= Solution
Suppose first that $L=K(\alpha)$ and the degree-$n$ minimal polynomial of $\alpha$ is an <Eisenstein polynomial>. It is irreducible, $\alpha$ is a <uniformizer> of $L$, and its valuation shows that $e(L/K)\geq n=[L:K]$. Equality follows, so $L/K$ is a <totally ramified extension>.
Conversely, suppose $L/K$ is totally ramified of degree $n$ and choose a uniformizer $\alpha$ of $L$. The field $K(\alpha)$ already has ramification index at least $n$, so it equals $L$. Let
$$
f(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_0
$$
be the minimal polynomial. All conjugates of $\alpha$ have $L$-valuation one. Each $a_i$ with $i<n$ is an elementary symmetric polynomial in products of at least one conjugate and therefore has positive $L$-valuation. Since valuations of elements of $K$ are multiples of $n$, every $a_i$ lies in the maximal ideal of $\mathcal O_K$. Moreover
$$
v_K(a_0)=v_K(N_{L/K}(\alpha))=1,
$$
so $a_0$ is not divisible by the square of that ideal. Thus $f$ is Eisenstein, proving the <Eisenstein generator of a totally ramified extension> criterion.
Back to article page
