# Counting tamely ramified Galois extensions of $\mathbb{Q}_p$ with a given Galois group.

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.

• 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 $\mathbb Q_3$. Commented Nov 30, 2014 at 23:16
• @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. Commented Jan 9, 2019 at 16:33
• Thanks, @GiovanniDiMatteo. Commented Jan 9, 2019 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