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Let $p \equiv 5 \pmod{8}, q \equiv 7 \pmod{8}$ be primes and $N = pq$. I want to show that the class number $n$ of $\mathbb{Q}(\sqrt{-N})$ satisfies $n \equiv 2 \pmod{4}$ if $\left(\frac{q}{p}\right) = -1$ and $n \equiv 0 \pmod{4}$ otherwise.

This comes from my trying to understand the answer to question 28462 (Why are there usually an even number of representations as a sum of 11 squares).

In a comment, Paul Monsky wrote: "To see this note that the non-trivial ambiguous ideal class is represented by an ideal of norm q, and so is not in the principal genus."

Can someone explain more fully how to show this?

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Cox's book "primes of the form $x^2+ny^2$" is a good read for this sort of question. I think it defines all the terms in Paul's quote, for example, and you can go on from there. – user30035 Mar 29 '13 at 7:33
On a side note, I think it's better to think of this as a statement about the 2-divisibility of the class number than it is to think about it as the value of the class number mod 4. – Cam McLeman Mar 29 '13 at 12:48
@Sarah--In the answer to your earlier question I could have used the seventh power of g= x+x^4+x^9+x^16+... rather than the eleventh. Then it would turn out that the coefficient of x^pq in the expansion is odd or even according as the class number of Q(root(-2pq)) is 4 mod 8 or 0 mod 8. And the Gauss theory of forms axx+bxy+cyy with b^2-4ac=-8pq would show that the class number is 4 mod 8 when (q/p)=-1 and 0 mod 8 when (q/p)=1. So my argument would show that g^7 isn't the reduction of the expansion at infinity of any modular form for any gamma_0. – paul Monsky Mar 29 '13 at 13:37
up vote 6 down vote accepted

The way Gauss did things was in terms of SL(2,Z) equivalence classes of (primitive) binary quadratic forms over Z. So lets consider such definite forms, axx+bxy+cyy with b^2-4ac=-N. For simplicity suppose N is squarefree and odd. Gauss defines a composition on the classes, making the set of classes into a finite group. For each prime dividing N he defines a genus character from this group to Z/2. The product of these characters is the trivial map, and the joint kernel of them all consists of the squares, a subgroup that he calls the principal genus. In the case N=pq, the character attached to p maps a form to (M/p) where M is any integer prime to p represented by the form. So the squares form, in your case, a subgroup of index 2, and there is a unique non-trivial class of order 2 in the group. Gauss calls the classes of order 2 the "ambiguous classes". They're easily written down in general and are represented by "reduced" forms axx+bxy+cyy with b=0 or with a=b (or c). So in your case the non-trivial ambiguous class is represented by pxx+qyy. The genus character attached to p maps this class to (q/p). So the class is a square if and only if (q/p)=1, giving the result you want. The theory also works for even N not necessarily square-free, though there are 2 genus characters attached to the prime 2 when 32 divides N, and it works for indefinite forms as well.

Oops--I should have said that the non-trivial ambiguous class is represented by qxx+qxy+(1/4)(p+q)yy.

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This really clarifies things for me Paul -- thanks. I couldn't see how these sorts of arguments could say anything about the class group other than the size of the 2-torsion -- I knew they could access assertions about squares, but now I see that "I have order 2" and "I am a square" implies "some element has order 4"; I hadn't ever internalised the fact that these arguments could go this deep into the class group. – user30035 Mar 29 '13 at 17:43

Your conjecture is true, it follows from Corollary 1 to Theorem 39 (Page 181) combined with Theorem 41 (Page 190) in Fröhlich-Taylor: Algebraic number theory (Cambridge University Press, 1991).

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Thanks! I'm going to read that section now. – stl Mar 29 '13 at 15:33

I come later than GH and paul Monsky and with an intricate and way too long answer, but since I had fun in solving the exercice, I post it anyhow. Most probably (as paul Monsky suggested) there are easier solutions; I should also say that it was Monsky's comment about ambiguous classes who made me take the following route. Finally, let me say that I think what follows works for general $p\equiv 1\pmod{4}$ and $q\equiv 3\pmod{4}$.

