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Let $K$ be a number field and let $p$ denote an odd prime. We let $H(K)$ and $M(K)$ denote the maximal $p$-abelian everywhere unramified extension of $K$ and the maximal $p$-abelian $p$-ramified extension of $K$, respectively. Let $\mathfrak{p}_1, \ldots, \mathfrak{p}_s$ denote the primes in $K$ above $p$, and for each $i=1,\ldots, s$, let $U_i$ denote the local units in $K_{\mathfrak{p}_i}$ which are one modulo $\mathfrak{p}_i$. We also let $$U(K)=\prod\limits_{i=1}^s U_i$$ and we denote by $E(K)$ the units of $K$ which embed diagonally in $U(K)$. Class field theory gives an isomorphism $$Gal(M(K)/H(K)) \cong U(K)/\overline{E(K)}.$$

My questions is: if we let $K_{\infty}$ denote the cyclotomic $Z_p$-extension of $K$, is there a similarly nice description of the Galois group $Gal(M(K)/K_{\infty}\cdot H(K))$?

When $K=\mathbb{Q}(\zeta_p)$, I found some references which prove that $$Gal(M(K)/K_{\infty}\cdot H(K)) \cong U'(K)/\overline{E(K)},$$ where $U'(K)$ is the subgroup of elements in $U(K)$ which have norm $1$ to $\mathbb{Q}_p$. Does this result generalize to arbitrary number fields, or at least those which contain $\zeta_p$?

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Your notations are not very good because it is not visible at first sight that you deal only with $\mathbf Z_p$-modules. Anyway, keep $E(K)$ but change the notation $U(K)$, which will now be the product over all $p$-places of the local units, and introduce $U^1(K)$ = the product of the principal local units. Moreover, for simplification, put $Y(K) = Gal(M(K)/K)$ and $X(K)=Gal(H(K)/K)$. Instead of the isomorphism that you recalled, I'd rather write down the so called "CFT exact sequence relative to inertia":

(In) $ 1\to \Delta(K)\to E(K) \otimes \mathbf Z_p \to U(K) \otimes \mathbf Z_p = U^1(K) \to Y(K) \to X(K) \to 1$ ,

where the global-local map $E(K) \otimes \mathbf Z_p \to U(K) \otimes \mathbf Z_p $ is the diagonal embedding (its image is your $\bar E{(K)}$) , the local-global map $U^1(K) \to Y(K)$ is CFT, and the map $Y(K) \to X(K)$ is the natural surjection; the kernel $\Delta(K)$ is the so called Leopoldt kernel, whose vanishing constitutes the famous Leopoldt conjecture. Note that $X(K)$ is finite, but $Y(K)$ is not; actually, if $c$ is the number of complex primes of $K$, one can derive from (In) that $rank_{\mathbf Z_p} Y(K)=1 + c +rank_{\mathbf Z_p} \Delta(K)$. In this setting, it is more natural to consider the "CFT exact sequence relative to decomposition" : let $E'(K)$ be the group of global $p$-units ( = units outside the $p$-primes), $X'(K)=Gal(H'(K)/K)$, where $H'(K)$ is the subfield of $H(K)$ in which all the $p$-places $v$ of $K$ are totally decomposed; for a local field $K_v$, introduce the pro-$p$-completion $\hat K_{v}= \varprojlim K^*_v/(K^*_v)^{p^n}$, and denote by $\hat U(K)$ the product over all $p$-places of the $\hat K_{v}$'s ; then we have the exact sequence :

(Dec) $ 1\to \Delta(K)\to E'(K) \otimes \mathbf Z_p \to \hat U(K) \to Y(K) \to X'(K) \to 1$

NB. There is a subtlety here. The pro-$p$-completions of $E(K), E'(K)$ and $U(K)$ are $E(K) \otimes \mathbf Z_p$, $E'(K) \otimes \mathbf Z_p$ and $U(K) \otimes \mathbf Z_p$ because $E(K), E'(K)$ and $U(K)$ are $\mathbf Z$-modules of finite type, which not the case of $K^*_v$.

