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$N_chi（\alpha，T）$上的估计如何导致算术序列的Dirichlet素数定理？

设$N_chi（\alpha，T）$是$L（s=\sigma+it，\chi）=\sum\frac{\chi（N）}{N^s}$的零个数，其中$c>0$和$（\sigma，T）$位于矩形$[\阿尔法，1]\times[-T，T]$中。

在各种论文中，人们可以阅读所有L函数的零点数估计：

$$\sum_{\chi\mod q}N_chi（\alpha，T）\ll T^{c（1-\alpha）}$$

这如何暗示算术级数的素数定理？

$$\sum_{1\leq\leqQ}\sum_{chi}\left|\sum_x^{x+h}\chi（p）\log p\right|\ll h\cdot e^{large-a\frac{\log x}{\log-q}}$$

我几乎认不出这是声明任何算术序列中都有无限多个素数

由于这是一个关于所有字符$\chi$的语句，我们如何得到一个算术序列$n=aq+b$？

我是一个不专业的人，我怀疑这样的结果对该领域以外的人来说并不是100%清楚的。

精神上类似，Polya-Vinogradov不等式是否意味着算术序列的PNT？这可能是一个单独的问题。

答案

<trans data-src="Firstly, a general comment: as understanding of a mathematical problem deepens, it is common (and even expected) for the most mathematically natural formulation of a given problem (or class of problems) to become "hardly recognisable" as arising from the formulation of the problem that historically motivated work in this area.  ">首先，一般性评论：随着对数学问题的理解加深，一个给定问题（或一类问题）的数学上最自然的形式通常（甚至可以预期）会变得“难以识别”，因为这是历史驱动的问题在这一领域发挥作用的形式。</trans><trans data-src="For instance, the classical Greek problems of trisecting the angle or doubling the cube by straightedge and compass became transformed into part of the Galois theory of field extensions,">例如，经典的希腊问题——用直尺和指南针将角度三等分或将立方体加倍——变成了伽罗瓦场扩展理论的一部分，</trans><trans data-src=" whereas the superficially similar classical Greek question of squaring the circle by straightedge and compass has now instead become part of transcendence theory.  ">然而，用直尺和指南针将圆平方的表面上类似的古希腊问题，现在却成为了超越理论的一部分。</trans><trans data-src="These fields have since transformed further, for instance in modern algebraic geometry one can think of Galois theory as a special case of the theory of schemes (and their fundamental groups).  ">这些领域已经发生了进一步的转变，例如在现代代数几何中，人们可以将伽罗瓦理论视为图式理论（及其基本群）的特例。</trans><trans data-src="These transformations arise because the mathematical tools and insights that one slowly acquires to attack these problems are often naturally abstracted using a formalism which often has no direct connection to the historical motivations of the problem,">这些转变之所以出现，是因为人们为解决这些问题而慢慢获得的数学工具和见解往往是使用形式主义自然抽象出来的，而形式主义通常与问题的历史动机没有直接联系，</trans><trans data-src=" other than that the historical problem happens to be transformable to a special case of a much broader (and often much more interesting, ultimately) class of questions that the formalism is well equipped to handle.">除此之外，历史问题恰好可以转化为形式主义能够处理的更广泛（最终往往更有趣）的一类问题的特殊情况。</trans><trans data-src="Now for the specific question at hand.  ">现在谈谈手头的具体问题。</trans><trans data-src="The first insight, fundamental to modern analytic number theory, is that historical number-theoretic questions such as "how can I show that there infinitely many primes of the form $X$?" should be thought of as special cases of the much more general problem "How can I obtain good asymptotics or bounds for sums of the form $\sum_p f(p)">第一个洞见是现代分析数论的基础，它是历史数论问题，例如“我如何证明$X$形式有无穷多个素数？”应被视为更一般问题的特例“我如何获得$\sum_p f（p）形式和的良好渐近性或界</trans><trans data-src="$, where $p$ ranges over primes?">$，$p$在哪里超过素数？</trans><trans data-src="".  The original problem can be viewed as the special case where $f(p)$ is the indicator function $1_X$ associated to the property $X$;">“。原始问题可以视为特殊情况，其中$f（p）$是与属性$X$；关联的指示符函数$1_X$；</trans><trans data-src=" however the latter problem is in a much more general, flexible, and powerful framework, because of all the tools we have to compare one sum with another.  ">然而，后一个问题是在一个更通用、更灵活、更强大的框架中，因为我们必须使用所有工具来比较一个和另一个。