Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 28 / 4

Normalize $v_K$ to have value group $\mathbb{Z}$, and extend it to an algebraic closure. Suppose $\alpha$ is a root of a monic Eisenstein polynomial of degree $n$. A root cannot have nonpositive valuation: the leading term would then have uniquely smallest valuation. If $w=v_K(\alpha )>0$, the constant term has valuation one, all the other nonleading terms have valuation strictly greater than one, and the leading term has valuation $nw$. Cancellation forces $nw=1$. The value group of $K(\alpha )$ is $(1/e)\mathbb{Z}$, so $n\mid e$. Since $e\le [K(\alpha ):K]\le n$, equality holds throughout. Thus the polynomial is irreducible and the extension is totally ramified.

Conversely, let $L/K$ be totally ramified of degree $n$, and choose a uniformizer ${\pi }_L$. Its extended $v_K$ value is $1/n$. The same denominator argument gives $[K({\pi }_L):K]\ge n$, so $K({\pi }_L)=L$. Every conjugate has valuation $1/n$, by the unique extension of an absolute value to a finite extension of a complete field. The nonleading coefficients of its minimal polynomial are elementary symmetric expressions in these conjugates, so their valuations are positive, hence at least one because the coefficients lie in $K$. The constant coefficient is their product and has valuation exactly one. Its minimal polynomial is an Eisenstein polynomial, and $\boxed{L=K({\pi }_L)}$.

One form of Krasner's lemma is: if $K$ is a complete non-Archimedean field, $\alpha$ is separable over $K$, and $\beta$ is algebraic withwhere ${\alpha }^{\prime }$ ranges over the other $K$-conjugates, then $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is immediate.

To prove the nontrivial case, let $\sigma$ be an automorphism of an algebraic closure fixing $K(\beta )$. The unique extension of an absolute value to a finite extension implies that $\sigma$ preserves the extended absolute value. ThereforeThis is strictly less than the distance to every other conjugate, so $\sigma (\alpha )=\alpha$. As $\alpha$ is separable over $K(\beta )$, being fixed by all these automorphisms puts it in $K(\beta )$. This proves $\boxed{K(\alpha )\subseteq K(\beta )}$. The strict inequality is what prevents switching to another conjugate.

We count extensions up to $K$-isomorphism. An extension of degree $n$ has a maximal unramified extension $E/K$ inside it, with $[E:K]=f\mid n$, and $L/E$ is totally ramified of degree $e=n/f$. The unramified extension of each degree is unique, so there are only finitely many possibilities for $E$.

Fix $E$ and $e$. The coefficient space of monic degree-$e$ Eisenstein polynomials is compact: its nonconstant nonleading coefficients lie in the compact maximal ideal, and its constant coefficient lies in the compact set ${\mathfrak{m}}_E\setminus {\mathfrak{m}}_E^2$. Each such polynomial $F$ has a root $\alpha$ generating a degree-$e$ extension. For sufficiently close coefficients of a polynomial $G$, the numbers $G(\alpha )$ and $G^{\prime }(\alpha )-F^{\prime }(\alpha )$ are arbitrarily small, while $F^{\prime }(\alpha )\ne 0$ because the characteristic is zero. The strong form of Hensel lemma, applied in $E(\alpha )$, supplies a root $\beta$ of $G$ as close to $\alpha$ as desired. Choose the neighborhood so that the Krasner's lemma inequality holds. Then $E(\alpha )\subseteq E(\beta )$, and both degrees are $e$, so the extensions are equal. Compactness gives a finite cover by such neighborhoods, proving finiteness of degree-$n$ extensions. If extensions are realized as subfields of a fixed algebraic closure rather than counted up to isomorphism, each has only finitely many embeddings, so that count is finite too.

For odd $p$, write $K^{\times }={\pi }^{\mathbb{Z}}{\mathcal{O}}_K^{\times }$. The square map is bijective on $1+{\mathfrak{m}}_K$ by Hensel lemma, and the finite residue field has cyclic multiplicative group of even order. ThusThe quadratic extensions from square classes correspond to its three nontrivial classes. Representatives are $u,\pi ,u\pi$, where $u$ is a unit with nonsquare residue. There are $\boxed{3}$ quadratic extensions: $K(\sqrt{u})$ is unramified, while $K(\sqrt{\pi })$ and $K(\sqrt{u\pi })$ are totally ramified. For completeness, two nonsquare parameters give the same extension only if their ratio is a square: writing $\sqrt{b}=x+y\sqrt{a}$ and squaring forces $x=0$, hence $b/a=y^2$.