I'm trying to understand the proof of the following theorem.
A local field $K$ of characteristic $0$ admits (inside a fixed algebraic closure) only finitely many extensions of a given degree.
The proof runs something like this:
Since the case of $\Bbb R$ and $\Bbb C$ is trivial, we may assume $K$ is non-archimedean. Then $K$ has only one unramified extension of a given degree, so by taking the maximal unramified extension in $L/K$, we may assume $L/K$ is totally ramified and the theorem reduces to proving there are only finitely many totally ramified extensions of given degree. It is known that $L/K$ totally ramified $\implies L=K(\alpha)$ with $\alpha$ a root of an Eisenstein polynomial over $K$, i.e. one of the form $f(x)=x^n+\pi c_{n-1}x^{n-1}+\cdots+\pi c_0$ with $\pi$ a uniformiser of $K$, $c_j\in R$ (the valuation ring of $K$), and $c_0\in U=R^\times$. By taking the tuple $c=(c_0,\dots,c_{n-1})$, the set of Eisenstein polynomials of degree $n$ may be identified with $U\times R^{n-1}$, a compact space. According to Theorem 3, each tuple $c$ has a small neighbourhood all of whose elements determine the same (finitely many) extensions as $c$. By compactness, it follows there are only finitely many totally ramified extensions of degree $n$.
The Theorem 3 quoted here is the following:
Let $K$ be a complete non-archimedean field and let $f\in K[x]$ be monic irreducible separable of degree $n$. There exists a constant $\delta>0$ with the following property: if $g\in K[x]$ is monic of degree $n$ and $|g-f|<\delta$ (where $|\sum_ja_jx^j|=\max_j(|a_i|)$), then $g$ is irreducible and for any root $\alpha$ of $f$ there is a root $\beta$ of $g$ with $K(\alpha)=K(\beta)$ (inside a fixed algebraic closure of $K$).
It's not clear to me how Theorem 3 is used to derive the highlighted part. If $c$ is a tuple and $f_c$ is the associated Eisenstein polynomial, by the theorem there is a $\delta_c$ (depending on $f_c$ and hence on $c$) such that $g$'s $\delta_c$-close to $f_c$ have certain properties. If I take as $V_c$ the product of open intervals of radius $\delta_c$ around each coordinate $c_j$, I get an open neighbourhood of $c$ in $U\times R^{n-1}$ and if $d\in V_c$, then $|f_d-f_c|<\delta_c$ and extensions by any roots of $f_c$ are equal to extensions by certain roots of $f_d$ (statement of the theorem). But I'm looking for precisely the opposite relation, i.e. I want to describe the extensions by roots of $f_d$ for $d\in V_c$ in relation to extensions by roots of $f_c$. What if some $d$ has some root $\gamma$ such that $K(\gamma)$ is not equal to $K(\alpha)$ for a root of $f_c$?
Additionally, the book states the following after the proof (of the theorem about extensions):
It is easy to see that the statement also holds in nonzero characteristic if only those extensions $L/K$ such that $e(L/K)$ is not divisible by $\operatorname{char}K$ are allowed.
How does the characteristic of $K$ enter into the proof and why shouldn't it hold verbatim in characteristic $p$? At first I thought that maybe it has something to do with Eisenstein polynomials in positive characteristic not being separable, but something like $x^p+tx^{p-1}+tx^{p-2}+\cdots+t$ over $\Bbb F_p((t))$ looks perfectly alright to me.
Edit: The source is Lorenz's "Algebra II" (available here), page $83$ for the theorem in question and page $79$ for Theorem $3$.
Edit2: The secondary question (positive characteristic) has since been clarified, the primary is remains open.
Best Answer
As Torsten suggested, it is possible to sharpen the theorem to have it say
but even that's unnecessary, as in the proof the original relation is "for every root $\beta$ of $g$ there is a root $\alpha$ of $f$ with $K(\alpha)=K(\beta)$" (it is then modified using $\operatorname{Aut}(C/K)$ to mean the opposite), and this is sufficient for my purposes: cover $U\times R^{n-1}$ by finitely many neighbourhoods of the form $V_c$, then $d\in V_c\implies|f_c-f_d|<\delta_c\implies$ extensions by roots of $f_d$ classified by extensions by roots of $f_c\implies$ finitely many totally ramified extensions of degree $n$.