For a homework exercise, I'm to determine for each $p$ the number of non-isomorphic tamely ramified Galois extensions $K/\mathbb{Q}_p$ such that $\operatorname{Gal}(K/\mathbb{Q}_p) \cong \mathcal{Q}_8$ (the quaternion group of order $8$).

I was out of class the day we covered this, and I don't see the topic discussed in Gouvêa's book, so all I have is a lone equation from a classmate's notes:

$$\#\{\text{tame Galois extensions } K/\mathbb{Q}_p \, : \, \operatorname{Gal}(K/\mathbb{Q}_p) \cong G\} = \frac{\#\{(a,b) \in G \times G \, : \, aba^{-1}=b^{p},\, \langle a,b \rangle = G\}}{\left|\operatorname{Aut}(G)\right|}$$

I know that $\left|\operatorname{Aut}(\mathcal{Q}_8)\right| = \!\left|\mathcal{S}_4\right| = 24$, so when $p\equiv 1,2 \!\pmod{4}$, I get that there are $0$ such Galois extensions, and when $p\equiv 3 \!\pmod{4}$, I get that there is only $1$. This seems odd to me.

Where does this equation come from? Is it even correct? I have tried to Google it, but have thus far been unsuccessful.

  • 1
    $\begingroup$ I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $\mathbb Q_3$. $\endgroup$ – Lubin Nov 30 '14 at 23:16
  • $\begingroup$ @Lubin The construction of the unique $Q_8$-extension of $\mathbb{Q}_p$ in the $p\equiv\, 3\, (\text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui. $\endgroup$ – Giovanni Di Matteo Jan 9 '19 at 16:33
  • $\begingroup$ Thanks, @GiovanniDiMatteo. $\endgroup$ – Lubin Jan 9 '19 at 21:18

I think the formula you cite is true and is a consequence of the following observations :

Let $F/\mathbf{Q}_p$ be a finite sub-extension of $\overline{\mathbf{Q}}_p/\mathbf{Q}_p$, let $G_F = \text{Gal}(\overline{\mathbf{Q}}_p/F)$, and let $k_F$ denote the residual field of $F$. Let $G$ denote a fixed finite group. Let's say that a $G$-extension of $F$ is a finite Galois sub-extension $K/F$ of $\overline{\mathbf{Q}}_p/F$ with $\text{Gal}(K/F)\simeq G$.

Assertion 1 : (see section 1.1 of [1]) The set of $G$-extensions of $F$ is in bijection with set of surjective homomorphisms $f:G_F\to G$, modulo $\text{Aut}(G)$.

Assertion 2 : Under this correspondence, the tamely ramified $G$-extensions correspond to homomorphisms $f$ which factor through the quotient $G_F\twoheadrightarrow\text{Gal}(F^{\text{tame}}/F)$, where $F^{\text{tame}}$ is the maximal tamely ramified extension of $F$.

Assertion 3 : We have $ \text{Gal}(F^{\text{tame}}/F) = \langle x, y\, |\, yxy^{-1} = x^q\rangle$ as a profinite group (i.e. $x$ and $y$ are topological generators), where $q = |k_F|$. This is theorem 2.(i) of Iwasawa 's paper [2].

Putting these things together, we get the following formula : if $F/\mathbb{Q}_p$ is a finite extension and $q = |k_F|$, then the number of tamely ramified $G$-extensions of $F$ is equal to $$\frac{\#\{ (a,b) \in G\times G | aba^{-1} = b^q \text{ and }\langle a,b \rangle = G\}}{|\text{Aut}(G)|}$$ Note that this number depends only on the group $G$ and the integer $q$. The interpretation of the exponent $q$ coming from Iwasawa's theorem allows us to deduce a few interesting consequences (see remarks 2 and 3 below).

Specializing to the case when $F=\mathbf{Q}_p$, we have $q=p$ and I think we recover the formula you stated, which is valid for any finite group $G$.

Remarks :

  1. If $|G|$ is prime to the residue characteristic of $F$, then all $G$-extensions of $F$ are tamely ramified (see section 3.2 of [1]). Therefore, if $G$ is a finite group with $|G|$ prime to $p$, then the above formula for $F/\mathbf{Q}_p$ finite counts the total number of all $G$-extensions of $F$. In particular, the unique tamely ramified $Q_8$-extension you find in the $F=\mathbf{Q}_p$ and $p\equiv 3 \,(\text{mod} 4)$ case is the only $Q_8$-extension.
  2. If $F/\mathbf{Q}_p$ is totally ramified, then the residual field of $F$ is equal to $\mathbf{F}_p$; in particular, in light of Iwasawa's result and OP's calculation, we get the following for free : if $p\equiv 3 (\text{mod} 4)$ and if $F/\mathbf{Q}_p$ is totally ramified, then there is a unique octic tamely ramified Galois extension $K/F$ with $\text{Gal}(K/F)\simeq Q_8$, and this extension is necessarily the unique $Q_8$-extension of $F$ by remark (1).
  3. More generally, if $F/\mathbf{Q}_p$ is finite and if $F'/F$ is totally ramified (so that $k_{F'} = k_{F}$), then the number of tamely ramified $G$-extensions of $F$ is equal to the number of tamely ramified $G$-extensions of $F'$. And again, when $|G|$ is prime to $p$, all of the $G$-extensions of $F$ (resp. $F'$) are tamely ramified.

    [1]: M. Yamagishi, On the number of Galois p-extensions of a local field, Proceedings of the AMS (1995). http://www.ams.org/journals/proc/1995-123-08/S0002-9939-1995-1264832-0/S0002-9939-1995-1264832-0.pdf

    [2]: K. Iwasawa, On Galois groups of local fields, Trans. Amer. Math. Soc. 80 ( 1955), 448-469. http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/S0002-9947-1955-0075239-5.pdf


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.