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This question is inspired from the post linked below:

Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?

What I am curious about is the following: let $\alpha$ be an algebraic integer lying in a kroneckerian Galois extension $K/\mathbb Q$ (i.e. wherein complex conjugation commutes with the rest of the Galois group, more specifically, in a CM-field). If $|\alpha|<1$, must all the other conjugates of $\alpha$ also lie inside the unit circle?

It seems I am getting heurstic arguments both for and against the statement I am trying to investigate! (Of course, one of them is more naive than the other.)

On one hand, the answer would be positive if I can show that for any $\alpha$ lying in a CM field all conjugates of $\alpha$ must lie on the same circle as $\alpha$. I did manage to show that if $\alpha$ lies on a circle of radius $r$ for some $r>0$ such that $r^n \in \mathbb Q \text{ for some }n$ $\in \mathbb N$ (i.e. $|\alpha| \in \sqrt{\mathbb Q}$), then all conjugates of $\alpha$ must also lie on the same circle, simply because

\begin{multline*} |\sigma\alpha|^2 = \sigma\alpha \cdot \overline{\sigma\alpha}= \sigma(\alpha)\sigma(\overline{\alpha}) = \sigma(\alpha\overline{\alpha}) = \sigma(|\alpha|^2) \\ = \sigma(r^2) = \sigma(r^{2n})^{1/n} = (r^{2n})^{1/n} = r^2 = |\alpha|^2 \end{multline*}

for every $\sigma \in \text{Gal}(K/\mathbb Q)$, hence all conjugates lie on the same circle. So things look somewhat promising, though I am stuck at this point. I am not sure if there is some continuity argument involved either. (Are Galois elements in CM fields continuous maps with respect to archimedean absolute values?). Also, it is to be noted that I haven't used the assumption that $\alpha$ is an algebraic integer, yet.

On the other hand, the Weak Approximation Theorem yields an $\alpha \in K$ for which $|\alpha|>1$ but $|\alpha|_v<1$ for every other archimedean place $v$ of $K$. This would give us an $\alpha$ which serves as a counterexample ... but $\alpha$ need not be an algebraic integer. I tried using the Strong Approximation Theorem (the only version of which I know being Bombieri-Gubler's Theorem 1.4.5.) to choose $\beta$ sufficiently close to $\alpha^{-1}$ at those nonarchimedean places $w$ for which $|\alpha|_w>1$, which gave me an algebraic integer $\beta$ but I lose control over archimedean values of $\beta$. Maybe I'm missing something really simple... In this argument, I haven't gotten any obstruction from the "CM-ness" of $K$ either.

Having said all that, it is possible that there are easy proofs/counterexamples to the statement that I am missing, although I haven't been come up with one of the latter either. I would really appreciate any suggestion or help. Thanks.

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2 Answers 2

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Zero is the only algebraic integer which has all its conjugates strictly inside the complex unit circle. (Look at the norm.)

For explicit examples with conjugates on either side of the unit circle, you can start with a real quadratic field with a totally positive unit that isn't already a square in this field, such as $\varepsilon = 2+\sqrt{3}$. Then take $\alpha=\sqrt{-\varepsilon}$ to obtain a CM extension and an algebraic integer (even an algebraic unit) with a pair of conjugates of absolute value less than $1$ and another pair of absolute value greater than $1$.

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  • $\begingroup$ Thank you, I was having this misconception that the norms of algebraic integers in CM fields should be close to radicals of rational numbers. $\endgroup$
    – asrxiiviii
    Jul 20, 2020 at 19:00
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To expand on GNiklasch's answer, and analyse what you write as well: we always have (when complex conjugation is central in the Galois group) have $\overline{\alpha^{\sigma}} = {\bar \alpha}^{\sigma}$ when $\alpha$ is an algebraic integer in $K$ and $\sigma$ is an element of the Galois group of $K$.

In general, there is no reason why $|\alpha|^{2}$ should be rational for an algebraic integer $\alpha$ in your field $K$. It is true that all algebraic conjugates of $|\alpha|^{2}$ are positive when $\alpha \neq 0$. Then a standard argument (just by the arithmetic-geometric mean inequality) yields that the arithmetic mean of the algebraic conjugate of $|\alpha|^{2}$ is at least one (since the product of all these algebraic conjugates is a positive rational integer). Hence if some algebraic conjugate of $\alpha (\neq 0)$ has absolute value strictly less than one, there must be another algebraic conjugate of $\alpha$ with absolute value greater than one.

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  • $\begingroup$ Thank you for your answer, I realized the error in my first thought process. $\endgroup$
    – asrxiiviii
    Jul 20, 2020 at 19:01

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