Does $\mathbb{Z}[\zeta_n]$ contain $|x|=1$ with $x$ not a power of $\zeta_n$?

Question

Let $\zeta_n=e^\frac{{2i\pi}}{n}$. Suppose $n$ is even. Then $\mathbb{Z}[\zeta_n]$ contains no roots of unity that are not powers of $\zeta_n$ because this is true of $\mathbb{Q}[\zeta_n]$ which is a way to show that these fields are non-isomorphic for different values of even $n$. If $n$ is odd, then $\mathbb{Z}[\zeta_n]=\mathbb{Z}[\zeta_{2n}]$. I was wondering if this extends to all $x$ with $|x|=1$. I know this is false for the fields. However, are there are any $x$ with $|x|=1$ in $\mathbb{Z}[\zeta_n]$ that are not $n$th roots of unity? Clearly there are none for $n=2,4,6$.

Answer 1

Let $K$ be an abelian Galois extension of $\mathbf Q$, such as a cyclotomic field. View it as a subfield of $\mathbf C$. Since $K/\mathbf Q$ is Galois and complex conjugation fixes $\mathbf Q$, complex conjugation is a field automorphism of $K$. It would be the identity automorphism if $K \subset \mathbf R$, but also your question trivially has the answer "no" for $K$ if it is a subfield of $\mathbf R$, so you could assume complex conjugation is not the identity on $K$ below. But I actually won't need to know whether or not complex conjugation is the identity on $K$ below.

Theorem. With notation as above, if $\alpha \in K$ has absolute value $1$ then every $\mathbf Q$-conjugate of $\alpha$ has absolute value $1$.

Proof. We assume $\alpha\overline{\alpha} = 1$. Let $\alpha' \in K$ be a $\mathbf Q$-conjugate of $\alpha$, so $\alpha' = \sigma(\alpha)$ for some $\sigma \in {\rm Gal}(K/\mathbf Q)$. Then $\overline{\alpha'} = \overline{\sigma(\alpha)} = \sigma(\overline{\alpha})$ since elements of ${\rm Gal}(K/\mathbf Q)$ commute. Then $$ |\alpha'|^2 = \alpha'\overline{\alpha'} = \sigma(\alpha)\sigma(\overline{\alpha}) = \sigma(\alpha\overline{\alpha}) = \sigma(1) = 1, $$ so $|\alpha'| = 1$. $\Box$

Note. The idea for the above argument comes from the answer by Cam McLeman here.

Corollary. With notation as above, an algebraic integer in $K$ with absolute value $1$ must be a root of unity.

Proof. If an algebraic integer in $K$ has absolute value $1$, then by the theorem all of its $\mathbf Q$-conjugates have absolute value $1$. A famous theorem of Kronecker says that an algebraic integer whose $\mathbf Q$-conjugates all have absolute value $1$ must be a root of unity. $\Box$