5
$\begingroup$

Let $K$ be a local field (of characteristic 0) with (finite) residue field of characteristic $l$ and let $p$ be a prime.

Considering the cases, whether the $p$-th roots of unity are in $K$ and whether $l$ equals $p$ (and maybe whether $p=2$) or not, my question is:

How many Galois extensions of $K$ of degree $p$ exist?

$\endgroup$
1
9
$\begingroup$

If $K$ contains the $p$-th roots of unity, then Kummer theory tells us that the degree $p$ Galois extensions of $K$ are in bijective correspondence with the subgroups of $K^{\times}/(K^{\times})^p$ of order $p$. The structure of $K^{\times}$ is well-known; see http://en.wikipedia.org/wiki/Local_fields or any decent book on local fields. So you can work out the answer in this case.

If $K$ doesn't contain the $p$-th roots of unity, then it becomes harder. Here are some special cases. For every $d \in \mathbb{N}$, $K$ has a unique unramified extension of degree $d$, which is necessarily cyclic - see Corollary 4.4 of these nice notes: http://websites.math.leidenuniv.nl/algebra/localfields.pdf So in particular there is a unique unramified degree p Galois extension of $K$.

If $l \neq p$, then any ramified extension of $K$ must be totally and tamely ramified. But then by 5.3 and 5.4 of the above notes, $\mathbb{Z}/p\mathbb{Z}$ must embed into the unit group of the residue field of $K$. Then by Hensel's Lemma, $K$ must contain $p$-th roots of unity and so we are reduced to the Kummer case above.

So we are left with the case $l=p$ and $K$ not containing $p$-th roots of unity. I'll think about this some more, but you should be able to use class field theory as mentioned above.

$\endgroup$
2
  • $\begingroup$ Just for clarity (I just confused myself for a minute): What you are saying in your third paragraph is that either $K$ contains all $p$-th roots of unity (so we are in the Kummer case), or there are no ramified Galois extensions of degree $p$ at all (so the unramified is the only one). $\endgroup$ May 8 '14 at 18:41
  • $\begingroup$ Yes (assuming $l \neq p$, of course). $\endgroup$ May 8 '14 at 22:38
10
$\begingroup$

I am sorry if I see this question only now, but since no one gave the following answer, it seems worth posting it.

There is a general formula for the number of extensions of degree $d$ of a $p$-adic field $K$ contained inside a fixed algebraic closure $\overline{K}$, which is given by $$ \# \{ L \mid K \subseteq L \subseteq \overline{K}, \, [L \colon K] = d \} = \sigma(h) \cdot \sum_{j = 0}^m \frac{p^{m+j+1} - p^{2j}}{p - 1} \cdot (p^{\varepsilon_p(j) \cdot d \cdot d_0} - p^{\varepsilon_p(j - 1) \cdot d \cdot d_0}) $$ where:

  • $\sigma$ denotes the sum of divisors function;
  • $h, m \in \mathbb{N}$ are the unique natural numbers such that $p \nmid h$ and $d = h \cdot p^m$;
  • $\varepsilon_p(j) := \sum_{k=1}^j p^{-k}$ if $j \geq 1$, $\varepsilon_p(0) := 0$ and $\varepsilon_p(-1) := -\infty$, i.e. $p^{\varepsilon_p(-1) \cdot d} = 0$. In particular, $p^{\varepsilon_p(j) \cdot d} \in \mathbb{N}$ if $-1 \leq j \leq m$;
  • $d_0 := [K \colon \mathbb{Q}_p]$.

This formula is due to Krasner, and has been proved in the paper "Nombre des extensions d'un degré donné d'un corps $\mathfrak{p}$-adique". The proof uses the same analytic techniques that go into the proof of the (much more famous) Krasner lemma.

Observe that this number is different from the number of $K$-isomorphism classes of extensions of $K$ having a given degree. This is of course due to the presence of non-Galois extensions. Here are two examples of this phenomenon:

  • if $q \neq p$ is a prime then there are $q + 1$ fields $K \subseteq L \subseteq \overline{K}$ having degree $[L \colon K] = q$, but there are only two isomorphism classes of these fields: one containing the only unramified extension, and the other containing the tamely and totally ramified extensions;
  • if $p \geq 3$ there are $1 + p + (p^2 - p) \cdot p$ extensions $\mathbb{Q}_p \subseteq L \subseteq \overline{\mathbb{Q}_p}$ such that $[L \colon \mathbb{Q}_p] = p$, but they form $1 + p + p^2 - p = p^2 + 1$ isomorphism classes. $p + 1$ of these contain a unique extension (which is Galois over $\mathbb{Q}_p$) and every other isomorphism class contains $p$ extensions (see for example Proposition 2.3.1 in the paper "A database of local fields" by Jones and Roberts).

