As is likely the case with many math bloggers, I’ve been looking quite a bit at MO and haven’t updated on some of the previous series in a while.

Back to ANT. Today, we tackle the case {e=1}. We work in the local case where all our DVRs are complete, and all our residue fields are perfect (e.g. finite) (EDIT: I don’t think this works out in the non-local case). I’ll just state these assumptions at the outset. Then, unramified extensions can be described fairly explicitly.

So fix DVRs {R, S} with quotient fields {K,L} and residue fields {\overline{K}, \overline{L}}. Recall that since {ef=n}, unramifiedness is equivalent to {f=n}, i.e.

\displaystyle [\overline{L}:\overline{K}] = [L:K].

Now by the primitive element theorem (recall we assumed perfection of {\overline{K}}), we can write {\overline{L} = \overline{K}(\overline{\alpha})} for some {\overline{\alpha} \in \overline{L}}. The goal is to lift {\overline{\alpha}} to a generator of {S} over {R}. Well, there is a polynomial {\overline{P}(X) \in \overline{K}[X]} with {\overline{P}(\overline{a}) = 0}; we can choose {\overline{P}} irreducible and thus of degree {n}. Lift {\overline{P}} to {P(X) \in R[X]} and {\overline{a}} to {a' \in R}; then of course {P(a') \neq 0} in general, but {P(a) \equiv 0 \mod \mathfrak{m}'} if {\mathfrak{m}'} is the maximal ideal in {S}, say lying over {\mathfrak{m} \subset R}. So, we use Hensel’s lemma to find {a} reducing to {\overline{a}} with {P(a)=0}—indeed {P'(a')} is a unit by separability of {[\overline{L}:\overline{K}]}.

I claim that {S = R[a]}. Indeed, let {T=R[a]}; this is an {R}-submodule of {S}, and

\displaystyle \mathfrak{m} S + T = S

because of the fact that {S/\mathfrak{m'}} is generated by {\alpha} as a field over {k}. Now Nakayama’s lemma implies that {S=T}.  

Proposition 1 Notation as above, if {L/K} is unramified, then we can write {L=K(\alpha)} for some {\alpha \in S} with {S=R[\alpha]}; the irreducible monic polynomial {P} satisfied by {\alpha} remains irreducible upon reduction to {k} 

There is a converse as well: 

Proposition 2 If {L=K(\alpha)} for {\alpha \in S} whose monic irreducible {P} remains irreducible upon reduction to {k}, then {L/K} is unramified, and {S=R[\alpha]} 

Consider {T := R[X]/(P(X))}. I claim that {T \simeq S}. First, {T} is a DVR. Now {T} is a finitely generated {R}-module, so any maximal ideal of {T} must contain {\mathfrak{m}T} by the same Nakayama-type argument. In particular, a maximal ideal of {T} can be obtained as an inverse image of a maximal ideal in

\displaystyle T \otimes_R k = k[X]/(\overline{P(X)})

by right-exactness of the tensor product. But this is a field by the assumptions, so {\mathfrak{m}T} is the only maximal ideal of {T}. This is principal so {T} is a DVR and thus must be the integral closure {S}, since the field of fractions of {T} is {L}.

Now {[L:K] = \deg P(X) = \deg \overline{P}(X) = [\overline{L}:\overline{K}]}, so unramifiedness follows.

Next up: totally ramified extensions, differents, and discriminants.