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Let $F$ be the field obtained by adjoining to $\mathbb{Q}_l$ a $p$-th root of unity, with $p \not = l$. Denote by $F(p)$ the maximal $p$-extension of $F$, i.e. the maximal extension $L:F$ such that the degree is a power of $p$.

Question: what is the structure of the Galois group $G(F(p):F)$?

I found a reference in the literature stating that it is equal to the semidirect product of two copies of $\mathbb{Z}_p$, but I don't know why, and have been unable to track down any references. Any help would be appreciated.

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up vote 3 down vote accepted

Let $K$ be the maximal unramified p-extension of $F=\mathbb{Q}_l(\zeta_p)$. Note that the valuation on $K$ is discrete (since $K/F$ is unramified), so many of the standard results on local fields apply to $K$. Let $k$ be the residue field of $K$, so $k$ is the maximal p-extension of $\mathbb{F}_l(\zeta_p)$. Then $k(\sqrt[p]{u})/k$ is a p-extension for all $u\in k$ (this uses the fact that $\zeta_p\in k$). Hence by maximality, $k$ contains $p$-th roots of all its elements. By Hensel's lemma, one can show that if $u\in \mathcal{O}_K^\times$, then $\sqrt[p]{u}\in K$. One easy consequence is that $\zeta_{p^n}\in K$ for all natural numbers $n$.

Let $L/K$ be a finite p-extension. Then we may split $L/K$ into an unramified extension $K_0/K$ and a totally ramified one $L/K_0$, both of which are p-extensions. By the definition of $K$, we must have $K_0=K$ so $L/K$ is totally ramified. $L/K$ is tamely ramified because it is a p-extension and $p\neq l$. Let $p^n$ be its degree. I claim that $L=K(\sqrt[p^n]{l})$. The classification of totally tamely ramified extensions gives $L=K(\sqrt[p^n]{\pi})$ for some generator $\pi$ of $\mathfrak{m}_K$. But $\pi=ul$ for some $u\in\mathcal{O}_K^\times$. Since $\sqrt[p^n]{u}\in K$, $K(\sqrt[p^n]{\pi})=K(\sqrt[p^n]{l})$, proving the claim.

Since $K/F$ is a p-extension, the maximal p-extension $F(p)$ of $F$ equals the maximal p-extension of $K$, which by the above is the extension $K(\sqrt[p^\infty]{l})$ obtained by adjoining all $p$-th power roots of $l$ to $K$. By the Galois theory of finite fields, we have $\mathrm{Gal}(K/F)\cong\mathrm{Gal}(k/\mathbb{F}_l(\zeta_p))\cong \mathbb{Z}_p$. Since $K$ contains $p^n$-th roots of unity, $\mathrm{Gal}(K(\sqrt[p^k]{l})/K)\cong\mathbb{Z}/p^k\mathbb{Z}$. By taking limits we have $\mathrm{Gal}(F(p)/K)=\mathrm{Gal}(K(\sqrt[p^\infty]{l})/K)\cong \mathbb{Z}_p$. In particular $\mathrm{Gal}(F(p)/F)$ is an extension of $\mathbb{Z}_p$ by $\mathbb{Z}_p$.

To show it is a semidirect product, we need to construct a splitting of $\mathrm{Gal}(F(p)/F)\rightarrow \mathrm{Gal}(K/F)\cong \mathbb{Z}_p$. I suspect such a splitting can be obtained by sending $1$ to the Frobenius automorphism of $F(p)/F$.

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Thanks, I'm happy with this argument. – KristianJS Nov 19 '12 at 14:21
The splitting is automatic, since Z_p is a free object ( = topologically free on one generator) in the category of pro-p-groups. – nguyen quang do Feb 18 at 20:33

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