\input amstex
\documentstyle{amsppt}
\magnification=\magstep1
\NoBlackBoxes
\pagewidth{6.5truein}
\pageheight{8.5truein}
\loadbold
\def\A{{\Cal A}}
\def\B{{\Cal B}}
\def\L{{\Cal L}}
\def\M{{\Cal M}}
%\def\S{{\Cal S}}
\def\D{{\Cal D}}
\def\d{{\delta}}
\def\cee{{\Bbb C}}
\def\que{{\Bbb Q}}            
\def\real{{\Bbb R}}
\def\zed{{\Bbb Z}}
\def\bm{\text{\bf m}}
\def\bt{\text{\bf t}}
\def\bv{\text{\bf v}}
\def\bx{\text{\bf x}}
\def\bwy{\text{\bf y}}
\def\bL{\text{\bf L}}
\def\ba{{\boldsymbol\alpha}}
\def\bb{{\boldsymbol\beta}}
\def\bxi{{\boldsymbol\xi}}
\def\bl{{\boldsymbol\lambda}}
\def\bo{{\boldkey 0}}
\def\bu{{\boldkey 1}}
\def\ep{\varepsilon}
\def\nrm{\text{Norm}_{k/{\Bbb Q}}}
\def\sgn{\operatorname{sgn}}
\def\trc{\operatorname{Trace}_{k/\Bbb Q}}
\def\arcsinh{\operatorname{arcsinh}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\topmatter
\title The least nonsplit prime in Galois extensions of $\que$ \endtitle
\author Jeffrey D. Vaaler*\\\ Jos\'e Felipe Voloch**\endauthor
\leftheadtext{JEFFREY D. VAALER, JOS\'E FELIPE VOLOCH}
\address Jeffrey D. Vaaler,
Department of Mathematics, University of Texas, 
Austin, TX 78712, {\rm e-mail:} {\tt vaaler\@math.utexas.edu} 
\endaddress 
\address Jos\'e Felipe Voloch, 
Department of Mathematics, University of Texas, 
Austin, TX 78712, {\rm e-mail:} {\tt voloch\@math.utexas.edu} 
\endaddress  
\thanks * Research of the first author was supported in part by the National 
Science Foundation (DMS-9622556) and the Texas Advanced Research 
Project, \endthanks
\thanks ** Research of the second author was supported in part by the
National Security Agency (MDA904-97-1-0037). \endthanks
\endtopmatter

\document 
\baselineskip=18pt		%% DOUBLE-SPACED DRAFT MODE 

\head 1. Introduction\endhead%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $k$ be a Galois extension of $\que$ with $[k:\que]=d\ge 2$. The purpose of this paper is
to give an upper bound for the least prime which does not split completely in $k$ in terms of
the degree $d$ and the discriminant $\Delta_k$. Our estimate
improves on the bound given by Lagarias, Montgomery and Odlyzko \cite{2}.  We note, however, that 
with the assumption of the generalized Riemann hypothesis much stronger bounds have been
obtained by Murty \cite{6}.  In fact the analytic method employed in \cite{6} can be used to 
produce an unconditional bound of the same general type as ours.  Our method is essentially
elementary.  It is based on an application of the product formula to the binomial coefficient
${\alpha\choose N}$, where $\alpha$ is an irrational algebraic integer in $k$ 
and $\trc(\alpha)=0$. A similar idea has been used in \cite{7} to give a lower bound on the
number of primes that do split completely in $k$.  In the special case $k=\que(\sqrt{p})$
our argument differs insignificantly from that used by Gauss in the course of his first
proof of quadratic reciprocity \cite{1, art. 129}.

\proclaim{THEOREM 1} There exists a constant $c_1>0$ with the following property: if 
$$\exp\{c_1d(\log d)^2\} \le |\Delta_k|,\eqno(1.1)$$
then there exists a prime $p$ such that $p$ does not split completely in $k$ and
$$p\le 48d^2|\Delta_k|^{1/2(d-1)}\eqno(1.2)$$
\endproclaim

Here the constant $c_1$ is readily computed from constants which occur in the error
term in the prime number theorem.
 
\head 2. Nonarchimedean estimates\endhead%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Throughout this section we assume 
that all primes $p\le dN(d-1)^{-1}$ split in $k$, where $N\ge d$ is a positive integer parameter.  We
further assume that $\alpha$ is a nonzero algebraic integer in $k$ with $[\que(\alpha):\que]=\d$
and $\trc(\alpha)=0$.  Then we write $\alpha=\alpha_1, \alpha_2, \dots , \alpha_{\d}$ for the {\it distinct} 
conjugates of $\alpha$ in $k$ and $$f(x)= \prod_{i=1}^{\d} (x-\alpha_i)$$
for the minimal polynomial of $\alpha$ over $\que$.  Obviously $f$ is a monic, irreducible
polynomial in $\zed[x]$ with $2\le \d =\deg(f)$ and $\d~|~d$.  We also define
$$A_N(\alpha)=\nrm\Bigl\{{\alpha\choose N}\Bigr\} 
= (-1)^{dN}(N!)^{-d}\Bigl\{\prod_{n=0}^{N-1} f(n)\Bigr\}^{d/\d},\eqno(2.1)$$
where ${x\choose N}$ is the binomial coefficient.  Clearly $A_N(\alpha)$ is a nonzero rational number.