Now, given any $\mathbf Z_p$-extension $K_\infty /K$, you can take the proj. lim. relative to norms of (In) and (Dec) along the finite subextensions $K_n$ of $K_\infty /K$ to get analogous exact sequences at infinite level. Because of the functoriality of CFT w.r.t. norm maps, the proj. lim. $Y(K_\infty), X(K_\infty), X'(K_\infty)$ are subject to the same Galois theoretic interpretation over $K_\infty$ as previously $Y(K), X(K), X'(K)$ over $K$. If $K_\infty$ is specifically the cyclotomic $\mathbf Z_p$-extension, it is known that the Leopoldt defects $\Delta(K_n)$ are bounded, and their proj. lim. is null ./.

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  • $\begingroup$ Maybe I am unable to see it from your answer, but it seems that you are showing me how to find the description of the Galois group of the maximal $p$-abelian $p$-ramified extension of $K_{\infty}$, whereas I would like to know how one finds the Galois group of $M(K)/K_{\infty}$ (M(K) being the maximal $p$-abelian $p$-ramified of $K$, rather than $K_{\infty}$ (I am happy if instead we have $M(K)/K_{\infty}\cdot H(K)$. $\endgroup$
    – F. Holz
    Jan 26 '18 at 13:35
  • $\begingroup$ Ah, I misread... As precised, your problem is much more interesting, it requires making descent as is currently done in Iwasawa theory. I must add an answer since this is not so straightforward. $\endgroup$ Jan 26 '18 at 13:51
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1) After going up, we must now go down to retrieve possible information at the level of $K$. We stick to the case where $K_{\infty}$ is the cyclotomic $\mathbf Z_p$-extension and denote $\Gamma=Gal(K_{\infty}/K)$ throughout. Write $Z^{\Gamma}$ (resp. $Z_{\Gamma}$) for the module of invariants (resp. co-invariants) of $Z$ under the action of $\Gamma$, i.e. the maximal submodule (resp. quotient module) of $Z$ on which $\Gamma$ acts trivially. The following are classically known for the module $Y(K_{\infty})$ : (i) $Y(K_{\infty})^{\Gamma}$ is null ; (ii) there is a natural exact sequence $1\to Y(K_{\infty})_{\Gamma}\to Y(K)\to \Gamma=\mathbf Z_p \to 1$ : this follows right away from the inflation-restriction sequence in Galois cohomology, using the fact that $\Gamma$ is a free pro-$p$-group; but a computation "by hand", using commutators, can be found in any textbook in Iwasawa theory, e.g. Washington's "Introduction to Cyclotomic Fields".

2) Things are much less simple for $X(K_{\infty})$ : there is of course a natural map $Y(K_{\infty})_{\Gamma}\to X(K_{\infty})_{\Gamma} (*)$, and what you want to know is $Gal (M/K_{\infty}.H(K))$, or $Gal(K_{\infty}.H(K)/K_{\infty})$ which is the image of $Y(K_{\infty})_{\Gamma}\to X(K)$. One can get a certain hold of the kernel/cokernel of $(*)$ by cutting the exact sequence $(In_{\infty}$) in two, and playing with the snake lemma (relative to the action of $\Gamma$). But this is not a strong hold, essentially because descent does not work smoothly with units.

3) In fact, as stressed before, it is more natural to study $X'(K_{\infty})$ because here we deal with restricted ramification (i.e. in the setting of "étale cohomology"). The relationship between $Y(K_{\infty})$ and $X'(K_{\infty})$ belongs to the so called "Spiegelung" phenomenom, or "mirror relations". At finite level, these are a combination of isomorphism (CFT) and duality (Kummer theory), so the the best setting for "Spiegelung" is at infinite level, above $K({\mu_p}^{\infty})$, where ${\mu_p}^{\infty}$ is the group of all $p$-th power roots of unity. Suppose $p$ odd for simplicity, and put $\Delta=Gal(K({\mu_p})/K)$, $G=Gal(K({\mu_p}^{\infty}/K))$, so that $G\cong \Delta \times \Gamma$. Then we can study our problem over $K({\mu_p})$ and descend to $K$ via $\Delta$ without any problem since $p$ does not divide the order of $\Delta$. The result is the following : the natural surjection $Y(K({\mu_p}^{\infty}))\to X'(K({\mu_p}^{\infty}))$ factors through an Iwasawa adjoint $\alpha (X'(K({\mu_p}^{\infty}))(-1))$ (I don't explain what it is exactly, just view it as a "twisted" dual related to "Spiegelung"), and its image in $X'(K)$ is isomorphic to $Hom (X'(K({\mu_p}^{\infty})), {\mu_p}^{\infty})^G$ . For details, see my article "Sur la $\mathbf Z_p$-torsion de certains modules galoisiens", Ann. Inst. Fourier, 36, 2 (1986), 27-46, thm. 1.1 and propos. 3.1 .

