Classifying algebraic integers satisfying a positivity condition Let $a $ be an algebraic integer such that $1/a$ is also an algebraic integer belonging to the ring of integers of $\mathbb {Q}(a) $. Then, what is the condition for $a $ to satisfy:
For any integer-coefficient polynomial $f (x) $ and any Galois conjugate $a'$ of $a $, $f (a')f (1/a')\ge 0. $
?
From the condition, we can deduce that $a'+1/a' $ is real and $2 \ge a'+1/a' \ge -2$ for each conjugate $a'$, at least. To clarify what is the point, consider the roots of unity $\xi_n $. By the reflection principle, we have $f (a')f (1/a') = |f (a')|^2$ in this case. Therefore, the real question is whether there is a non-cyclotomic example. I guess the answer is no, but I'm not sure.
+
What if I just require $f (a)f (1/a)\ge 0 $?(not all of the Galois conjugates)
 A: Proffering the following counterexample to the wish that the conditions


*

*$a$ is a unit of (the ring of integers of) an algebraic number field such that $a+\dfrac1a\in\Bbb{R}$, and

*$f(a)f(\dfrac1a)\ge0$ for all $f\in\Bbb{Z}[x]$


would force $a$ to be a root of unity.
Let 
$$
u=\sqrt2-1+i\sqrt{2\sqrt2-2}.
$$
We easily see that $u$ lies on the unit circle of the complex plane. Furthermore,
$$
(x-u)(x-\overline{u})=x^2-2\sqrt2 x+2x+1.
$$
Therefore $u$ is a zero of the polynomial
$$
p(x)=(x^2-2\sqrt2 x+2x+1)(x^2+2\sqrt2 x+2x+1)=x^4+4x^3-2x^2+4x+1.
$$
In particular $u$ and $1/u$ are both algebraic integers.
I claim that $p(x)$ is irreducible over $\Bbb{Q}$. The latter quadratic factor of $p(x)$ has two real zeros, $u_{3,4}=-1-\sqrt{2}\pm\sqrt{2(1+\sqrt2)}$. So over the reals $p(x)$ has one quadratic factor and two linear factors. We immediately see that the product of the quadratic and one of the linear factors has irrational coefficients, so $p(x)$ must be the minimal polynomial of $u$.
Because $|u|=1$ we have that for all $f\in\Bbb{Z}[x]$
$$
f(u)f(\frac1u)=f(u)f(\overline{u})=|f(u)|^2\ge0.
$$
The last claim is that $u$ is not a root of unity. This follows from the irreducibility of $p(x)$. For if $u$ were a root of unity, so would all the other zeros of its minimal polynomial $p(x)$. This is not the case, because $u_3\approx-0.217$ and $u_4\approx-4.612$ are clearly not roots of unity.
However, this is not a counterexample to the conjecture, where you include the requirement that conditions 1 & 2 should hold for all the conjugates of $a$ as well. For if $f(x)=x+3$ then $f(u_3)$ and $f(u_4)=f(1/u_3)$ have opposite signs.
A: Proving the conjecture that the conditions


*

*$a$ is unit of an algebraic number field such that $a'+1/a'\in\Bbb{R}$ for all conjugates $a'$ of $a$ (needed to ensure that $f(a')f(1/a')$ is real for all $f$ and the question about its sign is meaningful), and

*$f(a')f(1/a')\ge0$ for all conjugates $a'$ and all $f\in\Bbb{Z}[x]$
together imply that $a$ is a root of unity.


We begin by observing that for a complex number $z$ the sum $z+1/z$ is real, if and only if $z$ is real itself or $z$ is on the unit circle. For if $z$ is not real, then $z$ and $1/z$ have arguments with opposite signs. Therefore their imaginary parts can cancel only if $|z|=1$.
Next I claim that condition 2 implies that all the conjugates of $a$ are on the unit circle of the complex plane. The alternative is that at least one of them, say $a_1$, is a real number $\neq\pm1$. Then there exists a ratioanl number $q=m/n$ strictly between $a_1$ and $1/a_1$. But this implies that for $f(x)=nx-m\in\Bbb{Z}[x]$ the values $f(a_1)$ and $f(1/a_1)$ have opposite signs contradicting 2.
So we can assume that all the conjugates of $a$ are on the unit circle. The following standard argument then shows that $a$ is a root of unity. A key observation is that the powers $a^k$ then also have all their conjugates on the unit circle for all (infinitely many) positive integers $k$. The Vieta relations then imply that the coefficients of the minimal polynomials of $a^k$ are all bounded from above by a number that depends on the degree $[\Bbb{Q}(a):\Bbb{Q}]$, but does not depend on $k$. There are only finitely many such polynomials (the coefficients are integers!). Therefore there are only finitely many distinct powers $a^k$. The claim follows.