\proclaim{LEMMA 2} For each prime $p$ with $p\le dN(d-1)^{-1}$, the congruence 
$$f(x)\equiv 0 \mod p$$ has at least one root in the set $\{0, 1, 2, \dots ,[{(\d-1)p}/{\d}]\}$.
\endproclaim

\demo{Proof} Let $p$ be a prime with $p\le dN(d-1)^{-1}$.  All embeddings of $k$ in an algebraic closure
$\overline{\que_p}$ are contained in $\que_p$.  Hence all roots of $f$ occur in $\zed_p$ and therefore 
$f$ splits in $\zed/p\zed[x]$.  Let $a_1, a_2, \dots ,a_{\d}$ be elements of $\{0, 1, 2, \dots ,p-1\}$ 
such that $$f(x)\equiv \prod_{i=1}^{\d} (x-a_i) \mod p.$$  If $p\le \d$ then the result is trivial,
so we may assume that $\d < p$.  Because $\trc(\alpha)=0$ we have 
$$\sum_{i=1}^{\d} a_i \equiv 0 \mod p.\eqno(2.2)$$
Now assume, contrary to the statement of the lemma, that ${(\d-1)p}/{\d} < a_i \le p-1$ for each
$i=1, 2, \dots ,\d$. Then we get
$$(\d-1)p < \sum_{i=1}^{\d} a_i \le {\d}(p-1),$$
which contradicts $(2.2)$
\enddemo


\proclaim{LEMMA 3} The number $A_N(\alpha)$ is a nonzero integer.  
\endproclaim

\demo{Proof}  Let $p$ be a prime with $p\le N$.  As before all roots of $f$ occur in $\zed_p$.  It follows that
${\alpha_i\choose N}$ belongs to $\zed_p$ for each $i=1, 2, \dots ,\d$.  In particular the $p$-adic
absolute value of $A_N(\alpha)$ satisfies $|A_N(\alpha)|_p\le 1$ for each $p\le N$, and of
course for $p>N$ the bound $|A_N(\alpha)|_p\le 1$ is trivial.  Thus $A_N(\alpha)$ is a nonzero integer.
\enddemo

\proclaim{LEMMA 4} For each prime $p$ such that $N<p\le dN(d-1)^{-1}$, we have
$$\log|A_N(\alpha)|_p \le d{\d}^{-1}(-\log p).\eqno(2.3)$$
\endproclaim

\demo{Proof}  As $\log|N!|_p = 0$ for $N<p$, we find that
$$\eqalign{\log|A_N(\alpha)|_p&=d{\d}^{-1}\Big\{\sum_{n=0}^{N-1}\log|f(n)|_p\Big\} -d\log|N!|_p\cr
&=d{\d}^{-1}\sum_{n=0}^{N-1}\sum_{\scriptstyle m=1\atop\scriptstyle p^m|f(n)}^{\infty}(-\log p)\cr
&\le d{\d}^{-1}(-\log p)\sum_{\scriptstyle n=0\atop\scriptstyle p|f(n)}^{N-1} 1.\cr}\eqno(2.4)$$
By hypothesis we have $d\le N < p\le dN(d-1)^{-1}$, and therefore
$$[{(\d-1)p}/{\d}]\le [(d-1)p/d] < (d-1)p/d \le N.$$  By Lemma 2 the sum on the right 
of $(2.4)$ contains at least one nonzero term and this verifies $(2.3)$.  
\enddemo

\proclaim{THEOREM 5} There exists a constant $c_2\ge 20$ such that
$$\sum_{p} \log|A_N(\alpha)|_p \le -N{\d}^{-1}\eqno(2.5)$$
whenever $\exp\{c_2(\log d)^2\}\le N$.
\endproclaim

\demo{Proof} From the error term in the prime number theorem there exist constants $c_3>0$ 
and $c_4>0$ such that
$$|\sum_{p\le X} \log p \ - X| \le c_3X\exp\{-c_4\sqrt{\log X}\}$$
for all $X\ge 2$.  Hence there exists $c_2 \ge 20$ such that
$$|\sum_{p\le X} \log p \ - X| \le {{X}\over{d(2d-1)}}\eqno(2.6)$$
whenever $\exp\{c_2(\log d)^2\}\le X$.  Let $\exp\{c_2(\log d)^2\}\le N$,
set $X=dN(d-1)^{-1}$ and $\epsilon=(d(2d-1))^{-1}$.  Then $(2.3)$ and $(2.6)$ imply that
$$\eqalign{\sum_{p} \log|A_N(\alpha)|_p &\le d{\d}^{-1}\sum_{N<p\le X}(-\log p)\cr
	&=d{\d}^{-1}\big\{(N-X)+(\sum_{p\le N}\log p -N)+(X-\sum_{p\le X}\log p)\big\}\cr	
	&\le d{\d}^{-1}\{N-X + \epsilon(N+X)\}\cr
	&=-N{\d}^{-1}.\cr}$$
\enddemo

