Totally ramified extensions of local fields

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I'm trying to solve two exercises from a former course on Local Fields & Class Field Theory. Here is the original exercise sheet.

Definition a field extension $L/K$ of non-archimedian local fields $(L, \pi)$ is totally ramified iff $L$ and $F$ have the same residue fields $O_L/(\pi)=k_L=k_K=O_K/(t)$; this is equivalent to existence of a $c \in O_L^*$ with $t= c \cdot \pi^d$ with $d=[L:K]$.

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Questions:

In don't know how solve 2.(a) & (b).

Below I would like present my "approach" in order to solve (a). The main problem in (a) for me is that I don't know how to use the condition that the degree $d$ of the extension is coprime to characteristic $p$ of the residue fields $k_L=k_F$.

My approach on (a): We have to show the existence of a $a \in K$ with $L= K(\sqrt[d]{a})$.

We know by assumption that $O_L/(\pi)=k_L=k= \mathbb{F}_q$ with $q=p^n$ and that we can find a a $c \in O_L^*$ with uniformizer $t= c \cdot \pi^d$. Again, by assumption, $d$ is coprime to $p$. How does it help to find a appropriate root? Again, as $L/K$ is t.r. we know that $L=K(\pi)$ and thus there exist a irreducible polynomial $f(X) \in K[X]$ of degree $d$ with $f(\pi)=0$. If $f$ has the shape $f= X^d-r, r \in K$ we are done.

The key seems to find a $s \in L$ such that $(\pi \cdot s)^d \cdot c= \pi^d \cdot (s^d \cdot c)= r \cdot t$ such that $s^d \cdot c, r \in O_K^*$ that is is seems to be reasonable to try to modify $\pi$ and $c$. Does anybody see how this can be done or how to continue? Possibly by another approach?

On (b) I have no idea, especially the part that is $L/K$ is cyclic extension.

Best Answer

For $L/K$ totally tamely ramified of degree $d$

$$O_K/(\pi_K) \cong \Bbb{F}_q,\qquad \pi_L^d= \pi_K u,\qquad u \in O_L^\times,\qquad u = \zeta_{q-1}^l(1+a\pi_L), \qquad a\in O_L$$

$$\varpi_L = \pi_L (1+a\pi_L)^{-1/d}\in O_L,\qquad \varpi_L^d = \pi_K\zeta_{q-1}^l=\varpi_K \in O_K$$ $$O_L = O_K[\varpi_K^{1/d}]$$

Iff $d | q-1$ then $\zeta_d\in O_L$ so $O_L/O_K$ is Galois and since $\zeta_d\in O_K$ it is abelian.

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