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Let $\chi$ be a Dirichlet character and $L(1,\chi)$ the associated L-function evaluated at $s=1$. What would be the 'shortest' proof of the non-vanishing of $L(1,\chi)$?

Background: The non-vanishing of $L(1,\chi)$ plays an essential role in the proof of Dirichlet´s theorem on primes in arithmetic progressions. In his "Introduction to analytic number theory", T. M. Apostol gives an elementary proof of the above fact estimating various sums in a few lemmas in the context of a proof of the aforementioned Dirichlet theorem. While his approach has the advantage of being self-contained and not requiring much of a background, it is quite lenghty. In their "Analytic number theory", H. Iwaniec and E. Kowalski remark that in Dirichlet´s original proof the non-vanishing of $L(1,\chi)$ for real Dirichlet characters is a simple consequence of Dirichlet´s class number formula. However, in both approaches it is necessary to distinguish between real and complex Dirichlet characters. Hence my two "sub"-questions:

1) Is there a proof that avoids the distinction between the complex and real case?

2) Are there in general other proof strategies for $L(1,\chi)\neq 0$ that can be considered shorter and/or more elegant than the two mentioned above?

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As for (1), I think it is a bad idea. There is a fundamental distinction between the two cases. The idea is, that if $L(1,\chi)=0$ for $\chi$ complex, then so does its conjugate, which gives a double zero in the product that David Speyer gives below. Enlarging this idea, we get a fairly decent zero-free region for $L(1,\chi)=0$ for $\chi$ complex (I don't recall, but $1/\log D$ maybe). But for real characters this is not true, and we have to worry about the so-called Siegel zeros, and the zero-free region is much worse (as $1/\sqrt D$). The difference is the single vs double effect. – Junkie May 25 '10 at 5:03
Also, this is longer not shorter, but the use of Eisenstein series to prove non-vanishing on the 1-line was in vogue (originally Jacquet-Shalika in the 70s ), due to Sarnak's reworking a few years ago. See (and the Sarnak ref of there), which does a zero-free region. The proof that symmetric square $L$-functions don't vanish at the edge, and indeed lack Siegel zeros (unless induced) by Goldfeld, Hoffstein, Lieman followed the product idea. – Junkie May 25 '10 at 5:52
In terms of "clearest causation", I still do think that the spectral argument using the constant term of Eisenstein series for $GL_2$ is the most memorable, the most explanatory, and the most suggestive of the broader situation... – paul garrett Jan 25 '15 at 0:29

15 Answers 15

up vote 27 down vote accepted

I like the proof by Paul Monsky: 'Simplifying the Proof of Dirichlet's Theorem' American Mathematical Monthly, Vol. 100 (1993), pp. 861-862.

Naturally this does maintain the distinction between real and complex as whatever you do, the complex case always seems to be easier as one would have two vanishing L-functions for the price of one.

I incorporated this argument into my note on a "real-variable" proof of Dirichlet's theorem at .

There are proofs, notably in Serre's Course in Arithmetic which claim to treat the real and complex case on the same footing. But this is an illusion; it pretends the complex case is as hard as the real case. Serre considers the product $\zeta_m(s)=\prod L(s,\chi)$ where $\chi$ ranges over the modulo $m$ Dirichlet characters. If one of the $L(1,\chi)$ vanishes then $\zeta_m(s)$ is bounded as $s\to 1$ and Serre obtains a contradiction by using Landau's theorem on the abscissa of convergence of a positive Dirichlet series. But all this subtlety is only needed for the case of real $\chi$. In the non-real case, at least two of the $L(1,\chi)$ vanish so that $\zeta_m(s)\to0$ as $s\to1$. But it's elementary that $\zeta_m(s)>1$ for real $s>1$ and the contradiction is immediate, without the need of Landau's subtle result.

Added (25/5/2010) I like the Ingham/Bateman method. It is superficially elegant, but as I said in the comments, it makes the complex case as hard as the real. Again it reduces to using Landau's result or a choice of other trickery. What one should look at is not $\zeta(s)^2L(s,\chi)L(s,\overline\chi)$ but $$G(s)=\zeta(s)^6 L(s,\chi)^4 L(s,\overline\chi)^4 L(s,\chi^2)L(s,\overline\chi^2)$$ (cf the famous proof of nonvanishing of $\zeta$ on $s=1+it$ by Mertens). Unless $\chi$ is real-valued this function will vanish at $s=1$ if $L(1,\chi)=0$. But one shows that $\log G(s)$ is a Dirichlet series with nonnegative coefficients and we get an immediate contradiction without any subtle lemmas. Again it shows that the real case is the hard one. For real $\chi$ then $G(s)=[\zeta(s)L(s,\chi)]^8$ while Ingham/Bateman would have us consider $[\zeta(s)L(s,\chi)]^2$. This leads us to the realization that for real $\chi$ we should look at $\zeta(s)L(s,\chi)$ which is the Dedekind zeta function of a quadratic field. (So if one is minded to prove the nonvanishing by showing that a Dedekind zeta function has a pole, quadratic fields suffice, and one needn't bother with cyclotomic fields).

