Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In his answer to my question The Green-Tao theorem and positive binary quadratic forms
Kevin Ventullo answers my initial question in the affirmative. What remains is the title question here, of separate interest to me.

Any integral positive binary quadratic form integrally represents a set of primes with known Dirichlet density. This is an application of the Cebotarev density theorem (or Chebotarev or Tchebotarev), in particular it is Theorem 9.12 in David A. Cox, Primes of the form $x^2 + n y^2,$ with the example $\Delta = -56$ on page 190. I typed this out in the previous question.

Now, Jurgen Neukirch "Class Field Theory" points to Serre "A Course in Arithmetic," and on page 76 Serre says that the set of primes p such that a fixed polynomial has a root $\pmod p$ has a natural density, and refers to K. Prachar "Primzahlverteilung" chapter 5 section 7. By results (theorem 9.2, page 180) in the Cox book, this means that the principal form $x^2+ny^2$ or $x^2+xy+ky^2$ does represent a set of primes with a natural density, therefore equalling the Dirichlet density. And by the result on arithmetic progressions, a full genus of forms has a natural density.

Combining observations, the principal form always has a natural density of primes, any full genus does, therefore we are done for one class per genus, and in the case of two classes per genus we are done with the principal genus and any genus with two distinct opposite forms. So we have natural densities for Cox's $\Delta = -56$ example. We are also done with the principal genus when it has three classes.

So, (and I would love a reference), does every positive binary quadratic form represent a set of primes for which the natural density exists?

share|improve this question
    
Will, You should look at "Sets of primes determined by systems of polynomial congruences" , J. C. Lagarias, Illinois J. Math. 27 (1983), pp. 224-239. –  Victor Miller Jun 16 '10 at 2:59
    
Thanks, Victor. Robin is quite clear in his answer, but after Cox's book I am left with this sort of problem: given $\Delta = -47,$ I can tell if a prime with $ ( -47 | p) = 1 $ is represented by $ x^2 + x y + 12 y^2 $ by whether a certain $ f_{-47}(t)$ has a root $\pmod p,$ and I would love to know the monic degree 5 $ f_{-47}(t)!$ If not by the principal form, how can I tell whether $p$ is represented by $ 2 x^2 \pm x y + 6 y^2 $ or by $ 3 x^2 \pm x y + 4 y^2 ?$ –  Will Jagy Jun 16 '10 at 17:20
    
Will, That's exactly what Jeff covers in his paper. –  Victor Miller Jun 19 '10 at 3:07
add comment

2 Answers

up vote 7 down vote accepted

Accoring to H. Lenstra, Chebotarev's theorem holds both for Dirichlet and for natural density (but he doesn't give a reference in this document). Applying Chebotarev to the extension $H/\mathbb{Q}$ where $H$ is the Hilbert class field of $\mathbb{Q}(\sqrt{-n})$ gives the result you want. (At least for primitive discriminants; for non-primitive discriminants you need an appropriate generalization of the Hilbert class field).

Added in response to Will's comment There is always a suitable field. Let $-D$ be a primitive negative discriminant and let $d=-m^2D$ be a general discriminant. Let $H_n$ be the abelian extension of $K=\mathbb{Q}{\sqrt{-D}}$ where a non-ramified prime splits iff its is principal and generated by an element of $\mathbb{Z}+m\mathcal{O}_K$. Such a field $H_n$ is called a ring class field and exists by class field theory. It also is an extension of $K$ by a singular value of the $j$-function.

Then $G_m=\mathrm{Gal}(H_n/\mathbb{Q})$ is a generalized dihedral group. There is a correspondence between conjugacy classes in $G_m$ and pairs of equivalence classes of $ax^2\pm bxy+cy^2$ of discriminant $d$, such that an unramified prime has its Frobenius in a conjugacy class iff it it represented by the corresponding form. This is why we can apply Chebotarev.

