If $p$ is a prime of the form $4n+3$, the class number $h$ of $Q[\sqrt{-p}]$ can be expressed using the number $V$ of quadratic residues and $N$ nonresidues in the interval $[1,\frac{p-1}{2}]$:

  • If $p=8n+7$ then $h=V-N$
  • If $p=8n+3$ then $h=\frac{1}{3}(V-N)$

This result seems so simple and elegant, but its proof (which I saw in Number Theory by Borevich & Shafarevich - p. 346), while definitely beautiful, is not very short and is based on the analytic class number formula. And so I didn't feel that the proof really helped "demystify" the result.

This leads me to ask: Can someone see intuition for this result? Any short heuristic argument that would lead to it? Any obvious meaning of $h=V-N$? In short, "why" is it true?


I'll have to be brief; I can think of two reasons "why":

  1. Cauchy and Jacobi proved that for a prime ideal ${\mathfrak p}$ in a complex quadratic number field with prime discriminant, the h-th power of ${\mathfrak p}$ (with $h$ as in your question) is principal. Their technique was what we nowadays know as the Stickelberger ideal. You should find references in Ireland-Rosen.

  2. Venkov, in the 1920s, gave an arithmetic proof of a large part of Dirichlet's class number formulas based on Gauss's work on ternary quadratic forms and the arithmetic of quaternions. There are modern accounts floating around, and I remember the names Rehm and Shemanske in this connection (if google doesn't help, I'll provide you with references when I'm back from the holidays -). This approach is probably more involved than Dirichlet's, so I don't know whether it explains anything.

EDIT: Let me first give you the references for 2.

  • B.A. Venkov, On the arithmetic of quaternion algebras (Russ.), Izv. Akad. Nauk (1922), 205--220, 221--246; ibid. (1929), 489--509, 532--562, 607--622

  • H.P. Rehm, On a theorem of Gauss concerning the number of solutions of the equation $x^2 + y^2 + z^2 = m$, in "Selected topics on ternary forms and norms" (O. Taussky, ed.), Dekker 1976

  • Th.R. Shemanske, Representations of ternary quadratic forms and the class number of imaginary quadratic fields, Pac. J. Math. {\bf 122} (1986), 223--250

As for 1, let $K/{\mathbb Q}$ be a finite abelian extension with Galois group $G$ and conductor $m$. Let $\sigma_a$ denote the restriction of the automorphism $\zeta_m \to \zeta_m^a$ of ${\mathbb Q}(\zeta_m)$ to $K$. Then $$ \theta(K) = \frac{1}{m} \sum a\sigma_a^{-1} \in {\mathbb Q}[G], $$ where the sum is over all $0 < a < m$ with $(a,m) = 1$, is called the Stickelberger element corresponding to $K$. The fact that $(b-\sigma_b)\theta \in {\mathbb Z}[G]$ for integers $b$ coprime to $m$ allows us to define the Stickelberger ideal $I_0(K)$ as the ideal in ${\mathbb Z}[G]$ generated by elements of the form $(b-\sigma_b)\theta$. Set $I(K) = {\mathbb Z}[G] \cap \theta {\mathbb Z}[G]$. If $K = {\mathbb Q}(\zeta_m)$ is a full cyclotomic field, then $I(K) = I_0(K)$.

Stickelberger's Theorem says that if $K/{\mathbb Q}$ is an abelian extension, then the Stickelberger ideal $I(K)$ annihilates the class group $Cl(K)$.

Applied to quadratic extensions, you get the theorem first proved by Cauchy and Jacobi mentioned above. See my Reciprocity Laws, Chapter 11, for details.

  • $\begingroup$ Thanks! Could you please explain more about #1? I can't seem to see exactly how this relates my question. $\endgroup$ – danseetea May 24 '10 at 0:37
  • $\begingroup$ That's great, thanks! This answer combined with more information from your Reciprocity Laws has now finally convinced me of "why" this result makes sense. $\endgroup$ – danseetea Jun 2 '10 at 12:36

Update: more thoughts, including a shorter and more nonsensical "proof", on my blog.

Orde's paper, "On Dirichlet's Class number formula", gives a beautiful nonsense proof, before giving a rigorous one. Thanks to KConrad for pointing out Orde's paper to me. Orde is a little terse, so let me expand:

Let $R$ be the ring of integers in $\mathbb{Q}[\sqrt{-p}]$. For simplicity, take $p>3$. For any positive integer $N$, let $S(N)$ be the number of ideals in $R$ with norm $N$. By unique factorization into prime ideals and a little thought, we have $$S(N) = \sum_{d|N} \left( \frac{-p}{d} \right). \quad (*)$$

This formula is correct for $N>0$. Orde explains how to extend this formula to be correct for $N \neq 0$. The formula at $N=0$ will then be the class number formula!

Let $C$ be the class group of $R$. Let $Q$ be the set of integral quadratic forms with discriminant $-p$, modulo equivalence. Note that $Q$ is the disjoint union of $Q^{+}$ and $Q^{-}$; the positive definite forms and the negative definite ones. For most purposes, we discard $Q^{-}$, but today we want it around. There is a standard bijection between $C$ and $Q^{+}$.

