Let $K$ be a non archimedian local field of characteristic $p>0$ with residue field $\mathbb{F}_p$ and $l\neq p$ be a prime.
It is known by local classfieldtheory that any abelian Galois extension $L|K$ lies in $L\subset K^{nr}K_\infty$, where $K^{nr}|K$ denotes the maximally unramified extension and $K_\infty|K$ is a fixed Lubin-Tate extension, which is a tower of fields, which are totally ramified over $K$ with galois group $Gal(K_\infty|K)\cong\mathcal{O}_K^\times$.
Can this fact be used to show that any cyclic Galois extension $L|K$ with degree $l^2$ is either unramified or totally ramified, or are there examples of $L|K$ with $Gal(L|K)=\mathbb{Z}/l^2\mathbb{Z}$, such that the inertia degree $f(L|K)=l$ and the ramification index $e(L|K)=l$?
I am searching explicitly for the case that $p=3$ and $l=2$.
Best Answer
I think the following is an example of a cyclic extension of degree $4$ over the field $K=\Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $\Bbb{F}_9$. Let $$ u=\sqrt{(1+i)x},\quad v=\sqrt{(1-i)x}. $$ It follows that $uv=\pm ix\notin K$. Similarly $u^2,v^2\notin K$, but $u^2+v^2=-x\in K$. This implies that $$ p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2\in K[T] $$ is irreducible. None of the zeros $\pm u,\pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$ is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$ is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1\in L$ as is $v=ix/u$. So $L$ has a subfield isomorphic to $\Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $\sqrt{1+i}\notin\Bbb{F}_9$. The explanation is surely that $\sqrt x\notin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,\ell)$ pairs? A key ingredient seems to be that $L=\Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=\ell$ whenever $\ell\mid p-1$, but that needs a bit more work.