# To show that if all roots of an integer monic polynomial have norm 1, then they are roots of 1

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

• 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". Dec 25, 2017 at 16:14
• 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 Dec 25, 2017 at 16:15
• 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 $k\in\mathbb{N}$. Dec 25, 2017 at 16:23
• 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...) Dec 25, 2017 at 16:23
• Ah! Yes. I think I understand my folly. How do I approach this problem then? Dec 25, 2017 at 16:39

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.

• The use of the Lindeman-Weierstrass Theorem is very clever (+1). Dec 25, 2017 at 17:18

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.

• (+1) I started writing my answer before you, and I finished later. We used similar ideas, but your proof is cleaner and more elementary. Dec 25, 2017 at 17:17
• Probably I'm just being stupid again: Why may we suppose the $\alpha_j$ are distinct? Dec 25, 2017 at 17:23
• @David C Ullrich We have only to show that the different values of the $\alpha_j$ are roots of unity, Dec 25, 2017 at 17:26
• ??? 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 $\alpha_1^k+\alpha_2^k\in\mathbb Z$? Dec 25, 2017 at 17:34
• 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. Dec 25, 2017 at 20:09

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 $1$ .

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 $$\alpha_i^{k^t}= \alpha_{i}^{l^t }$$

• What's $t$? Have you used that in place of $k$? Dec 27, 2017 at 18:37
• @junkquill $t$ is a natural number , arbitrary Dec 27, 2017 at 19:54
• Right, right. Of course. I had failed to notice the tiny $t$ power on $k$ and $l$. Dec 27, 2017 at 19:57
• There are algebraic numbers on the unit circle which aren't roots of unity. E.g. take $3z^2-2z+3=0$. Oct 23, 2023 at 23:37
• 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.) Oct 24, 2023 at 0:04