But we can do more. Let $t$ be real and consider $$G_t(s)= \zeta(s)^6 L(s+it,\chi)^4 L(s-it,\overline\chi)^4 L(s+2it,\chi^2)L(s-2it,\overline\chi^2).$$ Unless both $t=0$ and $\chi$ is real, if $L(1+it,\chi)=0$ one gets a contradiction just as before. So the nonvanishing of any $L(s,\chi)$ on the line $1+it$ is easy except at $1$ for real $\chi$. This special case really does seem to be deeper!

Added (26/5/2010) The argument I outlined with the function $G_t(s)$ is well-known to extend to a proof for a zero-free region of the L-function to the left of the line $1+it$. At least it does when unless $t=0$ and $\chi$ is real-valued. In that case it breaks down and we get the phenomenon of the Siegel zero; the possible zero of $L(s,\chi)$ for $\chi$ real-valued, just to the left of $1$ on the real line. So the extra difficulty of proving $L(1,\chi)\ne0$ for $\chi$ real-valued is liked to the persistent intractability of showing that Siegel zeroes never exist.

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You mean at the end that zeta_m(s) > 1 for real s > 1, not real s > 0. (In fact the Riemann zeta-function is negative for 0 < s < 1.) – KConrad May 24 '10 at 22:59
Robin, what are your thoughts about the uniform treatment by Bateman in the answer I wrote (I edited it in after J.H.S. pointed it out in his answer)? Note there are two factors of the zeta-function in the product considered there, so the common treatment of the real and complex cases doesn't seem like an illusion (to me). – KConrad May 24 '10 at 23:28
I definitely agree that it is an illusion for the lack of real-complex distinction. I really like your re-phrasing as "pretends the complex case is as hard as the real case", which can be said about almost the attempts given here. – Junkie May 25 '10 at 5:05
Keith, thanks for the correction. Maybe with the Ingham/Bateman proof we finally have a proof for the complex case as difficult as the real case :-) – Robin Chapman May 25 '10 at 6:28
Cute. Actually, if you wanted to show nonvanishing of L(s,chi) everywhere on the line Re(s) = 1, and not just at s = 1, do you still think the complex case should be considered easier than the real case? – KConrad May 25 '10 at 6:41

My preferred proof is to consider $Z(s) := \prod_{\chi} L(\chi, s)$, where the product is over all characters of $\mathbb{Z}/n$, including the trivial character. It is easy to see that $L(s, \chi)$ is meromorphic near $s=1$, and analytic for $\chi$ nontrivial. So, if any of the $L(1,\chi)$'s were zero, then $Z(s)$ would be analytic near $1$.

But $Z(s)$ is the $\zeta$-function of $\mathbb{Z}[\zeta_n]$. If you know the class number formula for $\mathbb{Z}[\zeta_n]$, then you immediately know that $Z(s)$ has a pole at $s=1$. I think the best approach is to prove this class number formula.

If you want, you can also use the idea of the class number proof but cut some corners. Namely:

  1. Discard primes dividing $n$ from the Euler products. The modified $Z(s)$ is the generating function for ideals $\mathbb{Z}[\zeta_n]$ which are relatively prime to $n$, and that also works. This means you only have to analyze how unrammified primes factor.

  2. We have $Z(s) = \sum_I |I|^{-s}$, where the sum is over all nonzero ideals. We can bound this below by the sum over principal ideals. That already blows up as $s \to 1^{+}$.

  3. You only need to prove that the unit group of $\mathbb{Z}[\zeta_n]$ is discrete, which is the easy part of the Dirichlet unit theorem. If there were fewer units than the unit theorem predicts, the sum would blow up even faster.

I've cut these corners when presenting the idea of the class number proof over tea; but I would probably give the right proof if I were teaching a class.

