%plain TeX \normalbaselineskip=1.6\normalbaselineskip\normalbaselines \magnification=1200 \def\max{\mathop{\rm max}} \def\min{\mathop{\rm min}} \def\rk{\mathop{\rm rk}} \def\Z{{\bf Z}} \def\Q{{\bf Q}} \def\C{{\bf C}} \def\F{{\bf F}} \def\N{{\bf N}} \def\zmp{\Z/p\Z} \def\O{{\cal O}} \def\isom{\cong} \def\Ext{\mathop{\hbox{Ext}}} \def\mod{\mathop{\rm mod}\nolimits} \def\Qp{{\bf Q}_p} \def\Qpb{\bar{\bf Q}_p} \def\Cp{{\bf C}_p} \def\P{{\bf P}} \def\Hom{\mathop{\rm Hom}\nolimits} \def \H{{\cal H}} \def \X{{\cal X}} \def\Spec{\mathop{\rm Spec}\nolimits} \def \bs{\bigskip} \def\Gal{\mathop{\rm Gal}\nolimits} \def\End{\mathop{\rm End}\nolimits} \def \L{\Lambda} \def\bk{\bar{k}} \def \ra{\rightarrow} \def \op{\frac{1}{p}} \def \st{\stackrel} \def \da{\downarrow} \def \R{{\bf Rings}} \def \d{\delta} \def \s{\sigma} \def \a{\alpha} \def \b{\beta} \def \x{\chi} \def \t{\tau} \def \z{\zeta} \def \con{\equiv } \def \e{\epsilon} \def \bC{\bar{C}} \def \bX{\bar{X}} \def \k{\kappa} \centerline{\bf Chebyshev's method for number fields} \medskip \centerline{\bf Jos\'e Felipe Voloch} \bs Chebyshev [C] proved that the number of primes up to $x$ was between multiples of $x/\log x$. In the simplified form of Chebyshev's method worked out by Erd\"os this can be obtained by looking at the prime factorization of the binomial coefficients $2n \choose n$. Note that ${2n \choose n} = (-4)^n{-1/2 \choose n}$ is the $n$-th coefficient of the Taylor expansion of $(1-4x)^{-1/2}$. In the course of their work on Grothendieck's conjecture on differential equations, by considering Pad\'e approximations to $(1+x)^{i\a}, i=1,2,\ldots$ where $\a$ is an irrational algebraic integer, D. and G. Chudnovsky proved that there are infinitely many primes which do not split in $\Q(\a)$ (which is of course a special case of Chebotarev's density theorem). The proof requires estimating complicated expressions in various binomial coefficients $i\a+j \choose n$. In this paper we show that, at least for $\Q(\a)/\Q$ Galois, we can very easily obtain not only that there are infinitely many primes which split completely in $\Q(\a)$, but that there are at least a multiple of $x^{1/d}/\log x$ ($d=[\Q(\a):\Q]$) such primes up to $x$, by simple estimates of $\a \choose n$. We will also study the prime factorization of $\a \choose n$ in the light of the present knowledge about the distribution of primes to analyse the scope of this method. \proclaim Lemma 1. Let $\a \in \C, \a \notin \N$, then $\log |{\a \choose n}| = o(n)$. {\it Proof:} This is of course elementary and well-known. The function $(1+x)^{\a}$ is holomorphic on the unit disk and has a singularity at $x=1$ so its Taylor expansion about zero has radius of convergence $1$, hence the result. J. Vaaler pointed out that the $o(n)$ can be improved to $O(\log n)$ but we won't need this sharper result. For now on let $\a$ be an irrational algebraic integer such that $\Q(\a)/\Q$ is Galois and denote by $\a=\a_1,\ldots,\a_d$ the conjugates of $\a$, so $d=[\Q(\a):\Q]$. Put $f(x) = \prod (x-\a_i)$, the minimal polynomial of $\a$ over $\Q$. Under our assumptions, $f(x) \in \Z[x]$. Let $A_n$ be the absolute norm of $\a \choose n$, which is a non-zero rational number. We denote by $v_p$ the $p$-adic valuation