Let $K$ be a finite extension of $\mathbf{Q}_p$. Let $F_d$ be the unramified extension of $\mathbf{Q}_p$ of degree $d$. I would like to know whether there exists some $d \geq 1$ and some $L \subset K \cdot F_d$ such that $L/\mathbf{Q}_p$ is totally ramified and $K \cdot F_d / L$ is unramified.

If $K/\mathbf{Q}_p$ is Galois, this can be done as follows. Take $d=[K:\mathbf{Q}_p]$. Write $Gal( F_d / \mathbf{Q}_p) = \langle \sigma \rangle$ and take any $\tilde{\sigma} \in Gal(K \cdot F_d / \mathbf{Q}_p)$ that lifts $\sigma$. Since $\tilde{\sigma}^d$ is the identity on both $K$ and $F_d$, it is the identity in $Gal(K \cdot F_d / \mathbf{Q}_p)$. We can then take $L = (K \cdot F_d)^{\langle \tilde{\sigma} \rangle}$.

Is the result still true if $K/\mathbf{Q}_p$ is not Galois?


This is not a complete answer, but perhaps it's a roadmap to a counterexample.

My strategy is to consider some non-Galois $K/\mathbf{Q}_p$ for which the result is true, and let's make some deductions about what $K$ must look like in this situation. Then let's find a $K$ that doesn't look like this.

Let $M/\mathbf{Q}_p$ denote the Galois closure of $K/\mathbf{Q}_p$. We're assuming the result is true, so let's choose $d$ and $L$ as in the question. Every field in the question is a subfield of $M.F_d$. Note also that the extension $K.F_d/L$ is unramified, and a composite of unramified extensions is unramified, so I am free to change $d$ to any multiple of $d$ and in particular I can assume that $d$ is a multiple of the degree of the maximal unramified subextension $M_0/\mathbf{Q}_p$ of $M/\mathbf{Q}_p$. In fact I really want to take $d=\infty$ and let $F_\infty$ denote the maximal unramified extension of $\mathbf{Q}_p$ in a fixed algebraic closure of $M$. So let's consider $M.F_\infty$, an infinite Galois extension of $\mathbf{Q}_p$, and let's see what the assertion that $L$ exists tells us about this situation.

We now have $F_\infty\subseteq L.F_\infty\subseteq K.F_\infty\subseteq M.F_\infty$. Recall that $M/\mathbf{Q}_p$ is Galois, with group $D$ say, and say $I$ is the inertia subgroup of $D$. Then $M.F_\infty/F_\infty$ is Galois with group $I$, and $M.F_\infty/\mathbf{Q}_p$ is Galois with group $=:D_M$, a semidirect product of $I$ and $\widehat{\mathbf{Z}}$ (with $I$ normal). The subfields $L.F_\infty$ and $K.F_\infty$ of the extension $M.F_\infty/F_\infty$ correspond to subgroups $I_L$ and $I_K$ of $I$, with $I_K\subseteq I_L$.

The tension in the situation is that $L/\mathbf{Q}_p$ is totally ramified -- we've not used this yet. Let's work in $Gal(M.F_\infty/\mathbf{Q}_p)$; recall this is a group with a finite subgroup $I$ and quotient $\widehat{\mathbf{Z}}$. The existence of $L$ from a Galois theory point of view tells us that there's a subgroup $D_L$ of this group with the property that $I_L$ is normal in $D_L$ (as $L.F_\infty/L$ is Galois) and such that the natural map from $Gal(L.F_\infty/L)$ to $Gal(F_\infty/\mathbf{Q}_p)$ is an isomorphism (here is where we use $L/\mathbf{Q}_p$ totally ramified).

In particular the normaliser of $I_L$ in $D_M$ must be quite big -- and I think that this is too much to ask. If $\sigma$ is a lift of Frobenius in $D_M$ then I think that $\sigma$ must normalise $I_L$. Note that $I_K$ and $I_L$ are both subgroups of $I$ and we know exactly how $\sigma$ acts on this, it's just the action coming from $D=Gal(M/\mathbf{Q_p})$.

So now let me imagine that I can choose $M/\mathbf{Q}_p$ such that $I=(\mathbf{Z}/p\mathbf{Z})^2$ with Frobenius acting as an automorphism of $I$ of order $p+1$ which cycles around all of the subgroups of order $p$. I think that I can explicitly build such an extension when $p=2$ by using Kummer theory on the unramified degree 3 extension of $\mathbf{Q}_2$.

Then if $I_K$ corresponds to a degree $p$ subgroup of $I$, we cannot have $I_L=I_K$ as the normaliser isn't big enough to contain $\sigma$, so we must have $I_L=I$. But this is bad because now $L$ corresponds to a subgroup of $D_M$ that contains $I$ and some lift of $\sigma$.

I have to see a student right now but what do you think? Sorry to be so sketchy!

Addition after seeing Doris' answer; So I think I now have an explicit counterexample.

There's a unique $A_4$ extension $M$ of $\mathbf{Q}_2$ apparently, with inertia the Sylow 2-subgroup and $f=3$. It's the splitting field of $x^4 - 2x^3 + 2x^2 + 2$. Take an element of order 2 in inertia and let $K$ be the corresponding degree 6 extension of $\mathbf{Q}_2$. We have $e=2$ for $K$ so if $L$ existed it would be a quadratic extension. Rather than count masses like Doris suggests, I am just going to do something far more moronic -- I will check using a computer that if $L/\mathbf{Q}_2$ is ramified and quadratic, then $ML/M$ is ramified and hence $KL/K$ must be too. There is probably a sensible way to do this but I just checked all 6 cases explicitly on a computer.

