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Let $p$ be a prime number and $K$ be a finite extension of $\mathbb{Q}_p$. Denote by $\zeta_{p^n}$ a primitive $p^n$-th root of unity (where $n$ is a positive integer). Assume that $K$ contains $\zeta_{p^m}$ for some $m$.

Is it possible to find a subextension $F \subset K$ such that $K$ is abelian over $F$ of degree prime to $p$ and such that $F$ does not contain $\zeta_p$ (and I mean $\zeta_p$ not $\zeta_{p^n}$) ?

This is true when $K = \mathbb{Q}_p (\zeta_{p^m})$. What about the general case ?

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1 Answer 1

up vote 4 down vote accepted

It's not true for any $K$ if $p = 2$, for obvious reasons.

It's not true for $p = 3$; every subfield of $K = \mathbf{Q}_3(\sqrt[4]{-3})$ contains $\zeta_3$ except $F = \mathbf{Q}_3$, and $K/F$ is not abelian (nor even Galois).

It's not true for any $p$. To be explicit, let $E = \mathbf{Q}_p(\zeta_p)$, let $\pi = 1 - \zeta_p$, and let $K = E(\pi^{1/q})$, where $q$ is a prime not dividing $p(p-1)$. Suppose that $F$ is a subfield of $K$ such that $K/F$ is Galois. If $F$ contains $\alpha^{1/q}$ then $F = K$, which is impossible. Hence $K$ must contain the Galois conjugate $\alpha^{1/q} \zeta_q$, and thus also $\zeta_q$. Since $q$ does not divide $p(p-1)$, the extension $\mathbf{Q}_p(\zeta_q)$ is a non-trivial unramified extension contained in $K$. This contradicts the fact that $K$ is totally ramified. (In this case, there are no subfields $F$ of $K$ such that $K/F$ is Galois except $F = K$.)

It is true if $K/\mathbf{Q}_p$ is Galois and $p > 2$. Let $E$ denote the maximal unramified extension of $\mathbf{Q}_p$ inside $K$. Let $D = \mathrm{Gal}(K/\mathbf{Q}_p)$, let $I \subset D = \mathrm{Gal}(K/E)$ denote the inertia subgroup, and $I^{w} \subset I$ the wild inertia subgroup. The group $I^w$ is pro-$p$, and normal in $I$; the quotient $I/I^w$ has order prime to $p$ and is cyclic. In particular, there exists an element $\sigma \in I \subset D$ such that $\sigma$ has order prime to $p$ and generates $I/I^w$. Let $F = K^{\sigma}$ be the fixed field of $\sigma$. Then $\mathrm{Gal}(K/F) = \langle \sigma \rangle$ is Galois and abelian of order prime to $p$. Moreover, $[F:E] = |I^w|$ has $p$-power order, and $E/\mathbf{Q}_p$ is unramified, so the ramification degree $e_F$ of $F/\mathbf{Q}_p$ is a power of $p$. In particular, $F$ does not contain $\zeta_p$ as long as $p > 2$ (since the ramification degree of $\mathbf{Q}_p(\zeta_p)$ is $p-1$).

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Dear Bullwinkle, In para. 3, does $\alpha = \pi$? Regards, –  Matt E Dec 18 '12 at 3:13

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