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First a quibble: is it common to take the $\zeta$-function of a cyclotomic ring, rather than a cyclotomic field? And #1 should say "ideals of Z[zeta_n]"? Is it that easy to declare (to a student) that $Z(s)$ is indeed the $\zeta$-function of the cyclotomic field -- coming from an analytic number theory background (Ayoub's book from the 60s for me actually), Kronecker-Weber, or whatever Lang would have in his book, might end up being a side-trip if this is not obvious. For that matter, discreteness of the unit group (as compared to class no formula for quadratic chars) is also overhead. – Junkie May 25 '10 at 5:10
I guess I can re-phrase my comment as: if you are going to use the class number formula in the end, it seems an excess to do so with cyclotomic fields. As Robin Chapman noted, you can just pair off complex conjugate characters quite easily, and then are left with just quadratics for the class number formula (which are easier from a ground-up viewpoint, though for someone facile in number fields, it matters not I guess). Of course, this is notably opposite to the what the OP wanted. – Junkie May 25 '10 at 5:24
I'm kind of surprised that this was the highest rated answer, since it is more of an answer to the most elegant proof than the shortest proof. But I will reply to some of your comments: – David Speyer May 25 '10 at 10:53
I'm not sure why the class number formula for arbitrary number fields is harder than the quadratic case. The presence of an infinite unit group makes the proof harder, but you already have to deal with real quadratic fields. The differences between a rank one unit group and a general one are only notational. (Although, depending on your audience, notation can be a big savings.) – David Speyer May 25 '10 at 10:56
Of course, it is true that the quadratic case is harder than the others. But I think that, pedagogically and aesthetically, it is better to give a proof which handles the hardest case in such a way that the other cases are covered at the same time. When I am trying to learn a proof, I find that every case is another burden on my understanding. – David Speyer May 25 '10 at 11:00

It's easy to show that one cannot have two real characters $\chi, \chi'$ with $L(1,\chi)=L(1,\chi')=0$, as then $\zeta(s) L(s,\chi) L(s,\chi') L(s,\chi \chi')$ would have a simple zero at s=1 while having non-negative coefficients, which is absurd. This is of course a minor variant of the argument that rules out vanishing for a complex character (and its complex conjugate). The intuition here is that the primes can almost conceivably conspire to be almost totally correlated (or more precisely, anti-correlated) with one character, but not with two; see this blog post of mine.

In some sense, the fact that $L(1,\chi) \neq 0$ with at most one exception is the best one can do using without introducing algebraic number theory methods (class number formula) or moving away from s=1 (in particular, using s=1/2 information); this is discussed in this paper of Granville.

The fact that nobody knows how to make the constants in Siegel's theorem (the one on Siegel zeroes) effective seems to me to be a significant upper bound as to how short or elegant a known proof of $L(1,\chi) \neq 0$ can be; at some point one must perform some maneuvre that is very expensive with regards to effective bounds (e.g. moving one's attention from s=1 to s=1/2, or bounding the class number from below by 1).

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Variants of the proof I am about to give have already been given; here's "the rest of the story".

Let $\chi$ be a quadratic character with conductor $m$. Then $L(s,\chi) \zeta(s) = \zeta_K(s)$ is the Dedekind zeta function of the quadratic number field $K$ whose decomposition law is determined by $\chi$ (we have $K = {\mathbb Q}(\sqrt{ \pm m})$ for a suitable choice of the sign), hence $\zeta_K(s) = \sum c_n n^{-s}$, where $c_n$ denotes the number of ideals with norm $n$. In particular, $c_n \ge 0$ and $c_{n^2} \ge 1$ since $(n)$ has norm $n$. If we had $L(1,\chi) = 0$, this zero would cancel the pole of $\zeta(s)$, and $\zeta_K(s)$ would be analytic for all $s$ with real part $> 0$; on the other hand, the divergence of the harmonic series implies that $\zeta_K(s)$ diverges at $s = \frac12$.

Now we can strip off the background by observing that $c_n = \sum_{d \mid n} \chi(d)$. We can show directly that $c_n \ge 0$ and $c_{n^2} \ge 1$ and show that $\sum c_n n^{-1/2}$ diverges. A slightly technical but elementary manipulation shows that $$ \sigma = \sum_{t = 1}^{n^2} \frac{c_t}{\sqrt{t}} = 2n L(1,\chi) + O(1). $$ If we had $L(1,\chi) = 0$, this sum would be bounded: contradiction.