For $c \in C$ and $N>0$, let $S_c(N)$ be the number of ideals of $R$ of class $c$ and norm $N$. So $S(N)=\sum_{c \in C} S_c(N)$. Let $q$ be the corresponding positive definite form and let $T_q(N)$ be the number of representations of $N$ be the form $q$. By the standard relationship between quadratic forms and ideals, $T_q(N)=2 S_c(N)$. (That $2$ is because $R$ has $2$ units.) Also, since $N>0$, we have $T_{-q}(N)=0$. So $$\frac{1}{2} \sum_{q \in Q} T_q(N) = S(N) = \sum_{d|N} \left( \frac{-p}{d} \right) \quad (**).$$

The left and right hand sides of $(**)$ are symmetric in exchanging $N$ and $-N$, so $(**)$ is also valid for $N<0$.

Now, consider $(**)$ for $N=0$. For any $q \in Q$, we have $T_q(0)=1$, since $q$ is either positive or negative definite. So the left hand side is $(1/2) |Q|=|C|$.

Everything divides $0$, so the right hand side is $\sum_{d>0} \left( \frac{-p}{d} \right)$. That doesn't converge, but its Cesaro sum is $(1/p) \sum_{d=1}^{p} (p-d) \left( \frac{-p}{d} \right).$ (If we were doing the case that $p \equiv 1 \mod 4$, that average would be over $4p$ terms, instead of just $p$ of them.) So we "derive" that $$|C| = (1/p) \sum_{d=1}^{p} (p-d) \left( \frac{-p}{d} \right).$$

This is easily shown to be equivalent to the class number formula.

  • 1
    $\begingroup$ Let me add a vague comment, which maybe someone else can build on. Most proofs take some sort of weighted average of $S(N)$, and take the limits of both sides of (**). The standard approach is to work with $Z(s) := \sum S(N)/N^s$; Orde uses $\Theta(x) := \sum S(N) x^N$. What is easy to see is that the asymptotics of $Z(s)$ (resp $\Theta(x)$) as $s \to 1^+$ (resp $x \to 1^{-}$) are related to $h$, with some extra junk terms. It is also easy to see that these limits are related to $L(1, \chi)$, times some junk terms. (Continued) $\endgroup$ – David E Speyer May 25 '10 at 15:29
  • $\begingroup$ The Cesaro sum is easily related to $L(0, \chi)$. One can use the functional equation, or explicit computations with Gauss sums, to relate $L(0, \chi)$ and $L(1, \chi)$. When one does, all the junk terms cancel! So the challenge is to give a proof which never produces those junk terms in the first place. $\endgroup$ – David E Speyer May 25 '10 at 15:30
  • $\begingroup$ Orde's trick (in the real proof) is to write $\Theta(x) = a/(1-x) + b + O(1-x)$ and consider the expressions for $b$, rather than for $a$. Somehow, this makes none of the junk terms come up any more. Is this part of some general phenomenon? And why did Orde's nonsensical limit, above, wind up with exactly the same sums which are involved in computing this term? $\endgroup$ – David E Speyer May 25 '10 at 15:33

This question was asked recently. Take a look at

Most squares in the first half-interval

for some further references. In Ireland & Rosen p. 225 (notes to Chapter 14) they say that the difference formula annihilating the class group (as an exponent) goes back to Kummer.

  • $\begingroup$ Thanks! That is very useful and interesting (funny I missed that thread being that it was asked only a few days ago). Do you have more direct intuition for 'why' the result is true? $\endgroup$ – danseetea May 24 '10 at 0:40
  • 1
    $\begingroup$ Nope, sorry. Class number formulas don't look intuitive to me. The basic idea of what Franz is saying is that if you have a formula for an integer which kills everything in a group, you might guess that formula is in fact the size of the group. $\endgroup$ – KConrad May 24 '10 at 1:18
  • 1
    $\begingroup$ By "doesn't look intuitive" I mean I can't pin down a specific intuition from each part of the formula. In the same way I can't see intuitively why the alternating harmonic series should be log(2), although of course I can explain it. There is no intuition as far as I know about why 1 - 1/2 + 1/3 - ... should be log(2) just by staring at the individual terms in the series. $\endgroup$ – KConrad May 24 '10 at 1:20
  • $\begingroup$ I see what you mean. Lots of thanks for your answers. $\endgroup$ – danseetea May 24 '10 at 1:29
  • $\begingroup$ @KConrad Regroup $1-1/2+1/3-1/4+\cdots - 1/(2n)$ as $(1+1/2+1/3+\cdots + 1/(2n)) - 2 (1/2+1/4+\cdots + 1/(2n))$ = $(1+1/2+1/3+\cdots + 1/(2n)) - (1/1+1/2+\cdots + 1/n)$ = $(1/(n+1) + 1/(n+2) + ... + 1/(2n))$ = $(1/n) (1/(1+1/n) + 1/(1+2/n) + ... + 1/(1+1))$. Note that, as $n \to \infty$, this approaches a Riemann sum for $\int_1^2 dt/t$. It is cute to note that the usual false proof that $1-1/2+1/3-\cdots =0$ is doing this rearragnement and then acting as if $\lim_{n \to \infty} ( 1/(n+1) + \cdots 1/(2n)) = 0+0+\cdots+0=0$. $\endgroup$ – David E Speyer May 8 '14 at 18:47

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