3
$\begingroup$

Characterize the polynomials $f\in\mathbb{Z}[x]$ with the following property:

There is $M\in\mathbb{N}$ such that for all $k\in\mathbb{N}^*$ there is an irreducible factor (over $\mathbb{Q}$) of $f(x^k)$ that has degree at most $M$.

If $x\mid f(x)$ then $x\mid f(x^k)$. Also if $x-1\mid f(x)$, then $x-1\mid f(x^k)$.

Question: Is it true that these are all such polynomials?

We could assume $f$ irreducible with $f(0)$ and $f(1)$ different from zero.

The linear case we can always make $f(x^k)$ itself irreducible for arbitrarily large $k$.


Maybe they end up not very related, but at least aesthetically this question has the same flavor as this other question [A square integer matrix with $k$-th roots for all $k$ is a projection]. Well, I invented this problem out of the linked one.

$\endgroup$
1
  • 1
    $\begingroup$ A cool question. The cyclotomic polynomials would otherwise come close, but fail when $k$ is divisible by a key prime. For example, if $f(x)=x^4+x^3+x^2+x+1$ is the fifth cyclotomic polynomial, we have $f(x)\mid f(x^k)$ whenever $5\nmid k$. But $f(x^{5^\ell})$ is irreducible for all $\ell>0$, and therefore $f$ fails to qualify. $\endgroup$ Sep 21, 2022 at 4:51

1 Answer 1

4
$\begingroup$

These polynomials (i.e those for which either $0$ or $1$ is a root) are the only such polynomials.

There exists some $c\in\mathbb R$ for which every root $z$ of $f$ satisfies $|z|\leq c$, and so every root $z_k$ of $f(x^k)$ satisfies $|z|\leq c^{1/k}$. Suppose $g(x)=\sum_{i=0}^d a_ix^i$ is an irreducible factor of some $f(x^k)$, with $a_d\neq 0$. Then $g$ has $d$ roots $\lambda_1,\dots,\lambda_d$ in $\mathbb C$, and for every $0\leq j\leq d$ $$\left|\frac{a_j}{a_d}\right|=\left|\sum_{\substack{S\subset\{1,2,\dots,d\}\\|S|=j}}\prod_{i\in S}\lambda_i\right|\leq \binom djc^{j/k}.$$ Note that $a_d$ divides the leading coefficient $C$ of $f$, since $g\in\mathbb Z[x]$ is a factor, so $$|a_j|\leq \binom djc^{j/k}|a_d|\leq C2^dc^{d/k}.$$ Now, suppose that $f$ satisfies the condition, and let $g_k\mid f(x^k)$ be a factor of degree at most $k$. Every coefficient of $g_k$ is at most $$C2^{\deg g_k}c^{\frac{\deg g_k}k}\leq C2^Mc^{M/k}\leq Cc2^M$$ in magnitude. So, there are only finitely many possible polynomials $g$ of degree at most $M$ which may divide $f(x^k)$ for any $k$. In particular, there exists some $g$ which divides $f(x^k)$ for infinitely many $k$.

Fix such a $g$ irreducible, and let $z$ be any root of $g$. If $g(x)$ divides $f(x^k)$, then $z$ is a root of $f(x^k)$, and so $z^k$ is a root of $f$. Since there are finitely many roots of $f$, one must have $z^k=z^\ell$ for some $k<\ell$ (as there are infinitely many $k$ for which $z^k$ is a root of $f$). This means that $z$ is either $0$ or a root of unity. This implies that $g(x)$ is either $x$ (in which case $x\mid f(x)$ and we are done) or some cyclotomic polynomial $\Phi_n(x)$. This means that for all sufficiently large $k$, some cyclotomic polynomial of degree at most $M$ divides $f(x^k)$.

Let $N$ be the largest integer such that $\deg \Phi_N\leq M$, and consider $f(x^{N!})$. For every root $z$ of $f(x^{N!})$, $z^{N!}$ is a root of $f$. If some $\Phi_n(x)$ with $n\leq N$ divides $f(x^{N!})$, then $z^{N!}$ is a root of $f$ for every $n$th root of unity, but $z^{N!}=1$ for these roots. So, $1$ must be a root of $f$, as desired.

$\endgroup$
1
  • $\begingroup$ Very nice! I tried to churn through the Galois theory but it looked messy. Your observation that the roots getting smaller means there are only finitely many possible $g$ is much better. $\endgroup$ Sep 22, 2022 at 17:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .