Order of $\operatorname{Gal}(K_s/K_\ell)$

Question

I am reading the proof of Grothendieck’s proposition about $\ell$-adic representations of the decomposition group of some discretely valued field, the proposition in the appendix of Serre and Tate’s Good Reduction of Abelian Varieties, and have a question.

The setting is that $K$ is a field complete with respect to a discrete valuation, and $K_\ell$ is the $\ell$-part of the maximal tamely ramified extension of $K_{nr},$ which itself is the maximal nonramified extension of $K$; i.e. $K_\ell$ is generated over $K_{nr}$ by the $\ell^{n\text{th}}$ roots of a uniformizer ($\ell$ is a prime distinct from $p$, the characteristic of the residue field — $p$ may be $0$). The proof goes on to say that one sees easily that, if $L$ is a finite extension of $K_\ell$, every element of $L$ is an $\ell^{\text{th}}$ power, hence the order of $\operatorname{Gal}(K_s/K_\ell)$ is prime to $\ell$. In other words, there is no finite Galois extension of $K_\ell$ of order divisible by $\ell$.

I don’t understand this reasoning. My understanding of the situation is that $\operatorname{Gal}(K_s/K_\ell)$ is an extension of a group isomorphic to $$\prod_{q\text{ prime}\ne p,\ell}\mathbf{Z}_q$$ by a pro-$p$ group (where $p$ is the characteristic of the residue field). This should imply that no element of a finite quotient of $\operatorname{Gal}(K_s/K_\ell)$ has order divisible by $\ell$, and in turn this can be used to see that every element of $L$ as above is an $\ell^\text{th}$ power.

Can someone explain to me the reasoning in the original proof?

Answer 1

I think this is the whole idea, adapting to other local fields shouldn't be very different :

Let $K = \mathbf{Q}_p$, $K^{nr} = K(\zeta_{p^\infty-1})$, $K_\ell = K^{nr}(p^{1/\ell^\infty})$ ($\ell$ prime $\ne p$) and $L/K_\ell$ a finite extension.

Let $f(x) = (1+x)^{1/\ell} = \sum_{n=0}^\infty {1/\ell \choose n} x^n,x \in \overline{\mathbf{Q}_p}$. Since $|{1/\ell \choose n}|_p \le 1$ the series converges for $|x|_p < 1$ in which case $f(x) \in \mathbf{Q}_p(x)$ (as $\mathbf{Q}_p(x)$ is complete for $|.|_p$)

If $y \in L$ then $|y|_p = p^{\ell^r u/v }$ with $\ell \nmid v \in \mathbb{Z}$ so $|y^v (p^u)^{\ell^r}|_p = 1$ and for some $\zeta^{p^m-1}=1$ and $|x|_p < 1$ : $y^v = (p^u)^{-\ell^r} \zeta\, (1+x)$ whence $y^{v/\ell} \in L$ and $vw = 1 + \ell t \implies y^{1/\ell}= y^{-t} (y^{v/\ell})^w \in L$.

If $L/K_\ell$ is Galois of degree $N \equiv 0 \bmod \ell$ then $Gal(L/K_\ell)$ contains a cyclic subgroup $H$ of order $\ell$ so $L/L^H$ is cyclic of order $\ell$ so (*) $L= L^H(a^{1/\ell})$ which is a contradiction.

Whence finite separable extensions of $K_\ell$ are of degree not divisible by $\ell$.

(*) Since $L/L^H$ is cyclic of degree $\ell$ with Galois group generated by $\sigma$ and $\zeta_\ell \in L^H$, letting $b \in L, b \not \in L^H$ then $c= \sum_{m=1}^{\ell} \zeta_\ell^m \sigma^m(b)$ satisfies $c=\zeta_{\ell} \sigma(c)$ whence $c^\ell=\prod_{m=1}^{\ell} \sigma^m(c) \in L^H$, letting $a = c^\ell$ then $L^H(a^{1/\ell})/L^H$ is a non-trivial subextension which has to be $L/L^H$ since $[L:L^H]$ is prime.

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In characteristic $p$ the same idea works since $\sum_{n=0}^\infty {1/\ell \choose n} x^n \in \mathbf{Z}_p[[x]]$ so it can be reduced $\bmod p$ to obtain $(1+x)^{1/\ell} \in \mathbf{F}_p[[x]]$.

Answer 2

To see that every element of $L$ is an $\ell^\text{th}$ power, let $L=K_l[t]/a(t)$ for $a(t)$ an irreducible separable polynomial $a(t)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0$, and suppose $t$ is not an $l^\text{th}$ power in $L$. This implies that the polynomial $a_l(t)=a_nx^{ln}+a_{n-1}x^{l(n-1)}+\ldots+a_0$ is irreducible and separable over $K_l$. Let $K'/K$ be a finite Galois extension containing all $l^{\text{th}}$ roots of unity and the $a_i$, and contained in $K_l$. The extension $K'[t]/a_l(t)$ is finite and separable and is contained in a finite Galois extension $K''$ of $K'$ with $l$ dividing $[K'':K']$, so $l$ divides the ramification index or the residual degree. If the latter, making a finite unramified extension of $K'$ produces a contradition on the irreducibility of $a_l(t)$ over $K_l$. If the former, replacing $K''$ by $K'[\pi^{1/l}]\subset K_l$ similarly yields a contradiction.