I am reading A Classical introduction to Number Theory by Ireland Rosen. There, they have the following statement and proof:
Let $K/\mathbb{Q}$ be an algebraic number field and let $\sigma_1, \sigma_2, \ldots, \sigma_n$ be the $n$ embeddings of $K$ into $\mathbb{C}$. If $\alpha \in K$ such that $\mid \sigma_i(\alpha)\mid <1$ for all $i=1,\ldots, n$, then $\alpha$ isa root of unity.
I have doubts in the proof which goes as follows:
$\alpha$ is a root of \begin{align*}\prod_{i=1}^{n}(x-\sigma_i(\alpha)) \in \mathbb{Z}[x].\end{align*} The hypothesis of the lemma implies that the coefficient of $x^m$ is bounded by the integer $\binom{n}{m}$. Thus, only finitely many polynomials of degree $n$ in $\mathbb{Z}[x]$ can arise this way. If $\alpha$ satisfies the hypothesis of the lemma, so do all powers of $\alpha$. Since finitely many polynomials can have only finitely many roots, it follows that two distinct powers of $\alpha$ must be equal. Thus, $\alpha$ is a root of unity.
I don't understand two things:
i) Why does the product of the polynomials lie in $\mathbb{Z}[x]$ and not $\mathbb{Q}[x]$? What guarantees them to be integers because $\sigma_i(\alpha)$ could be a rational as well, right?
ii) How exactly does the last line imply that $\alpha$ must be a root of unity? Can two distinct powers might be same, but that doesn't mean they must also satisfy $x^k=1$ for some $k$?
Thank you so much in advance!!
It should be $|\sigma_i(\alpha)| \leqslant 1$ instead of $<$ in the proposition and in this case, $\alpha$ is a root of unity or $0$. If it is a root of unity we obviously have $|\sigma_i(\alpha)| = 1$ for all $i$ by the way. To conlude from $\alpha^n = \alpha^m$ for some $n \neq m$ that $\alpha = 0$ or is a root of unity, notice that when $\alpha \neq 0$, it implies $\alpha^{n - m} = 1$ and $n - m \neq 0$, as said @red_trumpet in your comments.
Now, an other hypothesis misses to get that $P(x) = \prod_{i = 1}^n (x - \sigma_i(\alpha))$ has coefficients in $\mathbb{Z}$, which is that $\alpha$ must be integral over $K$. Indeed, if $\alpha$ is integral over $K$, its characteristic polynomial in $K/\mathbb{Q}$ (which is $P$) has integer coefficients (and the reciprocal is true by the way).
This hypothesis is necessary. Take for example $\frac{1}{2} \in \mathbb{Q}$. $\mathbb{Q}/\mathbb{Q}$ has degree $1$ hence $\sigma = \mathbb{id}_\mathbb{Q}$ and $P(x) = x - \frac{1}{2}$ in this case. Moreover, $|\sigma(1/2)| = \frac{1}{2} \leqslant 1$ but $\frac{1}{2}$ is not a root of unity.