Finally, let me remark that this formula is related to Serre's "mass formula", which is valid in any characteristic. This formula says that a certain "count" of totally ramified extensions of a local, non-Archimedean field $K$ of degree $d$ is equal to $d$. More precisely, $$ \sum_{L \in \Sigma_d} (\# \kappa)^{d - 1 - \mathrm{v}_K(\mathrm{disc}(L/K))} = d $$ where $\Sigma_d$ denotes the set of totally ramified extensions of $K$ which have degree $d$, and $\kappa$ is the residue field of $K$. Observe that if $p \nmid d$ then the formula can be written simply as $\# \Sigma_d = d$. Two useful references for this are:

$\endgroup$
9
$\begingroup$

A Galois extension of degree $p$ has Galois group $\mathbb{Z}/p\mathbb{Z}$, so you are asking about abelian extensions of your local field. Thus the answer to your question can be obtained explicitly via local class field theory -- you can get the needed results out of Serre's Local Fields or many other books.

$\endgroup$
8
$\begingroup$

Here is a belated answer (I came across this question only today) which doesn't require anything more than Kummer theory --- or Artin-Schreier theory, if you want to allow $K$ to be a finite extension of ${\mathbf F}_l((t))$. I will confine myself to the more interesting case of degree-$l$ extensions (where $l$ is the residual characteristic).

If $K$ contains a primitive $l$-th root of $1$, then there is a natural bijection between the set of degree-$l$ cyclic extensions of $K$ and the set of ${\mathbf F}_l$-lines in $K^\times/K^{\times l}$ (Kummer theory). The structure of this filtered ${\mathbf F}_l$-space is completely known; see for example Section V of arXiv:0711.3878 (where your $l$ is called $p$).

If $K$ does not contain a primitive $l$-th root $\zeta$ of $1$, then put $K'=K(\zeta)$, $\Delta={\rm Gal}(K'|K)$ and $\omega:\Delta\to{\mathbf F}_l^\times$ the cyclotomic character giving the action of $G$ on the $l$-th roots of $1$. Then there is a natural bijection between the set of degree-$l$ cyclic extensions of $K$ and ${\mathbf F}_l$-lines in the $\omega$-eigenspace for the action of $\Delta$ on $K^{\prime\times}/K^{\prime\times l}$. The structure of this filtered ${\mathbf F}_l[\Delta]$-module can be completely determined; see for example arXiv:0912.2829.

If you are interested more generally in all degree-$l$ (separable) extensions of $K$ (and not just the cyclic ones), then something similar can be done. Put $L=K(\root{l-1}\of{K^\times})$ and $G={\rm Gal}(L|K)$. There is a natural bijection between the set of (isomorphism classes of) degree-$l$ (separable) extensions of $K$ and the set of $G$-stable lines in the ${\mathbf F}_l$-space $L^\times/L^{\times l}$. Again, the structure of this filtered ${\mathbf F}_l[G]$-module is completely known: see for example arXiv:1005.2016.

Finally, if you allow $K$ to be a finite extension of ${\mathbf F}_l((t))$, there are similar results using Artin-Schreier theory instead of Kummer theory. See for example arXiv:0909.2541 for degree-$l$ cyclic extensions (which correspond to ${\mathbf F}_l$-lines in $K^+/\wp(K^+)$, where $\wp(x)=x^l-x$) and arXiv:1005.2016 for degree-$l$ separable extensions, which correspond to $G$-stable ${\mathbf F}_l$-lines in $L^+/\wp(L^+)$, where $L$ is still $K(\root{l-1}\of{K^\times})$ and $G={\rm Gal}(L|K)$. The filtered ${\mathbf F}_l$-space (resp. ${\mathbf F}_l[G]$-module) $K^+/\wp(K^+)$ (resp. $L^+/\wp(L^+)$) has been completely determined therein. These results allow you in particular to count the number of extensions with bounded ramification, of which there are only finitely many.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.