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\begin{document}
\vspace*{-50pt}
\centerline{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7
(2007), \#A18}
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\begin{center}
\uppercase{\bf A remark on the Chebotarev theorem \\about roots of unity}
\vskip 20pt
{\bf F. Pakovich\footnote{Research supported by the ISF, Grant No.
979/05}}\\ {\smallit Department of Mathematics, Ben Gurion University,
P.O.B. 653, Beer-Sheva 84105, Israel}\\ {\tt pakovich@math.bgu.ac.il}\\
\end{center}
\vskip 20pt
\centerline{\smallit Received: 2/21/07,
Accepted: 4/8/07, Published: 4/12/07}
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\centerline{\bf Abstract}

\noindent
Let $\Omega$ be a matrix 
with entries 
$a_{i,j}=\omega^{ij},$ $1\leq i,j \leq n,$ where $\omega=e^{2\pi \sqrt{-1}/n},$ $n\in \mathbb N.$ 
The Chebotarev theorem states that if $n$ 
is a prime then any minor of $\Omega$ is non-zero. 
In this note we provide an analogue of this statement for composite $n.$ 

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\markright{\smalltt INTEGERS: \smallrm ELECTRONIC
 JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A18\hfill}

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Let $\Omega$ be a matrix 
with entries 
$a_{i,j}=\omega^{ij},$ $1\leq i,j \leq n,$ where $\omega=e^{2\pi \sqrt{-1}/n},$ $n\in \mathbb N.$ 
The Chebotarev theorem states that if $n$ 
is a prime then any minor of $\Omega$ is non-zero. Chebotarev's proof 
of this theorem and the references to other proofs can be found in \cite{l}. Yet other 
proofs can be found in recent papers \cite{g} and \cite{t}.

For a complex polynomial $P(z)$ denote by $w(P)$ the number of non-zero coefficients of $P(z).$
It is easy to see that the Chebotarev theorem is equivalent to the following statement: if a non-zero polynomial $P(z),$ $\deg P(z) \leq n-1,$  
has $k$ different roots which are $n$-roots of unity 
then $w(P)> k$ whenever $n$ is a prime. 

A natural question is: How small can  $w(P)$ be if $n$ is a composite
number ? The example %of the polynomial 
$D_{n,r,l}(z)=z^l(1+z^r+z^{2r}+ \cdots + z^{(\frac{n}{r}-1)r}),$ where
$r\vert n$, $0\leq l \leq r-1,$ shows that $w(P)$ could be as small as
$n/(n-k).$   In this note we show that actually it is the ``worst''
possible case.


\noindent{\bf Theorem.} {\it Let $n$ be a composite number and $P(z)$ be a non-zero complex polynomial, $\deg P(z) \leq n-1.$ 
Suppose that $P(z)$ has
exactly   
$k$ different roots which are $n$-roots of unity. Then the inequality 
$$ w(P)\geq  \frac{n}{n-k}\eqno(*)$$ holds. Furthermore, the equality attains if and only 
if $P(z)$ up to a multiplication by a complex number coincides with
$D_{n,r,l}(\omega^j z)$ 
%$$P(z)=\alpha z^l(1+(\nu^j)z^r+ (\nu^j)^{2}z^{2r}+ ... + (\nu^j)^{(\frac{n}{r}-1)}z^{(\frac{n}{r}-1)r}),$$ 
for some $j,$ $0\leq j \leq n-1,$ and $r,l$ as above.  
}

