2
$\begingroup$

Let $S$ be a set of prime numbers and $K$ be a number field. What is the definition of maximal abelian extension over $K$ unramified outside $S$? Why does it exist and why if $S=\{p\}$, where $p$ is a prime number, it is equal to $$\bigcup_{n\ge 1}K_{p^n}$$ where $K_{q^n}$ is the ray class field modulo $q^n\mathcal{O}_K$? Thanks.

$\endgroup$
  • $\begingroup$ It is an abelian Galois extension $L/K$ which is unramified at all primes not in $S$, such that if $L'$ is another field with this property, then $L\supset L'$. Its existence follows from class field theory, and is highly non-trivial. Are you familiar with the statements of class field theory? $\endgroup$ – Mathmo123 Aug 19 '16 at 14:20
  • $\begingroup$ @Mathmo123: But this extension can be infinite, right? So, what does it mean "unramified" for an infinite extension? $\endgroup$ – user72870 Aug 19 '16 at 22:29
  • $\begingroup$ Are you saying $p$ is a prime number, or $p$ is a prime ideal of $K$? $\endgroup$ – D_S Aug 20 '16 at 15:58
  • $\begingroup$ @D_S: a prime number $\endgroup$ – user72870 Aug 20 '16 at 18:51
  • 1
    $\begingroup$ @RaviFernando Yes of course. I'm confusing myself with the Hilbert class field. $\endgroup$ – Mathmo123 Aug 24 '16 at 7:54
4
$\begingroup$

Here are some answers to your questions. Throughout, fix a number field K and a finite S of primes of K (one could also take infinite sets of primes by inductively enlarging finite S ’s, but this is a detail).

1) For short, a finite extension L/K is called S-ramified if it is unramified outside S. The composite of two such extensions is again S-ramified, so it makes sense to consider the composite $K_S$ of all the finite S-ramified extensions of K. Because of maximality, $K_S/K$ is obviously Galois ; its Galois group $G_S (K)$ is profinite. The quotient $G_S (K)^{ab}$ of $G_S (K)$ modulo the closure of its commutator subgroup fixes the maximal S-ramified abelian extension $K_S^{ab}$ of K. This $K_S^{ab}$ can also obviously be defined as the composite of all the finite abelian S-ramified extensions of K. For more details on ramification in infinite Galois theory (although there is no mystery), see e.g. §2 of the appendix of Washington’s « Introduction to cyclotomic fields ».

2) Actually, class field theory (CFT) also takes into account the ramification at infinite primes (i.e. embeddings into an algebraic closure) in a finite extension L/K : by definition, a complex prime is always unramified ; a real prime of K is ramified iff it becomes complex in L. A « K-modulus » M is a formal product of an integral ideal of K and a finite number of distinct real embeddings. The classical formulation of CFT in terms of ideals and modulii is not very suggestive at first sight. Let us refer for this to the excellent survey by D. Garbannati, « CFT summarized », Rocky Mountain J. of Math., 11, 2 (1981), 195-225. Then the ray class field modulo M over $K$ is the largest finite abelian M-ramified extension of K. If $S = S_p$ := the set of primes of $K$ above an odd prime $p$ (to avoid petty details on ramification at infinite primes), indeed $K_S^{ab}$ is the direct limit of the ray class fields mod $p^n$ as you suspected.

3) The nowadays prevailing formulation of CFT in terms of idèles is well-adapted to the machinery of Galois cohomology and Poitou-Tate duality. But even so, the description of $G_S (K)^{ab}$ is explicit in particular cases (if $K = \mathbf Q$ , see the Kronecker-Weber theorem), but not in general. Actually many deep conjectures on the determination of $G_S (K)^{ab}$ and far reaching applications remain unsolved to this day. Let us for instance take the p-adic point of view and study the maximal pro-p-quotient $X_S (K)$ of $G_S (K)^{ab}$ . If S contains $S_p$, CFT tells us that $X_S (K)$ is a $\mathbf Z_p$-module of finite type, hence it is a priori of the form $T_S (K)$ x $\mathbf Z_p^{1+c+d}$, where $T_S (K)$ is a finite abelian p-group, $2c$ is the number of complex primes of $K$ and $d$ is a natural integer. The celebrated Leopoldt conjecture (up to now proved only for abelian $K$) predicts the nullity of $d$. If $K$ is totally real, then $T_S (K)$ contains, in a precise sense, the theory of $p$-adic $L$-functions : this is the « Iwasawa main conjecture », proved by Mazur-Wiles and Wiles. If $S$ does not contain $S_p$, much less is known ./.

$\endgroup$
  • $\begingroup$ A very nice exposition. One slip: “ a real prime of $K$ is unramified iff it becomes complex in $L$” — I think that should be “ramified”. $\endgroup$ – Lubin Aug 20 '16 at 22:27
  • $\begingroup$ Right, thank you. I'll edit that. $\endgroup$ – nguyen quang do Aug 21 '16 at 6:17
  • $\begingroup$ Thank you! I have some questions: 1) Does your definition coincide with that given by Mathmo123 in the comments? (I think so) 2) Why $K_S^{ab}$ is finite? In particular, I'm trying to show (without success) that $K_{p^n}=K_{p^{n+1}}$ when $n$ is big. 3) Why when you define $S_p$ in (2), you assume $p$ to be odd? $\endgroup$ – user72870 Aug 21 '16 at 18:07
  • 1
    $\begingroup$ 1) Yes for the definition. But the existence does not depend on CFT; 2) It is NOT finite. See the subject of Leopoldt's conjecture ; 3) No $\endgroup$ – nguyen quang do Aug 21 '16 at 19:30
  • $\begingroup$ The existence of $K_S^{ab}$ does non depend on CFT, but to show that $K_S^{ab}$ is nontrivial, i.e. there exists an abelian extension of $K$ that is $S-$ramified, we need CFT, right? $\endgroup$ – user72870 Aug 21 '16 at 21:47

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.