With slightly more effort, but still completely elementary means, we can prove that
$$ \log n < \sigma < 2n L(1,\chi) + 6\phi(m) + 1, $$ from which we can easily deduce that $$ L(1,\chi) > \frac12 e^{-6 \phi(m) - 2}. $$ This is a lousy lower bound, but it gives us what we want.

The basic idea behind this proof is due to Mertens in the late 1890s. It was the first proof of Dirichlet's theorem that did not use the quadratic reciprocity law (Dirichlet showed that quadratic characters $\chi$ could be written in the form $\chi(a) = (\frac{m}{a})$, and then showed that $L(1,\chi)$ is a product of nonzero terms involving the class number of the quadratic number field $K$), and hence it was possible to use Dirichlet's theorem for closing the gap in Legendre's proof (the fact that this had already achieved by Kummer had not been noticed by then). Mertens [Wiener Berichte 106 (1897), 254-286; J. Reine Angew. Math. 117 (1897), 169-184] did not start with the Dedekind zeta function but rather with Dirichlet's proof that $$\sum_{n=1}^N d(n) = N \log N + (2\gamma - 1) + O(\sqrt{N}), $$ where $d(n)$ denotes the sum of divisors of $n$. Dirichlet proved this result about the same time he proved his theorem on primes in arithmetic progression; he admitted that his first proof of the nonvanishing of his L-series was complicated, so he dropped it altogether and replaced it by his proof of the class number formula. It is not unlikely that his lost first proof of the nonvanishing contained elements of Mertens's idea.

Pieces of this story occur in Hasse's Vorlesungen über Zahlentheorie, Siegel's lectures in analytic number theory (in German; have these been translated into English?) and Zagier's book on quadratic fields and forms (also in German).

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I like very much the proof in Serre's A Course in Arithmetic, which I have rewritten slightly as Section 3 of

It turns on a simple analytic fact about convergence of Dirichlet series with non-negative coefficients due to Landau: see Theorem 18 on p. 13 of

In particular, this proof does not require separate consideration of real and complex characters.

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I guess I don't understand all the fuss about the real versus complex case. Agreed, the complex case has a two-line proof: that leaves you with the case of real characters. I see that this is a more special situation to which you could profitably apply other tools like the analytic class number formula for real quadratic fields. I don't agree that this is necessary or even "simpler": the proof I enclosed of Landau's theorem is only a single paragraph and uses only standard undergraduate real and complex analysis. Surely this is simple enough, anyway? – Pete L. Clark May 26 '10 at 2:01
The other tool of the class number formula is in fact exactly how Dirichlet first handled the real case! – KConrad May 26 '10 at 2:46
@KC: I heard that. If you know the class number formula very well -- e.g., if you were the one to prove it! -- this is certainly a natural way to go. However, the analytic class number formula is another big result in its own right. (Maybe I should admit that although I wrote up my notes for an under/grad course on number theory, in the two times I taught the course I never found time / dared to cover it. But it is in the range of things I feel like I could cover, whereas the analytic class number formula feels beyond that.) Landau's Theorem is essentially a brilliant remark. – Pete L. Clark May 26 '10 at 3:42
I submit that it's always a mess, to balance between the proof that generalizes the most, and the proof that is the most direct, possibly using special tricks. I think that is perhaps the underlying issue in most of the discussion. – Junkie May 26 '10 at 7:51

There are proofs that treat the cases of real and non-real $\chi$ on an equal footing. One proof is in Serre's Course in Arithmetic, which the answers by Pete and David are basically about. That method is using the (hidden) fact that the zeta-function of the $m$-th cyclotomic field has a simple pole at $s = 1$, just like the Riemann zeta-function. Here is another proof which focuses only on the $L$-function of the character $\chi$ under discussion, the $L$-function of the conjugate character, and the Riemann zeta-function.

Consider the product $$ H(s) = \zeta(s)^2L(s,\chi)L(s,\overline{\chi}). $$ This function is analytic for $\sigma > 0$, with the possible exception of a pole at $s = 1$. (As usual I write $s = \sigma + it$.)

Assume $L(1,\chi) = 0$. Then also $L(1,\overline{\chi}) = 0$. So in the product defining $H(s)$, the double pole of $\zeta(s)^2$ at $s = 1$ is cancelled and $H(s)$ is therefore analytic throughout the half-plane $\sigma > 0$.

