$ are the roots of unity in ${\bf Q}(\omega_l)$. More generally, if $K_+$ is any totally real field, and $K$ is a totally imaginary quadratic extension of $K_+$, then the only algebraic integers in $K$ satisfying $|z|=1$ are the roots of unity in $K$.

This result is surely well-known/standard, but it's easier to recite or reconstruct a proof than to track down a reference. Let $d = [K_+:{\bf Q}]$, so that $2d = [K:{\bf Q}]$; for example $d = \varphi(l)/2$ when $K = {\bf Q}[\omega_l]$. Now $|z|=1$ iff $z$ is in the kernel of the map ${\rm Nm} : z \mapsto z \bar z$ from $K^*$ to $K_+^*$. In our setting $z$ is assumed to be an algebraic integer in the kernel, so it is in the unit group of [the algebraic integers in] $K$, with inverse $\bar z$. But by the Dirichlet unit theorem both $K$ and $K_+$ have unit groups of rank $d-1$. Moreover the image of ${\rm Nm}$ has rank at least $d-1$ because it contains the squares of all units in $K_+$. Hence its rank is exactly $d-1$, and the kernel of ${\rm Nm}$ has rank zero, and therefore consists only of roots of unity, as desired.

EDIT: Thanks to David Speyer for the link to his answer to the same question on Stackexchange (http://math.stackexchange.com/questions/39856), using an alternative route via Kronecker's theorem (an algebraic integer $z$ is a root of unity iff every Galois conjugate of $z$ has absolute value $1$ in ${\bf C}$) instead of the Dirichlet unit theorem. This method too works in the general "CM" setting of a totally imaginary quadratic extension of a totally real field.

2 Generalization to arbitrary CM fields

A.Quas already noted in his comment that ${\bf Z}[\omega_l]$ might contain a $2l$-th root of unity (he gave $l=3$ but even $l=1$ works...). But that's the only possibility: we show that the only algebraic integers $z$ in ${\bf Q}(\omega_l)$ satisfying $|z| = 1$ are the roots of unity in ${\bf Q}(\omega_l)$. More generally, if $K_+$ is any totally real field, and $K$ is a totally imaginary quadratic extension of $K_+$, then the only algebraic integers in $K$ satisfying $|z|=1$ are the roots of unity in $K$.

This result is surely well-known/standard, but it's easier to recite or reconstruct a proof than to track down a reference. Let $d = \phi(l)/2$ [K_+:{\bf Q}]$, so that$2d = [K:{\bf Q}]$; for example$d = \varphi(l)/2$when$K := {\bf Q}[\omega_l]$has degree$2d$over${\bf Q}$. Then the real subfield$K_+$has degree$d$. Now$|z|=1$iff$z$is in the kernel of the map$N {\rm Nm} : z \mapsto z \bar z$from$K^*$to$K_+^*$. In our setting$z$is also assumed to be an algebraic integer in the kernel, so a it is in the unit group of [the algebraic integers in]$K$. K$, with inverse $\bar z$. But by the Dirichlet unit theorem , both $K$ and $K_+$ have unit groups of rank $d-1$. Moreover the image of $N$ {\rm Nm}$has rank at least$d-1$because it contains the squares of all units in$K_+$. Hence its rank is exactly$d-1$, and the kernel of${\rm Nm}$has rank zero, and therefore consists only of roots of unity, as desired. 1 A.Quas already noted in his comment that${\bf Z}[\omega_l]$might contain a$2l$-th root of unity (he gave$l=3$but even$l=1$works...). But that's the only possibility. Let$d = \phi(l)/2$so that$K := {\bf Q}[\omega_l]$has degree$2d$over${\bf Q}$. Then the real subfield$K_+$has degree$d$. Now$|z|=1$iff$z$is in the kernel of the map$N : z \mapsto z \bar z$from$K^*$to$K_+^*$. In our setting$z$is also an algebraic integer, so a unit of$K$. But by the Dirichlet unit theorem, both$K$and$K_+$have unit groups of rank$d-1$. Moreover the image of$N$has rank$d-1$because it contains the squares of all units in$K_+\$. Hence the kernel has rank zero, and consists only of roots of unity, as desired.