Salem Numbers, roots on the unit circle

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.