## Unramified extensions October 20, 2009

Posted by Akhil Mathew in algebra, algebraic number theory, number theory.
Tags: , ,

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.