  • $\begingroup$ Nice counterexample! Nevertheless, I'll point out that my mass counting argument will show that there are counterexamples for pretty much all $e(K/\mathbb{Q}_p)>1$ and $f(K/\mathbb{Q}_p)>1$, and indeed it applies over any base field too. $\endgroup$ – Doris Feb 14 '17 at 14:36
  • $\begingroup$ @Doris: I'm sure you're right and I'm sure you're thinking about things in the right way. The reason I answered this question at all was because by chance I was thinking about a closely related problem on the way to work today. I wanted to find some explicit local field $K$ with the property that the absolute Frobenius automorphism on the residue field did not lift to $K$ (just to prove that such a thing can happen -- I am giving a talk on Monsky-Washnitzer cohomology tomorrow and one is always lifting Frobenius; I wanted to show that this was an assumption). $\endgroup$ – Kevin Buzzard Feb 14 '17 at 14:38

Note that the question is equivalent to the following: Given $K/\mathbb{Q}_p$, is there $L/\mathbb{Q}_p$ totally ramified so that $KL/K$ and $KL/L$ are unramified?

You note that it is true for $K/\mathbb{Q}_p$ Galois. It is also trivially true for $K/\mathbb{Q}_p$ totally ramified.

It is also true for $K/\mathbb{Q}_p$ tamely ramified: in this case you can write $K=\mathbb{Q}_p(\zeta,\sqrt[e]{\zeta^r p})$ where $\zeta$ is a primitive $p^f-1$th root of unity, so taking $L=\mathbb{Q}_p(\sqrt[e]{p})$ then $KL=K(\sqrt[e]{\zeta})=L(\sqrt[e]{\zeta})$ are unramified, as required.

I think that in general however, the answer is no. Fix the parameters $e=e(K/\mathbb{Q}_p)$ and $f=f(K/\mathbb{Q}_p)$, then we are looking for $L/\mathbb{Q}_p$ totally ramified of degree $e$ so that $KL/K$ and $KL/L$ are unramified. Assume $f>1$ and $e>1$ (because we know $f=1$ or $e=1$ works), then by Serre's mass formula (counting the number of totally ramified extensions of an extension of a p-adic field, which essentially is exponential in the degree of the base field over $\mathbb{Q}_p$) there are loads of possible $K$ ($n_K$ say), but not many possible $L$ ($n_L$ say). Let $d=(KL:K)$, then $(KL:\mathbb{Q}_p)=def=(KL:L)e$ so $(KL:L)=df$ is determined by $d$, and $d \leq e$. Hence the number of possible $KL$ by counting the $L/\mathbb{Q}_p$ is $n_Le$, whereas the number of different $K$ to consider is at least $n_K$, and assuming $n_Le < n_K$ (which is probably true if you look up the statement of the mass formula) then there are fewer solutions than questions, and hence some questions have no solution.

  • $\begingroup$ I thought a bit about $p=2$ so let me feed that into your answer. There is a unique $A_4$ extension of $\mathbf{Q}_2$ with inertia the Sylow 2-subgroup, so $e=4$ and $f=3$. Corresponding to a cyclic subgroup of order 2 in $I$ is a degree 6 extension $K/\mathbf{Q}_2$ with $e=2$ and $f=3$. My answer is a sketch of why I believe this should be a counterexample. With your eyes we see that we are now asking if there is $L/\mathbf{Q}_2$ ramified quadratic such that $LK/L$ and $LK/K$ are unramified. This looks unlikely to me and we can just check on a computer :-) $\endgroup$ – Kevin Buzzard Feb 14 '17 at 13:53
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    $\begingroup$ I just did this -- I'll add the details to my answer. $\endgroup$ – Kevin Buzzard Feb 14 '17 at 14:25

I'm a bit late but here are a few remarks on Kevin's example :

  1. There is a unique $\mathfrak{A}_4$-extension of $\mathbf{Q}_2$ because there is a unique cyclic cubic extension $C$ (namely the unramified one), the group $G=\mathrm{Gal}(C|\mathbf{Q}_2)=\mathbf{Z}/3\mathbf{Z}$ has a unique irreducible degree-$2$ $\mathbf{F}_2$-representation $\rho$, and $\rho$ occurs with multiplicity $1$ in $C^\times/C^{\times2}$. If $G$ acts on $D\subset C^\times/C^{\times2}$ through $\rho$, then Kevin's $M$ is $M=C(\sqrt D)$.

  2. Kevin's $K$ is not galoisian over $\mathbf{Q}_2$ (its galoisian closure is $M$) and nor is any of its unramified extensions, so no such $L$ exists (because an unramified extension of a quadratic extension is always galoisian).

  3. You can find $\mathfrak{A}_4$-extensions of ramification index $4$ and residual degree $3$ over every local field $F$ with finite residue field of characteristic $2$ (although they might not be unique, and in fact there are infinitely many of them if $F$ has characteristic $2$), so the argument can be made to work over every such $F$.

  4. $\mathfrak{A}_4$-extensions are galoisian closures of primitive quartic extensions (as are $\mathfrak{S}_4$-extensions, and these are the only two possibilities). I'm confident that the same trick can be applied over local fields with finite residue field of characteristic $p$ (arbitrary prime) by working with primitive extensions of degree $p^2$ or perhaps $p^n$ for some $n$. How does one find such extensions ? See https://arxiv.org/abs/1608.04183.


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