</trans><trans data-src="For instance, rather than doing a naive unweighted count $\sum_p 1_X(p)$, one could work with more general weighted sums such as $\sum_p \frac{1_X(p)}{p}$, $\sum_p \frac{1_X(p)}{p^s}$, $\sum_p 1_X(p) \log p$, etc. which may be more advantageous to work with.  ">例如，与其进行简单的未加权计数$\sum_p1_X（p）$，还不如使用更一般的加权和，例如$\sum_p\frac{1_X。</trans><trans data-src="In the case of Dirichlet's theorem on arithmetic progressions, a key insight of Dirichlet himself was that the multiplicative Fourier expansion">在Dirichlet关于算术级数的定理的例子中，Dirichllet自己的一个关键见解是乘法傅里叶展开</trans><trans data-src="$$ 1_{a \hbox{ mod } q}(n) = \frac{1}{\phi(q)} \sum_\chi \overline{\chi(a)} \chi(n)$$">$$1_{a\hbox{mod}q}（n）=\frac{1}{\phi（q）}\sum_\chi\上划线{\chi（a）}\chi$$</trans><trans data-src="allowed one to convert the original question of primes into arithmetic progressions $a \hbox{ mod } q$ into what turned out to ultimately be a much more interesting question, namely the estimation of Dirichlet character sums over primes, such as $\sum_{x \leq p \leq x+h} \chi(p) \log p$.  ">允许人们将素数的原始问题转换为算术级数$a\hbox{mod}q$，最终变成一个更有趣的问题，即素数上Dirichlet字符和的估计，例如$\sum{x\leqp\leqx+h}\chi（p）\logp$。</trans><trans data-src="Basically, the more upper bounds one has on these sums for non-principal Dirichlet characters, the more easily one is able to control historically interesting statistics such as $\sum_{x \leq p \leq x+h: p = a \hbox{ mod } q} 1$ in terms of the principal character $\chi_0(n) := 1_{(n,q)=1}$, which relatively easy to work with.">基本上，非主Dirichlet字符的这些和的上界越大，就越容易控制历史上有趣的统计数据，例如以主字符$\chi_0（n）：=1_{（n，q）=1}$表示的$\sum_{x\leqp\leqx+h:p=a\hbox{mod}q}1$，这相对容易处理。</trans><trans data-src="The main reason that one prefers to work with characters over arithmetic progressions is because the former has much better multiplicative structure than the latter, allowing the full power of multiplicative number theory to be brought into play.  ">相比算术级数，人们更喜欢使用字符的主要原因是前者的乘法结构比后者好得多，从而可以充分发挥乘法数论的威力。</trans><trans data-src="Indeed, the next great insight of Dirichlet was that character sums and products over primes were closely tied to the associated Dirichlet $L$-function, through identities such the Euler product formula">事实上，Dirichlet的下一个伟大见解是，素数上的特征和和乘积通过Euler乘积公式等恒等式与相关的Dirichlet$L$-函数紧密相连</trans><trans data-src="$$ L(s,\chi) = \prod_p (1 - \frac{\chi(p)}{p^s})^{-1}$$">$$L（s，\chi）=\prod_p（1-\frac{\chi（p）}{p^s}）^{-1}$$</trans><trans data-src="(which encodes the fundamental theorem of arithmetic and the multiplicative nature of the Dirichlet character $\chi$) or the log-derivative of this formula,">（编码算术基本定理和Dirichlet字符$\chi$的乘法性质）或此公式的对数导数，</trans><trans data-src="$$ -\frac{L'(s,\chi)}{L(s,\chi)} = \sum_{j=1}^\infty \sum_p \frac{\chi(p)^j}{p^{sj}} \log p.$$">$$-\frac{L'（s，\chi）}{L（s，\ chi）{=\sum_{j=1}^\infty\sum_p\frac{\chi（p）^j}{p^{sj}}\log p$$</trans><trans data-src="In the latter formula one can already begin seeing why it is in fact natural to weight primes by characters and by the logarithmic weight $\log p$.  ">在后一个公式中，人们已经开始明白为什么用字符和对数权重$\log p$来衡量素数实际上是很自然的。</trans><trans data-src="(Actually, the most convenient way to package all this information is to replace sums over primes with sums over natural numbers weighted by the [von Mangoldt function][1].)">（实际上，打包所有这些信息的最方便的方法是将素数和替换为按[von Mangoldt函数][1]加权的自然数和。）</trans><trans data-src="Through complex analysis tools such as the residue theorem, we know that the behaviour of the meromorphic function $-\frac{L'(s,\chi)}{L(s,\chi)}$ is controlled by the location of the zeroes of the Dirichlet $L$-function $L(s,\chi)$.  ">通过余数定理等复杂分析工具，我们知道亚纯函数$-\frac{L'（s，\chi）}{L（s，\chi）}$的行为由狄利克雷$L$-函数$L（s，\chi）$的零的位置控制。</trans><trans data-src="Putting all this together, we now see that the key question one needs to resolve to understand primes in arithmetic progressions is where the zeroes of $L(s,\chi)$ are.  ">综上所述，我们现在看到，要理解算术级数中的素数，需要解决的关键问题是$L（s，\chi）$的零在哪里。</trans><trans data-src="For instance, Dirichlet's original proof of his theorem ultimately reduced matters to showing that $s=1$ is not a zero of this function.  ">例如，Dirichlet对其定理的原始证明最终将问题简化为表明$s=1$不是该函数的零。</trans><trans data-src="Through various "explicit formulae", one can make this connection much more manifest, for instance the von Mangoldt explicit formula for the Dirichlet L-function basically reads ">通过各种“显式公式”，可以使这种联系更加明显，例如，Dirichlet L函数的von Mangoldt显式公式基本上如下所示</trans><trans data-src="$$ \sum_{p \leq X} \chi(p) \log p + \dots = - \sum_\rho \frac{X^\rho}{\rho} + \dots$$">$$\sum_{p\leq X}\chi（p）\log p+\dotes=-\sum_\rho\frac{X^\rho}{\rho{+\dots$$</trans><trans data-src="where $\rho$ ranges over zeroes of $L(s,\chi)$, and the $\dots$ conceal some lower order terms which I will omit here (together with the issue of how to interpret the infinite sum on the RHS, or whether one should truncate it to a finite sum) for simplicity.  ">其中$\rho$的范围超过$L（s，\chi）$的零，而$\dots$隐藏了一些低阶项，为了简单起见，我将在这里省略这些项（以及如何解释RHS上的无穷和，或者是否应将其截断为有限和的问题）。</trans><trans data-src="Combining such formulae with the multiplicative Fourier expansion mentioned earlier gives some useful direct connection between primes in arithmetic progressions and zeroes of $L$-functions.  ">将这些公式与前面提到的乘法傅里叶展开相结合，可以在算术级数中的素数和$L$-函数的零点之间提供一些有用的直接联系。</trans><trans data-src="(This connection is sometimes popularly referred to as "the music of the primes", particularly in the $q=1$ case.  Actually, with an appropriately "adelic" mindset, the $q=1$ case and $q>1$ cases can be unified into what is arguably an even more natural, albeit abstract, formalism, but that is perhaps a story for another time.)">（这种联系有时被普遍称为“素数的音乐”，尤其是在$q=1$的情况下。实际上，通过适当的“专业”思维方式，$q=1$case和$q>1$的情况可以统一成一种更自然、更抽象的形式主义，但这可能是另一个时代的故事。）</trans><trans data-src="One thing that the explicit formula reveals is that the zeroes $\rho$ that are of large real part, $\operatorname{Re}(s) > \alpha$, will have an outsize impact on the distribution of the primes.  ">显式公式揭示的一件事是，大实数部分的零$\rho$，即$\operatorname{Re}（s）>\alpha$，将对素数的分布产生巨大影响。</trans><trans data-src="Ideally, to get the most out of the explicit formula, we would like no zeroes whatsoever to the right of the critical line $\operatorname{Re}(s)=\frac{1}{2}$ (which is where all known zeroes of Dirichlet L-functions reside), and this is the main reason why the generalised Riemann hypothesis (GRH) is such a prized objective.  ">理想情况下，为了最大限度地利用显式公式，我们希望在临界线$\operatorname｛Re｝（s）=\frac｛1｝｛2｝$（狄利克雷L函数的所有已知零都位于此处）的右边没有任何零，这就是广义黎曼假设（GRH）是如此重要的目标的主要原因。</trans><trans data-src="Of course, we can't prove GRH, but in some applications we can make do with weaker <I>density theorems</I> that don't entirely prohibit zeroes from appearing to the right of the critical line, but at least limit how many of them can do so,">当然，我们不能证明GRH，但在某些应用中，我们可以使用较弱的密度定理来解决这个问题，这些密度定理并不完全禁止零出现在临界线的右侧，但至少限制了它们中有多少个可以这样做，</trans><trans data-src=" which can still lead to reasonable upper bounds on various error terms that arise when using the explicit formula.  ">这仍然可以导致在使用显式公式时出现的各种误差项的合理上限。