# Show that all the nonzero roots of $f(x)$ are roots of unity. (A theorem of Kronecker)

Let $$f(x)$$ be a monic polynomial with integer coefficients all of whose roots have absolute values at most 1. Show that all the nonzero roots of $$f(x)$$ are roots of unity. A root of unity $$\xi$$ is a complex number such that there exists a positive integer $$n$$ so that $$\xi^n = 1$$.

I think it might be useful to let $$a_1,\cdots, a_n$$ denote the roots of $$f(x)$$. Then define for each positive integer $$r, f_r(x)=\prod_{i=1}^n (x-a_i^r)$$. The coefficients of $$f_r(x)$$ are bounded, and to see why, observe that the coefficient of $$x^{n-j}$$ has absolute value equal to $$|\sigma_j (a_1^r,\cdots, a_n^r)|$$. The latter expression is at most $${n\choose j}$$ since each product of $$j$$ terms of $$a_1^r,\cdots, a_n^r$$ has absolute value at most 1. I think one can then apply the pigeonhole principle somehow.

• You need first to divide through by $x^r$ to make the constant term non-zero (eliminating all the zero roots). And think about why that might be significant. Oct 10, 2022 at 13:43
• Note that it's not enough for the roots to be on the unit circle. Oct 10, 2022 at 13:55
• Look at the constant term Oct 10, 2022 at 14:03
• Relevant MO post: mathoverflow.net/questions/10911
– Sil
Oct 10, 2022 at 14:10
• I suggest that you re-read the comments of @TheBestMagician, since these comments immediately conquer the problem. Oct 10, 2022 at 14:11

Since a good proof of this on MathOverflow has already been linked in the comments, I will provide an alternate and original proof.

Firstly, note that we may write $$f(x)=x^r g(x)$$ for some monic $$g(x)\in\mathbb{Z}[x]$$ with no roots at $$x=0$$.

Let $$n=\deg(g)$$, and let $$\alpha_1,\cdots,\alpha_n$$ be the roots of $$g(x)$$. Define the sequence

$$A_k=\alpha_1^k+\alpha_2^k+\cdots+\alpha_n^k$$

It is a fact that the fixed field of the Galois group of a splitting field of a polynomial in $$\mathbb{Q}[x]$$ is $$\mathbb{Q}$$, so since $$A_k$$ is fixed by every automorphism in $$\text{Gal}(\mathbb{Q}(\alpha_1,\cdots,\alpha_n)/\mathbb{Q})$$, then $$A_k\in\mathbb{Q}$$. Furthermore, since $$\alpha_i$$ are algebraic integers, then so is $$A_k$$, and since $$A_k$$ is also rational, then it must be an integer. (note that this fact can also be proven by combining of Newton's identities and Vieta's formulas)

Now, note that since $$|\alpha_i|\leq 1$$, then

$$|A_k|\leq |\alpha_1|^k+|\alpha_2|^k+\cdots+|\alpha_n|^k\leq n$$

Combined with the fact that $$A_k$$ is an integer, this means that $$A_k$$ only takes on finitely many values. Writing, $$g(x)=x^n-b_{n-1}x^{n-1}-\cdots-b_1x-b_0$$, we know that $$A_k$$ satisfies the linear recurrence

$$A_k=b_{n-1}A_{k-1}+\cdots+b_1A_{k-n+1}+b_0A_{k-n}$$

Since $$A_k$$ satisfies a linear recurrance and takes on only finitely many values, then $$A_k$$ is periodic with some period $$m$$. This means that $$A_{mk}=A_0=n$$ for all $$k\geq 0$$. Therefore, for $$|x|<1$$, we have that $$$$\begin{split} \frac{n}{1-x}&=\sum_{k=0}^\infty nx^k\\ &=\sum_{k=0}^\infty A_{mk}x^k\\ &=\sum_{k=0}^\infty \sum_{i=1}^n \alpha_i^{mk}x^k\\ &=\sum_{i=1}^n\sum_{k=0}^\infty\alpha_i^{mk}x^k\\ &=\sum_{i=1}^n\frac{1}{1-\alpha_i^mx}\\ \end{split}$$$$ and therefore, multiplying both sides by $$1-x$$, and taking the limit as $$x\rightarrow 1^-$$, we have that $$$$n=\lim_{x\rightarrow 1^-}\sum_{i=1}^n\frac{1-x}{1-\alpha_i^mx}\\$$$$ Notice that if $$\alpha_i$$ is an $$m$$-th root of unity, then $$\lim\limits_{x\rightarrow 1^-}\frac{1-x}{1-\alpha_i^mx}=1$$ , and otherwise $$\lim\limits_{x\rightarrow 1^-}\frac{1-x}{1-\alpha_i^mx}=0$$. Combining this with the formula above tells us that $$g(x)$$ has precisely $$n$$ roots $$\alpha_i$$ which are $$m$$-th roots of unity (this is every root of $$g(x)$$).

Equivalently, every root of $$f(x)=x^r g(x)$$ is either $$0$$ or a root of unity, as desired.

• Thanks for your answer. How would one prove in an elementary way that $A_k$ is always an integer (e.g. using Newton's identities as shown here and induction?) Oct 22, 2022 at 23:19
• Glad you asked. Letting $g(x)=x^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0$ and letting $s_i=s_i(\alpha_1,\cdots,\alpha_n)$ be the elementary symmetric polynomials of the roots of $g(x)$, Vieta's formulas tell us that $s_i=(-1)^ib_{n-i}$ (so they are integers since $b_i$ are integers). Newton's identities tell us that $$A_k=-(ks_k+\sum_{i=1}^{k-1}s_iA_{k-i})$$ We see that for any $k\geq 0$, if $A_i$ is an integer for all $0\leq i<k$, then $A_k$ is equal to a sum of integers, and is thus an integer. Noting that $A_0=n$ is an integer, then by strong induction, $A_k$ is an integer for all $k\geq 0$. Oct 22, 2022 at 23:45
• I actually expected a proof along the lines you typed up but I didn't have time to type out the proof when I asked it. The equality $\dfrac{1}{1-x} = \sum_{i=0}^\infty x^i$ only holds for complex numbers $x$ when $|x| < 1$. So it seems all the given equalities actually hold when $|x| < 1$. Then all your results would be valid if $|x| < 1$ so you can just take the left limit as $x\to 1^-$ to get the required result since equality holds for each such x. Oct 23, 2022 at 12:28
• However, my question is, in general, is it okay to use formal power series and then take limits like you did, or does there need to be some interval of convergence that'll allow you to take the limit (in this case $x\in (-1,1)$, but what if $x\in (-1/2, 1/2)$ due to convergence requirements)? Correct me if I'm wrong, but I think the validity of taking limits depends on the range over which the formal power series is actually a power series. Oct 23, 2022 at 12:29
• @user3379 You are right that those particular manipulations only hold when $|x|<1$. Originaly, I did phrase things this way. However, in the context of formal power series, we can discuss such series while ignoring convergence. What I really proved was that $$\frac{n}{1-x}=\sum_{i=1}^n\frac{1}{1-\alpha_i^m x}$$ as an equality of elements of $\mathbb{C}(x)$ (rational functions of $x$ with coefficients in $\mathbb{C}$). Notice that this equality features no power series. Once we have equality in $\mathbb{C}(x)$, we can manipulate expressions in all the usual ways, taking limits without issue. Oct 23, 2022 at 12:44