Suppose first that $L/K$ is a <totally ramified extension> of degree $n$, and let $\alpha$ be a <uniformizer> of $L$. If the valuation on $L$ is normalized by $v_L(\alpha)=1$, then $v_L(K^\times)=n\mathbb Z$. The value group of $K(\alpha)$ already contains both $n\mathbb Z$ and $1$, so its <ramification index> over $K$ is at least $n$. Hence $[K(\alpha):K]\geq n$, and therefore $L=K(\alpha)$.

Let
$$
f(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_0
$$
be the <minimal polynomial> of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of $\mathcal O_K$. Moreover
$$
v_L(a_0)=v_L\bigl(N_{L/K}(\alpha)\bigr)=n,
$$
which means $v_K(a_0)=1$. Thus $f$ is an <Eisenstein polynomial>.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the <Eisenstein criterion> makes that polynomial irreducible and its <Newton polygon> gives $v_L(\alpha)=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha)/K)\geq n$; equality with the field degree forces $e=n$ and <residue-field degree> one. Thus $K(\alpha)/K$ is totally ramified.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 1 / a / Solution

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For a <discrete valuation ring> with <uniformizer> $\pi$, an Eisenstein polynomial is a monic polynomial whose nonleading coefficients are divisible by $\pi$ and whose constant coefficient is not divisible by $\pi^2$. It is <irreducible polynomial>[irreducible], and adjoining one of its roots gives a totally ramified extension.

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 Incoming links: Eisenstein polynomial