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4 Proof of Dirichlet's theorem
4 狄利克雷定理的证明
We need the notion of the Dirichlet density of sets of prime numbers.
首先引入素数集合的狄利克雷密度概念。
Theorem 13 Let
定理 13 记
ψ ( s ) = ∑ p 1 p s \psi(s)=\sum_{p} \frac{1}{p^{s}} ψ(s)=p∑ps1
where the sum is over all primes p . Then this series is convergent for s > 1 s>1 s>1 and as s → 1 + s \to 1^{+} s→1+
求和遍历全体素数 p p p。该级数在 s > 1 s>1 s>1 收敛，且当 s → 1 + s\to1^+ s→1+ 时满足
ψ ( s ) = − l o g ( s − 1 ) + O ( 1 ) \psi (s)=-log (s-1)+O(1) ψ(s)=−log(s−1)+O(1)
Proof Let s > 1 s>1 s>1 . The series for ψ ( s ) \psi(s) ψ(s) ) converges by comparison with that for ζ ( s ) \zeta(s) ζ(s) ). From the Euler product for ζ ( s ) \zeta(s) ζ(s) ), we have
证明 取 s > 1 s>1 s>1，与 ζ ( s ) \zeta(s) ζ(s) 级数比较即得 ψ ( s ) \psi(s) ψ(s) 收敛。对 ζ ( s ) \zeta(s) ζ(s) 的欧拉乘积取对数：
l o g ζ ( s ) = − ∑ p l o g ( 1 − 1 p s ) = ∑ p ∑ m = 1 ∞ 1 m p m s = ψ ( s ) + ω ( s ) log \zeta(s)=-\sum_{p} log \left(1-\frac{1}{p^{s}}\right)=\sum_{p} \sum_{m=1}^{\infty} \frac{1}{m p^{m s}}=\psi(s)+\omega(s) logζ(s)=−p∑log(1−ps1)=p∑m=1∑∞mpms1=ψ(s)+ω(s)
where
其中余项
ω ( s ) = ∑ p ∑ m = 2 ∞ 1 m p m s . \omega(s)=\sum_{p} \sum_{m=2}^{\infty} \frac{1}{m p^{m s}} . ω(s)=p∑m=2∑∞mpms1.
But
余项满足一致有界估计：
0 < ω ( s ) < ∑ p ∑ m = 2 ∞ 1 2 p m s = 1 2 ∑ p 1 p 2 s ( 1 − 1 p s ) − 1 = 1 2 ∑ p 1 ( p s − 1 ) p s < 1 2 ∑ p 1 ( p − 1 ) p < 1 2 ∑ m = 1 ∞ 1 m ( m + 1 ) = 1 2 . \begin{aligned} 0<\omega(s) & <\sum_{p} \sum_{m=2}^{\infty} \frac{1}{2 p^{m s}}=\frac{1}{2} \sum_{p} \frac{1}{p^{2 s}}\left(1-\frac{1}{p^{s}}\right)^{-1}=\frac{1}{2} \sum_{p} \frac{1}{\left(p^{s}-1\right) p^{s}} \\ & <\frac{1}{2} \sum_{p} \frac{1}{(p-1) p}<\frac{1}{2} \sum_{m=1}^{\infty} \frac{1}{m(m+1)}=\frac{1}{2} . \end{aligned} 0<ω(s)<p∑m=2∑∞2pms1=21p∑p2s1(1−ps1)−1=21p∑(ps−1)ps1<21p∑(p−1)p1<21m=1∑∞m(m+1)1=21.
It follows that
因此当 s → 1 + s\to1^+ s→1+ 时
ψ ( s ) = l o g ζ ( s ) + O ( 1 ) \psi (s)=log \zeta (s)+O(1) ψ(s)=logζ(s)+O(1)
as s → 1 + s \to 1^{+} s→1+ . But
已知 s > 1 s>1 s>1 时
1 s − 1 < ζ ( s ) < s s − 1 \frac{1}{s-1}<\zeta(s)<\frac{s}{s-1} s−11<ζ(s)<s−1s
for s > 1 s>1 s>1 , and so
取对数得
− l o g ( s − 1 ) < l o g ζ ( s ) < l o g s − l o g ( s − 1 ) -log (s-1)<log \zeta(s)<log s-log (s-1) −log(s−1)<logζ(s)<logs−log(s−1)
and hence
即
l o g ζ ( s ) = − l o g ( s − 1 ) + O ( 1 ) log \zeta (s)=-log (s-1)+O(1) logζ(s)=−log(s−1)+O(1)
as s → 1 + s \to 1^{+} s→1+ . Consequently
代入即证
ψ ( s ) = − l o g ( s − 1 ) + O ( 1 ) \psi (s)=-log (s-1)+O(1) ψ(s)=−log(s−1)+O(1)
Let A denote a subset of the set of primes. The Dirichlet density of A is
l i m s → 1 + − 1 l o g ( s − 1 ) ∑ p ∈ A 1 p s lim {s \to 1^{+}} \frac{-1}{log (s-1)} \sum{p \in A} \frac{1}{p^{s}} lims→1+log(s−1)−1p∈A∑ps1
provided this limit exists. By Theorem 13, the set of all primes has Dirichlet density 1. Evidently any finite set of primes has Dirichlet density 0. Any set of primes with nonzero Dirichlet density is infinite.
设 A A A 为全体素数集合的子集。若下述极限存在，则称其为 A A A 的狄利克雷密度：
lim s → 1 + − 1 log ( s − 1 ) ∑ p ∈ A 1 p s . \lim_{s\to1^+}\frac{-1}{\log(s-1)}\sum_{p\in A}\frac{1}{p^s} . s→1+limlog(s−1)−1p∈A∑ps1.
由定理 13，全体素数的狄利克雷密度为 1；任意有限素数集的狄利克雷密度为 0；狄利克雷密度非零的素数集合必为无限集。
We need to show that L ( 1 , χ ) L(1, \chi) L(1,χ) ) is nonzero whenever x ≠ ε x ≠\varepsilon x=ε . The next result is a start in this direction.