Complement. I forgot to give a link with the special case $K= \mathbf Q (\mu_p)$ which you cite at the end of your post. In this case things simplify greatly because $\alpha (X'(K({\mu_p}^{\infty}))(-1))$ coincides with the $\Lambda$-torsion of $Y(K_{\infty})$. This gives me the opportunity to explain (in the general case) the last isomorphism that you allude to. Actually, in accordance with the approach in section 1) above, but replacing $X(K_{\infty})$ by $X'(K_{\infty})$, we are led to study the map $X'(K_{\infty})_{\Gamma} \to X'(K)$, but in view of determining its kernel. This is achieved by the so called Sinnott exact sequence $1 \to \tilde E'(K) \otimes \mathbf Z_p \to E'(K) \otimes \mathbf Z_p \to {\oplus}' \hat{K_v}^*/\tilde {K_v^*} \to X'(K_{\infty})_{\Gamma} \to X'(K)$ (see Appendix to L. Federer, B. Gross, "Regulators and Iwasawa modules", Invent. Math., 62 (1981), 443-457), where $\tilde {K_v^*}$ is the subgroup of $\hat{K_v}^*$ (notation of my previous post) consisting of the universal norms in ${K_v}_{,\infty}/K_v$, the leftmost term is tautologically defined as the kernel of the 2nd map (which is just induced by the diagonal), and the 3rd map is CFT. In the case $K=\mathbf Q (\mu_p)$, the 3rd term is the $U'(K)$ that you mentioned ./.

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  • $\begingroup$ Thank you very much for this, I will read your article to try to understand all the details. Just a quick question: can one find a satisfactory description for the subgroup of $U^{1}(K)$ (your notation from previous answer) which quotiented by $E(K) \otimes \mathbb{Z}_p$ will correspond to the Galois group in question ($Gal(M(K)/K_{\infty}\cdot H(K))$)? $\endgroup$
    – F. Holz
    Jan 26 '18 at 19:13
  • $\begingroup$ Unfortunately, no. You could start from a "satisfying" answer to the question for $K_{\infty}.H'(K)$ and then try to come back to $K_{\infty}.H(K)$ by diagram chasing, but you'll get into quite a few uncontrolled parameters. For instance try between the map $(X'(K_{\infty}))_{\Gamma} \to X'(K)$ (the end of Sinnott's sequence) and the analogous map for $X(.)$, and you'll see. The case $K=Q(\mu_p)$ is very particular, because there is only one prime ideal above $p$ and it is principal, so that X(.) = X'(.) $\endgroup$ Jan 27 '18 at 7:12
  • $\begingroup$ I see. According to your answer to a previous question I posted (math.stackexchange.com/questions/2420504/…), we understand quite well the universal norms on the cyclotomic extension of local fields. Wouldn't (at least in the case $\zeta_p \in K$) then the term $\oplus ' \hat{K}_v^{\ast}/\tilde{K}_v^{\ast}$ become again just the subgroup $U^{1}(K)$ consisting of $u$ such that $N_{K/\mathbb{Q}}(u)=1$ (the usual action of global norm on ideles) ? $\endgroup$
    – F. Holz
    Jan 27 '18 at 8:28
  • $\begingroup$ Perhaps, in order not to impede the columns of SE, we should pursue this discussion privately. Here is my e-mail address tguyenq@univ-fcomte.fr . Just send me yours, and we'll go on. $\endgroup$ Jan 27 '18 at 9:51

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