# Number of extensions of a local field of fixed degree

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.

• @hardmath I did not mention it, but it's Lorenz's Algebra II. I'll put it in the body. Commented Jul 25, 2023 at 14:42
• Sorry, $q$ should be a prime dividing $p-1$ such that there's a primitive $q$-th root of unity. Kummer theory provides a criterion for when $K(\sqrt[q]{a})=K(\sqrt[q]{b})$: this is the case iff $\langle aK^{q}\rangle =\langle bK^{q} \rangle$. So if $K^\times/q$ is infinite, then there are infinitely many distinct Kummer extensions. Commented Jul 25, 2023 at 15:13
• Fair enough. -- Regarding the open question, my first thought would be: Can the theorem be sharpened (or just phrased better) as saying that: If $\alpha_1, ..., \alpha_n$ are the roots of $f$, then there is an ordering $\beta_1, ..., \beta_n$ of the roots of $g$ so that $K(\alpha_i)= K(\beta_i)$ for all $1\le i \le n$? That would give what we need, right? Commented Jul 26, 2023 at 16:20
• @TorstenSchoeneberg I think you may be right, looking through the proof now it looks like the root $\beta$ of $g$ one starts with is arbitrary and one finds a root of $f$ tailored to it (then later one argues with $K$-automorphisms to extend it to all roots of $f$). Commented Jul 30, 2023 at 10:46
• @TorstenSchoeneberg I'm almost sure what you say is true, I'd just have to show $g$ is necessarily separable (which I'm having trouble with). Regardless, the actual proof phrases the relationship conversely to the statement (roots of $g$ to roots of $f$) and that's good enough for my purposes. Commented Jul 30, 2023 at 11:44

As Torsten suggested, it is possible to sharpen the theorem to have it say

If $$|f-g|<\delta$$, then $$g$$ is irreducible and separable and there is an ordering of the roots $$\{\alpha_1,\dots,\alpha_n\}$$ of $$f$$ and $$\{\beta_1,\dots,\beta_n\}$$ of $$g$$ so that $$K(\alpha_i)=K(\beta_i)$$.

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 $$n$$.

In finite characteristic, Eisenstein polynomials are not necessarily separable. Consider $$X^p-\pi_K$$, where $$p=\mathrm{char}(K)$$ and $$\pi_K$$ is a uniformiser. Thus Theorem 3 is not applicable.

I think the critical point (perhaps misunderstanding on your side) is that theorem 3 is not just applied to some Eisenstein polynomials of degree $$n$$, but to all of them. The argument breaks down as soon as one is not separable.

• That makes sense, thanks. What about the proof itself, the part I'm struggling with, any ideas? Commented Jul 25, 2023 at 15:17
• Just to comment on what you said about interchanging $g$ and $f$: that is the crux of my problem. The way I understand the proof of Theorem $3$, $\delta$ depends on $f$ (even if it's not explicitly said in the statement), so you can't just interchange $f$ and $g$. Commented Jul 25, 2023 at 15:30
• @V.Ch. yeah I realized, that's why I deleted that part. Commented Jul 25, 2023 at 15:48