\head 3. Archimedean estimates\endhead%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In this section we assume that $f$ is a polynomial in $\real[x]$ with $\deg(f)=M\ge 1$.  Then 
we write $\alpha_1, \alpha_2, \dots ,\alpha_L$
for the {\it distinct} roots of $f$ in $\cee$ so that
$$f(x)=c_0\prod_{l=1}^L(x-\alpha_i)^{e(l)}\quad\text{with}\quad e(l)\in \{1, 2, \dots \}\quad\text{and}\quad c_0\not=0.$$
It follows that the Mahler measure $\mu(f)$ is given by
$$\log \mu(f) = \log |c_0| + \sum_{l=1}^L e(l)\log^+|\alpha_l|.$$
We also require the norm
$$\|f\|_{\infty} = \sup\lbrace |f(z)|: z\in \cee,\ |z| \le 1\rbrace.$$
And by the square-free kernel of $f$ we understand the polynomial
$$q(x)=\prod_{l=1}^L(x-\alpha_l).$$

\proclaim{LEMMA 6} Let $f$ be a polynomial in $\real[x]$, $q$ the square free kernel of $f$, and
$$B_f=\{\beta\in \real: f'(\beta)=0\ \text{and}\ f(\beta)\not=0\}.$$
Then we have 
$$|B_f|\le L-1\quad\text{and}\quad\sum_{\beta \in B_f} \log^+|\beta| \le \log \|q\|_{\infty}.\eqno(3.1)$$
\endproclaim

\demo{Proof} There exists a polynomial $p(x)$ in $\real[x]$ uniquely determined by the 
identity
$${{f'(x)}\over{f(x)}}=\sum_{l=1}^L{{e(l)}\over{x - \alpha_l}}={{p(x)}\over{q(x)}}.$$
Clearly we have
$$B_f=\{\beta\in \real: p(\beta)=0\}.$$ 
>From the identity $f'(x)q(x) = f(x)p(x)$ we find that $\deg(p)=L-1$ and
the leading coefficient of $p$ is $M$.  Therefore $|B_f|\le L-1$ and
$$\log M + \sum_{\beta\in B_f} \log^+|\beta| \le \log \mu(p).\eqno(3.2)$$
By a basic inequality for $\mu$ (see \cite{5}) we get
$$\log \mu(p)+\log \mu(f)=\log \mu(q)+\log \mu(f') \le \log \mu(q)+\log M +\log \mu(f).\eqno(3.3)$$
The remaining inequality in $(3.1)$ follows from $(3.2)$ and $(3.3)$. 
\enddemo

\proclaim{LEMMA 7} Let $f$ be a polynomial in $\real[x]$ and $q$ the square free kernel of $f$.  
Then we have
$$\int_U^V \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx 
\le M\big\{\log^+|U| + \log^+|V| + 2\log^+ \|q\|_{\infty}\big\}+2L\log^+\|f\|_{\infty}.\eqno(3.4)$$
\endproclaim

\demo{Proof}  Write $B_f(U,V)$ for the set of distinct roots of $f'$ which occur
in the interval $(U,V)$ and which are not roots of $f$.  To begin with we will establish the inequality 
$$\int_U^V \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx 
\le \log^+|f(U)| + \log^+|f(V)| + 2\sum_{\beta \in B_f(U,V)} \log^+|f(\beta)|.\eqno(3.5)$$
Suppose that $(u,v)\subseteq \real$ is a bounded open interval
such that $$1 < f(x)\quad\text{whenever}\quad u<x<v.\eqno(3.6)$$  Then let
$$B_f(u,v) = \lbrace \beta_1, \beta_2, \dots \beta_J \rbrace,$$ and write
$$u=\beta_0 < \beta_1 < \beta_2 < \dots < \beta_J < \beta_{J+1} = v.$$
Obviously $J=0$ in case $B_f(u,v)$ is empty.  In view of $(3.6)$ we have
$$\eqalign{\int_u^v \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx 
&= \sum_{j=0}^J \int_{\beta_j}^{\beta_{j+1}} \bigl|{f'(x)\over f(x)}\bigr| \ dx\cr
&= \sum_{j=0}^J \bigl| \int_{\beta_j}^{\beta_{j+1}} {f'(x)\over f(x)} \ dx \bigr|\cr
&= \sum_{j=0}^J \bigl| \log f(\beta_{j+1}) - \log f(\beta_j) \bigr| \cr
&\le \sum_{j=0}^J \max \bigl( \log^+|f(\beta_{j+1})|,\ \log^+|f(\beta_j)| \bigr)\cr
&\le \log^+|f(u)| + \log^+|f(v)| + 2\sum_{\beta \in B_f(u,v)} \log^+|f(\beta)|.\cr}\eqno(3.7)$$
It is clear that $(3.7)$ continues to hold if
$$f(x) < -1\quad\text{whenever}\quad u<x<v.\eqno(3.8)$$

