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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)

D. Lee
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  • I may be wrong but I would think that "more often than not" $f(a')f(1/a')$ might not be real, so comparing it with $0$ is already a bit problematic. +1 anyway - an interesting question somewhere here :-) – Jyrki Lahtonen Jul 25 '15 at 06:02
  • For example: Let $\phi=(1+\sqrt5)/2$ be the golden ratio. Then $a=i\phi$ is a unit in the ring of integers of $\Bbb{Q}(a)$. With $f(x)=x+1$ we get $$ f(a)f(1/a)=(1+i\phi)(1-i/\phi)=2+i(\phi-1/\phi)=2+i.$$ – Jyrki Lahtonen Jul 25 '15 at 06:14
  • Ok. So are we further restricting to such numbers $a$ that $f(a)f(1/a)$ is always real? – Jyrki Lahtonen Jul 25 '15 at 06:19
  • Yes. um.. I think it suffices for us to consider the cases $a+1/a\in\mathbb{R} $. – D. Lee Jul 25 '15 at 06:23
  • Ok. I think that doesn't imply that $a'+1/a'$ would be real for all the conjugates, so we may need to drop the conjugate? For example with $a=\root3\of2-1$ we clearly have real $$a+1/a=(\root3\of2-1)+(\root3\of4+\root3\of2+1)\approx 4.10724,$$ but with the conjugate $a'=\omega\root3\of2-1$ we get $$a'+1/a'\approx-2.0536+i0.8075.$$ – Jyrki Lahtonen Jul 25 '15 at 06:25
  • If you assume that $a'+1/a'$ is real for all the conjugates $a'$, then I think the assumption $f(a')f(1/a')\ge0$ for all $f$ does imply that $a$ is a root of unity. The assumption implies that all those $a'$s are either real or on the unit circle. If one of $a'$s is real, then it seems to be nearly certain that we can find an $f$ such that $f(a')f(1/a')<0$. OTOH, if all the conjugates are on the unit circle, it is well known that $a$ must be a root of unity. Have you seen that result? Thinking about this part as well. – Jyrki Lahtonen Jul 25 '15 at 11:52
  • I'm almost a newbie of algebraic number theory. But the result you refer to looks interesting.. Is it a canonical result in a graduate textbook? I guess I can find the proof. – D. Lee Jul 25 '15 at 12:04

2 Answers2

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Proffering the following counterexample to the wish that the conditions

  1. $a$ is a unit of (the ring of integers of) an algebraic number field such that $a+\dfrac1a\in\Bbb{R}$, and
  2. $f(a)f(\dfrac1a)\ge0$ 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.

Jyrki Lahtonen
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  • Wow. I couldn't even guess the form of this counterexample. You're a genius! :o – D. Lee Jul 25 '15 at 11:51
  • Hmm.. then, any modulus 1 element, which is not a root of unity, is a counterexample of this type. Are there infinitely many such elements? – D. Lee Jul 25 '15 at 12:00
  • Anyway, thank you very much! :-D – D. Lee Jul 25 '15 at 12:10
  • Yes. There are infinitely many such numbers. Fix a real part to be some algebraic integer $\in(0,1)$, and find the matching imaginary part (which is then also an algebraic integer). The number $u$ get is then an algebraic integer also. The polynomial $(x-u)(x-\overline{u})$ has a messy linear term and constant term $=1$. Find all the conjugates of that linear term and start multiplying... – Jyrki Lahtonen Jul 25 '15 at 12:49
  • More related discussion in this thread. – Jyrki Lahtonen Sep 16 '15 at 19:50
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Proving the conjecture that the conditions

  1. $a$ 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
  2. $f(a')f(1/a')\ge0$ 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.

Jyrki Lahtonen
  • 133,153