接下来证明：所有非平凡特征标 χ \chi χ 均满足 L ( 1 , χ ) ≠ 0 L(1,\chi)\neq0 L(1,χ)=0，先给出辅助定理。
Theorem 14 Let N be a positive integer. Then
定理 14 设 N N N 为正整数，则对任意 s > 1 s>1 s>1，有
∏ χ ∈ X N L ( s , χ ) > 1 \prod_{\chi \in X_{N}} L(s, \chi)>1 χ∈XN∏L(s,χ)>1
for all s > 1 s>1 s>1 .
Proof Using Euler's product we have
证明 代入欧拉乘积公式，交换乘积顺序得
∏ χ ∈ X N L ( s , χ ) = ∏ p ∏ χ ∈ X N ( 1 − χ ( p ) p s ) − 1 . \prod_{\chi \in X_{N}} L(s, \chi)=\prod_{p} \prod_{\chi \in X_{N}}\left(1-\frac{\chi(p)}{p^{s}}\right)^{-1} . χ∈XN∏L(s,χ)=p∏χ∈XN∏(1−psχ(p))−1.
If p ∣ N p | N p∣N then
若素数 p ∣ N p\mid N p∣N，则 χ ( p ) = 0 \chi(p)=0 χ(p)=0，于是
( 1 − χ ( p ) p s ) − 1 = 1. \left(1-\frac{\chi(p)}{p^s}\right)^{-1}=1 . (1−psχ(p))−1=1.
Otherwise χ ( p ) ≠ 0 \chi(p) ≠0 χ(p)=0 for all χ ∈ X N \chi \in X_{N} χ∈XN .Let H be the subgroup of Z N ∗ Z_{N}^{*} ZN∗ generated by the residue of p , and let r = ∣ H ∣ r=|H| r=∣H∣ . By Theorem 5
若 p ∤ N p\nmid N p∤N，则对任意 χ ∈ X N \chi\in X_N χ∈XN， χ ( p ) ≠ 0 \chi(p)\neq0 χ(p)=0。记 H H H 为 Z N ∗ \mathbb{Z}_N^* ZN∗ 中由 p p p 的剩余类生成的循环子群， r = ∣ H ∣ r=|H| r=∣H∣。由定理 5，
∏ χ ∈ X N ( 1 − χ ( p ) p s ) − 1 = ∏ ψ ∈ H \^ ( 1 − ψ ( p ) p s ) − ϕ ( N ) / r . \prod_{\chi\in X_N}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}=\left\\prod_{\\psi\\in\\widehat{H}}\\left(1-\\frac{\\psi(p)}{p\^s}\\right)\\right^{-\phi(N)/r} . χ∈XN∏(1−psχ(p))−1= ψ∈H ∏(1−psψ(p)) −ϕ(N)/r.
As H is cyclic of order r , generated by p , its characters are ψ j \psi_{j} ψj where ψ j ( p ) = \psi_{j}(p)= ψj(p)= exp ( 2 π i j / r ) (2 \pi i j / r) (2πij/r) ). Thus
H H H 是阶为 r r r 的循环群，生成元为 p p p，其特征标满足 ψ j ( p ) = exp ( 2 π i j / r ) \psi_j(p)=\exp(2\pi i j/r) ψj(p)=exp(2πij/r)，因此
∏ ψ ∈ H ^ ( 1 − ψ ( p ) p s ) = ∏ j = 0 r − 1 ( 1 − exp ( 2 π i j / r ) p s ) = 1 − 1 p r s . \prod_{\psi\in\widehat{H}}\left(1-\frac{\psi(p)}{p^s}\right)=\prod_{j=0}^{r-1}\left(1-\frac{\exp(2\pi i j/r)}{p^s}\right)=1-\frac{1}{p^{rs}} . ψ∈H ∏(1−psψ(p))=j=0∏r−1(1−psexp(2πij/r))=1−prs1.
Hence
于是
∏ χ ∈ X N ( 1 − χ ( p ) p s ) − 1 = ( 1 − 1 p r s ) − ϕ ( N ) / r > 1 , s > 1. \prod_{\chi\in X_N}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}=\left(1-\frac{1}{p^{rs}}\right)^{-\phi(N)/r}>1,\quad s>1 . χ∈XN∏(1−psχ(p))−1=(1−prs1)−ϕ(N)/r>1,s>1.
It follows that
对全体素数取乘积，即得
∏ χ ∈ X N L ( s , χ ) > 1 , s > 1. \prod_{\chi\in X_N}L(s,\chi)>1,\quad s>1 . χ∈XN∏L(s,χ)>1,s>1.
We can now prove that L ( 1 , χ ) ≠ 0 L(1, \chi) ≠0 L(1,χ)=0 = 0 for all non-real valued Dirichlet characters x
下面先证明：非实值狄利克雷特征标 χ \chi χ 满足 L ( 1 , χ ) ≠ 0 L(1,\chi)\neq0 L(1,χ)=0。
Theorem 15 Let χ ∈ X N \chi \in X_{N} χ∈XN , and suppose that x ≠ x ˉ x ≠\bar{x} x=xˉ . Then
定理 15 设 χ ∈ X N \chi\in X_N χ∈XN，且 χ ≠ χ ˉ \chi\neq\bar{\chi} χ=χˉ（非实特征标），则
L ( 1 , χ ) ≠ 0. L(1,\chi)\neq0 . L(1,χ)=0.
Proof If χ ≠ χ ˉ \chi ≠\bar{\chi} χ=χˉ the three characters ε, x and χ ˉ ∈ X N \bar{\chi} \in X_{N} χˉ∈XN are all distinct. Suppose that L ( 1 , χ ) = 0 L(1, \chi)=0 L(1,χ)=0 .Then
证明 若 χ ≠ χ ˉ \chi\neq\bar{\chi} χ=χˉ，则 ε , χ , χ ˉ \varepsilon,\chi,\bar{\chi} ε,χ,χˉ 是三个两两不同的特征标。反设 L ( 1 , χ ) = 0 L(1,\chi)=0 L(1,χ)=0，则
L ( 1 , χ ˉ ) = ∑ n = 1 ∞ χ ( n ) ‾ n = ∑ n = 1 ∞ χ ( n ) n ‾ = L ( 1 , χ ) ‾ = 0. L(1,\bar{\chi})=\sum_{n=1}^\infty \frac{\overline{\chi(n)}}{n}=\overline{\sum_{n=1}^\infty \frac{\chi(n)}{n}}=\overline{L(1,\chi)}=0 . L(1,χˉ)=n=1∑∞nχ(n)=n=1∑∞nχ(n)=L(1,χ)=0.
