I'm looking for a reference or proof of the following. Let $K/\mathbb{Q}$ be a finite Galois extension of degree $n$. Let $a\_1,\ldots,a\_n$ be Galois conjugate elements in the ring of integers of $K$ with $a\_1, \ldots, a\_k$ ($1 \le k < n$) lying on the unit circle (hence are not roots of unity). Show there are infinitely many positive integers $m$ such that $a\_1^m, \ldots, a\_k^m$ are all simultaneously far from $1$ (where ``far" needs to be that the angle is at least $\pm\pi/6$).


It seems that Kroenecker's generalization of Dirichlet's simultaneous approximation should work as a hammer for this nail (see: [Simultaneous diophantine approximation](https://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation)) but I've been unable to finish it.


Thank you,


Ben



Simultaneous Powers Far From 1
There is no proof because the desired result is false!
Indeed, for any $\theta \gt 0$ there exists an algebraic integer
of degree $n$ with $k \lt n$ conjugates $a\_1,\ldots,a\_k$ on the unit circle
such that for each $m$ at least one of $a\_1^m,\ldots,a\_k^m$ 
is within $\theta$ of $1$.


Suppose $\theta \geq \pi/r$ for some integer $r$.
Note that an algebraic integer with a conjugate $x$ on the unit circle
must be a unit, because $\bar x$ is an algebraic conjugate of $x$
(and thus also an algebraic integer) such that $x \bar x = 1$.
Let $b \in {\bf R}$, then, be an algebraic unit of degree $2r+4$,
with $2r+2$ conjugates $b\_1,\ldots,b\_{2r+2}$ on the unit circle,
and $b$ the unique conjugate such that $|b| \gt 1$;
and let $K$ be the splitting field of ${\bf Q}$.
Assume that the subgroup of ${\rm Gal}(K/{\bf Q})$
that fixes $b$ acts transitively on $b\_1,\ldots,b\_{2r+2}$.
Then take $a = b/b\_1$.
[We shall see later how to construct such $b$; that's where I needed
the clarification on whether ${\bf Q}(a)$ is allowed to be a
non-Galois extension of ${\bf Q}$, though it may be possible to have
${\bf Q}(a)/{\bf Q}$ Galois.]
The conjugates of $a$ are the quotients $\beta/\beta'$
where $\beta,\beta'$ are conjugates of $b$ with $\beta' \neq \beta^{\pm 1}$.
I claim that for each $m$ at least one of these conjugates is
within $\pi/r$ of $1$, and thus *a fortiori* within $\theta$ of $1$.
Indeed the $2r+2$ numbers $b\_j^m$ come in $r+1$ conjugate pairs,
and none equals $\pm 1$. Therefore $r+1$ of the $b\_j^m$ are on
the open arc $\lbrace e^{i\psi}: 0 \lt \psi \lt \pi \rbrace$
of length $\pi$. We conclude that two of them are within $\pi/r$
of each other, and their ratio is a conjugate of $a^m$
whose angular distance from $1$ is less than $\pi/r \leq \theta$,
as claimed.


[Remark: $a$ has $n = (r+1)(2r+4)$ conjugates. Indeed there are
$2r+4$ choices of $\beta$, and for each one $2r+2$ choices of $\beta'$;
but $\beta / \beta' = ({\beta'}^{-1}) / (\beta^{-1})$, so each conjugate
arises at least twice. But $a = \beta/\beta'$ only for
$(\beta,\beta') = (b,b\_1)$ and $(b\_1^{-1},b^{-1})$, because
$|b| = |a| = |\beta|/|\beta'|$, and $|\beta|=1$ for all
$\beta \neq b^{\pm 1}$. It follows that the number $k$ of
conjugates of norm $1$ is $\frac12(2r+2)2r = 2(r^2+r)$.]


It remains to find our unit $b$. Let $F \subset {\bf R}$ be any totally real
number field of degree $r+2$ whose normal closure has Galois group $S\_{r+2}$.
Choose positive $c \in F$ all of whose other embeddings are negative,
and assume that $F' := F(c^{1/2})$ has ${\rm Gal}(F'/{\bf Q})$ the full
hyperoctahedral group $\lbrace \pm 1 \rbrace^{r+2} \rtimes S\_{r+2}$
(which is the usual case).
Let $\sigma$ be the Galois involution of $F'/F$,
which permutes the two real embeddings of $F'$
and acts as Galois conjugation on the $r+1$ complex embeddings.
By Dirichlet, the group of units of $F'$ has rank $r$,
and its $\sigma$-invariant subgroup has rank $r-1$.
Hence there is a rank-$1$ subgroup of units inverted by $\sigma$.
Let $b \in F'$, then, be a unit $b$ of infinite order
such that $b^{\sigma} = b^{-1}$. Then all conjugates of $b$
other than $b^{\pm 1}$ lie on the unit circle, and are permuted
transitively by ${\rm Gal}(K/{\bf Q}(b))$ because
${\rm Gal}(K/{\bf Q}))$ is as large as possible.
Thus $a = b/b\_1$ works as claimed, **QED**.


To get explicit examples for small $r$, we can let $b$ be a
[Salem number](http://en.wikipedia.org/wiki/Salem_number)
(which usually has hyperoctahedral Galois group, though that's
not guaranteed). For example, for $r=3$ we can use for $b$
Lehmer's number, the larger real root of
$y^{10} + y^9 - y^7 - y^6 - y^5 - y^4 - y^3 + y + 1$.
This makes $a$ a unit of degree $n=40$ with $k=24$
conjugates on the unit circle such that for each $m$
at least one conjugate pair of conjugates $a\_k$ satisfies
$|a\_k^m - 1| \lt 1$. The supremum over $m$ of $\prod\_{j=1}^{24} |a\_j^m - 1|$
still exceeds $1$, though probably not by as much as it would for a
typical unit with $24$ conjugates on the unit circle:
the value is apparently $2^{24}/5^5 = 5368.70912$,
nearly attained when the eight $b\_k^m$ are approximately at
$1$, $1$, $-1$, $-1$, and the four roots of $5z^4 + 6z^2 + 5$.
Numerically, for $m \leq 10^7$ the largest product observed is
$5359.938\ldots$ for $m=953110$.


