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I am interested to know whether there is an intuitive/straightforward proof of the following result.

There is no monic polynomial $p\in\mathbb{Z}[x]$ with roots $re^{\textrm{i} \theta}, re^{-\textrm{i} \theta}, r^{-1}e^{\textrm{i} \theta}$, and $r^{-1}e^{-\textrm{i} \theta}$ such that

  • $|r|>1$, and
  • no quotient of two distinct roots of $p$ is a root of unity.

I have a sketch proof along the following lines. We can setup a sequence of quadratic extensions that ultimately shows that $e^{\textrm{i}\theta}$ lies in an imaginary quadratic extension of a totally real field that is closed under complex conjugation. Thus $e^{\textrm{i}\theta}$ is a unimodular algebraic integer in a CM-field closed under complex conjugation and so a root of unity (see, for instance, https://math.stackexchange.com/a/4343/95131). It follows that the second hypothesis does not hold as $re^{\textrm{i} \theta} / re^{-\textrm{i} \theta} = e^{2\textrm{i}\theta}$ is then also a root of unity.

The point of this question is to ask whether there is a straightforward proof.

Zephos
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1 Answers1

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I confess it took me a while trying to prove this before realizing it was false.

Consider the roots of the polynomial

$$P(x) = x^4 - x^3 + 3 x^2 - x + 1 = 0.$$

Note that $ x^4 P(1/x) = P(x)$, so if $\alpha$ is a root then so is $\alpha^{-1}$. Hence we can write the four roots as $\alpha, \alpha^{-1}, \beta, \beta^{-1}$.

Some easy calculus shows that (for some choice) that $r:=|\alpha| > 1$. Since the complex conjugate of $\alpha$ is also a root, and the complex conjugate has absolute value $|\overline{\alpha}| = |\alpha| = r > 1$, it follows that the complex conjugate is not $\alpha^{-1}$ and so (after relabeling) it is $\beta$. But now writing $\alpha = r e^{i \theta}$, we see the other roots are

$$\beta = r e^{- i \theta}, \alpha^{-1} = r^{-1} e^{-i \theta}, \beta^{-1} = r^{-1} e^{i \theta}.$$

Hence the polynomial is of the desired form. So it suffices to show that $\alpha/\beta = e^{2 i \theta}$ is not a root of unity. But it's easy enough to compute that the minimal polynomial of $\alpha/\beta$ is

$$Q(x) = x^4 - x^3 - 3 x^2 - x + 1,$$

which has roots $\alpha \beta$, $\alpha^{-1} \beta^{-1}$, $\alpha \beta^{-1}$, and $\alpha^{-1} \beta$. The first two are real, the second are complex with absolute value one but are not roots of unity. To see this, the polynomial is irreducible, so if $\alpha/\beta$ was a root of unity then all the other roots would also be roots of unity, but the first two roots are real numbers different from $\pm 1$ and so definitely not roots of unity.

If I try to imagine where you made a mistake, it might have been that you imagined the number $r^2 = \alpha \beta \in \mathbf{R}$ was totally real? But it doesn't have to be (and is not in this case), and indeed $\mathbf{Q}[x]/Q(x)$ has signature $(2,1)$.

user952367
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