By Theorems 11 and 12, L ( s , ε ) = O ( 1 / ( s − 1 ) ) L(s, \varepsilon)=O(1 /(s-1)) L(s,ε)=O(1/(s−1)) , L ( s , χ ) = O ( s − 1 ) L(s, \chi)=O(s-1) L(s,χ)=O(s−1) and L ( s , χ ˉ ) = O ( s − 1 ) L(s, \bar{\chi})=O(s-1) L(s,χˉ)=O(s−1) as s → 1 + s \to 1^{+} s→1+ . Thus
由定理 11、12，当 s → 1 + s\to1^+ s→1+ 时： L ( s , ε ) = O ( 1 s − 1 ) L(s,\varepsilon)=O\left(\frac{1}{s-1}\right) L(s,ε)=O(s−11)， L ( s , χ ) = O ( s − 1 ) L(s,\chi)=O(s-1) L(s,χ)=O(s−1)， L ( s , χ ˉ ) = O ( s − 1 ) L(s,\bar{\chi})=O(s-1) L(s,χˉ)=O(s−1)，三者乘积满足
lim s → 1 + L ( s , ε ) L ( s , χ ) L ( s , χ ˉ ) = 0. \lim_{s\to1^+}L(s,\varepsilon)L(s,\chi)L(s,\bar{\chi})=0 . s→1+limL(s,ε)L(s,χ)L(s,χˉ)=0.
Also
其余特征标对应的 L ( s , ψ ) L(s,\psi) L(s,ψ) 在 s → 1 + s\to1^+ s→1+ 极限存在，故
lim s → 1 + ∏ ψ ∈ X N ψ ≠ ε , χ , χ ˉ L ( s , ψ ) = ∏ ψ ∈ X N ψ ≠ ε , χ , χ ˉ L ( 1 , ψ ) . \lim_{s\to1^+}\prod_{\substack{\psi\in X_N\\\psi\neq\varepsilon,\chi,\bar{\chi}}}L(s,\psi)=\prod_{\substack{\psi\in X_N\\\psi\neq\varepsilon,\chi,\bar{\chi}}}L(1,\psi). s→1+limψ∈XNψ=ε,χ,χˉ∏L(s,ψ)=ψ∈XNψ=ε,χ,χˉ∏L(1,ψ).
It follows that
由此可得
lim s → 1 + ∏ ψ ∈ X N L ( s , ψ ) = 0. \lim_{s\to1^+}\prod_{\psi\in X_N}L(s,\psi)=0 . s→1+limψ∈XN∏L(s,ψ)=0.
This is impossible, as
这是不可能的，因为
∏ ψ ∈ X N L ( s , ψ ) > 1 \prod_{\psi \in X_{N}} L(s, \psi)>1 ψ∈XN∏L(s,ψ)>1
for all s > 1 s>1 s>1 , by Theorem 14. This contradiction shows that it is impossible for L ( 1 , χ ) L(1, \chi) L(1,χ) ) to equal zero.
由定理 14，对任意 s > 1 s>1 s>1 恒有 ∏ ψ ∈ X N L ( s , ψ ) > 1 \displaystyle\prod_{\psi \in X_{N}} L(s, \psi)>1 ψ∈XN∏L(s,ψ)>1，上述极限为 0 与之矛盾。故假设不成立，必有 L ( 1 , χ ) ≠ 0 L(1,\chi)\neq0 L(1,χ)=0。
The most difficult part in the proof of Dirichlet's theorem is the proof that L ( 1 , χ ) ≠ 0 L(1, \chi) ≠0 L(1,χ)=0 when x is a nontrivial real-valued Dirichlet character. To achieve this some trick or other is needed. We shall employ an elegant device due to Monsky 1.
狄利克雷定理证明中最难的环节，是证明非平凡实值狄利克雷特征标 χ \chi χ 满足 L ( 1 , χ ) ≠ 0 L(1,\chi)\neq0 L(1,χ)=0，这一步需要特殊技巧，本文采用蒙斯基 1 提出的简洁证法。
Theorem 16 Let χ ∈ X N \chi \in X_{N} χ∈XN , and suppose that x ≠ ε x ≠\varepsilon x=ε and x = x ˉ x=\bar{x} x=xˉ . Then
定理 16 设 χ ∈ X N \chi\in X_N χ∈XN， χ ≠ ε \chi\neq\varepsilon χ=ε 且 χ = χ ˉ \chi=\bar{\chi} χ=χˉ（非平凡实特征标），则
L ( 1 , χ ) ≠ 0. L(1,\chi)\neq0 . L(1,χ)=0.
Proof We introduce a Lambert series
证明 引入朗伯级数
f ( x ) = ∑ d = 1 ∞ χ ( d ) x d 1 − x d . f(x)=\sum_{d=1}^{\infty} \frac{\chi(d) x^{d}}{1-x^{d}} . f(x)=d=1∑∞1−xdχ(d)xd.
This series converges for 0 < x < 1 0<x<1 0<x<1 by comparison with
该级数在 0 < x < 1 0<x<1 0<x<1 上收敛；可与级数 ∑ d = 1 ∞ x d 1 − x d \displaystyle\sum_{d=1}^{\infty}\frac{x^d}{1-x^d} d=1∑∞1−xdxd 比较判别，
∑ d = 1 ∞ x d 1 − x d \sum_{d=1}^{\infty} \frac{x^{d}}{1-x^{d}} d=1∑∞1−xdxd
which converges by the ratio test. We calculate
后者由比值判别法收敛。展开重排求和：
f ( x ) = ∑ d = 1 ∞ χ ( d ) ∑ t = 1 ∞ x d t = ∑ n = 1 ∞ x n ∑ d ∣ n χ ( d ) = ∑ n = 1 ∞ c n x n f(x)=\sum_{d=1}^{\infty} \chi(d) \sum_{t=1}^{\infty} x^{d t}=\sum_{n=1}^{\infty} x^{n} \sum_{d | n} \chi(d)=\sum_{n=1}^{\infty} c_{n} x^{n} f(x)=d=1∑∞χ(d)t=1∑∞xdt=n=1∑∞xnd∣n∑χ(d)=n=1∑∞cnxn
where
其中系数定义为
c n = ∑ d ∣ n χ ( d ) . c_n=\sum_{d\mid n}\chi(d) . cn=d∣n∑χ(d).
