7
$\begingroup$

Some background: when I first heard about ray class fields a year or so ago, I was told that the Hilbert class field of a global field $K$ is the maximal abelian extension of $K$ that is unramified at all (finite and infinite) primes of $K$, and that ray class fields are similar except allowing some ramification at finitely many primes. This eventually crystallized in my head as a quasi-definition of ray class fields: "the ray class field of $K$ with modulus $\mathfrak m$ is the maximal abelian extension of $K$ that is unramified away from $\mathfrak m$, and is allowed to have some restricted ramification at primes dividing $\mathfrak m$." Now that I'm learning about ray class fields properly, I'm struggling to formulate precisely what this "restricted ramification" is, and I haven't found a reference that does so explicitly.

The definition of ray class fields that I'm working with is summarized in section 2.9 of these wonderfully concise notes by Bjorn Poonen: given a global field $K$ and a modulus $\mathfrak m$, we construct a particular open subgroup $U_{\mathfrak m}$ in the idele group $\mathbb A_K^{\times}$, and let $U_{\mathfrak m}'$ be its image in the quotient $\mathbb A_K^{\times}/K^{\times} = C_K$. Then this is a finite-index open subgroup, so it corresponds to a finite-index open subgroup of $\widehat{C_K}$, isomorphic to $\mathrm{Gal}(K^{ab}/K)$ via the global Artin homomorphism, which fixes a finite extension $K_{\mathfrak m}/K$ that we call the ray class field.

I would like the following to be true: if $\mathfrak{m} = \prod_{\mathfrak p} \mathfrak p^{a_p}$, the ray class field $K_\mathfrak{m}$ is the maximal abelian extension of $K$ that is unramified away from $\mathfrak m$, and that has trivial higher ramification group $G^{a_{\mathfrak p}}$ at finite primes $\mathfrak p$ dividing $\mathfrak m$. (Note the upper numbering on the ramification groups; I previously thought I had a counterexample to this, but I was using lower numbering.) A friend and I have more or less worked out why this should be true: from section 1.3 in Poonen's notes, the local Artin map for $K_{\mathfrak p}$ maps the filtration $\mathcal O_{K_{\mathfrak p}}^{\times} \supset 1 + \mathfrak p \supset 1 + \mathfrak p^2 \supset \cdots$ isomorphically onto the higher ramification groups $G^0 \supset G^1 \supset G^2 \supset \cdots$ in $\mathrm{Gal}(K_{\mathfrak p}^{ab}/K_{\mathfrak p})$, so (after using local-global compatibility) the fact that $U_{\mathfrak m}$ contains $1 + \mathfrak p^{a_{\mathfrak p}}$ should be exactly what we need to force the triviality of $G^{a_{\mathfrak p}}$ in $\mathrm{Gal}(K_{\mathfrak m}/K)$.

Can someone confirm that this statement is correct, or fix it if not?

$\endgroup$
1
$\begingroup$

B. Poonen’s summary misses a few important theorems on ramification, which you can find e.g. in D. Garbanati, « CFT summarized », Rocky Mountain J. M., 11, 2 (1981), 195-225 . I adopt Garbanati’s notations (which are slightly different from Poonen’s). For an abelian finite extension $L/K$, one can define a conductor $\mathcal F_{L/K}$ which verifies the following conditions : (i) $\mathcal F_{L/K}$ is the smallest (w.r.t. division) $K $-modulus $\mathcal M$ such that $L$ is contained in $K(R_\mathcal M)$, the ray-class field mod $\mathcal M$ (ii) $\mathcal P$ ramifies in $L/K$ iff $\mathcal P$ divides $\mathcal F_{L/K}$. When taking $L = K(R_\mathcal M)$, these readily imply that the ray class-field $K(R_\mathcal M)$ is the maximal abelian extension of $K$ which is unramified outside $\mathcal M$.

As for your question concerning local ramification groups in upper numbering, I have difficulty to understand what you mean precisely by « using local-global compatibility ». Does this mean that you consider the local extension at $\mathcal P$ and want to study its ramification (compute the last jump of the filtration) in upper numbering ?

$\endgroup$
  • $\begingroup$ "the ray class-field K(RM) is the maximal abelian extension of K which is unramified outside M" This seems immediately odd to me because it only takes into account which primes occur in M, and not their multiplicities. And yet the ray class field depends on the multiplicities: the ray class field mod $(n)$ of $\mathbb{Q}$ is the cyclotomic field $\mathbb{Q}(\mu_n)$, even though $\mathbb{Q}(\mu_{n^2})$ is also unramified outside $(n)$. Am I missing something? $\endgroup$ – Tom Price Jul 7 '17 at 20:29
  • $\begingroup$ @ Tom Price No, you missed nothing. It was my error, I got carried away by the terminology of the OP : « the ray class field $K_M$ is the maximal abelian extension of K unramified away from $M$ ». In the usual terminology of « ramification outside S » or « S-ramification », S is a set of given primes, so the maximal abelian S-ramified extension of K is generally infinite (start e.g. from your example and take inductive limits). I guess that in order to make sense, the OP should read : « … maximal abelian extension of K whose conductor divides M », or something like that. $\endgroup$ – nguyen quang do Jul 8 '17 at 14:09
  • $\begingroup$ But being unramified away from M is only part of the condition. Maybe the second part, about the trivial higher ramification groups, guarantees that it's a finite extension. $\endgroup$ – Tom Price Jul 8 '17 at 17:13
  • $\begingroup$ We must first agree on the terminology "unramified outside M". According to the theorem: In an abelian extension L/K, a prime P ramifies iff P divides the conductor, the only sensible definition should be "uramified outside S (= the support of M)". But then, in the case where the abelian extension L/K unramified outside S is infinite, what does the additional condition on the triviality of certain higher ramification groups bring ? First of all, to make sense, it should not contradict the infiniteness of L/K. But consider the following example. $\endgroup$ – nguyen quang do Jul 9 '17 at 6:45
  • 1
    $\begingroup$ I think you are interpreting the author's condition as "the maximal abelian extension that is unramified away from M, which I claim also has trivial higher ramification groups", when he actually meant "the abelian extension that is maximal among the ones that are both unramified away from M and have higher trivial higher ramification groups". So the higher ramification groups are trivial by definition. $\endgroup$ – Tom Price Jul 9 '17 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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