Let $p$ be a prime number and $K$ a finite extension of $\mathbb{Q}_p$. Put $K_\infty = K(\mu_{p^\infty})$, the field extension obtained by adjoining all $p$-power roots of unity to $K$.

I want to prove that: if $u \in {\mathcal{O}_K^\times}$ is not a root of unity, then the field extension $K_u$ of $K_\infty$ generated by the $p$-power roots of $u$ is non-abelian over $K$.

Here is a vague idea that I believe will allow us to prove the statement:

First, note that $\mathcal{O}_K^\times = F \times (1 + m_K)$, where $F$ is a finite group and $1 + m_K$ is the group of principal units. We may then reduce the problem to the case when $u$ is a principal unit. Kummer theory provides us with an isomorphism $$ K^\times / (K^\times)^{p^n} \cong H^1 (G_K, \mu_{p^n}), $$ for all $n \in \mathbb{Z}_{\geq 1}$. Taking the projective limits for $n$, we get $$\widehat{K^\times} \cong H^1 (G_K, \mathbb{Z}_p(1)).$$ We consider $u$ now as an element of $\widehat{K^\times}$. I saw in a recent paper by Khare and Wintenberger entitled, Ramification in Iwasawa theory and Splitting Conjectures, that

  • $K_u$ is the extension of $K_\infty$ corresponding to the fixed field of the kernel of the homomorphism arising from the image of $u$ under the map $$ f: \widehat{K^\times} \cong H^1 (G_K, \mathbb{Z}_p(1)) \rightarrow \text{Hom} (G_{K_\infty}, \mathbb{Z}_p)(1)^{\text{Gal}(K_\infty/K)}.$$

Let $\rho = f(u)$. Then, to prove that $K_u$ is non-abelian over $K$, it is enough to verify that the image of $\rho$ is non-abelian.

Now, with this line of argument,

(1) I do not know how the assumption that $u$ is a principal unit and is not a root of unity comes to use;

(2) I am having difficulty in understanding the statement in bullet, so I would be glad if someone can explain to me how $f$ was constructed and how come $K_u$ can be interpreted in the manner described above;

(3) I wonder if there is some effect to the proof if $K$ is assumed to contain some $p$-power roots of unity.

Of course, I also welcome other ideas that can lead to the proof of the above statement. Thanks in advance.

  • Is $K$ local here, or at least a non-Archimedean valued field? If not, what is $\mathscr{O}_K$? – Keenan Kidwell Jul 17 '13 at 12:44
  • Apologies for the confusion. Although I am more interested in a more general case, I will now consider $K$ to be a finite extension of $\mathbb{Q}_p$. This will also enable me to understand what happens in a basic case. – Octobris Jul 17 '13 at 12:51
  • 2
    I think $Hom(G_{K_\infty},\mathbb{Z}_p)(1)^{Gal(K_\infty/K)}$ should actually be $Hom(G_{K_\infty}, \mathbb{Z}_p(1))^{Gal(K_\infty/K)}$. The map comes from inflation-restriction. Any nonzero element of the image gives a non-abelian extension as long as $K\neq K_\infty$. I will try to write a more complete answer a bit later if no one else does. – Kevin Ventullo Jul 17 '13 at 20:55
up vote 2 down vote accepted

Well, let me try in an elementary way. Pick any field $K$ of characteristic prime to $p$ and suppose it does not contain any $p^n$-th root of unity. Let $a\in K^\times\setminus (K^\times)^p$. Then, the extension $K(\sqrt[p^n]{a},\mu_{p^n})/K$ is normal, not abelian and its Galois group is isomorphic to $\Delta_n\ltimes\mathbb{Z}/p^n$ where $\Delta_n=\mathrm{Gal}(K_n(\zeta_{p^n})/K)$. To see this, start by observing that it is clearly Galois (it contains all roots $\xi\;\sqrt[p^n]{a}$ of $X^{p^n}-a$, for $\xi$ running in $\mu_{p^n}$), so the point is to study its Galois group. Set $F_n=K(\zeta_{p^n})$ and let $H=\mathrm{Gal}\big(F_n(\sqrt[p^n]{a})/F_n)\big)$ and $G=\mathrm{Gal}\big(F_n(\sqrt[p^n]{a})/K(\sqrt[p^n]{a})\big)$. I claim that

(*) the order of $a$ in the quotient $F_n^\times/(F_n^\times)^{p^n}$ is exactly $p^n$ (at least if $K/\mathbb{Q}_p$ is finite)

