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?



Totally ramified subextension in a finite extension of $\mathbf{Q}_p$
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!




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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.