First of all, since $p\equiv 1\pmod{4}$, Gauss' reciprocity tells you that $$ \Big(\frac{p}{q}\Big)=1 $$ if and only if $p$ splits in $\mathbb{Q}(\sqrt{-q})$, if and only if $q$ splits in $\mathbb{Q}(\sqrt{p})$. Moreover, $p$ is the unique ramified prime in $\mathbb{Q}(\sqrt{p})/\mathbb{Q}$ and this implies that the narrow class group of $\mathbb{Q}(\sqrt{p})$ is trivial (there is a general discussion about this kind of results in L. Washington's ''Introduction to Cyclotomic Fields'', Theorem 10.4; we will use this later on, too: the point is that the Galois group $\mathrm{Gal}(\mathbb{Q}(\sqrt{p})/\mathbb{Q})$ is a $2$-group, hence acts non-trivially on the $2$-Sylow of the class group of $\mathbb{Q}(\sqrt{p})$ and this allows you to ''descend'' to $\mathbb{Q}$ a piece of the would-be narrow Hilbert class field, contradicting non-existence of extensions of the rationals unramified outside infinity). And now a final remark before starting: denote by $K$ your field $\mathbb{Q}(\sqrt{-N})$, call $L=\mathbb{Q}(\sqrt{p},\sqrt{-q})$ and $\Delta$ the Galois group of $K/\mathbb{Q}$. Clearly $L/K$ is everywhere unramified, thus $\mathrm{Cl}_K$ has non-trivial $2$-Sylow: in particular, $\Delta$ must act on this $2$-Sylow with non-trivial fixed points (and, a fortiori, it has non-trivial fixed points when acting on the whole class group). I claim that there is a surjection $$\tag{$\dagger$} H^0(\Delta,\mathrm{Id}_K)\twoheadrightarrow H^0(\Delta,\mathrm{Cl}_K)\neq 0 $$ from the fractional ideals of $K$ which are fixed by $\Delta$ to the $2$-Sylow of ideal classes which are fixed by $\Delta$ (such fixed classes are normally referred to as ``ambigous classes'', and here is where paul Monsky's quote enters (my version of) the game). Let me postpone the proof of ($\dagger$) and draw a consequence of it: pick a class of order a power of $2$ which is fixed by $\Delta$ and chose an ideal $I$ representing it which is fixed by $\Delta$. Then the primes occurring in the decomposition of $I$ either ramify or are inert or come into conjugate pairs. But inert primes are principal, and so are products of conjugate split primes, so you can throw them away and you are left with ramified primes. The moral is that all ambiguous classes are represented by ramified primes: call $\mathfrak{p},\mathfrak{q}$, respectively, the ideal of $K$ above $p,q$: then both $\mathfrak{p}^2,\mathfrak{q}^2$ are principal, so $H^0(\Delta,\mathrm{Cl}_K)=\mathbb{Z}/2$ or $=\mathbb{Z}/2\times\mathbb{Z}/2$.

We have seen that the class group you are interested in has even order and I call it $n$ as you do in your question. Then

1) Case $n\equiv 2\pmod{4}$. By computing inertia degrees, $\mathfrak{p}$ and $\mathfrak{q}$ have the same splitting behaviour in $L/K$: they split if and only if $$ \Big(\frac{p}{q}\Big)=1\;. $$ Suppose that they split in $L/K$: since $n\equiv 2\pmod{4}$, $L$ is the $2$-Hilbert class field of $K$ and the splitting means that they are both principal, contradicting the fact that they generate a non-trivial subgroup of $\mathrm{Cl}_K$, so $$ \Big(\frac{p}{q}\Big)=-1\;. $$ 2) Suppose now that $n\equiv 0\pmod{4}$ and call $H$ the $2$-Hilbert class field of $K$: then the class group of $L$ is divisible by $2$. Suppose that there is a unique prime in $\mathbb{Q}(\sqrt{p})$ above $q$, call it $\wp$: then $L/\mathbb{Q}(\sqrt{p})$ is a $2$-extension which is only ramified at $\wp\cdot\infty$. It follows that if $2$ divides the class number of $K$, then $2$ must divide the narrow class number of $\mathbb{Q}(\sqrt{p})$ by the same argument as above (I mean: as in Washington's Theorem 10.4), leading to a contradiction. Therefore $q$ must split in $\mathbb{Q}(\sqrt{p})$ and $$ \Big(\frac{p}{q}\Big)=1\;. $$

I am left with the proof of the surjection in $(\dagger)$. For this, observe that $-1$ is not a norm from $\mathcal{O}_K^\times$ to $\mathbb{Q}^\times$, since $\mathcal{O}_K$ has only two units $\{\pm 1\}$ and both have norm $1$. It is also not a local norm at the archimedean place, so it does not belong to the kernel of $$ \hat{H}^0(\Delta,\mathcal{O}_K^\times)\to\hat{H}^0(\Delta,\prod_v\mathcal{O}_v^\times) $$ where the right hand side is the product of all local units over all places of $K$ (I call it $\mathcal{U})$. Since the left-hand side has order $2$, the above map is injective. Consider then the diagram, in which the horizontal injectivity on top comes from Hilbert 90 and the vertical injectivity on the right we have just explained, $$\left.\begin{array}{ccc} \hat{H}^{-1}(\Delta,\mathrm{Pic}_K)&\hookrightarrow&\hat{H}^0(\Delta,\mathcal{O}_K^\times)\\ \downarrow&&\stackrel{\;\;\cap}{\downarrow}\\ \hat{H}^{-1}(\Delta,\mathrm{Id}_K)&\rightarrow&\hat{H}^0(\Delta,\mathcal{U}) \end{array}\right. $$ induced by taking $\Delta$ Tate cohomologies of the sequences $$ 1\rightarrow \mathcal{O}_K^\times\rightarrow K^\times\rightarrow \mathrm{Pic}_K\rightarrow 1 $$ and $$ 1\rightarrow \mathcal{U}\rightarrow \mathbb{A}_K^\times\rightarrow\mathrm{Id}_K\rightarrow 1 $$ respectively (I hope notation in self-explaining...). Since going right-down in the diagram is injective, so is going down-right, i. e. $H^1(\Delta,\mathrm{Pic}_K)$ injects into $H^1(\Delta,\mathrm{Id}_K)$ ($\Delta$ being cyclic, I shift without warnings Tate cohomology by $2$). Considering now the long exact sequence of $\Delta$-cohomology attached to $$ 1\rightarrow\mathrm{Pic}_K\rightarrow\mathrm{Id}_K\rightarrow\mathrm{Cl}_K\rightarrow 1 $$ gives $(\dagger)$.

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