Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $p=37$. Since $p$ divides the numerator of $B_{32}$, by Ribet's proof of the converse of Herbrand's theorem, we know that the class group of ${\bf Q}(\mu_p)$ has size divisible by $p$. More specifically, the $p$-part of the class group decomposes under the action of $\Delta = {\rm Gal}({\bf Q}(\mu_p)/{\bf Q})$, and we know exactly which eigenspace is non-trivial; in this case, the $\omega^5$-eigenspace is non-trivial where $\omega$ is the mod $p$ cyclotomic character.

Now let $X$ denote the $p$-Hilbert class field of ${\bf Q}(\mu_p)$. As above, $X$ decomposes under the action of $\Delta$, and we have that $X^{(\omega^5)}$ is 1-dimensional over ${\bf F}_p$. Thus, there is some abelian everywhere unramified extension $H_5$ of ${\bf Q}(\mu_p)$ such that $\Delta$ acts on ${\rm Gal}(H_5/{\bf Q}(\mu_p))$ by $\omega^5$.

Let $Y$ denote the Galois group of the maximal abelian $p$-extension of ${\bf Q}(\mu_p)$ unramified outside of $p$. By Leopoldt, the ${\bf Z}_p$-rank of $Y$ (modulo its torsion) equals $(p+1)/2$. We can also decomposing $Y$ under the action of $\Delta$; I presume what happens is that eigenspace of $\omega^0$ is rank 1 over ${\bf Z}_p$ (corresponding to the cyclotomic ${\bf Z}_p$-extension), and the eigenspace of $\omega^i$ is rank 1 over ${\bf Z}_p$ for $i$ odd. In particular, there is some ${\bf Z}_p$-extension $M_i$ over ${\bf Q}(\mu_p)$ such that $\Delta$ acts on ${\rm Gal}(M_i/{\bf Q}(\mu_p))$ by $\omega^i$ for each odd $i$.

My question: is $H_5$ the first layer of the ${\bf Z}_p$-extension $M_5/{\bf Q}(\mu_p)$?

share|improve this question
5  
Dear Rob: I think so. The following works for all but a few small odd $p$. Let $K = \mathbf{Q}(\zeta_p)$, $S$ the places over $p$ and $\infty$ (for any number field). If negative answer, get 2-dim'l subspace of $H^1_S(\mathbf{Q},\omega^5)$ (in fact, 2 classes with distinct splitting fields). By global Euler char., has dim. 1 more than $H^2_S(\mathbf{Q},\omega^5)$, so want this $H^2$ to vanish. By local duality at $p$ and oddness for $\infty$, this $H^2$ is Sha$^2$, so injects into Sha$^2(K,\mathbf{F}_p)$, which is dual to Sha$^1(K,\omega)$, which vanishes by Grunwald-Wang (as $p \ne 2$). QED –  BCnrd Dec 11 '10 at 0:38
7  
Dear BCnrd, could you please once in a while foresake your beloved telegraphic style and flesh out the argument for the rest of humanity ? $$ $$ –  Chandan Singh Dalawat Dec 11 '10 at 4:08
2  
Dear Chandan: Let me clarify some things which may have been unclear. When writing $H^i_S(F,M)$ for a number field $F$, finite set of places $S$, and finite discrete $G_F$-module $M$ unramified outside $S$, I mean $H^i(G_{F,S},M)$. For $i = 1$ this is the same as the subgroup of classes in $H^i(G_F,M)$ unramified outside $S$ (due to infl.-restr.), but perhaps not for $i > 1$. For $v \not\in S$, the image of $H^2(G_{F,S},M)$ in $H^2(F_v,M)$ factors through inflation of $H^2(F_v^{\rm{un}}/F_v,M)$. This latter $H^2$ vanishes since cd($\widehat{Z})=1$! That's why Sha$^2$ intervenes. –  BCnrd Dec 11 '10 at 18:58
1  
ERROR: the actual flavor of Sha$^2$ which comes up above is not quite dual to the Sha$^1$ that I was thinking, due to an error that I explain in comments to Rassilon's answer below (caused by using notation that was too concise). So my suggested argument does not yield the conclusion that I expected. (Presumably when fleshed out correctly, it boils down to the same Vandivier issue that Rassilon gets by the Wiles formula, as the proof of that formula uses related global results of Tate.) Sorry about that. –  BCnrd Dec 12 '10 at 4:10
    
Many thanks for your insights, BCnrd. –  Chandan Singh Dalawat Dec 12 '10 at 4:38
show 1 more comment

1 Answer 1

up vote 21 down vote accepted

$\newcommand{\L}{\mathcal{L}}$ $\newcommand{\Q}{\mathbf{Q}}$ $\newcommand{\Z}{\mathbf{Z}}$ $\newcommand{\F}{\mathbf{F}}$

The answer is yes. It suffices (as Brian mentions) to show that the corresponding space of extensions is one dimensional.

