In relation to the (possible non-)existence of non-contracting ergodic endomorphisms on the torus, I have the following question on polynomials: If the polynomial is monic, with integer coefficients and has no root strictly inside the unit circle, is it then also true that roots on the unit circle are roots of unity? For polynomials up to degree 4 this seems to be the case, but I'd be happy with a higher degree counter-example. I am aware that Kronecker's Theorem says the answer is "yes" if all roots are on the unit circle.


1 Answer 1


Let $f(z)$ be a degree-$n$ irreducible polynomial with integer coefficients and roots on the unit circle but no roots strictly within it. Then $g(z) := z^nf(1/z)$ is also a degree-$n$ irreducible polynomial, whose roots on the unit circle are the same as those of $f$. Since $f$ is the minimal polynomial of those unit roots and $g$ has the same degree as $f$, $g$ must also have the same roots as $f$.

Now since $f$ has no roots within the unit circle, $g$ has by construction no roots outside the unit circle, so all roots of $f$ lie on the unit circle. The rest follows from Kronecker's theorem.

  • $\begingroup$ I’m a bit thick today, but have you shown that $g$ is monic? Is that necessary to your argument? $\endgroup$
    – Lubin
    Jan 1 at 2:01
  • $\begingroup$ Kronecker's theorem as OP describes it only requires $f$ to be monic integer, not $g$. $\endgroup$
    – Magma
    Jan 1 at 2:13
  • 1
    $\begingroup$ I thought of this too, but g is really not monic if f has roots outside the unit circle, so I couldn't see how this would work. I don't think Kronecker's theorem applies to g, or at least I don't see how. $\endgroup$
    – Henk Bruin
    Jan 1 at 16:54
  • $\begingroup$ I just showed that if $f$ has roots on the unit circle but not inside, then it doesn't have roots outside either by showing that $f$ is equal to $g$ up to scale. Once you see that, assuming $f$ to be monic you would instantly obtain that $g$ is either monic or $1-z$, because the $z^n$ coefficient of $g$ is integer and the product of the roots of $f$, but it is not necessary to do so since you can immediately apply Kronecker's result to a monic $f$ instead. $\endgroup$
    – Magma
    Jan 1 at 17:09
  • $\begingroup$ Unfortunately, I still don't understand your argument. You may well be right; the only polynomials with roots on and inside the unit circle I could find are reducible. I also checked Kronecker's original publication: "monic" is quite essential. I think we agree that the leading coefficient of g(z) = z^n f(1/z) is the constant coefficient of f which is the product of the roots, and therefore greater than 1 in absolute value. I don't see what you mean by "𝑓 is equal to 𝑔 up to scale". Certainly not f = c g for some scalar c, because I don't see why that should be true. $\endgroup$
    – Henk Bruin
    Jan 4 at 8:44

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