For $\sigma > 1$, we have the exponential representation $$ H(s) = \exp\left(\sum_{p, k} \frac{2 + \chi(p^k) + \overline{\chi}(p^k)} {kp^{ks}}\right), $$ where the sum is over $k \geq 1$ and primes $p$. If $p$ does not divide $m$, then we write $\chi(p) = e^{i\theta_p}$ and find
$$ \frac{2 + \chi(p^k) + \overline{\chi}(p^k)}{k} = \frac{2(1 + \cos(k\theta_p))}{k} \geq 0. $$
If $p$ divides $m$ then this sum is $2/k > 0$.
Either way, inside that exponential is a Dirichlet series with nonnegative coefficients, so when we exponentiate and rearrange terms (on the half-plane of abs. convergence, namely where $\sigma > 1$), we see that $H(s)$ is a Dirichlet series with nonnegative coefficients. A lemma of Landau on Dirichlet series with nonnegative coefficients then assures us that the Dirichlet series representation of $H(s)$ is valid on any half-plane where $H(s)$ can be analytically continued.

To get a contradiction at this point, here are several methods.

[Edit: In the answer by J.H.S., and due to Bateman, is the slickest argument I have seen, so let me put it here. The idea is to look at the coefficient of $1/p^{2s}$ in the Dirichlet series for $H(s)$. By multiplying out the $p$-part of the Euler product, the coefficient of $1/p^s$ is $2 + \chi(p) + \overline{\chi}(p)$, which is nonnegative, but the coefficient of $1/p^{2s}$ is $(\chi(p) + \overline{\chi}(p) + 1)^2 + 1$, which is not only nonnegative but in fact is greater than or equal to 1. Therefore if $H(s)$ has an analytic continuation along the real line out to the number $\sigma$, then for real $s \geq \sigma$ we have $H(s) \geq \sum_{p} 1/p^{2s}$. The hypothesis that $L(1,\chi) = 0$ makes $H(s)$ analytic for all complex numbers with positive real part, so we can take $s = 1/2$ and get $H(1/2) \geq \sum_{p} 1/p$, which is absurd since that series over the primes diverges. QED!]

  1. If you are willing to accept that $L(s,\chi)$ (and therefore $L(s,\overline{\chi})$) has an analytic continuation to the whole plane, or at least out to the point $s = -2$, then $H(s)$ extends to $s = -2$. The Dirichlet series representation of $H(s)$ is convergent at $s = -2$ by our analytic continuation hypothesis and it shows $H(-2) > 1$, or the exponential representation implies that at least $H(-2) \not= 0$. But $\zeta(-2) = 0$, so $H(-2) = 0$. Either way, we have a contradiction.

  2. There is a similar argument, pointed out to me by Adrian Barbu, that does not require analytic continuation of $L(s,\chi)$ beyond the half-plane $\sigma > 0$. If you are willing to accept that $\zeta(s)$ has zeros in the critical strip $0 < \sigma < 1$ (which is a region that the Dirichlet series and exponential representations of $H(s)$ are both valid since $H(s)$ is analytic on $\sigma > 0$), we can evaluate the exponential representation of $H(s)$ at such a zero to get a contradiction. Of course the amount of analysis that lies behind this is more substantial than what is used to continue $L(s,\chi)$ out to $s = -2$.

  3. We consider $H(s)$ as $s \rightarrow 0^{+}$. We need to accept that $H$ is bounded as $s \rightarrow 0^{+}$. (It's even holomorphic there, but we don't quite need that.) For real $s > 0$ and a fixed prime $p_0$ (not dividing $m$, say), we can bound $H(s)$ from below by the sum of the $p_0$-power terms in its Dirichlet series. The sum of these terms is exactly the $p_0$-Euler factor of $H(s)$, so we have the lower bound $$ H(s) > \frac{1}{(1 - p_0^{-s})^2(1 - \chi(p_0)p_0^{-s})(1 - \overline{\chi}(p_0)p_0^{-s})} = \frac{1}{(1 - p_0^{-s})^2(1 - (\chi(p_0)+ \overline{\chi}(p_0))p_{0}^{-s} + p_0^{-2s})} $$ for real $s > 0$. The right side tends to $\infty$ as $s \rightarrow 0^{+}$. We have a contradiction. QED

These three arguments at some point use knowledge beyond the half-plane $\sigma > 0$ or a nontrivial zero of the zeta-function. Granting any of those lets you see easily that $H(s)$ can't vanish at $s = 1$, but that "granting" may seem overly technical. If you want a proof for the real and complex cases uniformly which does not go outside the region $\sigma > 0$, use the method in the answer by Pete or David [edit: or use the method I edited in as the first one in this answer].