Next we write
$$\lbrace x\in \real : U<x<V,\ 1 < |f(x)|\rbrace = \bigcup_{k=1}^K (u_k, v_k),$$
where $\lbrace (u_k, v_k) : k=1, 2, \dots K\rbrace$ is a finite, disjoint collection of open
intervals. Then we have
$$\int_U^V \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx = 
\sum_{k=1}^K \int_{u_k}^{v_k} \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx\eqno(3.9)$$
and so we can apply the estimate in $(3.7)$ to each term in the sum on the right
of $(3.9)$.  In doing so we note that if $U<u_k<V$ then, since $x\rightarrow \log^+|f(x)|$
is continuous, we have $\log^+|f(u_k)| = 0$, and similarly if $U<v_k<V$.  It follows then
that 
$$\eqalign{\int_U^V \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx 
&=\sum_{k=1}^K \int_{u_k}^{v_k} \bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx\cr
&\le \log^+|f(U)| + \log^+|f(V)| + 2\sum_{k=1}^K \sum_{\beta\in B(u_k,v_k)} \log^+|f(\beta)|\cr
&=\log^+|f(U)| + \log^+|f(V)| + 2\sum_{\beta\in B_f(U,V)} \log^+|f(\beta)|,\cr}$$
and this verifies the inequality $(3.5)$.

In order to further estimate the terms on the right of $(3.5)$ we employ the inequality
$$|f(z)| \le \|f\|_{\infty} \max(1,|z|)^M$$
which follows easily from the maximum modulus theorem.  
Also, we have $$\log^+|w_1w_2| \le \log^+|w_1| + \log^+|w_2|$$
for all pairs of complex numbers $w_1$ and $w_2$.  Combining these observations
we find that 
$$\log^+|f(z)| \le \log^+\|f\|_{\infty} + M\log^+|z|.\eqno(3.10)$$
Now  we can combine $(3.1)$, $(3.5)$, $(3.10)$ and so establish the bound:
$$\eqalign{\int_U^V &\bigl|{d\over dx} \log^+|f(x)|\bigr| \ dx\cr 
&\le M\log^+|U|+M\log^+|V|+(2|B_f|+2)\log^+\|f\|_{\infty}+2M\sum_{\beta\in B_f}\log^+|\beta|.\cr
&\le M\log^+|U|+M\log^+|V|+2L\log^+\|f\|_{\infty}+2M\log^+\|q\|_{\infty}}\eqno(3.11)$$
This proves the lemma.
\enddemo

\proclaim{LEMMA 8} Let $f$ be a polynomial in $\zed[x]$, $q$ the square-free kernel of $f$, $\deg(f)=M$
and $\deg(q)=L$. Let $N$ be a positive integer and assume that that $f$ has no roots in the set 
$\lbrace 0, 1, 2, \dots , N-1\rbrace.$  Then we have
$$\eqalign{\bigl|\sum_{n=0}^{N-1}\log|f(n)| - &\int_0^N \log|f(x)|\ dx\bigr|\cr
&\le M(2 +\log N) + (L+{\textstyle{1\over 2}})\log\|f\|_{\infty}+M\log\|q\|_{\infty}.\cr}\eqno(3.12)$$
\endproclaim 