Even more added A good reference for ring class fields is Cox's book Primes of the Form $x^2+ny^2$.

share|improve this answer
    
I used Google with "chebotarev density" "natural density". Our own Pete L. Clark, in some lecture notes, attributes the result for natural density to Hecke. www.math.uga.edu/~pete/8430notes5.pdf –  Will Jagy Jun 15 '10 at 18:24
    
@Will: yes, that's what I say. I would be happy to have a precise reference though. –  Pete L. Clark Jun 15 '10 at 20:06
    
I hope that I have managed to make genuine links by putting the string "http://" in front. On the Chebotarev density theorem proved for natural density, Stevenhagen and Lenstra give a specific 1917 Hecke article math.leidenuniv.nl/~hwl/papers/cheb.pdf Then Anatoly Preygel modular.fas.harvard.edu/129-05/final_papers/Anatoly-Preygel.pdf refers to a modern book, Larry Joel Goldstein , Analytic Number Theory (1971) while I already mentioned K. Prachar "Primzahlverteilung" (1957) so I am planning to go borrow more books from campus. –  Will Jagy Jun 15 '10 at 20:20
    
There's an alternative to class-field theory here. For the most part the argument is like that used nowadays for showing primes in a progression have a relative density, but you have to use ray-class characters in the ring of integers of the quadratic field. The hard part is showing that the L-function attached to a real character doesn't vanish at 1. Once one can extend the L-function to all of C, one can give a Landau style argument. What Hecke did was show how to extend the L-function-- his argument is much like one of Riemann's; Tate later put it in the modern style. –  paul Monsky Jun 15 '10 at 21:21
    
Dear Prof. Monsky, Could you please tell me where the Tate article appeared? Also, would you say that the Hecke results you mean appear in the piece Stevenhagen and Lenstra mention, Uber die L-Funktionen und den Dirichletschen Primzahlsatz fur einen beliebigen Zahlkorper, Nachr. Akad. Wiss. Gottingen Math.-Phys. (1917) pages 299-318 or Mathematische Werke, pages 178-197. Thanks in advance, Will. –  Will Jagy Jun 16 '10 at 4:03
show 6 more comments

If $Q$ is a positive binary quadratic form whose discriminant $D$ is fundamental, then the number of primes $\leq X$ represented by $Q$ is given asymptotically as

$\pi_{Q}(X)=\frac{1}{2h(D)} \mathrm{Li}(X)+O(X \exp{(-c_{Q}\sqrt{\log{X}})})$.

Here $\mathrm{Li}(X)$ is the usual logarithmic integral. My reason for giving this as an answer is to point out that this was proven by de la Vallee Poussin himself, in the same paper where he proved the usual prime number theorem! Dating from the 1890s, this definitely predates Chebotarev, and de la V.P.'s contribution deserves (IMHO) to be better known than it is. Hadamard's work on the PNT is a little easier on the reader than de la V.P.'s, but de la V.P.'s results were both stronger and more general, which sometimes is forgotten.

share|improve this answer
    
That's really nice, David. I think you need to multiply it by 2 if the form is not equivalent to its opposite, the coarse relations being that the sum over all must be (1/2) Li(X), while the total per genus should be independent of genus. –  Will Jagy Jun 16 '10 at 19:09
    
Hi Will, could you give an example of what you mean? Does equivalent mean "represents the same integers"? –  David Hansen Jun 16 '10 at 19:22
    
Sure. This is in the book by Cox, page 190. For discriminant $-56,$ both $ x^2 + 14 y^2$ and $2 x^2 + 7 y^2,$ which are ambiguous forms, get Dirichlet density, and natural density of primes, $1/8,$ to total $1/4$ for the principal genus. In the other genus $ 3 x^2 + 2 x y + 5 y^2$ and its opposite $ 3 x^2 - 2 x y + 5 y^2$ represent exactly the same numbers, and in particular prime density $1/4.$ Entire discriminant $ 1/4 + 1/4 = 1/2.$ –  Will Jagy Jun 16 '10 at 19:42
    
@David Hansen: This is a very interesting answer, I happen to be looking for something exactly like this. Can you explain how one would go about proving this, or do you know any books or papers where I can read the proof? (Other then perhaps de la Vallee Poussin's original) Thanks a lot! –  Eric Naslund May 8 '12 at 18:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.