# Algebraic integers where all conjugates have absolute value at least 1

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 $$\pm 1$$.

• 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. – Alex J Best Dec 21 '18 at 20:16
• @AlexJBest There are number like $\sqrt{2}$ whose reciprocal is not an algebraic integer. – vukov Dec 21 '18 at 20:19
• cool I missed the intger bit thanks! – Alex J Best Dec 21 '18 at 20:22
• 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. – Jyrki Lahtonen Dec 21 '18 at 20:59