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With all due respect, it is J. H. S. and not J. S. E. Best of luck... – José Hdz. Stgo. May 25 '10 at 5:06
I can let Robin Chapman speak for himself for the illusion, but from the real-complex idea, you are using one extra $\zeta(s)$ when $\chi$ is complex (and so, overkill in pretending it is hard), and taking $(\zeta(s)L(s,\chi))^2$ when $\chi$ is real (and so, using the square of what can be used, and so for no reason I can determine, other than to uniformize the language). – Junkie May 25 '10 at 5:30
Junkie, when chi is real (and nontrivial), the product G(s) = zeta(s)L(s,chi) has a Dirichlet series with nonneg. coefficients. For p not dividing the modulus of chi, the coeff. of 1/p^{2s} is 2 + chi(p), which is at least 1, so if L(1,chi) = 0 then Bateman's argument applies to G(s) to get a contradiction using s = 1/2. However, from this point of view why should the real case be considered harder than the non-real case? That is, usually it's said that the real case is essentially harder than the non-real case. I don't think it looks that way with Bateman's argument. – KConrad May 25 '10 at 6:09
And sure, the argument in my answer was carried out in a common way for all characters, but I don't think Bateman's method is hiding some fundamental difficulty in the real case. (Unlike the method in Serre's book, say, I am not introducing lots of auxiliary characters to achieve the goal.) To the contrary, I think it makes the real and non-real cases equally accessible, although, sure, in the real case one can simplify a little by avoiding the excess multiplications that are used in the uniform argument. But it looks like a small difference between the two cases. – KConrad May 25 '10 at 6:13
By the way, the argument used here can be extended with essentially no extra effort to get nonvanishing of L(s,chi) everywhere on the line Re(s) = 1. (If L(1+ib,chi) = 0, consider the product H_b(s) = zeta(s)^2L(s,chi_b)L(s,chi_b*), where chi_b(n) = chi(n)|n|^{-ib}.) This can't be said of the argument in Serre's book. It doesn't give a zero-free region to the left of the line as far as I know, but the fact that it goes through with no changes to give nonvanishing everywhere on Re(s) = 1 is good. – KConrad May 25 '10 at 6:29

It is definitely not a general answer to the question, but I like a completely elementary approach for showing that $L(\chi,1)>0$ when $\chi$ is the real character modulo 3 and 4. One first computes $$ f(x)=\sum_{l=1}^\infty\chi(l)x^l=\sum_{m=1}^\infty\sum_{n=1}^{p-1}\chi(n)x^{mp+n} $$ $$ =\sum_{m=1}^\infty x^{mp}\sum_{n=1}^{p-1}\chi(n)x^n=\frac1{1-x^p}\sum_{n=1}^{p-1}\chi(n)x^n $$ to see that it is $$ f(x)=\frac{x-x^2}{1-x^3}=\frac x{1+x+x^2} \quad\text{or}\quad f(x)=\frac{x-x^3}{1-x^4}=\frac x{1+x^2}. $$ Then by Abel's theorem $$ L(\chi,1)=\int_0^1f(x)dx $$ and the integrand is a positive function on $(0,1)$ in both cases.

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This is essentially Dirichlet's proof. – Franz Lemmermeyer Aug 30 '10 at 12:02

Everyone seems to have their own favourite here. Mine is slightly similar to a couple of those mentioned earlier; it involves studying the properties of $\zeta(s) L(s,\chi)$ where $\chi$ is a real character (as mentioned earlier, it's quite simple to prove the case where $\chi$ is complex) --- it is the approach taken in Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory, for example.

It is in some sense quite natural to look at the product $\zeta(s) L(s,\chi)$, because $\zeta(s)$ is the "simplest" case of an $L$-series having a simple pole at $s=1$, while we are assuming that $L(s,\chi)$ has a zero at $s = 1$, and so these will cancel out and hence $\zeta(s) L(s,\chi)$ will extend holomorphically to $\Re(s) > 0$. A simple calculation shows that the coefficients $c(n) = \sum_{d \mid n}{\chi(d)}$ of the Dirichlet series for $\zeta(s) L(s,\chi)$ are positive whenever $c(n)$ is a perfect square. This Dirichlet series is valid only when $\zeta(s), L(s,\chi)$ are absolutely convergent (i.e. for $\Re(s) > 1$), but as the assumption $L(1,\chi) = 0$ implies that $\zeta(s) L(s,\chi)$ extends holomorphically to $\Re(s) > 0$, a theorem of Landau implies that this Dirichlet series is conditionally convergent for $\Re(s) > 0$. But $\sum^{\infty}_{n=1}{c(n)n^{-1/2}}$ diverges as $c(n^2) > 0$, which yields the result.

