How is $\mathbb{F}_p^*$ a subgroup of $\mathrm{Gal}(\mathbb{Q}_p)^{ab}$

Question

Let $p$ be a prime. How is $\mathbb{F}_p^*$ a subgroup of $\mathrm{Gal}(\mathbb{Q}_p)^{ab}$

Answer 1

$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}\def\Gal{\mathrm{Gal}}$ There are some shorter things to say here, which do not fully answer the question, and then some deep things to say which do. I don't think there is an easy way to fully answer the question. To be clear, there is more than one way to embed $\FF_p^{\times}$ into $\Gal(\QQ_p)^{ab}$; I am trying to give what I think most people would consider the "natural" embedding.

Throughout, let $\zeta_m$ denote a primitive $m$-th root of unity.

Shorter things: The extension $\QQ_p(\zeta_p)$ has Galois group naturally isomorphic to $\FF_p^{\times}$: The element $a \in \FF_p^{\times}$ corresponds to $\zeta_p \mapsto \zeta_p^a$. The proof that this is the Galois group is exactly the same as over $\QQ$. (Sketch: Every Galois symmetry must be of the form $\zeta_p \mapsto \zeta_p^a$, so this gives an embedding of the Galois group into $ \FF_p^{\times}$. To prove equality, we must show that the $p$-th cyclotomic polynomial, $x^{p-1} + \cdots + x + 1$, is irreducible in $\QQ_p$. Substitute $x = y+1$ and use Eisenstein's criterion.)

Thus, we have a surjection $\Gal(\QQ_p)^{ab} \to \Gal(\QQ_p(\zeta_p)/\QQ_p) \cong \FF_p^{\times}$. You are asking me to split this surjection.

By a similar argument we have $\Gal(\QQ_p(\zeta_{p^k})/\QQ_p) \cong (\ZZ/p^k \ZZ)^{\times}$. There is an obvious short exact sequence $$1 \to (1+p \ZZ/p^k \ZZ)/p^k \ZZ \to (\ZZ/p^k \ZZ)^{\times} \to \FF_p^{\times} \to 0. \quad (\ast)$$ The sequence $(\ast)$ has a unique splitting! Lift $a \in \FF_p^{\times}$ to the unique element $\tau(a) \in (\ZZ/p^k \ZZ)^{\times}$ which is $a \bmod p$ and obeys $a^p=a$. Taking the inverse limit on $k$, we obtain what is called the Teichmuller lift $\tau : \FF_p^{\times} \to \ZZ_p^{\times}$. Thus, $\FF_p^{\times}$ acts on $\bigcup_k \QQ_p(\zeta_{p^k})$ by $\zeta_{p^k} \mapsto \zeta_{p^k}^{\tau(a)}$ where $\tau(a)$ is the Tiechmuller lift of $a$.

Let $m$ be any positive integer and write $m = p^k n$ where $p$ does not divide $n$. Then $\QQ_p(\zeta_m) = \QQ_p(\zeta_n, \zeta_{p^k})$. The extension $\QQ_p(\zeta_{p^k})/\QQ_p$ is totally ramified and $\QQ_p(\zeta_n)/\QQ_p$ is unramified, so these extensions are disjoint and $\Gal(\QQ_p(\zeta_m)/\QQ_p) \cong \Gal(\QQ_p(\zeta_{p^k})/\QQ_p) \times \Gal(\QQ_p(\zeta_n)/\QQ_p)$. We can define an action of $\FF_p^{\times}$ on $\QQ_p(\zeta_m)$ by acting on $\QQ_p(\zeta_p)$ as described above and acting trivially on $\QQ_p(\zeta_n)$.

That is as far as I can get before I use deep tools.

Deeper things In fact, the local Kronecker-Weber theorem tells us that every abelian extension of $\QQ_p$ is contained in some $\QQ_p(\zeta_m)$. So I have described an action of $\FF_p^{\times}$ on the full $\QQ_p^{ab}$.

If you don't want to invoke local Kronecker-Weber, you can invoke local class field theory instead. Local class field theory tells us that $\Gal(\QQ_p)^{ab}$ is isomorphic to the profinite completion of $\QQ_p^{\times}$. Inside $\QQ_p^{\times}$, let $\ZZ_p^{\times}$ be $p$-adic integers which are nonzero $\bmod p$ and let $U_p$ be $p$-adic integers which are $1 \bmod p$. We have a short exact sequence $1 \to U_p \to \ZZ_p^{\times} \to \FF_p^{\times} \to 1$ which is split by the Teichmuller lift, so $\ZZ_p^{\times} \cong U_p \times \FF_p^{\times}$ and we get $\FF_p^{\times} \subset U_p \subset \QQ_p^{\times} \subset \Gal(\QQ_p)^{ab}$.

Presumably, if I understood local class field theory well enough, I could describe the explicit map $\QQ_p^{\times} \to \Gal(\QQ_p)^{ab}$ and restrict it to the Teichmuller representatives $\FF_p^{\times}$ sitting inside $\QQ_p^{\times}$. However, I do not.