Today we consider the case of a totally ramified extension of local fields {K \subset L}, with residue fields {\overline{K}, \overline{L}}—recall that this means {e=[L:K]=n,f=1}. It turns out that there is a similar characterization as for unramified extensions.

So, hypotheses as above, let {R \subset K,S \subset L} be the rings of integers and choose a uniformizer {\pi} of the DVR {S}. I claim that {S = R[\pi]}. This follows easily from:

Lemma 1 Let {A} be a DVR and let {S} be a set of elements of {A} whose image under reduction contains each element of the residue field of {A}. Let {\pi} be a uniformizer of {A}. Then each {a \in A} can be written 

\displaystyle a = \sum_{n=0}^{\infty} a_n \pi^n,

where each {a_n \in S}. 

This is a basic fact, which I discussed earlier, about systems of representatives.

Now, note that {L=K(\pi)} as a consequence. Consider the minimal polynomial of {\pi},

\displaystyle \pi^n + a_{n-1} \pi^{n-1} + \dots + a_0.

Because of the total ramification hypothesis, any two terms in the above sum must have different orders—except potentially the first and the last. Consequently the first and the last must have the same orders in {S} (that is, {n}) if the sum is to equal zero, so {a_0/\pi} is a unit, or {a_0} is a uniformizer in {R}. Moreover, it follows that none of the {a_i, i \neq 0}, can be a unit—otherwise the order of the term {a_i \pi^i} would be {i < n}, and the first such term would prevent the sum from being zero.

Note that I have repeatedly used the following fact: given a DVR and elements of pairwise distinct orders, the sum is nonzero.

In particular, what all this means is that {P} is an Eisenstein polynomial: 

Proposition 2 Given a totally ramified extension {K \subset L}, we can take {\pi \in S} with {S=R[\pi]} and such that the irreducible monic polynomial for {\pi} is an Eisenstein polynomial.  

Now we prove the converse: 

Proposition 3 If {K \subset L} is an extension with {L=K(\pi)} where {\pi \in S} satisfies an Eisenstein polynomial, then {S=R[\pi]} and {L/K} is totally ramified.  

Note first of all that the Eisenstein polynomial {P} mentioned in the statement is necessarily irreducible. As before, I claim that

\displaystyle T := R[X]/(P(X))

is a DVR, which will establish one claim. By the same Nakayama-type argument in the previous post, one can show that any maximal ideal in {T} contains the image of the maximal ideal {\mathfrak{m} \subset R}; in particular, it arises as the inverse image of an ideal in

\displaystyle T \otimes_R \overline{K} = \overline{K}[X]/(X^n);

this ideal must be {(X)}. In particular, the unique maximal ideal of {T} is generated by {(X,a_0)} for {a_0} a generator of {\mathfrak{m}}. But, since {P(X)} is Eisenstein and the leading term is {X^n}, it follows that {X^n \equiv a_0 \mod \mathfrak{m}^2}. This also implies that {X} is nonnilpotent in {T}.

Now any commutative ring {A} with a unique principal maximal ideal {\mathfrak{m}} generated by a nonilpotent element {\pi} is a DVR, and this implies that {S} is a DVR. This is a lemma in Serre, but we can take a slightly quicker approach to prove this. We can always write a nonzero {x \in A} as {x = \pi^n u} for {u} a unit, because {x \in \mathfrak{m}^n - \mathfrak{m}^{n-1}} for some {n}—this is the Krull intersection theorem (Cor. 6 here). Thus from this representation {A} is a domain, and the result is then clear.

Back to the proof of the second proposition. There is really only one more step, viz. to show that {L/K} is totally ramified. But this is straightforward, because {P} is Eisenstein, and if there was anything less than total ramification then one sees that {P(\pi)} would be nonzero—indeed, it would have the same order as the last constant coefficient {a_0}.