There are algebraic integers which are not roots of unity , for example consider the irreducible polynomial $ P(x)= x^4-2x^3-2x+1 $.

A computer software can show that this polynomial has two real roots outside the unit circle(one greater than one and the other less than one) and two roots on the unit circle. However I don't know how to prove this rigorously that there are two roots on the unit circle.

Usually when I wanted to prove some root of a polynomial is on the unit circle , I'd multiply that by some other polynomial to get something of the form $ x^n - 1 $ and it's obvious that every root of such expression has norm one, however , in this case this is not possible since none of roots of $ P(x) =0 $ are roots of unity.

Actually there is a whole lot of examples, called Salem numbers. It's an algebraic integer $\lambda > 1$ such that all of its Galois conjugates are on the unit circle except $\frac{1}{\lambda}$. The polynomial given above was an example of a minimal polynomial of a Salem number.

Does anyone have any idea how can I prove this, i.e. roots are on the unit circle except two of them?(I'm looking for a method that can be applied to more than just one example, hopefully lots of Salem numbers)


3 Answers 3


Factor the polynomial over $\mathbb{Q}(\sqrt{3})$ to get

$$(x^2+\alpha x + 1)(x^2 + \overline{\alpha} x + 1)$$

with $\alpha = -1+\sqrt{3}$ and $\overline{\alpha} = -1 - \sqrt{3}$. Let's give these polynomials names: $p = x^2 + \alpha x + 1$ and $q = x^2 + \overline{\alpha} x + 1$. I claim that the roots of $p$ are on the unit circle and the roots of $q$ are not.

Note first that the discriminant of $p$ is negative. Thus, $p$ has two complex roots, and since $p$ has real coefficients, these roots are complex conjugates of each other. The constant term of $p$ is the product of these roots, so we find that the product of each root with its complex conjugate is 1. In other words, each root lies on the unit circle.

On the other hand, because the discriminant of $q$ is positive, it has two real roots. For a real number to be on the unit circle, it must be $1$ or $-1$, and we see immediately that neither of those is a root of $q$. Thus, the roots of $q$ do not lie on the unit circle.

This points towards a way to generate lots of quartic examples. Construct a real quadratic field. Identify two quadratic extensions of it, one real and one complex, generated by roots of polynomials of the form $x^2 + \beta x + 1$ and $x^2 + \overline{\beta} x + 1$ where $\beta$ is an algebraic integer in the quadratic field, $\overline{\beta}$ is its algebraic conjugate, one polynomial has negative discriminant, and the other has positive. The the product of the polynomials is a Salem polynomial.


Here is the best way to count roots on the unit circle.

Your quartic polynomial has even degree and symmetric coefficients: $c_k = c_{4-k}$ for $0 \leq k \leq 4$. From the symmetry in the coefficients, your polynomial can be written in the form $x^2g(x + 1/x)$ for a polynomial $g(x)$, where in fact $g(x) = x^2 - 2x - 2$.

The mapping $z \mapsto z + 1/z$ sends the unit circle in a 2-to-1 way onto the interval $[-2,2]$ (except for being 1-to-1 at the endpoints). The roots of your polynomial on the unit circle (which don't include $\pm 1$) are in a 2-to-1 correspondence with roots of $g(x)$ in the interval $[-2,2]$. There is one such root for $g(x)$, so your polynomial has two roots on the unit circle.

Further examples are discussed at https://kconrad.math.uconn.edu/blurbs/galoistheory/numbersoncircle.pdf, including polynomials that don't have symmetric coefficients. Your example is treated at the very start by a more elementary method that does not easily generalize to higher degree and also in later examples by methods that do generalize to higher degree.


Here's an answer that my friend proposed: parametrize the unit circle by $e^{i \theta}$, to get one equation in terms of real parameter $\theta$, i.e. $|P(e^{i \theta})|^2 = 0$. This can be examined rigorously to have real roots using Intermediate Value Theorem.

  • 1
    $\begingroup$ Is it good to just solve a system of two equations for real and imaginary part of $P(e^{i \theta}) = 0$? They should be easier! Maybe it's easier your way in order to just predict the existence since we have just one equation! :) $\endgroup$
    – Maffred
    Feb 3, 2016 at 0:04
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    $\begingroup$ @Maffred The appearing difficulty with having two equations is that we need them to have the same number \theta as their root. I know how to prove a function has a real root using intermediate value theorem, but I didn't know how to make sure the two obtained roots for two equations are actually the same. So I mixed them into a single equation. $\endgroup$
    – Mehdi
    Feb 4, 2016 at 3:35
  • $\begingroup$ I was thinking exactely the same! Nice idea!!! $\endgroup$
    – Maffred
    Feb 4, 2016 at 3:38

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