# Unramified subextension of the Galois closure of a totaly ramified $p$-adic field

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.

• 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 – Lukas Nov 5 '18 at 10:08
• 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.) – dyf Nov 5 '18 at 14:28