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.

  • $\begingroup$ Do you mean "tamely ramified" and "wildly ramified"? $\endgroup$ – GH from MO Aug 9 '13 at 20:33
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    $\begingroup$ @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. $\endgroup$ – Chandan Singh Dalawat Aug 10 '13 at 1:27
  • $\begingroup$ @Chandan: What is the "unique ramification break"? $\endgroup$ – GH from MO Aug 10 '13 at 8:59
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    $\begingroup$ @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 $L|K$). $\endgroup$ – Chandan Singh Dalawat Aug 10 '13 at 9:11

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.

  • $\begingroup$ 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. $\endgroup$ – Chandan Singh Dalawat Aug 11 '13 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 '13 at 11:22
  • $\begingroup$ It will be nice if you clarify how you parametrize the set of degree-$p$ cyclic extensions of $\mathbf{Q}_p$. $\endgroup$ – Chandan Singh Dalawat Aug 11 '13 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 '13 at 13:38

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.

  • $\begingroup$ 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 '13 at 7:48
  • $\begingroup$ @AM : you should have objected to my comments under your question ! That would have have clarified your meaining. $\endgroup$ – Chandan Singh Dalawat Oct 5 '13 at 12:21
  • $\begingroup$ Indeed, sorry for the misunderstanding. $\endgroup$ – A M Oct 5 '13 at 19:27

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