We need to show that c n ≥ 0 c_{n} ≥0 cn≥0 for all n . As each χ ( d ) \chi(d) χ(d) ) is real and has modulus 0 or 1 then χ ( d ) ∈ − 1 , 0 , 1 \chi(d) \in{-1,0,1} χ(d)∈−1,0,1 for all d . Clearly c 1 = χ ( 1 ) = 1 c_{1}=\chi(1)=1 c1=χ(1)=1 . We use induction on n . If n > 1 n>1 n>1 then n = p k m n=p^{k} m n=pkm where p is prime, k > 0 k>0 k>0 and p ∤ m p \nmid m p∤m
需证对所有正整数 n n n， c n ≥ 0 c_n\geq0 cn≥0。 χ \chi χ 是实特征标，故对任意整数 d d d， χ ( d ) ∈ { − 1 , 0 , 1 } \chi(d)\in\{-1,0,1\} χ(d)∈{−1,0,1}。显然 c 1 = χ ( 1 ) = 1 c_1=\chi(1)=1 c1=χ(1)=1，对 n n n 作数学归纳法。若 n > 1 n>1 n>1，作素分解 n = p k m n=p^k m n=pkm，其中素数 p p p， k > 0 k>0 k>0， gcd ( p , m ) = 1 \gcd(p,m)=1 gcd(p,m)=1。
Then
于是
c n = ∑ d ∣ p k m χ ( d ) = ∑ j = 0 k ∑ r ∣ m χ ( p j r ) = ∑ j = 0 k ∑ r ∣ m χ ( p ) j χ ( r ) = c m ⋅ ∑ j = 0 k χ ( p ) j . c_n=\sum_{d\mid p^k m}\chi(d)=\sum_{j=0}^k\sum_{r\mid m}\chi(p^j r)=\sum_{j=0}^k\sum_{r\mid m}\chi(p)^j \chi(r)=c_m \cdot \sum_{j=0}^k \chi(p)^j . cn=d∣pkm∑χ(d)=j=0∑kr∣m∑χ(pjr)=j=0∑kr∣m∑χ(p)jχ(r)=cm⋅j=0∑kχ(p)j.
Thus
分情况讨论：
c n = { ( k + 1 ) c m if χ ( p ) = 1 , c m if χ ( p ) = 0 , c m if χ ( p ) = − 1 and k is even 偶 , 0 if χ ( p ) = − 1 and k is odd 奇 . c_{n}= \begin{cases} (k+1)c_{m} & \text{if } \chi(p)=1,\\ c_{m} & \text{if } \chi(p)=0,\\ c_{m} & \text{if } \chi(p)=-1 \text{ and } k \text{ is even 偶},\\ 0 & \text{if } \chi(p)=-1 \text{ and } k \text{ is odd 奇}. \end{cases} cn=⎩ ⎨ ⎧(k+1)cmcmcm0if χ(p)=1,if χ(p)=0,if χ(p)=−1 and k is even 偶,if χ(p)=−1 and k is odd 奇.
Inductively, as c m ≥ 0 c_{m} ≥0 cm≥0 , then c n ≥ 0 c_{n} ≥0 cn≥0 . Also c n k = 1 c_{n^{k}}=1 cnk=1 for ll k ≥ 0 k ≥0 k≥0 and so ∑ r = 1 ∞ c r \sum_{r=1}^{\infty} c_{r} ∑r=1∞cr diverges. For each M > 0 M>0 M>0 ,
由归纳假设 c m ≥ 0 c_m\geq0 cm≥0，故所有 c n ≥ 0 c_n\geq0 cn≥0。同时存在无穷多项 c n = 1 c_n=1 cn=1，因此级数 ∑ r = 1 ∞ c r \displaystyle\sum_{r=1}^{\infty}c_r r=1∑∞cr 发散。对任意固定 M > 0 M>0 M>0，
lim sup x → 1 − f ( x ) ≥ lim x → 1 − ∑ n = 1 M c n x n = ∑ n = 1 M c n . \limsup_{x\to1^-}f(x)\geq \lim_{x\to1^-}\sum_{n=1}^M c_n x^n=\sum_{n=1}^M c_n . x→1−limsupf(x)≥x→1−limn=1∑Mcnxn=n=1∑Mcn.
Hence f ( x ) → ∞ f(x) \to \infty f(x)→∞ as x → 1 − x \to 1^{-} x→1−
令 M → ∞ M\to\infty M→∞ 即得： x → 1 − x\to1^- x→1− 时 f ( x ) → + ∞ f(x)\to+\infty f(x)→+∞。
Suppose that L ( 1 , χ ) = 0 L(1, \chi)=0 L(1,χ)=0 . Then
反设 L ( 1 , χ ) = 0 L(1,\chi)=0 L(1,χ)=0，则
− f ( x ) = L ( 1 , χ ) 1 − x − f ( x ) = ∑ n = 1 ∞ χ ( n ) ( 1 n ( 1 − x ) − x n 1 − x n ) . -f(x)=\frac{L(1,\chi)}{1-x}-f(x)=\sum_{n=1}^{\infty}\chi(n)\left( \frac{1}{n(1-x)}-\frac{x^n}{1-x^n} \right) . −f(x)=1−xL(1,χ)−f(x)=n=1∑∞χ(n)(n(1−x)1−1−xnxn).
Let
记
b n ( x ) = 1 n ( 1 − x ) − x n 1 − x n . b_n(x)=\frac{1}{n(1-x)}-\frac{x^n}{1-x^n} . bn(x)=n(1−x)1−1−xnxn.
