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How does one show that there are only finitely many degree $n$ extensions of a local field? I understand how this follows from class field theory in the Abelian case but don't understand how to do the non-Abelian case.

Alexander
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2 Answers2

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In my comment, I said there was an argument from compactness, but on thinking a little more, I realized that since the set of irreducible polynomials over the ring of integers $\mathfrak o$ of your field is not closed, that approach would be a little trickier than I hoped.

Instead, let’s do it piecemeal; I’m sure there’s a quicker argument than what I give below, but I’m not recalling it now.

It’s not Class Field Theory, it’s Kummer Theory mostly. I’ll give a proof for extensions of $\Bbb Q_p$, and there will be necessary modifications for the function-field case. I’ll show that there are only finitely many Galois extensions of degree $n$; the general case follows from this. First, notice that there is only one unramified extension of degreee $n$, so I can concentrate on totally ramified extensions.

In case we’re talking about degree $n$ prime to $p$, then Ramification Theory says that the Galois group is injected into $\kappa^*$, where $\kappa$ is the residue field; in particular, the Galois group is cyclic. Now extend the base from your original $p$-adic field $k$ to the field $k'$ you get by adjoining the $n$-th roots of unity. Now you’re talking about a cyclic extension of degree $n'$ dividing $n$, and these are in one-to-one correspondence with the subgroups of ${k'}^*/({k'}^*)^{n'}$, only finitely many ’cause the $n$-th power map is open (here is one of the places where the fact that the field is local gets used).

Now for the case that the extension is of degree $p^m$ for some $m$. Just as before, I’ll pass to the situation where the base field $k$ has $p$-th roots of unity. You have a $p$-group, so a composition series where each factor is $C_p$, cyclic of order $p$. Kummer theory again, the number of extensions of degree $p$ is finite, each of these has only finitely many extensions of degree $p$, etc. Finitely many in all.

Those are the arguments for the separate layers. I only need to say that the maximal tamely ramified extension in your general extension of degree $n$ is Galois over the base, and it all works out.

Maybe somebody else has a slicker proof; I would welcome it.

Lubin
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  • Many thanks for this. One thing I am slightly confused about is this reduction to the Kummer case. Is it obvious that a tamely ramified extension of degree $n$ will contain the $n$th roots of unity? – Alexander Jan 25 '15 at 22:47
  • Not at all. I meant to make it clear that you’d have to adjoin these first. Then the extended field has only finitely many extensionf of degree $n$, and so… – Lubin Jan 26 '15 at 02:12
  • And of course, if you can find a compact set of monic polynomials of degree $n$ whose roots together generate all extensions of that degree, then Krasner gives you the result in a trice. – Lubin Jan 26 '15 at 02:17
  • Ok, this makes a ton of sense now. Thank you so much! Another approach I came across was to induct on the Abelian case using the solvability of the Galois group and then tackle the Abelian case with class field theory. – Alexander Jan 26 '15 at 23:51
  • Well, I just think of Kummer as more “basic” than CFT. – Lubin Jan 27 '15 at 03:18
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    Two comments. First, the original claim is not true for function fields. This is not an issue with Dr. Lubin's proof since he's working $p$-adically. Second, to fix your argument about compactness work with Eisenstein polynomials. Namely, since there are only finitely many unramified extensions, and every extension is an unramified followed by a totally ramified, we may as well assume we're totally ramified. But, all such extensions come from Eisenstein polynomials, and conversely all such Eisenstein polynomials give such extensions (we get around irreducibility since Eisenstein's are – Alex Youcis Aug 19 '16 at 05:39
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    automatically irreducible). This puts our polynomials in direct correspondence with $\mathcal{O}^\times\times\mathfrak{m}\times\cdots\times\mathfrak{m}\times(\mathfrak{m}-\mathfrak{m}^2))$. This is compact, and by Krasner's lemma sufficiently close polys give the same set of extensions (as you vary the roots). Since this compact set can be covered by finitely many there are finitely may extensions. – Alex Youcis Aug 19 '16 at 05:41
  • @AlexYoucis, thanks for this. I should absolutely have used an argument of this sort. And to complete your comment about char-$p$ local fields, here’s an infinite list of entirely different subfields of $k((t))$, all of degree $p$ beneath: $k((t^p+t^{np+1}))$. – Lubin Aug 19 '16 at 13:43
  • @Lubin No problem. I like your argument better to be frank--it's more 'obvious'.

    Also, just to remark to the OP (since I know you know) you don't have to guess about the extensions of $\mathbb{F}_p((T))$. Namely, by Kummer theory (since $\mathbb{F}_p((T))$ contains the $p^\text{th}$-roots of unity) we know that the cyclic $p$-extensions correspond to $\mathbb{F}_p((T))^\times/(\mathbb{F}_p((T))^\times)^p$. But, this is obviously infinite.

     – Alex Youcis Aug 22 '16 at 14:11
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    No, @AlexYoucis, Kummer Theory doesn’t apply when the characteristic divides the degree. In characteristic-$p$ Kummer theory, extensions of degree $p$ over $K$ are measured by $K^+/\wp(K^+)$, where here, $\wp(z)=z^p-z$. I’m under the impression that cyclic-$p^2$-extensions involve Witt vectors of length two, etc. (Over my head in dangerous waters here ! ) – Lubin Aug 22 '16 at 16:55
  • @Lubin Ah, of course, I'm being silly. Thank you! The issue, for those wondering, is that when one wants to apply the result $H^1(G_K,\mu_p)=K^\times/(K^\times)^p$ to Galois extensions one wants to identify the sheaves $\mu_p$ and $\Z/p\Z$ which requires a) your space contains the $p^\text{th}$ roots of unity in its global sections and b) that $p$ is invertible on your space (so that $\mu_p$ is smooth). Again, for those like me who were wondering, Lubin's other statement just follows from the Artin-Schrier sequence for $K$. I also don't know how the Length two Witt vectors come into play here. – Alex Youcis Aug 23 '16 at 09:21
  • The only place that I've seen them come up is in looking at non-trivial order $p^2$ unipotent algebraic groups over $\mathbb{F}_p$. Anyways, thanks again for the correction! – Alex Youcis Aug 23 '16 at 09:22
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About questions I have seen on the "explicit" number of extensions of fixed degree of a local field k : Krasner has shown complete (and ugly) formulae in the book "Tendances géométriques en Algèbre et Théorie des Nombres", ed. CNRS, Paris 1966, 143-169 ; a much more elegant "mass formula" has been given by Serre, Comptes Rendus Acad. Sciences Paris, 286 (1976), 1031-1036 . As for the same problem for Galois extensions with a given Galois group G, the answer is also known when G is a p-group: if k does not contain the group of p-th roots of 1, see Safarevic, Amer. Math. Soc. Transl., 4 (1956), 59-72 ; if k does, see Yamagishi, Proc. Amer. Math. Soc., 123 (1995), 2373-2380. ¤

nguyen quang do
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