Admitting the claim, $H$ is a cyclic group of order $p^n$ (let $\eta$ be a generator) while $G$ is a cyclic group of order $d>1$ for some $d\mid p^{n-1}(p-1)$ (this is classic, see for instance Lemma 1 in Birch's paper in Algebraic Number Theory by Cassels-Frölich). Define a $\mathbb{Z}_p^\times$-valued character $\omega:G\to\mathbb{Z}_p^\times$ by $g(\zeta)=\xi^{\omega(g)}$ for all $\xi\in \mu_{p^n}$ and $g\in G$: this implies $g(\xi\;\sqrt[p^n]{a})=\xi^{\omega(g)}\;\sqrt[p^n]{a}$. Finally, there is a $j$ such that $\eta(\sqrt[p^n]{a})=\zeta_{p^n}^j\;\sqrt[p^n]{a}$ where $\zeta_{p^n}$ is a fixed generator of $\mu_{p^n}$. You can readily compute that for each $g\in G$, $$ g\eta(\zeta_{p^n}\sqrt[p^n]{a})=\zeta_{p^n}^{(1+j)\omega(g)}\;\sqrt[p^n]{a} $$ while $$ \eta g(\zeta_{p^n}\sqrt[p^n]{a})=\zeta_{p^n}^{\omega(g)+j}\;\sqrt[p^n]{a} $$ which shows at once that $g$ and $\eta$ do not commute unless $g=1$ and that $H\cap G=\{1\}$, namely the extension $F_n(\sqrt[p^n]{a})/K$ is not abelian with Galois group $G\ltimes H$, isomorphic to $\Delta_n\ltimes \mathbb{Z}/p^n$.

We are left with my claim (*): this is where one needs some Kummer theory. Indeed, consider the inflation-restriction sequence $$ 1\to H^1(\Delta_n,\mu_p)\to H^1(K,\mu_p)\to H^0(\Delta_n,H^1(F_n,\mu_p))\to H^2(\Delta_n,\mu_p) $$ and observe that $\Delta_n$-cohomology of $\mu_p$ vanishes: this can be seen by computing $H^2=\hat{H}^0$ which is trivial because $\mu_p(K)=\{1\}$, and then using that the Herbrand quotient of a finite module is $1$: finally, identify the $H^1$'s with the quotients by $p$-th powers by Kummer theory to find $$ H^0(\Delta_n,F_n^\times/(F_n/^\times)^p)\cong K^\times/(K^\times)^p. $$ In particular, we see that $a$ does not become a $p$-th power in $F_n^\times$ and since $F_n/\mathbb{Q}_p$ is finite we know $F_n^\times/(F_n^\times)^{p^n}$ is isomorphic to finitely many copies of $\mathbb{Z}/p^n$, so not being a $p$-th power coincides with having order $p^n$.

Now, back to your situation, you simply observe that the extension you call $K_u$ is the direct limit of extensions $F_n(\sqrt[p^n]{u})$ so the Galois group $K_u/K$ is the inverse limit of $\Delta\ltimes(\mathbb{Z}/p^n)$ none of which is abelian, so it is a non-trivial semi-direct product $$ \Delta\ltimes\mathbb{Z}_p $$ for some finite-index subgroup $\Delta\subseteq\mathbb{Z}_p^\times$, all this provided that $u\notin (K^\times)^p$. This is certainly true if $u$ is a generator of principal units and certainly false if it is a root of unity of order prime to $p$. Of course, if $u$ is a non-trivial principal unit in $(K^\times)^{p^t}\setminus (K^\times)^{p^{t+1}}$, say $u=v^t$ you repeat the above argument with $v$ instead of $u$ (may be, getting some headache due to index-shifting), and similarly if $K$ contains some $p^k$-th root of unity.

As for Khare-Wintenberger's argument, as Kevin observed, they are simply restating the above elementary computation expressing it in terms of cohomology classes but I guess nothing new appears (observe we used Kummer isomorphism in proving (*) and that, is all is needed).

  • Ooops. Sorry about the first line. What I meant was that $K$ is a field of characteristic not equal to $p$. Thanks for your answer. So, upon reading your answer, it seems to me that it doesn't matter weather $u$ is a unit that is not a root of unity or not. The extension will be non-abelian as long as it is not a $p$th power. Moreover, what about when $K$ contains some $p^n$th root of unity? – Octobris Jul 17 '13 at 11:54

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.