</trans><trans data-src="As it turns out, the closer one gets to the line $\operatorname{Re}(s)=1$, the easier it is to exclude zeroes, and indeed it is known that there are no zeroes on or to the right of this line.  ">事实证明，越接近$\operatorname｛Re｝（s）=1$这一行，就越容易排除零，事实上，这一行上或右边都没有零。</trans><trans data-src="For $\alpha < 1$, we have thus far been unable to prevent an infinity of zeroes from lying in the infinite strip $\{ s: \operatorname{Re}(s) > \alpha \}$, but we have at least been able to get non-trivial bounds for the zeroes in rectangles such as $\{ s: \operatorname{Re}(s) > \alpha; ">对于$\alpha<1$，到目前为止，我们无法防止无穷多的零位于无限条带$\{s:\operatorname{Re}（s）>\alpha\}$中，但我们至少能够获得矩形中零的非平凡边界，例如$\{s:\operator name{Re{（s；</trans><trans data-src="0 \leq \operatorname{Im}(s) < T\}$, and this, together with suitably truncated versions of a suitable explicit formula, turns out to be good enough (after some calculation) to control primes in various arithmetic progressions, as is done in the paper of Gallagher you are citing.">0\leq\operatorname{Im}（s）<T\}$，这与适当的显式公式的适当截断版本一起，证明足够好（经过一些计算）来控制各种算术级数中的素数，正如您引用的Gallagher的论文中所做的那样。</trans><trans data-src="One might start with Davenport's multiplicative number theory book for an introduction to all this,">我们可以从达文波特的乘法数论书开始介绍这一切，</trans><trans data-src=" in particular to the proof of the prime number theorem in arithmetic progressions with classical error term which uses the same general strategy as in Gallagher's paper but is somewhat easier to execute ">特别是证明带有经典误差项的算术级数中的素数定理，它使用了与Gallagher论文中相同的一般策略，但更容易执行</trans><trans data-src="(one only needs the classical zero free region for the Dirichlet L-function, rather than the more difficult zero density estimates used by Gallagher).  ">（Dirichlet L函数只需要经典的零自由区，而不是Gallagher使用的更困难的零密度估计）。</trans><trans data-src="As mentioned in GH's answer, Iwaniec-Kowalski gives a more modern and advanced treatment of these topics.  ">正如GH的回答中所提到的，伊瓦尼埃克·科瓦尔斯基对这些主题进行了更现代、更高级的处理。</trans><trans data-src="I also like Montgomery's CBMS book.">我也喜欢蒙哥马利的CBMS书。</trans><trans data-src="As for the Polya-Vinogradov estimate, it is certainly of some relevance in bounding character sums, which in turn can be used to control L-functions,">至于Polya-Vinogradov估计，它在限定字符和方面肯定有一定的相关性，而这些字符和又可用于控制L函数，</trans><trans data-src=" but it is generally not the most convenient tool to use for this purpose (for instance, the original formulation of the inequality is concerned with sharply truncated character sums, whereas smoothed character sums are often more natural to work with for the purposes of understanding L-functions).  ">但它通常不是用于此目的的最方便的工具（例如，不等式的原始公式与急剧截断的字符和有关，而平滑的字符和通常更适合用于理解L函数）。</trans><trans data-src="Nevertheless there is certainly a close kinship.  ">然而，确实有一种亲密的亲属关系。</trans><trans data-src="For instance the Fourier inversion used to prove Polya-Vinogradov is closely tied to the Poisson formula proof of the functional equation for the Dirichlet L-function.">例如，用于证明Polya-Vinogradov的傅里叶反演与Dirichlet L函数的函数方程的泊松公式证明密切相关。</trans><trans data-src="[1]: ">[1]: </trans><trans data-src="https://en.wikipedia.org/wiki/Von_Mangoldt_function">https://en.wikipedia.org/wiki/Von_Mangoldt_function（维基百科网）</trans>

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$\开始组$ 总有一天这种风格会更自然。现在我将使用你提供的大纲。谢谢你写这篇文章！ $\端组$
– 约翰·曼古尔
2015年12月27日18:46
1

$\开始组$ “对于α<1，到目前为止，我们已经能够防止无穷多的零位于无限条{s:Re（s）>α}中。”这将是一个巨大的打击。等待你的论文。 $\端组$
– 范正
2015年12月28日3:24
1

$\开始组$ 更正了拼写错误。 $\端组$
– 陶哲轩
2015年12月28日3:49
$\开始组$ 解释得很好。我特别喜欢“把原来的问题[……]转化成最终更有趣的问题”。 $\端组$
– 来自MO的GH
2015年12月28日4:02

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