\noindent{\it Proof.} Let $P(z)=p_{0}+p_{1}z+...+p_{n-1}z^{n-1}$ and let 
$\displaystyle C=\begin{pmatrix}p_0 & p_1 & ... & p_{n-1}\\
p_{n-1} & p_0 & ...& p_{n-2}\\
... & ... & ... & ...\\
p_1& p_2 & ... & p_0\end{pmatrix}$ be the circulant matrix generated by
the coefficients  of $P(z).$ We will denote the row vectors of $C$ by
$\vec{t_j},$ $0\leq j \leq n-1$. Set $r={\rm rk}\, C.$
The key observation is that the number $k$ is equal to the number $n-r$. To establish it notice that eigenvectors of $C$ are
$$\vec{f_i}=((\omega^i)^0,(\omega^i)^1,...,(\omega^i)^{(n-1)}),\ \ \ \ \ 0\leq i \leq n-1,$$ and the corresponding eigenvalues are $P(\omega^{i}),$ $0\leq i \leq n-1.$ Furthermore, the vectors $\vec{f_i},$ $0\leq i \leq n-1,$ form a basis of $\mathbb C^n$. The matrix $C$ is diagonal with respect to this basis and therefore 
$k=n-r$.

It follows that in order to prove inequality (*) it is enough to establish
the inequality $$w(P)\hskip 0.05cm r\geq n. \eqno(**)$$ This inequality essentially is a particular case 
of Theorem B in \cite{g} and can be established easily as follows (\cite{g}).
Let $V$ be a vector space generated by the vectors $\vec{t}_j,$ $0\leq j \leq n-1$, and
$R \subseteq
\{\vec{t}_0, \vec{t}_1, ... , \vec{t}_{n-1}\}$
consisting of $r$ vectors which generate $V.$ Clearly, 
for any $i,$ $1\leq i \leq n,$ there exists a vector $\vec{v}\in V$ for which
its $i$-th coordinate is distinct from zero. Since
each vector from $R$ has exactly $w(P)$
non zero coordinates it follows that (**) holds.

For a vector $\vec{v}\in \mathbb C^n$ denote by ${\rm supp}\{\vec{v}\}$ the set consisting of numbers $i,$ 
$1\leq i \leq n,$ for which the $i^{\mathrm{th}}$ coordinate of $\vec{v}$
is non-zero. Observe now that the equality in (**) is attained only if
for any two vectors $\vec{v}_1,\vec{v}_2\in R$ we have ${\rm supp}\{\vec{v}_1\}\,\cap\,{\rm supp}\{\vec{v}_2\}=\emptyset$. 
This implies easily
that ${\rm supp}\{\vec{t}_0\}$ consists of numbers all 
congruent modulo $r$ to the same number $l,$ $0\leq l \leq r-1.$ 
Therefore, $P(z)=z^lQ(z^r)$ for some polynomial 
$Q(z)=q_{0}+q_{1}z+...+q_{(n/r)-1}z^{(n/r)-1}$
and number $l,$ $0\leq l \leq r-1.$

Furthermore, since the vectors $\vec{t}_0,\vec{t}_r,\vec{t}_{2r}, ... , 
\vec{t}_{(n/r)-1}$ have equal supports
the equality in (**) implies that any two of them are proportional. Therefore, the rank of the circulant matrix $W$ generated by the coefficients of $Q(z)$ equals 1. This implies that
the vector $\vec{q}=\{q_{0},q_{1},...,q_{(n/r)-1}\}$ is orthogonal to $(n/r)-1$ vectors from the   
collection $$\vec{g_j}=((\nu^j)^0,(\nu^j)^1,...,(\nu^j)^{(n/r)-1}),\ \ \ \ 0\leq j \leq (n/r)-1,$$ 
where $\nu=\omega^r.$ Since $\vec{g}_j,$ $0\leq j \leq (n/r)-1,$ are linearly independent
this implies that there exists $\alpha\in \mathbb C$ such that $\vec{q}=\alpha \vec{g}_j$ for some $0\leq j \leq (n/r)-1.$
\hfill $\Box$ 


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\bibitem {g} D. Goldstein, R. Guralnick, I. Isaacs,  \textit{
Inequalities for finite group permutation modules},
Trans. Am. Math. Soc. 357, No.10, 4017-4042 (2005)

\bibitem {l} P. Stevenhagen, H. Lenstra, \textit{
Chebotarev and his density theorem,}
Math. Intell. 18, No.2, 26-37 (1996)


\bibitem {t} T. Tao, \textit{
An uncertainty principle for cyclic groups of prime order},
Math. Res. Lett. 12, No.1, 121-127 (2005).

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