Let $\alpha$ be an algebraic integer with minimal polynomial $f$. Is there some natural condition on $f$ to guarantee that all Galois conjugates of $\alpha$ have absolute value at least $1$? Equivalently, under any embedding of $\alpha$ into $\mathbb{C}$, $\alpha$ has absolute value at least $1$?

The corresponding question with absolute values less than or equal to $1$ seems to be have been studied quite a bit ("Pisot numbers").

EDIT: In the definition of Pisot numbers, we actually require that all of the conjugates except for $1$ have absolute value less than or equal to $1$. Indeed, an algebraic integer whose conjugates all lie inside the unit disk is a root of unity.

I would also be interested in necessary conditions. For example, if one embedding of $\alpha$ has absolute value greater than $1$, then the constant term of the minimal polynomial cannot be $\pm 1$.

  • $\begingroup$ Is this problem equivalent to that for salem numbers, after all $|\sigma(1/\alpha)|=1/|\sigma(\alpha)|$. It sounds like you dont require one absolute value to equal 1 though. $\endgroup$ – Alex J Best Dec 21 '18 at 20:16
  • $\begingroup$ @AlexJBest There are number like $\sqrt{2}$ whose reciprocal is not an algebraic integer. $\endgroup$ – vukov Dec 21 '18 at 20:19
  • $\begingroup$ cool I missed the intger bit thanks! $\endgroup$ – Alex J Best Dec 21 '18 at 20:22
  • $\begingroup$ My gut feeling is that this is less interesting than, say, Pisot (or Salem).Basically because for any algebraic integer $\alpha$ we have that $\alpha+n$ and $n\alpha$ both have this property for large enough $n$ - there are plenty of large algebraic integers with this property. Making all but one of the conjugates small OTOH is much more difficult to arrange. $\endgroup$ – Jyrki Lahtonen Dec 21 '18 at 20:59

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