Let $L/L'/K/\mathbb{Q}_p$ be a tower of finite field extensions such that $L'/K$ is totally ramified and $L/K$ is its Galois closure. We suppose that $L/L'$ is unramified. Let $M/K$ be a Galois subextension of $L/K$ such that $L/M$ is unramified. Is necessarily $L=M$ ?

The answer seems to be yes if the ramification of $L'/K$ is tame. We then have $L'=K(\sqrt[e]{\pi_K})$, so the residual degree $f(L/K)=[L:L']$ is equal to the multiplicative order of $|k_K|\bmod e$, where $|k_K|$ is the order of the residue field. If we write $M/M_0/K$ with $M_0/K$ maximal unramified, then $M_0$ must contain the $e$-th roots of unity, so $|k_{M_0}|\equiv 1\bmod e$. We conclude that $f(L/K)$ divides $f(M_0/K)$, so they must we equal. Then $M=L$ by minimality of $L$.

In general, if $Gal(L/L')$ is generated by a Frobenius lift $\phi$, then $Gal(L/M)$ is generated by an element $i\phi^k$ with some $i\in I(L/K)$ and some $k$ dividing $f(L/K)$. The group $Gal(L/M)$ is normal in $Gal(L/K)$, so one can prove that $\phi$ commutes with $i$. Then, $L'\cap M$ is the subfield of $L$ fixed by $\phi$ and $i$ and $$Gal(L/L'\cap M)=\langle\phi\rangle\times\langle i\rangle.$$ Furtherfore, one can prove that the quotient $$Gal(M/L'\cap M)=(\langle\phi\rangle\times\langle i\rangle)/\langle i\phi^k\rangle$$ is cyclic generated by the image of the Frobenius $\phi$.

My idea was to try to prove that $M/L'\cap M$ is unramified, thus leading to a conclusion that $i$ is trivial, and then that $k=1$ by minimality of $L$, but I am not even sure if that's possible.


(edit: in hindsight, I think there is a flaw in my argument, so read if you please.)

Obviously if we know $M$ contains $L'$, by the definition of $L$ as the Galois closure of $L'$ we immediately get $L=M$.

Since your statement is about a bunch of finite extensions over $\mathbb{Q}_p$ and about unramifiedness, I figure it's best to think about it using finite fields.

Given this slogan, we know that $k_{L} / (k_{L'} = k_K) / \mathbb{F}_p$ are finite extensions. Let's think about where $k_M$ should sit in. Since $L/M/K$, we know that $k_L/k_M/k_K$, so now our tower of finite fields becomes $k_L/k_M/(k_K = k_{L'})$.

Theory about unramified extensions (and along with the fact that there is only one finite extension of a given degree of a finite field) now forces $M/L'$, so we are done.

  • $\begingroup$ for $k_{L'}\subseteq k_M$ to imply that $L' \subseteq M$ we need $L'$ to be unramified over some subfield of $M$, but I don't quite see how to prove the latter $\endgroup$ – Lukas Nov 5 '18 at 10:08
  • $\begingroup$ yes, that is indeed my flaw. (this prompts me now to guess this statement is incorrect in general (if it involves totally ramifiedness), but I haven't thought about any counterexamples yet.) $\endgroup$ – dyf Nov 5 '18 at 14:28

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