Of course, the idea behind this proof does not differ markedly from the proof in Serre's book, and even Monsky's proof uses some of the same ideas.

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Here is an elementary proof, the basic idea for which is in Selberg's 1949 paper "An elementary proof of Dirichlet's theorem about primes in an arithmetic progression" (Ann. Math., vol 2, 1949, pp. 297-304).

It is enough to prove the unboundedness $$ \sum_{\substack{ p < X \\ p \equiv 1 \mod{N}}} \frac{\log{p}}{p} \neq O_X(1). $$ This is just a paraphrasing of what David Speyer says in the first paragraph of his answer. The elementary reduction to this of $L(1,\chi) \neq 0$ is very easy, too: see (2.27) in Iwaniec and Kowalski's Analytic Number Theory.

Take $d := \phi(N)$ and $\zeta_1, \ldots,\zeta_d$ a basis for $A := \mathbb{Z}[\zeta_N]$, and let $S(X)$ be the primes $< X$ and $\equiv 1 \mod{N}$. With $Y := X^{1/d}$ (it is enough to vary $X$ through $d$-th powers), consider the non-zero rational integer $$ V = V(X):= \prod_{0 < n_i \leq Y} (n_1\zeta_1 + \cdots + n_d \zeta_d) \in \mathbb{Z} \setminus \{0\}. $$ There are $X$ terms in the product, and it is not too hard to prove that every non-zero ideal $I \subset A$ contains [correction!] $\frac{X}{\mathcal{N}(I)} + O_N\Big( (\frac{X}{\mathcal{N}(I)})^{1-\frac{1}{d}} \Big)$ of these terms. All asymptotics here are under $X \to \infty$ for a fixed $N$, uniformly in $I$.

Following Chebyshev and Mertens, let us then consider the prime factorization of $V$, which is a sort of a cyclotomic factorial. Applying the above with $I$ running through the powers of prime ideals of $A$, we get from the convergence of $\sum_p (\log{p}) / p^2$ and $\frac{1}{X} \sum_{p < X} (X/p)^{1-\frac{1}{d}}\log{p}$ the estimate $$ \frac{1}{X} \log{|V(X)|} = \sum_{p \in S(X)} \frac{\log{p}}{p} + O_N(1). $$ Here I have used the obvious fact that a prime $p$ splits the normal extension $\mathbb{Q}(\zeta_N)/\mathbb{Q}$ if and only if $p \equiv 1 \mod{N}$.

We are thus reduced to proving the unboundedness of $\frac{1}{X} \log{|V(X)|}$. But $$ \frac{1}{X} \log{|V(X)|} = \frac{1}{X} \sum_{\substack{0 < n_i < Y } } \max_i \log{n_i} + O_N(1), $$ and it is easy to see by partial summation that the last sum has an asymptotic development with leading term $Y^{d} \log{Y} + O(Y^{d}) = \frac{1}{d}X \log{X} + O(X)$.

This establishes the precise estimate $\sum_{p \in S(X)} \frac{\log{p}}{p} = \frac{1}{d} \log{X} + O_N(1)$, and thus completes the proof of Dirichlet's theorem on primes in arithmetic progressions.

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I like the proof in Newman's GTM book "Analytic number theory". He calls his presentation a natural (as opposed to "elementary" or "simple") proof. This is to say that the presentation emphasizes a motivated sequence of steps that illustrates how would might have discovered it. This proof (as well as many other "simple" and "elementary" proofs) proceeds using Landau's lemma on the abscissa of convergence of Dirichlet series with non-negative coefficients.

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I like a proof due to Ingham, found at (may require a university subscription to access, the title of the article is "A Note on Reimann's $\zeta$ function and Dirichlet's L-functions"). Ingham uses Landau's theorem and a slight generalization of Ramanujan's famous identity $\sum d(n) n^{-s} = \zeta^4(s)/\zeta(2s)$ as follows:

For completely multiplicative functions $a(n)$ and $b(n)$ which take on values of modulus only $0$ or $1$, if $g_a(s)$, $g_b(s)$, and $g_{ab}(s)$ are the Dirichlet series with coefficients $a(n), b(n),$ and $a(n)b(n)$, then

$$\frac{\zeta(s)g_a(s)g_b(s)g_{ab}(s)}{g_{ab}(2s)} = \sum\frac{1}{n^s}\left(\sum_{d|n}a(d)\right)\left(\sum_{\delta|n}b(\delta)\right),$$

where $\Re(s) >1$. One then specializes to the case that $a(n) = \chi(n)n^{-it}$, and $b(n)$ is the conjugate of $a(n)$.

