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In this question Hilbert class field of Quadratic fields it is mentioned that if $d\equiv 1 \mod 4$ then the Hilbert class field of $\mathbb{Q}(\sqrt{-d})$ contains $\mathbb{Q}(i,\sqrt{d})$.

Could someone point me to where the intersections of Hilbert class fields of imaginary quadratic fields is discussed in more detail?

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This is similar to a CM version of this question: mathoverflow.net/questions/74906/… –  Dror Speiser Sep 20 '12 at 18:56
@Dror: I think in the question you link to, as JSE comments, it is really a group theoretic question about $SL_2(\mathbb{Z}_p)\times SL_2(\mathbb{Z}_p)$. For an elliptic curve with CM by $\mathcal{O}$, a CM version of the other question would be talking about the extra structure on the Tate module as a rank 1 $\mathcal{O} \otimes \mathbb{Z}_p$-module. I see this question having more to do with the action of $Pic(\mathcal{O})$ on the Hilbert class field. I'm struggling to see the whole picture at the moment with CM curves, could you please expand a little as to the similarities you see? –  Adam Harris Sep 20 '12 at 21:58
Leaving the solution to the other question aside, torsion points of a CM elliptic curve generate the abelian extensions of the hilbert class field of the relevant quadratic field. So, this question is one level before the CM version of the linked question: here you want an intersection before taking torsion points, but just the defining field of the elliptic curve. So it's not exactly the CM version, but it's related. It's just a comment, and I'm not sure how to pursue it. (Also, I did not realise before that you asked the other question as well.) –  Dror Speiser Sep 21 '12 at 0:23
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2 Answers

up vote 7 down vote accepted

The generalization of the phenomena you see is genus theory. If $K = \mathbf{Q}(\sqrt{-d})$ and $H = K(j_d)$ then $H$ contains the Genus field $G$.

If $d = \prod_{i=1}^n p_i$ is squarefree (and odd for convenience's sake) then $G = K(\sqrt{p_i^*})$ where $p_i^* = (-1)^{(p_i-1)/2}p_i$. In particular $p_i^* = p_i$ if and only if $p_i \equiv 1\bmod 4$. Therefore if $d \equiv 1\bmod 4$ then $d = \prod_i p_i^*$. Therefore if $d \equiv 1\bmod 4$ then $G$ and thus $H$ contains $K(\sqrt d) = K(i)$.

In general, if $d_1$ and $d_2$ have lots of prime factors in common, then the genus fields of their corresponding imaginary quadratic fields will also have large intersections. Outside of the genus field, I don't know of any studies into the intersections of Hilbert Class Fields.

The standard reference for this is Cox's wonderful book "Primes of the form $x^2 + ny^2$."

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@stankewicz: Thanks! - I think you must mean $p_i^*=(-1)^{(p-1)/2}p_i$ though? –  Adam Harris Sep 20 '12 at 14:06
Yes, I always do that for some reason. Thank you very much. –  stankewicz Sep 20 '12 at 14:08
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It is not very clear to me what you mean by "intersection of Hilbert Class Fields [...] is discussed". The theory of Complex Multiplication (see, for instance, Serre's short note in Cassels and Frohlich's Algebraic Number Theory, or Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, or directly the bible from Shimura, Introduction to the Arithmetic Theory of Automorphic Functions) tells you that there is an explicit way, given an imaginary quadratic field $K=\mathbb{Q}(\sqrt{d})$ to construct not only its Hilbert class field, but all ray class fields of different conductors. For instance, for the Hilbert class field, you can first create the elliptic curve $\mathbb{C}/\mathcal{O}_K$: it is an elliptic curve with complex multiplication by $\mathcal{O}_K$. Then, the theory will tell you that the smallest extension of $\mathbb{Q}$ containing $\sqrt{d}$ over which this curve is defined is the Hilbert Class Field of $K$. Just to be convinced that what I say is believable (beside being true, for which you might look at the references), observe that there is an elliptic curve with CM by $\mathbb{Z}[i]$ which admits a Weierstrass equation $$ y^2=x^3+x\;. $$ This is defined over $\mathbb{Q}$, so the smallest field of definition which contains $\sqrt{-1}$ is $\mathbb{Q}(i)$, which is indeed its own Hilbert class field.

In more concrete terms, if you are given a squarefree $d<0$ then the Hilbert class field of $K=\mathbb{Q}(\sqrt{d})$ is$K(j(\sqrt{d})$, where $j$ is the modular function $$ j(q)=\frac{1}{q}+744+196884q+\dots $$ whose definition you'll find in the above references, or at http://en.wikipedia.org/wiki/J-invariant . All the fun is in proving that $j(\sqrt{d})$ is actually an algebraic number (it is even an algebraic integer!); then, of course, one proves the abelian+unramified property of the (now, finite!) extension $K(j(\sqrt{d}))/K$. Quiteremarkably, it is an algebraic number ''because" its conjugates are precisely the $j$-invariants of the elliptic curves $\mathbb{C}/\mathfrak{a}_i$ for $\mathfrak{a}_i$ running through a set of integral representatives of the ideal class. This, already, shows that the degree of the extension coincides with the class number (and rapidly leads to it being at least Galois).

In your case, life is easier: you only want to be sure that if $d\equiv 1\pmod{4}$ is negative and squarefree, then $K(i)/K$ is unramified (being abelian, this would force it to lie inside the Hilbert class field). For this, it is enough to observe that $K(i)/\mathbb{Q}$ is a biquadratic extension with Galois group $(\mathbb{Z}/2)^2$ and has therefore three quadratic subfields: $\mathbb{Q}(i),K$ and $F=\mathbb{Q}(\sqrt{-d})$. Now pick a prime $\ell$ dividing $d$: it is necessarily odd. Its ramification degree is $2$ both in $K$ and in $F$, while it is unramified in $\mathbb{Q}(i)$. Therefore it needs be unramified in $K(i)/K$, since ramification degrees are multiplicative in towers of extensions. Similarly, $2$ is unramified in $F/\mathbb{Q}$ and cannot ramify in $K(i)/K$. For what concerns the infinite primes, observe that $K$ is already totally complex, so no ramification can occur. Done!

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@Filippo: Thanks! I was thinking about intersections in general though, for example does a similar situation occur for $d \equiv 3 \mod 4$? Is there a standard reference for the statement for $d \equiv 1 \mod 4$? –  Adam Harris Sep 20 '12 at 11:37
The situation in the case $d\equiv 3\pmod{4}$ is slightly different because the extension $K(i)/K$ needs be ramified at $2$ for the same reason as before. It is then a little bit hard to understand what happens, but if you restrict to $\ell$ being a prime $\equiv 3\pmod{4}$ then $2$ cannot divide the class number of $K$ (see Theorem 10.4 (b) in Washington's Introduction to cyclotomic fields). I know of no standard references, and I would call this kind of results "small/clever/ingenous/trivial trick" according to your taste...Otherwise, Stankewicz' answer correctly pointing to genus theory. –  Filippo Alberto Edoardo Sep 20 '12 at 12:56
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