MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a complete characterization of those integer polynomials, that is $P\in{\mathbb Z}[X]$, such that $P(D)\subset D$, where $D$ is the unit disk ? At least, $P(X)=\pm X^k$ works, when $k\in\mathbb N$. But are there other ones, many other ones ?

share|cite|improve this question
I apologize for having posted that an easy question. – Denis Serre Apr 5 '12 at 9:08
up vote 18 down vote accepted

Since $P\in \mathbb{Z}[X]$, We have $P(0)\in \mathbb{Z}$. Suppose that $P(0)\neq 0$, then $|P(0)|\geq 1$. In that case $P(D)\subset D$ is not satisfied unless $P$ is constant by open mapping theorem. So, if $P$ is non-constant, then we must have $P(0)=0$.

Write $P(X)=XQ(X)$. Then $Q(X)=P(X)/X$. On a disk $D_r$ of radius $0< r<1$ centered at $0$, We have by Maximum modulus theorem that
$$|Q(X)|\leq 1/r$$

since $|P(X)|\leq 1$. Letting $r\rightarrow 1$, we have also that $|Q(X)|\leq 1$ whenever $|X|\leq 1$. By repeating the same argument for $Q(X)$, we obtain that $P(X)=\pm X^k$, when $k\in \mathbb{N}$ are the only polynomials in $\mathbb{Z}[X]$ satisfying the property.

share|cite|improve this answer
Well, and constant 0. – Harry Altman Apr 5 '12 at 9:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.