\demo{Proof}  From a standard application of Stieltjes integration we obtain the identity
$$\eqalign{\sum_{n=0}^{N-1}\log|f(n)| &= \sum_{n=0}^{N-1} \log^+|f(n)|\cr
&= \int_{0-}^{N-} \log^+|f(x)|\ d\lbrace[x]+{\textstyle{1\over 2}}\rbrace\cr
&= \int_0^N \log^+|f(x)|\ dx-{\textstyle{1\over 2}}\log^+|f(N)|
+{\textstyle{1\over 2}}\log^+|f(0)|\cr
&\qquad +\int_0^N B_1(x){d\over dx} \log^+|f(x)|\ dx.\cr}\eqno(3.13)$$ 
Here $$B_1(x)= x-[x]-{\textstyle{1\over 2}}\ \text{when}\ x\notin \zed,\ \text{and}
\ B_1(x)=0\ \text{when}\ x\in \zed,$$
is the first periodic Bernoulli polynomial.  We use the bound $|B_1(x)| \le {1\over 2}$,
the estimate $(3.4)$ from 
the previous lemma, $(3.10)$ and $(3.13)$.  In this way we arrive at the inequality
$$\eqalign{\bigl|\sum_{n=0}^{N-1}\log|f(n)| &- \int_0^N \log^+|f(x)|\ dx\bigr|\cr
&\le {\textstyle{1\over 2}} \max\big\{\log^+|f(0)|,\ \log^+|f(N)|\big\} 
+ {\textstyle{1\over 2}} \int_0^N \bigl|{d\over dx} \log^+|f(x)|\bigr|\ dx\cr
&\le M\log N + (L+{\textstyle{1\over 2}})\log^+\|f\|_{\infty}+M\log^+\|q\|_{\infty}.\cr}\eqno(3.14)$$
Next we observe that 
$$\eqalign{\bigl|\int_0^N \log|f(x)|\ dx-\int_0^N \log^+|f(x)|\ dx\bigr|
&=\int_0^N \log^-|f(x)|\ dx\cr
&\le\int_{\real} \log^-|f(x)|\ dx\cr
&\le 2M.\cr}\eqno(3.15)$$
Then we combine $(3.14)$ and $(3.15)$.  We find that 
$$\eqalign{\bigl|\sum_{n=0}^{N-1}\log|f(n)| - &\int_0^N \log|f(x)|\ dx\bigr|\cr
&\le M(2 +\log N) + (L+{\textstyle{1\over 2}})\log^+\|f\|_{\infty}+M\log^+\|q\|_{\infty}.\cr}\eqno(3.16)$$
To complete the proof we note that $1\le \|f\|_{\infty}$ and 
$1\le \|q\|_{\infty}$, because both $f$ and $q$ belong to $\zed[x]$. 
\enddemo

We define $\rho:\cee\rightarrow \real$ by
$$\rho(z) = {\textstyle{1\over 2}}\int_{-1}^1 \log|t-z|\ dt + 1.\eqno(3.17)$$
If follows from the basic theory of logarithmic potential 
functions (see \cite {3, Appendix 4, \S 1.}) that $\rho$ is nonnegative,
continuous, subharmonic, and the restriction of $\rho$ to $\cee\setminus[-1,1]$ is
harmonic.  For our purposes we require information about $\rho$ in the closed
unit disc.

\proclaim{LEMMA 9} For all complex $z=x+iy$ with $|z|\le 1$ we have 
$$\rho(z)\le{{\pi |y|}\over 2} + (\log 2)|z|^2.$$
\endproclaim

\demo{Proof}  Let $\psi(z)$ be defined in the closed unit disc by
$$\psi(z) = \sum_{n=1}^{\infty}{{z^{2n}}\over{2n(2n-1)}}.$$
In the upper half-disc $\{z\in \cee:0<\Im(z),\ |z|<1\}$ we find that
$$\eqalign{{\textstyle{1\over 2}}\int_{-1}^1 \log(z-t)\ dt + 1
&= {\textstyle{1\over 2}}(1-z)\log(z-1) + {\textstyle{1\over 2}}(1+z)\log(z+1)\cr
	&={{\pi i}\over 2} - {{\pi iz}\over 2} + \sum_{n=1}^{\infty}{{z^{2n}}\over{2n(2n-1)}}\cr
	&={{\pi i}\over 2} - {{\pi iz}\over 2} + \psi(z),\cr}\eqno(3.18)$$
where we have used the principal branch of the logarithm.  In 
the lower half-disc $\{z\in \cee:0>\Im(z),\ |z|<1\}$ the corresponding identity is 
$${\textstyle{1\over 2}}\int_{-1}^1 \log(z-t)\ dt + 1
={{-\pi i}\over 2} + {{\pi iz}\over 2} + \psi(z).\eqno(3.19)$$
Then $(3.18)$, $(3.19)$ and the continuity of $\rho$ imply that
$$\rho(z)=\Re\Big\{{\textstyle{1\over 2}}\int_{-1}^1 \log(z-t)\ dt + 1\Big\}
={{\pi |y|}\over 2} + \Re\big\{\psi(z)\big\},\eqno(3.20)$$
at all points $z$ in the closed unit disc.
By the maximum modulus theorem
$$|\psi(z)|=|z|^2\bigl|{{\psi(z)}\over{z^2}}\bigr|\le |z|^2\|\psi\|_{\infty}=(\log 2)|z|^2\eqno(3.21)$$ 
for all $z$ in the closed unit disc.  The lemma plainly follows from $(3.20)$ and $(3.21)$.
\enddemo

\head 4. The existence of special numbers\endhead%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Here we assume that $k$ is an algebraic number field having degree $d\ge 2$ over $\que$.  Let $\sigma_1, \sigma_2,
\dots ,\sigma_d$ be the distinct embeddings of $k$ into $\cee$.  We assume that $\sigma_1, \sigma_2,
\dots ,\sigma_r$ are real, that $\sigma_{r+1}, \sigma_{r+2}, \dots ,\sigma_{r+s}$ are complex and not real,
and that $\overline{\sigma}_{r+j}=\sigma_{r+s+j}$ for $j=1, 2, \dots ,s$.  We write $O_k$ for the ring of 
integers in $k$ and $\Delta_k$ for the discriminant. 

