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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) but I've been unable to finish it.

Thank you,

Ben

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Why do you say that roots of unity do not have all Galois conjugates on the unit circle? –  Jason Starr Jun 26 '13 at 0:38
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@Jason, I think what Ben is saying is that he wants numbers whose conjugates are not all on the unit circle, which implies that he wants numbers that are not roots of unity. –  Gerry Myerson Jun 26 '13 at 1:28
    
Clarification: The notation suggests that you might mean to require $a$ to be a primitive element of $K$, i.e. that ${\bf Q}(a)$ is Galois over ${\bf Q}$, or equivalently that the images $a_1,\ldots,a_n$ of $a$ under the $n$ elements of ${\rm Gal}(K/{\bf Q})$ are distinct. Is this the intention, or are some of the $a_j$ allowed to coincide? –  Noam D. Elkies Jun 30 '13 at 20:48
    
@Noam: I hadn't considered this. I don't believe I care if the $a_j$ coincide (as long as none of them are roots of unity). If this is easier to do assuming that all the $a_j$ are distinct, that that'd be good too. My hope was to bound away from zero the product over $1\le j\le n$ of $a_j^m - 1$ for some $m$. –  Ben Weiss Jul 1 '13 at 3:35
    
I guess you mean the product over $1 \leq j \leq k$, not all $j \leq n$ --- and that your desired separation from $1$ might be $\pi/3$, not $\pi/6$. I just posted an argument showing that no positive separation can be guaranteed for all $a$, at least if you don't require $a$ to be primitive (and probably not even then). Still, I think your desired result on $\prod_{j=1}^k |a_j^m-1|$ can be proved by an averaging argument, because the average of $\log|1-z|$ over the unit circle is zero. –  Noam D. Elkies Jul 3 '13 at 17:03

2 Answers 2

up vote 8 down vote accepted
+100

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 (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 $m=953110$.

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Edit: The following "proof" is wrong.

You can't use Kroenecker's theorem directly to $\alpha_1$, ..., $\alpha_k$, because the ratio of $\alpha_i$ and $\alpha_j$ can be root of unity. However it's not really an obstacle. Clearly there exists $m$ such that for all $i$ and $j$ either $\alpha_i^m=\alpha_j^m$, or the ratio $(\alpha_i/\alpha_j)^m$ is not a root of unity. You may use Kroenecker's theorem for numbers $\alpha_1^m$, ..., $\alpha_k^m$.

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Thanks, Oleg. Nice trick ``renaming" the $a_i$ to be an appropriate power of themselves. One question: doesn't Kroenecker's theorem require that we cannot generate any root of unity by multiplying any number of these together? That is, how can we get that their angles must be linearly independent over $\mathbb{Q}$ (as opposed to just pairwise independent). –  Ben Weiss Jun 26 '13 at 5:02
    
@BenWeiss Right. I was careless. The real question is "what are possible multiplicative relations between $\alpha_1$,...,$\alpha_k$". –  Oleg Eroshkin Jun 26 '13 at 13:10

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