associated to the prime $p$. We will often use that, since $\Q(\a)/\Q$ is Galois, for a prime $p$ of $\Q$, $p$ splits completely in $\Q(\a)$ if and only if there is a prime of $\Q(\a)$ above $p$ which is split over $\Q$. \proclaim Lemma 2. Assume that $p$ does not ramify in $\Q(\a)$. Then $v_p(A_n) \ge 0$, if $p$ splits completely in $\Q(\a)$ and $v_p(A_n) \le 0$ otherwise. {\it Proof:} If $x \in \Z_p$ then ${x \choose n} \in \Z_p$, so ${\a \choose n} \in \Z_p$ if $p$ splits, giving the first statement of the lemma. For the second statement we note that there is no root of the minimal polynomial of $\a$ in $\Z/p\Z$ in that case for that would force $p$ to split, so $p$ does not divide the numerator of $A_n$. \proclaim Lemma 3. Assume $p$ does not split in $\Q(\a)$. Then $v_p(A_n) \le -cn +O(\log n)$, for some $c>0$ depending on $p$. {\it Proof:} If $v_p$ denotes an extension of the $p$-adic valuation to $\Z_p[\a]$ then, since $p$ does not split, $\Z_p[\a]$ is strictly bigger than $\Z_p$ so there exists an integer $s \ge 0$ such that $v_p(\a_j -k) \le s, j=1,\ldots,d, k \in \Z$. Given $j, 1\le j \le d$, there are at most $n/p + O(1)$ integers $k, 0\le k < n$ with $v_p(\a_j -k) > 0$. At most $n/p^2 + O(1)$ of those satisfy $v_p(\a_j -k) > 1$ and so on, until at most $n/p^s + O(1)$ of those satisfy $v_p(\a_j -k) > s-1$ but none satisfy $v_p(\a_j -k) > s$. As $$A_n = {{\prod_{k=0}^{n-1} \prod_{j=1}^d(\a_j - k)}\over{n!^d}}$$ we get $v_p(A_n) \le d\sum_{i=1}^s n/p^i + O(1) - dv_p(n!) = -dn\sum_{i>s} 1/p^i + O(\log n) $, as was to be proved. \proclaim Lemma 4. If $p$ splits in $\Q(\a)$ then $v_p(A_n) \ll \log n/\log p$. {\it Proof:} Consider $A_n,\a$ as $p$-adic integers. We have that $$A_n = {{\prod_{k=0}^{n-1} \prod_{j=1}^d(\a_j - k)}\over{n!^d}}.$$ Given $j, 1 \le j \le d$, there are $n/p + O(1)$ integers $k, 0\le k < n$ with $k \con \a_j (\mod p)$ and $n/p^2 + O(1)$ of those satisfy $k \con \a_j (\mod p^2)$ and so on. However, if $p^r | (\a_j - k)$ then $p^r \ll k^d \le n^d$, so $r \ll \log n/\log p$. Therefore the $p$-adic valuation of the numerator of the above expression for $A_n$ is at most $dn/(p-1)+ O(\log n/\log p)$. But the last expression is also a lower bound for the $p$-adic valuation of the denominator of the above expression for $A_n$, namely $n!^d$. The lemma follows. \proclaim Theorem 1. If $S$ is the set of primes splitting in $\Q(\a)$ then $\#\{p \le x \mid p \in S\} \gg x^{1/d}/\log x$. {\it Proof:} For $x$ sufficiently large, let $y$ be the unique positive solution to $f(y)=x$ and put $n=[y]$. By lemma 1, $\log |A_n| = o(n)$. On the other hand, $\log |A_n| = \sum_{p \in S}v_p(A_n)\log p+ \sum_{p \notin S}v_p(A_n)\log p$. Clearly, if $v_p(A_n) > 0$ with $p \in S$, then $p$ divides $f(k)$ for some $k, 0 \le k 1$, $\sum_{p \in S, p \le cn} v_p(A_n)\log p = O(n)$. Therefore large primes must be contributing to the numerator of $A_n$. Note that $ v_p(A_n) > 0$ for $p>n$ if and only if there exists $a, 1\le a < n, f(a)\con 0 (\mod p)$. Thus, large primes $p$ which contribute to the numerator of $A_n$ are those for which there is a small solution to $f(x)\con 0 (\mod p)$. In the case that $\a$ is imaginary quadratic, Duke et al. [DFI] have shown that the values of $a/p$, where $0\le a