\proclaim{THEOREM 10} There exists a nonzero algebraic integer $\alpha$ in $k$ such that
$$\trc(\alpha)=0\quad\text{and}\quad \max_{1\le i \le d}|\sigma_i(\alpha)| \le 4|\Delta_k|^{1/{2(d-1)}}.\eqno(4.1)$$
Moreover, if $[\que(\alpha):\que]=\d$ and $f$ in $\zed[x]$ is the minimal polynomial of $\alpha$ over $\que$,
then 
$$\|f\|_{\infty} \le 8^{\d}|\Delta_k|^{{\d}/{2(d-1)}}.\eqno(4.2)$$
\endproclaim

\demo{Proof} To begin with we observe that the set of algebraic integers in $O_k$ which satisfy
$$\max_{1\le i \le d}|\sigma_i(\alpha)| \le T$$
is finite for every positive $T$.  Thus it suffices to prove that for every $\epsilon>0$ there exists
an algebraic integer $\alpha$ in $k$ such that
$$\trc(\alpha)=0\quad\text{and}\quad 
\max_{1\le i \le d}|\sigma_i(\alpha)| \le (4+\epsilon)|\Delta_k|^{1/{2(d-1)}}.$$

Let $\omega_1, \omega_2, \dots ,\omega_d$ be an integral basis for $O_k$.  We write
$\Omega$ for the $d\times d$ matrix $\Omega = (\sigma_i(\omega_j))$, where $i=1, 2, \dots ,d$ indexes
rows and $j=1, 2, \dots ,d$ indexes columns.  Then we define $W$ to be the $d\times d$ matrix which is
organized into blocks as
$$W=\pmatrix
\bu_r&\bo&\bo\cr
\bo&{\textstyle{1\over 2}}\bu_s&{\textstyle{1\over 2}}\bu_s\cr
\bo&{\textstyle{1\over {2i}}}\bu_s&{\textstyle{{-1}\over {2i}}}\bu_s\cr
\endpmatrix ,$$ where $\bu_r$ and $\bu_s$ are identity matrices.  We note 
that $(\det \Omega)^2 = \Delta_k$, $\det W = (-2i)^{-s}$ and the product
$W\Omega$ is a $d\times d$ matrix with real entries.  Next we define
$$\Lambda_k = \big\{\bl \in \zed^d: \sum_{j=1}^d \trc(\omega_j) \lambda_j \equiv 0 \mod d\big\}.$$
As $\Lambda_k$ is the kernel of the group homomorphism $\bl\rightarrow \sum_{j=1}^d \trc(\omega_j) \lambda_j$
from $\zed^d$ into $\zed/d\zed$, it follows that $\Lambda_k \subseteq \zed^d$ is a sublattice of index at
most $d$. 