Let $p$ be an odd prime, let $k$ be an integer, and let $\omega$ be the mod-$p$ cyclotomic character. The extensions we are interested in correspond to elements of the Selmer group $H^1_{\L}(\Q,\omega^k)$, where $\L$ is defined by the following local conditions: unramified away from $p$, and unrestricted at $p$ (so $\L_p = H^1(\Q_p,\omega^k)$). The dual Selmer group $\L^*$ corresponds to classes which are unramified outside $p$, and totally split at $p$. Let us further suppose that $k \not\equiv 0,1 \mod p-1$, so neither $\omega^k$ nor $\omega^{1-k}$ have invariants over $\Q_p$, and so $|\L_p| = p$. A formula of Wiles relating the Selmer group to the dual Selmer group implies (under the condition on $k$) that $$ \frac{|H^1_{\L}(\Q,\omega^k)|}{|H^1_{\L^*}(\Q,\omega^{1-k})|} = \frac{|\L_p|}{|H^0(G_{\infty},\omega^k)|} = \begin{cases} p, & k \equiv 1 \mod \ 2, \\\ 1, & k \equiv 0 \mod \ 2. \\ \end{cases}$$ (This result also follows more classicaly from so-called mirror theorems, and, phrased slightly differently, occurs in Washington's book on cyclotomic fields.) In your situation, $k$ is odd, and so the space $H^1_{\L}(\Q,\omega^k)$ is one dimensional if and only if $H^1_{\L^*}(\Q,\omega^{1-k})$ vanishes. Since $1-k$ is even, however, the vanishing of this latter group is essentially the same as Vandivier's conjecture (or rather, the $\omega^{1-k}$-part of Vandivier's conjecture). Since Vandiver's conjecture is true for $p = 37$, everything is ok in this case.

Update: Brian reconciles this answer with his previous comments in the comments to this answer below.

(If you don't assume Vandivier's conjecture, then it isn't clear what question to ask for general $p$, since $H^1_{\L^*}(\Q,\omega^k)$ could have dimension $\ge 2$.)


Dear Rob, For your second question (in the comments below), let $\Gamma$ denote the $\omega^{k}$-part of the maximal extension of $\Q(\zeta_p)$ unramified outside $p$. Then there is an exact sequence: $$0 \rightarrow \Z_p \rightarrow \Gamma \rightarrow C \rightarrow 0,$$ where $C$ is the $\omega^k$-part of the class group tensor $\Z_p$. You ask whether the $\Z_p$-extension is totally ramified. This boils down to the following question: Is the image of $\Z_p$ saturated in $\Gamma$? This is equivalent to asking that the sequence above remains exact after tensoring with $\F_p$. Equivalently, it is the same as asking that $$\Gamma/p \Gamma \simeq \F_p \oplus C/pC \simeq \F_p \oplus H^1_{\L^*}(\Q,\omega^k).$$ Yet $\Gamma/p \Gamma \simeq H^{1}_{\L^*}(\Q,\omega^k)$. Thus, using Wiles' formula again, this is the same as asking that: $$p = \frac{|H^1_{\L}(\Q,\omega^k)|}{|H^1_{\L^*}(\Q,\omega^k)|} = \frac{p |H^1_{\L^*}(\Q,\omega^{1-k})|}{|H^1_{\L}(\Q,\omega^{1-k}|},$$ or equivalently:

The $\Z_p$-extension is totally ramified if and only if there does not exist a $\omega^{1-k}$ extension of $\Q(\zeta_p)$ which is unramified outside $p$ but ramified at $p$.

Note that:

  1. If $p$ is regular, then there is no such extension, and so the $\Z_p$-extension is totally ramified. (This is obvious more directly.)

  2. If $p$ is irregular, then mirror theorems imply that there does exist a $\omega^{1-k}$-extension unramified outside $p$; if Vandiver's conjecture holds this extension is ramified, and so the $\Z_p$-extension is not totally ramified (this also follows from the first answer).

  3. If $p$ is irregular, and the $\omega^k$ part $C$ of the class group is cyclic (a consequence of Vandiver's conjecture), then $H^1_{\L}(\Q,\omega^{1-k})$ has order $p$. Thus, in this case, the $\Z_p$-extension is not totally ramified if and only if Vandiver's conjecture holds (for the $\omega^{1-k}$ part of the class group). This example shows that one can't really give a particularly clean answer, since cyclicity of the class group is seen as a strictly weaker property than Vandiver. This shows that you can't really expect a cleaner answer than I gave above (since proving Vandiver from cyclicity sounds very hard).

BTW, there is a little check mark button next to this answer, if you press it, it goes green, which lights up the dopamine receptors in my brain.

share|improve this answer
2  
And so...now I see my error: really the Sha$^2$ is defined as a Galois cohomology of $G_{\mathbf{Q},S}$ equipped with local triviality conditions along $S$. This is sensitive to the choice of $S$, and I erred by thinking it was the same as replacing $S$ with the set of all places since in degree 2 those unramified classes are automatically trivial at all places outside $S$ too. But such a switch can change things (and only for $i=1$ does it happen that the subgroup of $H^i(G_{F,S},M)$ trivial at all places is insensitive to changing $S$). Oops. –  BCnrd Dec 12 '10 at 4:29
    
@Rassilon: Many thanks for your answer. Regarding your last comment, the question I would want to know for general $p$ is whether or not the $Z_p$-extension of $Q_p(\mu_p)$ (in the $\omega^k$-eigenspace) is totally ramified. Any thoughts on this? –  Robert Pollack Dec 12 '10 at 21:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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