6
$\begingroup$

Let $p$ be a prime number. Let $\mathbb{F}$ be a finite extension of $\mathbb{F}_p$. Let $\omega$ be the mod $p$ cyclotomic character and let $V$ be a representation of $G_{p} = Gal(\bar{\mathbb{Q}}_p / \mathbb{Q}_p)$ over $\mathbb{F}$ which is a non-split extension of $\omega$ by $\omega$, namely we have a short exact sequence $1 \to \omega \to V \to \omega \to 1$.

We have a map $H^1 (G_p, V) \to H^1 (G_p, \omega)$ whose image is easily seen to be one dimensional over $\mathbb{F}$. Is it possible to describe it in terms of peu ramifiée or très ramifié extension ?

[EDIT] In order to be more precise, let me recall the definition of a peu ramifiée or très ramifiée extension in $H^1(G_p, \omega)$. (Those notions were introduced by Serre in Propriétés galoisiennes des points d'ordre fini des courbes elliptiques)

We have $H^1(G_p, \omega) \simeq \mathbb{Q}_p^{\times} / (\mathbb{Q}_p^{\times})^p \otimes_{\mathbb{F}_p} \mathbb{F}$ and the peu ramifiées extensions are the elements of the line $\mathbb{Z}_p^{\times} / (\mathbb{Z}_p^{\times})^p \otimes_{\mathbb{F}_p} \mathbb{F}$ and the très ramifiées ones are the complement of this set.

$\endgroup$
4
  • $\begingroup$ Do you mean "tamely ramified" and "wildly ramified"? $\endgroup$
    – GH from MO
    Aug 9, 2013 at 20:33
  • 4
    $\begingroup$ @GHfromMO The original title contained the French adjectives peu ramifiée and très ramifiée which have no equivalents in English. Applied to a (wildly) ramified degree-$p$ extension $L$ of a finite extension $K$ of $\mathbf{Q}_p$ or $\mathbf{F}_p((t))$, they tell you whether the unique ramification break of $\mathrm{Gal}(L|K)$ is prime to $p$ or divisible by $p$. This is quite different from the distinction between tamely or wildly ramified extensions. $\endgroup$ Aug 10, 2013 at 1:27
  • $\begingroup$ @Chandan: What is the "unique ramification break"? $\endgroup$
    – GH from MO
    Aug 10, 2013 at 8:59
  • 5
    $\begingroup$ @GHfromMO: [I should have included the hypothesis that the extension $L|K$ is cyclic.] The group $G=\mathrm{Gal}(L|K)$ comes with a (exhaustive and separated) decreasing filtration $\cdots\subset G_2\subset G_1\subset G_0\subset G$, called the ramification filtration (in the lower numbering). In our case, $G$ is cyclic of order $p$, so there is an integer $t$ such that $G_t=G$ and $G_{t+1}=1$~; this $t$ is called the unique ramification break of $G$ (or of $L|K$). $\endgroup$ Aug 10, 2013 at 9:11

2 Answers 2

3
$\begingroup$

I think I have the answer to the question.

By exactness, the image of $H^1(G_p, V) \to H^1(G_p, \omega)$ is the kernel of the cobord map $\delta : H^1(G_p, \omega) \to H^2(G_p, \omega)$.

As $V$ is a non split extension of $\omega$ by $\omega$, it defines a non zero element of $H^1(G_p, \mathbb{F})$, i.e. an additive character of $G_p$. Name this character $u$. Let $\eta$ be an element of $H^1(G_p, \omega)$. A direct computation (involving the definition of the map $\delta$) shows that $\delta(\eta) = u \cup \eta$ (the cup product of $u \in H^1(G_p, \mathbb{F})$ with $\eta \in H^1(G_p, \omega)$).

Now local class field theory tells us that this cup product $\cup : H^1(G_p, \mathbb{F}) \times H^1(G_p, \omega) \to H^2(G_p, \omega)$ is a perfect pairing and that the orthogonal of the line of unramified characters in $H^1(G_p, \mathbb{F})$ under this pairing is the line of peu ramifiées extensions.

Hence, if $V$ is defined (as an extension of $\omega$ by $\omega$) by an unramified character, the image of $H^1(G_p,V) \to H^1(G_p, \omega)$ consists of the peu ramifiées extensions whereas if $V$ is defined by a ramified character, then the image of $H^1(G_p, V) \to H^1(G_p, \omega)$ is generated by a très ramifiée extension.

$\endgroup$
4
  • $\begingroup$ I'm a bit worried that you seem to be thinking of degree-$p$ cyclic extensions of $\mathbf{Q}_p$ as elements of $\mathbf{Q}_p^\times/\mathbf{Q}_p^{\times p}$ rather than as lines in that $\mathbf{F}_p$-space. $\endgroup$ Aug 11, 2013 at 4:20
  • $\begingroup$ I'm not sure I understand completely your comment regarding my answer..., could you be more specific about your worries ? $\endgroup$
    – A M
    Aug 11, 2013 at 11:22
  • $\begingroup$ It will be nice if you clarify how you parametrize the set of degree-$p$ cyclic extensions of $\mathbf{Q}_p$. $\endgroup$ Aug 11, 2013 at 11:27
  • $\begingroup$ Well, I guess such an extension corresponds to a line in $H^1(G_p, \omega)$ but in my answer I'm only dealing with elements of $H^1$ and $H^2$ as cocycles. $\endgroup$
    – A M
    Aug 11, 2013 at 13:38
0
$\begingroup$

Let me just clarify the distinction between peu ramifiée and très ramifiée extensions of a local field $K$ with finite residue field of characteristic $p$. The reason for doing so is that I don't think $\mathbf{Q}_p$ ($p\neq2$) has any très ramifiées extensions, so it is not clear what the question is asking.

The distinction applies to cyclic extensions $L$ of $K$ of degree $p$. Let $t$ be the unique break in the ramification filtration on $\mathrm{Gal}(L|K)$ (as explained in my comments above). It can be shown that if $p\mid t$, then $K$ is a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root of $1$ and $L=K(\root p\of\pi)$ for some uniformiser $\pi$ of $K$. If so, $L$ is called très ramifiée; otherwise (when $t$ is not divisible by $p$), $L$ is called peu ramifiée.

The only case when $K=\mathbf{Q}_p$ contains a primitive $p$-th root of $1$ is when $p=2$, so the local fields $\mathbf{Q}_p$ have no très ramifiées extensions when $p\neq2$. In light of this, one should clarify what is being asked in the question.

$\endgroup$
3
  • $\begingroup$ There is a confusion here. A peu ramifiée extension in $H^1(G_p, \omega)$ is an extension of the trivial character by the $\omega$ (as Galois representations) which correspond via the Kummer isomorphism to an element in $\mathbb{Z}_p^{\times} / (\mathbb{Z}_p^{\times})^p$. So the term peu ramifiée (or très ramifiée) apply to an extension of Galois representations (and not an extension of a local field). $\endgroup$
    – A M
    Oct 5, 2013 at 7:48
  • $\begingroup$ @AM : you should have objected to my comments under your question ! That would have have clarified your meaining. $\endgroup$ Oct 5, 2013 at 12:21
  • $\begingroup$ Indeed, sorry for the misunderstanding. $\endgroup$
    – A M
    Oct 5, 2013 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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