Then for 0 < x < 1 0<x<1 0<x<1 ,
对 0 < x < 1 0<x<1 0<x<1，展开计算：
( 1 − x ) ( b n ( x ) − b n + 1 ( x ) ) = 1 n − 1 n + 1 − x n ( 1 − x ) 2 ( 1 − x n ) ( 1 − x n + 1 ) = 1 n ( n + 1 ) − x n ( 1 − x ) 2 ( 1 − x n ) ( 1 − x n + 1 ) . \begin{aligned} (1-x)\big(b_n(x)-b_{n+1}(x)\big) &=\frac{1}{n}-\frac{1}{n+1}-\frac{x^n(1-x)^2}{(1-x^n)(1-x^{n+1})}\\ &=\frac{1}{n(n+1)}-\frac{x^n(1-x)^2}{(1-x^n)(1-x^{n+1})}. \end{aligned} (1−x)(bn(x)−bn+1(x))=n1−n+11−(1−xn)(1−xn+1)xn(1−x)2=n(n+1)1−(1−xn)(1−xn+1)xn(1−x)2.
By the arithmetic-geometric mean inequality,
由均值不等式：
1 − x n 1 − x = ∑ j = 0 n − 1 x j = 1 2 ∑ j = 0 n − 1 ( x j + x n − 1 − j ) ≥ n x n − 1 2 . \frac{1-x^n}{1-x}=\sum_{j=0}^{n-1}x^j=\frac12\sum_{j=0}^{n-1}\big(x^j+x^{n-1-j}\big)\geq n x^{\frac{n-1}{2}} . 1−x1−xn=j=0∑n−1xj=21j=0∑n−1(xj+xn−1−j)≥nx2n−1.
Replacing n by n + 1 n+1 n+1 gives
将 n n n 替换为 n + 1 n+1 n+1 得
1 − x n + 1 1 − x ≥ ( n + 1 ) x n 2 . \frac{1-x^{n+1}}{1-x}\geq (n+1)x^{\frac{n}{2}} . 1−x1−xn+1≥(n+1)x2n.
Hence
代入化简可得
( 1 − x ) ( b n ( x ) − b n + 1 ( x ) ) ≥ 1 − x n ( n + 1 ) . (1-x)\big(b_n(x)-b_{n+1}(x)\big)\geq \frac{1-\sqrt{x}}{n(n+1)} . (1−x)(bn(x)−bn+1(x))≥n(n+1)1−x .
Thus ( ( b n ( x ) ) (b_{n}(x)) (bn(x)) )) is a decreasing sequence whenever 0 < x < 1 0<x<1 0<x<1
因此对固定 0 < x < 1 0<x<1 0<x<1，数列 { b n ( x ) } \{b_n(x)\} {bn(x)} 关于 n n n 单调递减。
Set a m = ∑ j = 1 m χ ( j ) a_{m}=\sum_{j=1}^{m} \chi(j) am=∑j=1mχ(j) . The sequence ( a m ) (a_{m}) (am) ) is bounded by Theorem 8; suppose that ∣ a m ∣ ≤ A |a_{m}| ≤A ∣am∣≤A for all m . Then
记部分和 a m = ∑ j = 1 m χ ( j ) a_m=\displaystyle\sum_{j=1}^m\chi(j) am=j=1∑mχ(j)，由定理 8， { a m } \{a_m\} {am} 有界，设 ∣ a m ∣ ≤ A |a_m|\leq A ∣am∣≤A。对有限和分部求和：
∑ n = 1 M χ ( n ) b n ( x ) = ∑ n = 1 M ( a n − a n − 1 ) b n ( x ) = a M b M ( x ) + ∑ n = 1 M − 1 a n ( b n ( x ) − b n + 1 ( x ) ) . \begin{aligned} \sum_{n=1}^M \chi(n)b_n(x) &=\sum_{n=1}^M (a_n-a_{n-1})b_n(x)\\ &=a_M b_M(x)+\sum_{n=1}^{M-1}a_n\big(b_n(x)-b_{n+1}(x)\big). \end{aligned} n=1∑Mχ(n)bn(x)=n=1∑M(an−an−1)bn(x)=aMbM(x)+n=1∑M−1an(bn(x)−bn+1(x)).
As M → ∞ M \to \infty M→∞ , b M ( x ) → 0 b_{M}(x) \to 0 bM(x)→0 and so
令 M → ∞ M\to\infty M→∞， b M ( x ) → 0 b_M(x)\to0 bM(x)→0，于是
− f ( x ) = ∑ n = 1 ∞ a n ( b n ( x ) − b n + 1 ( x ) ) . -f(x)=\sum_{n=1}^{\infty}a_n\big(b_n(x)-b_{n+1}(x)\big) . −f(x)=n=1∑∞an(bn(x)−bn+1(x)).
Hence
取绝对值放缩：
∣ f ( x ) ∣ ≤ ∑ n = 1 ∞ ∣ a n ∣ ( b n ( x ) − b n + 1 ( x ) ) ≤ A ∑ n = 1 ∞ ( b n ( x ) − b n + 1 ( x ) ) = A b 1 ( x ) . |f(x)|\leq \sum_{n=1}^{\infty}|a_n|\big(b_n(x)-b_{n+1}(x)\big)\leq A\sum_{n=1}^{\infty}\big(b_n(x)-b_{n+1}(x)\big)=A b_1(x) . ∣f(x)∣≤n=1∑∞∣an∣(bn(x)−bn+1(x))≤An=1∑∞(bn(x)−bn+1(x))=Ab1(x).
But
而直接计算
b 1 ( x ) = 1 1 − x − x 1 − x = 1. b_1(x)=\frac{1}{1-x}-\frac{x}{1-x}=1 . b1(x)=1−x1−1−xx=1.
It follows that f ( x ) f(x) f(x) ) is bounded for 0 < x < 1 0<x<1 0<x<1 . But this contradicts the fact that f ( x ) → ∞ f(x) \to \infty f(x)→∞ as x → 1 − x \to 1^{-} x→1− . Hence we cannot have L ( 1 , χ ) = 0 L(1, \chi)=0 L(1,χ)=0
由此推出 f ( x ) f(x) f(x) 在区间 0 < x < 1 0<x<1 0<x<1 上有界，与前文 x → 1 − x\to1^- x→1− 时 f ( x ) → + ∞ f(x)\to+\infty f(x)→+∞ 矛盾。故假设 L ( 1 , χ ) = 0 L(1,\chi)=0 L(1,χ)=0 不成立，证毕。
We now look at a sum over primes weighted by a nontrivial Dirichlet character.