I can't help but mention that one can generalize this identity even further, to the statement that where the sums converge absolutely,

$$\sum\frac{1}{n^s}\left(\sum_{d|n}a(d)\right)\left(\sum_{\delta|n}b(\delta)\right) = \zeta(s)\sum_{j,k=1}^\infty \frac{a(j)b(k)}{[j,k]^s},$$

(where $[j,k]$ is the least common multiple of $j$ and $k$) even when $a$ and $b$ are not completely multiplicative and can have any modulus. It is a fun exercise to derive this formula and then use it to derive Ingham's or Ramanujans.

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There is a simpler way to derive the result from the approach mentioned by K. Conrad. You can find it in the following 1997 note by Paul T. Bateman:

Bateman, Paul T.: A theorem of Ingham implying that Dirichlet's L-functions have no zeros with real part one. L'Enseignement Mathématique. 43 (1997), 281-284.

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A google search reveals it is " L'Enseignement Mathématique." I have to admit, I prefer Ingham's original proof, which I'm posting below (if no one beats me to it). – Brad Rodgers May 24 '10 at 23:01
Yes, Narasimhan's argument for showing the Riemann zeta function doesn't vanish on the line Re(s) = 1 is really nice. See L'Enseignement Math. (2) 14 1968 189--191 (1969). Or MathSciNet MR0249373. – KConrad May 24 '10 at 23:11
@KConrad: I cant find the link on MathSciNet. What is the name of the paper. – S.C. Mar 27 '12 at 21:40… – José Hdz. Stgo. Mar 30 '12 at 21:29

There is a beautiful proof on p. 36-37 of Iwaniec-Kowalski which uses nothing beyond basic manipulations of finite series.

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The proof you mentioned goes back to Mertens. – Micah Milinovich Mar 11 '11 at 3:32

I know it's not an answer to the question, but rather another (probably naive) question suggested by the original question.

It seems to me that proving the prime number theorem for arithmetic progressions isn't harder than proving the arithmetic progression theorem alone. In other words, you can get then prime number theorem for nothing.

I know that many people have already done that (see this text for references), but I'd expect almost everybody to do it, and I don't understand why they don't.

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The arithmetic progression theorem uses Dirichlet density, which involves behavior of L-functions near s = 1. The prime number theorem or its variant for arithmetic progressions uses natural density, and that involves behavior of L-functions everywhere along the line Re(s) = 1 and some extra analysis (to prove some Tauberian theorem). Proofs involving natural density do require more technique, so it is "harder" in a sense. Newman's ideas tack several pages to the proof to turn the Dirichlet density theorem into a natural density theorem. – KConrad May 25 '10 at 6:50
I know almost nothing on the subject, but I still feel that, to prove the prime number theorem for arithmetic progressions, the most delicate point remains, by far, the nonvanishing of $L(1,\chi)$. Would you agree with this formulation? – Pierre-Yves Gaillard May 25 '10 at 7:04
Sort of, but not quite. It's more than that: to prove the prime number theorem for arithmetic progressions you need to show nonvanishing of L(s,chi) on the whole line Re(s) = 1, not just at the point s = 1. (And then there is the extra Tauberian step to get a natural density asymptotic out of the nonvanishing, which isn't part of Dirichlet's theorem.) If you want to codify these theorems on primes in terms of nonvanishing theorems for L-functions, to prove anything for natural density you have to work with the whole line Re(s) = 1, but for Dirichlet density only s = 1 is what you need. – KConrad May 25 '10 at 7:15
I tried to make a careful comparison between Serre, Soprunov (, and Kedlaya ( It seemed to me that, if you want to prove the PNT for AP (as opposed to the AP Theorem alone) with the same amount of details as, say, Serre, you need to expand the text by about 10%. – Pierre-Yves Gaillard May 25 '10 at 9:24
I'm not sure that I agree with your claim that combining Dirichlet's theorem with PNT adds only 10% to the length of the proof, but it is nice to see some texts where the two are treated simultaneously. Thanks especially for the link to your notes. – Pete L. Clark May 25 '10 at 18:42

Yoshikatsu Yashiro's proof of the prime number theorem for the whole Selberg class, only available on Arxiv as far as I know. Elegant as a simple corollary of a far more general theorem.

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