Let $\zeta_p$ be a primitive $p$-th root of unity, and $f\in\Bbb Z[X]$. If $|f(\zeta_p)|=1,$ is $f(\zeta_p)$ a root of unity?

Question

Let $p$ be any prime. Let $K=\mathbb{Q}(\zeta_p)$, where $\zeta_p$ is primitive $p$-th root of unity. Suppose $f\in \mathbb{Z}[X]$. If $f(\zeta_p)$ has absolute value 1, is $f(\zeta_p)$ a root of unity?

I know the following two propositions:

1. The ring of integers in $K$ is $\mathbb{Z}[\zeta_p]$.
2. If $g(X)$ is a monic polynomial in $\mathbb Z[X]$ and all roots have absolute value 1, then all roots are roots of unity.

So, it will be sufficient to show that $f(\zeta_p^k)$ has absolute value 1 for all $k$ coprime with $p$. But how to do that?

Answer 1

Here is an easy answer:

We show for all $u\in K$ which has absolute value in $\mathbb{Q}$, $\vert \sigma(u)\vert =\vert u\vert$ for all $\sigma\in Gal(K/\mathbb{Q})$.

Let $\tau\in Gal(K/\mathbb{Q})$ be the conjugate map(i.e. maps $u$ to $\overline{u}$, its conjugate). Because $Gal(K/\mathbb{Q})$ is an abelian group, we have $\sigma(u)\overline{\sigma(u)}=\sigma(u\tau(u))=u\overline{u}$, therefore $\vert \sigma(u)\vert =\vert u\vert$.

So, all the conjugates of $f(\zeta_p)$ have absolute value 1. By proposition 2, it is a root of unity.

Answer 2

Yes this is true.

I checked out this related question: Are all algebraic integers with absolute value 1 roots of unity?

One of the answers gave a link to the following paper by Ryan Daileda: http://ramanujan.math.trinity.edu/rdaileda/research/papers/p1.pdf

On the first page it quotes a theorem by MacCluer and Parry which says that the statement you want is true for all fields that are Galois and CM, in particular for cyclotomic fields (and as a consequence all abelian fields).