接下来研究带非平凡狄利克雷特征标权重的素数求和。
Theorem 17 Let χ ∈ X N \chi \in X_{N} χ∈XN with x ≠ ε x ≠\varepsilon x=ε . Define
定理 17 设 χ ∈ X N \chi\in X_N χ∈XN， χ ≠ ε \chi\neq\varepsilon χ=ε，定义
ψ χ ( s ) = ∑ p χ ( p ) p s , \psi_\chi(s)=\sum_{p}\frac{\chi(p)}{p^s}, ψχ(s)=p∑psχ(p),
where the sum is over all primes p . Then this series is convergent for s > 1 s>1 s>1 and
求和遍历全体素数 p p p。该级数在 s > 1 s>1 s>1 收敛，且 s → 1 + s\to1^+ s→1+ 时满足
ψ χ ( s ) = O ( 1 ) \psi_{\chi}(s)=O(1) ψχ(s)=O(1)
as s → 1 + s \to 1^{+} s→1+ .
Proof By Theorem 13 the series
证明 由定理 13，级数 ψ ( s ) = ∑ p 1 p s \displaystyle\psi(s)=\sum_{p}\frac{1}{p^s} ψ(s)=p∑ps1 在 s > 1 s>1 s>1 收敛；
ψ ( s ) = ∑ p 1 p s \psi(s)=\sum_{p} \frac{1}{p^{s}} ψ(s)=p∑ps1
converges for s > 1 s>1 s>1 . By comparison the series for ψ χ ( s ) \psi_{\chi}(s) ψχ(s) ) also converges for s > 1 s>1 s>1
由比较判别法， ψ χ ( s ) \psi_\chi(s) ψχ(s) 的级数同样在 s > 1 s>1 s>1 收敛。
Let s > 1 s>1 s>1 . From the Euler product for L ( s , χ ) L(s, \chi) L(s,χ) ), we have
取 s > 1 s>1 s>1，对 L ( s , χ ) L(s,\chi) L(s,χ) 的欧拉乘积取对数展开：
log L ( s , χ ) = − ∑ p log ( 1 − χ ( p ) p s ) = ∑ p ∑ m = 1 ∞ χ ( p m ) m p m s = ψ χ ( s ) + ω χ ( s ) , \log L(s,\chi)=-\sum_{p}\log\left(1-\frac{\chi(p)}{p^s}\right)=\sum_{p}\sum_{m=1}^\infty \frac{\chi(p^m)}{m p^{ms}}=\psi_\chi(s)+\omega_\chi(s), logL(s,χ)=−p∑log(1−psχ(p))=p∑m=1∑∞mpmsχ(pm)=ψχ(s)+ωχ(s),
where
其中余项
ω χ ( s ) = ∑ p ∑ m = 2 ∞ χ ( p m ) m p m s . \omega_\chi(s)=\sum_{p}\sum_{m=2}^\infty \frac{\chi(p^m)}{m p^{ms}} . ωχ(s)=p∑m=2∑∞mpmsχ(pm).
But
余项满足估计
∣ ω χ ( s ) ∣ ≤ ω ( s ) = ∑ p ∑ m = 2 ∞ 1 m p m s , |\omega_\chi(s)|\leq \omega(s)=\sum_{p}\sum_{m=2}^\infty \frac{1}{m p^{ms}}, ∣ωχ(s)∣≤ω(s)=p∑m=2∑∞mpms1,
and from the proof of Theorem 13 the series ω ( s ) = O ( 1 ) \omega(s)=O(1) ω(s)=O(1) as s → 1 + s \to 1^{+} s→1+ . As L ( 1 , χ ) ≠ 0 L(1, \chi) ≠0 L(1,χ)=0 and
由定理 13 证明可知 s → 1 + s\to1^+ s→1+ 时 ω ( s ) = O ( 1 ) \omega(s)=O(1) ω(s)=O(1)。又前文已证 L ( 1 , χ ) ≠ 0 L(1,\chi)\neq0 L(1,χ)=0，且 lim s → 1 + L ( s , χ ) = L ( 1 , χ ) \displaystyle\lim_{s\to1^+}L(s,\chi)=L(1,\chi) s→1+limL(s,χ)=L(1,χ)，故
log L ( s , χ ) = O ( 1 ) , s → 1 + . \log L(s,\chi)=O(1),\quad s\to1^+ . logL(s,χ)=O(1),s→1+.
l i m s → 1 + L ( s , χ ) = L ( 1 , χ ) lim _{s \to 1^{+}} L(s, \chi )=L(1, \chi ) lims→1+L(s,χ)=L(1,χ)
then
于是
l o g L ( s , χ ) = O ( 1 ) log L(s, \chi)=O(1) logL(s,χ)=O(1)
as s → 1 + s \to 1^{+} s→1+ and so
ψ χ ( s ) = O ( 1 ) \psi_{\chi}(s)=O(1) ψχ(s)=O(1)
as s → 1 + s \to 1^{+} s→1+ .
于是 ψ χ ( s ) = log L ( s , χ ) − ω χ ( s ) = O ( 1 ) \psi_\chi(s)=\log L(s,\chi)-\omega_\chi(s)=O(1) ψχ(s)=logL(s,χ)−ωχ(s)=O(1)，定理证毕。
The proof of Dirichlet's theorem is now almost immediate.
至此，狄利克雷定理的证明可直接完成。
Theorem 18 (Dirichlet)
定理 18（狄利克雷算术级数素数定理）
Let N N N be a positive integer, and let a a a be an integer coprime to N N N. Then the set
设 N N N 为正整数，整数 a a a 满足 gcd ( a , N ) = 1 \gcd(a,N)=1 gcd(a,N)=1，记素数集合
P a , N = { p ∣ p prime 素数 , p ≡ a ( m o d N ) } , P_{a,N} = \{p \mid p \text{ prime 素数},\ p \equiv a \pmod{N}\}, Pa,N={p∣p prime 素数, p≡a(modN)},
has Dirichlet density 1 ϕ ( N ) \displaystyle\frac{1}{\phi(N)} ϕ(N)1. In particular, P a , N P_{a,N} Pa,N is infinite, that is, there are infinitely many primes congruent to a a a modulo N N N.
