Csdrhrt

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 $ell^{ntext{th}}$ 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
$$prod_{qtext{ prime}ne p,ell}mathbf{Z}_q$$
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?

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.

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 $(1+x)^{1/ell} in mathbf{F}_p[[x]]$.

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.