We now assume that $1\le r$ and let $t$ denote a positive parameter.  Then we define
$$\eqalign{C_{r,s}(t)=\Big\{\bwy\in \real^d:|y_1|<1&, |y_i|\le t\ \ \text{if}\ \ 2\le i\le r,\cr
&\text{and}\ \ (y_{r+j})^2+(y_{r+s+j})^2\le t^2\ \ \text{if}\ \ 1\le j\le s\Big\}.\cr}\eqno(4.3)$$
It is clear that $C_{r,s}(t)$ is a convex, symmetric subset of $\real^d$, and
a simple computation shows that
$$\text{Vol}_d\{C_{r,s}(t)\}=2^d(\pi/4)^s t^{d-1}.\eqno(4.4)$$
Hence the linear image
$$K_{r,s}(t) = (W\Omega)^{-1}C_{r,s}(t) = \big\{\bx\in\real^d:W\Omega\bx\in C_{r,s}(t)\big\},$$
is also a convex, symmetric subset.  And using $(4.4)$ we find that
$$\eqalign{\text{Vol}_d\big\{K_{r,s}(t)\big\}&=\text{Vol}_d\big\{(W\Omega)^{-1}C_{r,s}(t)\big\}\cr
&=|\det W\Omega|^{-1}\text{Vol}_d\{C_{r,s}(t)\}= 2^d(\pi/2)^s t^{d-1}|\Delta_k|^{-1/2}.\cr}\eqno(4.5)$$
Let $0<\eta$ and set $t=(2+\eta)|\Delta_k|^{1/2(d-1)}$.  Then
$$\text{Vol}_d\big\{K_{r,s}(t)\big\}= 2^d(2+\eta)^{d-1}(\pi/2)^s > [\zed^d:\Lambda_k]2^d,$$
and so by Minkowski's convex body theorem there exists a nonzero point $\bxi$ in 
$K_{r,s}(t)\cap\Lambda_k$.  Using $\bxi$ we
define $\beta = \sum_{j=1}^d \xi_j \omega_j$, so that $\beta$ is a nonzero point in $O_k$.  
>From the definitions of $W$, $\Omega$ and $C_{r,s}(t)$ we find that
$$|\sigma_1(\beta)|<1\ \ \text{and}\ \ |\sigma_i(\beta)|\le 
(2+\eta)|\Delta_k|^{1/2(d-1)}\ \ \text{for}\ \ i=2, 3, \dots ,d.\eqno(4.6)$$
It is clear from the first inequality on the left of $(4.6)$ that $\beta\in O_k\setminus \zed$.
>From the definition of $\Lambda_k$ we learn that
$$\trc(\beta)=md\quad \text{with}\quad m\in \zed.$$
We conclude that $\alpha = \beta - m$ also belongs to $O_k\setminus \zed$ and $\trc(\alpha)=0$.
We also get the estimate
$$|m|=\bigl|d^{-1}\sum_{i=1}^d \sigma_i(\beta)\bigr|
\le\max_{1\le i \le d}|\sigma_i(\beta)|\le (2+\eta)|\Delta_k|^{1/{2(d-1)}}$$
and therefore
$$\max_{1\le i \le d}|\sigma_i(\alpha)| \le (4+2\eta)|\Delta_k|^{1/{2(d-1)}}.$$
In view of our previous remarks, this verifies the inequality on the right of $(4.1)$.

Next we assume that $r=0$ and define
$$\eqalign{C_{0,s}(t)=\Big\{\bwy\in \real^d:(y_{1})^2&+t^{-2}(y_{s+1})^2<1,\cr 
&\text{and}\ \ (y_{j})^2+(y_{s+j})^2\le t^2\ \ \text{if}\ \ 2\le j\le s\Big\}.\cr}\eqno(4.7)$$
Clearly $C_{0,s}(t)$ is also a convex, symmetric subset of $\real^d$, and again we have
$$\text{Vol}_d\{C_{0,s}(t)\}=2^d(\pi/4)^s t^{d-1}.\eqno(4.8)$$
We set $t=(2+\eta)|\Delta_k|^{1/2(d-1)}$ and proceed as before to determine a nonzero 
point $\beta$ in $O_k$.  In this case we find that
$$\eqalign{(\Re(\sigma_{1}(\beta))^2&+t^{-2}(\Im(\sigma_{1}(\beta))^2<1,\cr
&\text{and}\ \ |\sigma_{j}(\beta)|\le (2+\eta)|\Delta_k|^{1/2(d-1)}\ \ \text{for}\ \ 2\le j\le s.\cr}\eqno(4.9)$$
The first inequality on the left of $(4.9)$ shows that $\beta\in O_k\setminus \zed$. The rest of the
argument verifying $(4.1)$ is essentially the same.

To complete the proof, let $f$ in $\zed[x]$ be the minimal polynomial of $\alpha$ over $\que$.  Then
$(4.1)$ implies that the Mahler measure $\mu(f)$ satisfies the bound
$$\mu(f) \le 4^{\d}|\Delta_k|^{{\d}/{2(d-1)}},\quad\text{where}\quad [\que(\alpha):\que]=\d.$$
And from a basic inequality for $\mu$ (see \cite {4, equation (4)}) we conclude that
$$\|f\|_{\infty} \le 2^{\d}\mu(f)\le 8^{\d}|\Delta_k|^{{\d}/{2(d-1)}}.$$ 
\enddemo