则 P a , N P_{a,N} Pa,N 的狄利克雷密度为 1 ϕ ( N ) \displaystyle\frac{1}{\phi(N)} ϕ(N)1。特别地， P a , N P_{a,N} Pa,N 是无限集，即存在无穷多素数模 N N N 同余于 a a a。
Proof
证明
For χ ∈ X N \chi \in X_{N} χ∈XN let
对任意 χ ∈ X N \chi\in X_{N} χ∈XN，记
ψ χ ( s ) = ∑ p χ ( p ) p s , s > 1. \psi_{\chi}(s) = \sum_{p} \frac{\chi(p)}{p^{s}},\quad s>1. ψχ(s)=p∑psχ(p),s>1.
If χ ≠ ε \chi \neq \varepsilon χ=ε, we have seen that ψ χ ( s ) = O ( 1 ) \psi_{\chi}(s) = O(1) ψχ(s)=O(1) as s → 1 + s \to 1^{+} s→1+. For χ = ε \chi = \varepsilon χ=ε we have
若 χ ≠ ε \chi \neq \varepsilon χ=ε，由前文结论， s → 1 + s\to1^+ s→1+ 时 ψ χ ( s ) = O ( 1 ) \psi_{\chi}(s)=O(1) ψχ(s)=O(1)；若 χ = ε \chi = \varepsilon χ=ε，拆分求和得
ψ ε ( s ) = − ∑ p ∣ N 1 p s + ∑ p 1 p s . \psi_{\varepsilon}(s) = -\sum_{p \mid N} \frac{1}{p^{s}} + \sum_{p} \frac{1}{p^{s}}. ψε(s)=−p∣N∑ps1+p∑ps1.
We have seen that
∑ p 1 p s = − log ( s − 1 ) + O ( 1 ) \sum_{p} \frac{1}{p^{s}} = -\log(s-1) + O(1) p∑ps1=−log(s−1)+O(1)
as s → 1 + s \to 1^{+} s→1+, and so
已知 s → 1 + s\to1^+ s→1+ 时
ψ ε ( s ) = − log ( s − 1 ) + O ( 1 ) , s → 1 + . \psi_{\varepsilon}(s) = -\log(s-1) + O(1),\quad s \to 1^{+}. ψε(s)=−log(s−1)+O(1),s→1+.
ψ χ ( s ) = ∑ p χ ( p ) p s , s > 1. \psi_{\chi}(s) = \sum_{p} \frac{\chi(p)}{p^{s}},\quad s>1. ψχ(s)=p∑psχ(p),s>1.
而有限和 ∑ p ∣ N 1 p s = O ( 1 ) \displaystyle\sum_{p\mid N}\frac{1}{p^{s}}=O(1) p∣N∑ps1=O(1)，因此
ψ ε ( s ) = − log ( s − 1 ) + O ( 1 ) , s → 1 + . \psi_{\varepsilon}(s) = -\log(s-1) + O(1),\quad s \to 1^{+}. ψε(s)=−log(s−1)+O(1),s→1+.
Let a a a be coprime to N N N and consider
取满足 gcd ( a , N ) = 1 \gcd(a,N)=1 gcd(a,N)=1 的整数 a a a，构造求和式：
θ a ( s ) = ∑ χ ∈ X N χ ( a ) ‾ ψ χ ( s ) = ∑ χ ∈ X N χ ( a ) ‾ ∑ p χ ( p ) p s = ∑ p 1 p s ∑ χ ∈ X N χ ( a ) ‾ χ ( p ) \begin{align*} \theta_{a}(s) &= \sum_{\chi \in X_{N}} \overline{\chi(a)} \psi_{\chi}(s) \\ &= \sum_{\chi \in X_{N}} \overline{\chi(a)} \sum_{p} \frac{\chi(p)}{p^{s}} \\ &= \sum_{p} \frac{1}{p^{s}} \sum_{\chi \in X_{N}} \overline{\chi(a)} \chi(p) \end{align*} θa(s)=χ∈XN∑χ(a)ψχ(s)=χ∈XN∑χ(a)p∑psχ(p)=p∑ps1χ∈XN∑χ(a)χ(p)
Let b b b be a positive integer with a b ≡ 1 ( m o d N ) ab \equiv 1 \pmod{N} ab≡1(modN). Then χ ( b ) = χ ( a ) ‾ \chi(b) = \overline{\chi(a)} χ(b)=χ(a), and so by Theorem 7
取正整数 b b b 满足 a b ≡ 1 ( m o d N ) ab \equiv 1 \pmod{N} ab≡1(modN)，则 χ ( b ) = χ ( a ) ‾ \chi(b) = \overline{\chi(a)} χ(b)=χ(a)。结合定理 7，
∑ χ ∈ X N χ ( a ) ‾ χ ( p ) = ∑ χ ∈ X N χ ( b p ) = { ∣ X N ∣ , b p ≡ 1 ( m o d N ) , 0 , b p ≢ 1 ( m o d N ) . \begin{align*} \sum_{\chi \in X_{N}} \overline{\chi(a)} \chi(p) &= \sum_{\chi \in X_{N}} \chi(bp) \\ &= \begin{cases} \left| X_{N} \right|, & bp \equiv 1 \pmod{N}, \\ 0, & bp \not\equiv 1 \pmod{N}. \end{cases} \end{align*} χ∈XN∑χ(a)χ(p)=χ∈XN∑χ(bp)={∣XN∣,0,bp≡1(modN),bp≡1(modN).
As b p ≡ 1 b p \equiv 1 bp≡1 (mod N ) if and only if p ≡ a p \equiv a p≡a (mod N ) then
注意 b p ≡ 1 ( m o d N ) bp\equiv1\pmod{N} bp≡1(modN) 等价于 p ≡ a ( m o d N ) p\equiv a\pmod{N} p≡a(modN)，因此
θ a ( s ) = ∣ X N ∣ ∑ p ∈ P a , N 1 p s . \theta_{a}(s) = \left| X_{N} \right| \sum_{p \in P_{a,N}} \frac{1}{p^{s}}. θa(s)=∣XN∣p∈Pa,N∑ps1.
