3
$\begingroup$

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.

$\endgroup$
10
  • 1
    $\begingroup$ @hardmath I did not mention it, but it's Lorenz's Algebra II. I'll put it in the body. $\endgroup$ Commented Jul 25, 2023 at 14:42
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jul 25, 2023 at 15:13
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Jul 26, 2023 at 16:20
  • 1
    $\begingroup$ @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 $f$). $\endgroup$ Commented Jul 30, 2023 at 10:46
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jul 30, 2023 at 11:44

2 Answers 2

2
$\begingroup$

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 $n$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ That makes sense, thanks. What about the proof itself, the part I'm struggling with, any ideas? $\endgroup$ Commented Jul 25, 2023 at 15:17
  • 1
    $\begingroup$ 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 $g$. $\endgroup$ Commented Jul 25, 2023 at 15:30
  • $\begingroup$ @V.Ch. yeah I realized, that's why I deleted that part. $\endgroup$ Commented Jul 25, 2023 at 15:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .