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Let $K$ be an imaginary quadratic field and $E$ be an elliptic curve with CM by $\mathcal{O}_K$. We know that the maximal unramified extension (Hilbert class field) $H/K$ is $K(j(E))$. Can we explicitly write down which are the primes that ramify in the ray class field ie $K(j(E), h(E[\mathfrak{p}]))$ for some $\mathfrak{p}$ where $h$ is the Weber function?

The only sort of information I have for now is the following: a prime $\mathfrak{q}$ splits completely in $K(j(E), h(E[\mathfrak{p}]))$ iff $\mathfrak{q}=(\alpha)$ with $\alpha\in \mathcal{O}_K$ and $\alpha\equiv 1 \mod \mathfrak{p}$.

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  • $\begingroup$ In Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem II.5.6 + Theorem II.3.2(c): The conductor of $K(j(E),h(E[\mathfrak{c}]))/K$ divides $\mathfrak{c}$. This gives a supset of the ramified primes. $\endgroup$
    – user19475
    Commented Jan 26, 2018 at 6:01
  • $\begingroup$ Probably one can show that if $\mathfrak p^v$ divides $\mathfrak c$ and $\left|\left( \mathcal O_K /\mathfrak p^v\right)^\times \right| > | \mathcal O_K^\times |$ then $\mathfrak p$ is in fact ramified, which combined with what TKe says handles everything except for very small primes, which can be done explicitly. $\endgroup$
    – Will Sawin
    Commented Jan 26, 2018 at 9:45
  • $\begingroup$ It would be interesting to know the (higher) ramification groups or at least the ramification index. $\endgroup$
    – user19475
    Commented Jan 26, 2018 at 11:05
  • $\begingroup$ Perhaps swc.math.arizona.edu/aws/1999/99RubinCM.pdf Corollary 5.20 helps (and 5.18). $\endgroup$
    – user19475
    Commented Jan 26, 2018 at 12:55

1 Answer 1

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All the information on the higher ramification groups can be derived from Theorem II.5.6, which TKe notes.

The Galois group of the ray class field is the group of fractional ideals relatively prime to $\mathfrak p$ modulo the principal ideals with generators $1$ mod $\mathfrak p$. This naturally maps to the class group, and the kernel consists of principal ideals, and is hence isomorphic to $(\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times$. The ramification group at $\mathfrak p$ in the abelianization of the Galois group is $\mathcal O_{K_p}^\times$, and the maps is $\mathcal O_{K_p}^\times \to (\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times$ is the obvious projection factoring through $(\mathcal O_K/\mathfrak p)^\times$.

Hence the ramification group of the extension is isomorphic to $(\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times $ and is entirely tame - there are no higher ramification groups.

$\mathfrak p$ is ramified if and only if the natural map $ \mathcal O_K^\times \to (\mathcal O_K/\mathfrak p)^\times$ is not surjective.

This map can only be surjective when $\left|(\mathcal O_K/\mathfrak p)^\times \right| \leq \left| \mathcal O_K^\times\right|$, which happens:

if $K\neq \mathbb Q(i), \mathbb Q(\mu_3)$, only when $\mathfrak p$ is a split or ramified prime lying over $2$ or $3$.

If $K = \mathbb Q(i)$, only when $\mathfrak p$ is the prime lying over $2$ or one of the two primes lying over $5$.

If $K = \mathbb Q(\mu_3)$, only when $\mathfrak p = 2$, the prime lying over $3$, or one of the two primes lying over $7$.

In all these cases, it is easy to see that the map is indeed surjective and so $\mathfrak p$ does not ramify.

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    $\begingroup$ Can one generalise this to $\mathfrak{p}^n$? $\endgroup$
    – user19475
    Commented Jan 27, 2018 at 11:56
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    $\begingroup$ Yes, and then the ramification map is $\mathcal O_{K_{\mathfrak p}}^\times \to (\mathcal O_K/\mathfrak p^n)^\times / \mathcal O_K^\times$. $\endgroup$
    – Will Sawin
    Commented Jan 27, 2018 at 13:40
  • $\begingroup$ So then there can be wild ramification if $n > 1$. $\endgroup$
    – user19475
    Commented Jan 27, 2018 at 14:10
  • $\begingroup$ @TKe And there always is, except for squares of split primes above $2$, the prime power $3$ in $\mathbb Q(\mu_3)$, and the second and third power of the prime above $2$ in $\mathbb Q(i)$. $\endgroup$
    – Will Sawin
    Commented Jan 27, 2018 at 16:36

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