But
另一方面
θ a ( s ) = − l o g ( s − 1 ) + O ( 1 ) \theta _{a}(s)=-log (s-1)+O(1) θa(s)=−log(s−1)+O(1)
and so P a , N P_{a, N} Pa,N has Dirichlet density
由狄利克雷密度定义， P a , N P_{a,N} Pa,N 的密度为
1 ∣ X N ∣ = 1 ∣ Z N ∗ ∣ = 1 ϕ ( N ) . \frac{1}{|X_N|}=\frac{1}{|\mathbb{Z}_N^*|}=\frac{1}{\phi(N)} . ∣XN∣1=∣ZN∗∣1=ϕ(N)1.
In particular P a , N P_{a, N} Pa,N is infinite.
密度非零，故集合 P a , N P_{a,N} Pa,N 有无穷多个素数，定理证毕。
5 Remarks on complex variables
5 复变方法补充说明
We briefly indicate on how this proof may be simplified when we use complex analysis.
简要说明若使用复分析，可大幅简化上述证明。
In accordance with the conventions of analytic number theory, let s = σ + i t s=\sigma+ it s=σ+it denote a complex variable in this section. Then the series for ζ ( s ) \zeta(s) ζ(s) ) converges for σ > 1 \sigma>1 σ>1 . The convergence is uniform on sets { s : σ > c } \{s: \sigma>c\} {s:σ>c} where c > 1 c>1 c>1 so that ζ ( s ) \zeta(s) ζ(s) ) is holomorphic for σ > 1 \sigma>1 σ>1 Similarly using partial summation, L ( s , χ ) L(s, \chi) L(s,χ) ) is holomorphic for σ > 0 \sigma>0 σ>0 whenever x is a nontrivial character. Then Theorem 12 become trivial (but also unnecessary). Also ζ ( s ) − 1 / ( s − 1 ) \zeta(s)-1 /(s-1) ζ(s)−1/(s−1) has an analytic continuation on the set { s : σ > 0 } \{s: \sigma>0\} {s:σ>0} . A corresponding result is true for L ( s , ε ) L(s, \varepsilon) L(s,ε) .
解析数论标准记号：本节记复变量 s = σ + i t s=\sigma+it s=σ+it。 ζ ( s ) \zeta(s) ζ(s) 的级数在 σ > 1 \sigma>1 σ>1 收敛，且在任意 σ > c ( c > 1 ) \sigma>c\ (c>1) σ>c (c>1) 上一致收敛，故 ζ ( s ) \zeta(s) ζ(s) 在半平面 σ > 1 \sigma>1 σ>1 全纯。同理通过分部求和可证：非平凡特征标对应的 L ( s , χ ) L(s,\chi) L(s,χ) 在 σ > 0 \sigma>0 σ>0 全纯。此时定理12的结论变得显然（甚至无需专门证明）。同时 ζ ( s ) − 1 s − 1 \zeta(s)-\dfrac{1}{s-1} ζ(s)−s−11 可解析延拓至 σ > 0 \sigma>0 σ>0，平凡特征标的 L ( s , ε ) L(s,\varepsilon) L(s,ε) 有完全相同的延拓性质。
If L ( 1 , χ ) L(1, \chi) L(1,χ) vanishes for some χ ∈ X N \chi \in X_{N} χ∈XN, then
若存在特征 χ ∈ X N \chi\in X_N χ∈XN 使得 L ( 1 , χ ) = 0 L(1,\chi)=0 L(1,χ)=0，则
f ( s ) = ∏ χ ∈ X N L ( s , χ ) f(s)=\prod_{\chi\in X_N}L(s,\chi) f(s)=χ∈XN∏L(s,χ)
is holomorphic on { s : σ > 0 } \{s: \sigma>0\} {s:σ>0}.
该函数在区域 { s : σ > 0 } \{s: \sigma>0\} {s:σ>0} 上全纯。
But for s > 1 s>1 s>1
而当实数 s > 1 s>1 s>1 时
f ( s ) = ∑ n = 1 ∞ a n n s f(s)=\sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} f(s)=n=1∑∞nsan
with all coefficients a n ≥ 0 a_n\geq0 an≥0.
其所有系数满足 a n ≥ 0 a_n\geq0 an≥0。
where the a n a_{n} an are nonnegative. A standard result on series of this form (Dirichlet series) shows that if all a n a_{n} an are nonnegative and f ( s ) f(s) f(s) ) analytically continues to { s : σ > 0 } \{s: \sigma>0\} {s:σ>0} , then the series is convergent there. However examination of our series shows that there is a real number s with 0 < s < 1 0<s<1 0<s<1 such that the series fails to converge. This gives the crucial non-vanishing result for the L ( 1 , χ ) L(1, \chi) L(1,χ) ). This approach is conceptually simpler, and much less ad hoc than the proof we give here.
狄利克雷级数经典结论：若系数全部非负，且函数可解析延拓到 σ > 0 \sigma>0 σ>0，则级数在 σ > 0 \sigma>0 σ>0 处处收敛。但本文构造的乘积级数在区间 0 < s < 1 0<s<1 0<s<1 存在发散点，导出矛盾，直接证得所有 L ( 1 , χ ) ≠ 0 L(1,\chi)\neq0 L(1,χ)=0。该复分析思路逻辑更简洁，相比本文实变量蒙斯基证法，无需构造特殊朗伯级数，更少依赖特殊技巧。
References
参考文献
1 Paul Monsky, 'Simplifying the proof of Dirichlet's theorem', American Mathematical Monthly, 100 (1993) 861--862.
1 保罗·蒙斯基，《简化狄利克雷定理的证明》，《美国数学月刊》，第100卷，1993年，861--862页。
2 Jean-Pierre Serre, A Course in Arithmetic, Springer-Verlag, 1973.
2 让-皮埃尔·塞尔，《算术教程》，施普林格出版社，1973年。