\head 5 Proof of Theorem 1\endhead%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $k$ be a Galois extension of $\que$ with $[k:\que]=d\ge 2$.  As in section 2 we assume 
that all primes $p\le dN(d-1)^{-1}$ split in $k$, where $N\ge d$ is a positive integer parameter.
By Theorem 10 there exists an algebraic integer 
$\alpha$ in $O_k\setminus \zed$ with $\trc(\alpha)=0$ and 
$$\max_{1\le i\le \d}|\alpha_i| \le 4|\Delta_k|^{1/2(d-1)},\eqno(5.1)$$ 
where $[\que(\alpha):\que]=\d$ and $\alpha=\alpha_1, \alpha_2, \dots ,\alpha_{\d}$ are the conjugates
of $\alpha$ in $k$. We assume that $$\exp\{c_2(\log d)^2\}\le 8|\Delta_k|^{1/2(d-1)}\le N.\eqno(5.2)$$
Then the minimal polynomial $f$ of $\alpha$ over $\que$ satisfies the bound
$(4.2)$ and so therefore $\|f\|_{\infty}\le N^{\d}$.  Let $A_N(\alpha)$ be defined as in $(2.1)$. 
Then Theorem 5 and the product formula imply that
$$0=\log|A_N(\alpha)| + \sum_{p}\log|A_N(\alpha)|_p \le \log|A_N(\alpha)| - N{\d}^{-1}.\eqno(5.3)$$
And from Lemma 9 we get the estimate
$$\eqalign{\log|A_N(\alpha)|&=d{\d}^{-1}\sum_{n=0}^{N-1}\log|f(n)|  - d\log N!\cr
&\le d{\d}^{-1}\Big\{\int_0^N\log|f(x)|\ dx + \d(2 + \log N) 
+ (2\d +{\textstyle{1\over 2}})\log \|f\|_{\infty}\Big\}\cr
&\phantom{..........................}- d\{N\log N - N\}\cr
&=dN{\d}^{-1}\Big\{\int_0^1\log|N^{-\d}f(Nx)|\ dx + \d \Big\} + 6d^2\log N \cr}\eqno(5.4)$$
Combining $(5.3)$ and $(5.4)$ we obtain the inequality
$$1\le d \Big\{\int_0^1\log|N^{-\d}f(Nx)|\ dx + \d \Big\} 
+ \Bigl({{6d^3 \log N}\over N}\Bigr).\eqno(5.5)$$
Next we derive $(5.5)$ again but with $-\alpha$ in place of $\alpha$, and then we combine the two 
bounds. In this way we establish the estimate
$$1 \le d \Big\{{\textstyle{1\over 2}}\int_{-1}^1\log|N^{-\d}f(Nx)|\ dx + \d \Big\} 
+ \Bigl({{6d^3 \log N}\over N}\Bigr).\eqno(5.6)$$
>From Lemma 9 and $(5.1)$ we get
$$\eqalign{\Big\{{\textstyle{1\over 2}}\int_{-1}^1\log|N^{-\d}f(Nx)|\ dx + \d \Big\} 
	&= \sum_{i=1}^{\d} \Big\{{\textstyle{1\over 2}}\int_{-1}^1 \log|t- N^{-1}\alpha_i|\ dt + 1 \Big\}\cr
	&= \sum_{i=1}^{\d} \rho(N^{-1}\alpha_i)\cr
	&\le 3N^{-1} \sum_{i=1}^{\d} |\alpha_i|\cr
	&\le 12\d N^{-1} |\Delta_k|^{1/2(d-1)}.\cr}\eqno(5.7)$$
Thus $(5.6)$ and $(5.7)$ lead to the bound
$$1\le \Bigl({{12d^2|\Delta_k|^{1/2(d-1)}}\over N}\Bigr) + \Bigl({{6d^3 \log N}\over N}\Bigr).\eqno(5.8)$$
We select
$$N=\bigl[24d^2|\Delta_k|^{1/2(d-1)}\bigr],$$
use the hypothesis $(5.2)$, and obtain a contradiction to $(5.8)$. We have shown that if 
$$\exp\{c_2(\log d)^2\}\le 8|\Delta_k|^{1/2(d-1)}\eqno(5.9),$$ then there exists a prime number $p$ with
$$p\le dN(d-1)^{-1}\le 48d^2|\Delta_k|^{1/2(d-1)}$$
such that $p$ does not split completely in $k$. Plainly there exists a constant $c_1>0$ such that $(1.1)$
implies $(5.9)$




\Refs

\ref
\no 1 
\by C. F. Gauss
\book Disquisitiones Arithmeticae  
\bookinfo (1801), English translation
\publ Springer-Verlag 
\publaddr New York
\yr 1986 
\endref 

\ref
\no 2 
\by J. C. Lagarias, H. L. Montgomery and A. M. Odlyzko \pages 271--296
\paper A bound for the least prime ideal in the Chebotarev density theorem 
\yr 1979 \vol 54
\jour Invent. Math. 
\endref

\ref
\no 3
\by G. G. Lorentz, M. v. Golitschek, Y. Makovoz 
\book Constructive Approximation \bookinfo 
\publ Springer-Verlag
\publaddr New York
\yr 1996
\endref

\ref
\no 4
\by K. Mahler \pages 98--100
\paper An application of Jensen's formula to polynomials
\yr 1960 \vol 7
\jour Mathematika
\endref

\ref
\no 5
\by K. Mahler \pages 145--154
\paper  On the zeros of the derivative of a polynomial
\yr 1961 \vol 264
\jour Proc. Roy. Soc. London, Ser. A
\endref

\ref
\no 6 
\by V. K. Murty \pages 555--565
\paper The least prime which does not split completely
\yr 1994 \vol 6
\jour Forum Math. 
\endref

\ref
\no 7 
\by J. F. Voloch
\paper Chebyshev's method for number fields
\yr 1999 
\endref

\endRefs

\enddocument