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We have, $f(x)=x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x]$, with $\alpha_i\in\mathbb{C},\ 1\leq i\leq n$ being all the roots of $f(x)$. If we have $|\alpha_i|=1$, for every $i$, then $\alpha_i$ is a root of 1.

Edit: As discussed in comments, I seem to have interpreted the problem incorrectly the first time around.

Essentially, we have to show that ${\alpha_i}^k =1$, for some $k>n$.

Note: This question was posed after a class on Units, including Dirichlet's Unit Theorem.

So I need some hint to help me get to the answer

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  • $\begingroup$ Are you supposed to show that $\alpha_j$ is an $n$th root of unity? That's not the same as just "root of unity". $\endgroup$ Dec 25, 2017 at 16:14
  • $\begingroup$ My bad, I'll edit the question to reflect the same. The question was framed using the word root of 1, instead of root of unity. I'll just write that $\endgroup$ Dec 25, 2017 at 16:15
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    $\begingroup$ It might be useful to notice that $f(x)\in\mathbb{Z}[x]$ and Newton's formulas imply that the power sum $$\alpha_1^k+\alpha_2^k+\ldots+\alpha_n^k$$ is an integer for every $k\in\mathbb{N}$. $\endgroup$ Dec 25, 2017 at 16:23
  • $\begingroup$ You missed my point. It looks like you're trying to show that $\alpha_j^n=1$. But that's not required by the problem as stated; you just have to show that $\alpha_j^k=1$ for some $k$. (The zeroes of $x^2+x+1$ are cube roots of $1$, not square roots...) $\endgroup$ Dec 25, 2017 at 16:23
  • $\begingroup$ Ah! Yes. I think I understand my folly. How do I approach this problem then? $\endgroup$ Dec 25, 2017 at 16:39

4 Answers 4

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Let us collect the comments into an answer. By Newton's identities the power sums $$ p(k) = \alpha_1^k+\ldots+\alpha_n^k $$ are integer numbers, and they belong to the $[-n,n]$ interval by the triangle inequality.

If $\alpha$ is a root of $f(x)$ then $\overline{\alpha}=\alpha^{-1}$ is also a root of $f(x)$ and by letting $\theta_i=\text{Arg}(\alpha_i)$ we have $$ \cos(k\theta_1)+\ldots+\cos(k\theta_n)\in [-n,n]\cap\mathbb{Z} $$ for any $k\in\mathbb{N}$. If all the angles $\theta_i$ are rational multiples of $\pi$ there is nothing to prove.
Let us assume that $\theta_1\not\in\pi\mathbb{Q}$. In such a case, by the Lindemann-Weierstrass theorem we have that $\cos(k\theta_1)$ is a trascendental number over $\mathbb{Q}$ for any $k\in\mathbb{N}^+$. On the other hand $p(k)$ may only take a finite number of values, so for some $j\in[-n,n]$ there are infinite $k\in\mathbb{N}$ such that $p(k)=j$. Additionally $\cos(k\theta)=T_k(\cos\theta)$, hence by elimination of variables we get that some algebraic combination (with coefficients in $\mathbb{Q}$) of $\cos(k_1\theta_1),\cos(k_2\theta_1),\ldots,\cos(k_M \theta_1)$ equals zero. This contradicts the fact that $\cos(\theta_1)$ is trascendental over $\mathbb{Q}$, proving the claim.

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  • $\begingroup$ The use of the Lindeman-Weierstrass Theorem is very clever (+1). $\endgroup$
    – Kelenner
    Dec 25, 2017 at 17:18
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We may suppose that all $\alpha_j$ are distincts. Put $b_k=\alpha_1^k+...+\alpha_n^k$. Then the $b_k$ are in $\mathbb{Z}$, as noticed by @Jack d'Aurizio, and clearly $|b_k|\leq n$ for all $k$. Now writing that the $\alpha_j$ are roots of $f$, multiplying by $\alpha_j^k$ and summing, we get that $$b_{k+n}+a_{n-1}b_{k+n-1}+...+a_0b_k=0$$ for all $k$. Put $w_k=(b_k,b_{k+1},..,b_{k+n-1})$. By the above, the $w_k$ take only a finite number of values. Hence there exists $m$ and $h\geq 1$ such that $w_m=w_{m+h}$. This imply by the recurrence relation that we have $b_{k}=b_{k+h}$ for all $k\geq m$. Hence for all $k$ we have $$(\alpha_1^h-1)\alpha_1^m\alpha_1^k+...+(\alpha_n^h-1)\alpha_n^m\alpha_n^k=0$$

Writing this for $k=0,...,n-1$ gives a linear system in the unknowns $x_j=(\alpha_j^h-1)\alpha_1^m$, with for determinant a Van der Monde determinant, hence non zero as the $\alpha_j$ are distincts. This imply that the $x_j$ are all $0$, and it is easy to finish.

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  • $\begingroup$ (+1) I started writing my answer before you, and I finished later. We used similar ideas, but your proof is cleaner and more elementary. $\endgroup$ Dec 25, 2017 at 17:17
  • $\begingroup$ Probably I'm just being stupid again: Why may we suppose the $\alpha_j$ are distinct? $\endgroup$ Dec 25, 2017 at 17:23
  • $\begingroup$ @David C Ullrich We have only to show that the different values of the $\alpha_j$ are roots of unity, $\endgroup$
    – Kelenner
    Dec 25, 2017 at 17:26
  • $\begingroup$ ??? Say $n=3$, $\alpha_1\ne\alpha_2=\alpha_3$. Then yes of course we need only show that $\alpha_1$ and $\alpha_2$ are roots of unity. But how do we know that for example $\alpha_1^k+\alpha_2^k\in\mathbb Z$? $\endgroup$ Dec 25, 2017 at 17:34
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    $\begingroup$ Oops. Since $\mathbb Z[x]$ is not a PID it's not clear what I mean by minimal polynomial. Ah - Gauss' Lemma: Irreducible in $\mathbb Z[x]$ implies irreducible in $\mathbb Q[x]$ - so I say what I said but for $\mathbb Q[x]$, fine. $\endgroup$ Dec 25, 2017 at 20:09
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Let me reformulate your question for security : If $\alpha$ is an algebraic integer all of whose conjugates have absolute value $1$, then $\alpha$ is a root of unity. Right ? Let us consider all the powers $\alpha^k$ for $k\in \mathbf N$. Such a power $\alpha^k$ is an algebraic integer, whose irreducible polynomial $f_k \in \mathbf Z[X]$ has degree less than the degree $n$ of $\alpha$ over $\mathbf Q$. But these coefficients are symmetric functions of the conjugates of $\alpha^k$, hence by the hypothesis and the triangle inequality, they are bounded by bounds depending only on $n$. It follows that there are only a finite number of possible $f_k$'s when $k$ varies. Therefore there are only finitely many distinct powers of $\alpha$, hence, after simplification, one of these powers must be $1$ .

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For $k$ natural number consider the monic polynomial $f_k$ with the roots $\alpha_i^k$, $i=1, \ldots, n$. The coefficients of $f_k$ are integers (as integral polynomials in the coefficients of $f$) and bounded in absolute value ( $\le \binom{n}{ [\frac{n}{2}]}$ ), since all the $\alpha_i$ have absolute value $\le 1$. Therefore, there exist $0\le k < l $ so that $f_k=f_l$. This means that the roots of these polynomials are equal, that, is, there exists a permutation $\sigma$ of $\{1, \ldots, n\}$ such that $\alpha_i^k= \alpha_{\sigma(i)}^l$ for $i=1, \ldots, n$. From here we get $$\alpha_i^{km}= \alpha_{\sigma(i)}^{lm}$$ for all $m$, and hence, by induction on $t$ $$\alpha_i^{k^tm}= \alpha_{\sigma^t(i)}^{l^t m}$$ for all possible $i$, $m$, $t$. Take now $t>1$ so that $\sigma^t=1$. We get $$\alpha_i^{k^t}= \alpha_{i}^{l^t }$$

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  • $\begingroup$ What's $t$? Have you used that in place of $k$? $\endgroup$ Dec 27, 2017 at 18:37
  • $\begingroup$ @junkquill $t$ is a natural number , arbitrary $\endgroup$
    – orangeskid
    Dec 27, 2017 at 19:54
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    $\begingroup$ Right, right. Of course. I had failed to notice the tiny $t$ power on $k$ and $l$. $\endgroup$ Dec 27, 2017 at 19:57
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    $\begingroup$ There are algebraic numbers on the unit circle which aren't roots of unity. E.g. take $3z^2-2z+3=0$. $\endgroup$
    – mr_e_man
    Oct 23, 2023 at 23:37
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    $\begingroup$ I think so. Try $z^2-2az+1=0$, where $a$ is a real algebraic number smaller than $1$, and $2a$ is not integral. (I suppose we should also require the algebraic conjugates of $a$, i.e. the other roots of its minimal polynomial, to have the same properties.) $\endgroup$
    – mr_e_man
    Oct 24, 2023 at 0:04

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