# Existence of intermediate field extensions for tamely ramified p-adic extensions

Let $$p$$ be a prime, and let $$K/\mathbb{Q}_p$$ be a tamely ramified finite extension of degree $$n$$. Let $$q$$ be a prime factor of $$n$$ with $$q\neq p$$. Must there exist an intermediate extension $$L$$ (with $$\mathbb{Q}_p\subset L\subset K$$) of degree $$q$$ over $$\mathbb{Q}_p$$?

Some partial progress, following Hasse's "Number Theory", Chapter 16:

We can decompose our field extension as $$\mathbb{Q}_p\subset K^{ur}\subset K$$, where $$K^{ur}$$ is the maximal unramified subextension. Let $$e=[K:K^{ur}]$$ and $$f=[K^{ur}:\mathbb{Q}_p]$$. If $$q$$ divides $$f$$, we are done, since $$K^{ur}/\mathbb{Q}_p$$ is Galois with cyclic Galois group.

Otherwise, write $$K^{ur}=\mathbb{Q}_p(\zeta)$$ and $$K=\mathbb{Q}_p(\zeta,(up\zeta^r)^{1/e})$$, where $$\zeta$$ is a primitive $$(p^f-1)st$$ root of unity, $$u$$ is a unit in the $$p$$-adic integers, and $$0\leq r<\gcd(p^f-1,e)$$. If $$\gcd(p^f-1,e)=1$$, then $$K$$ contains $$\mathbb{Q}_p((up)^{1/e})$$, and thus the intermediate field $$\mathbb{Q}_p((up)^{1/q})$$, which has degree $$q$$ over $$\mathbb{Q}_p$$.

We can also reduce to the case of $$e=q$$, by considering the subfield $$\mathbb{Q}_p(\zeta,(up\zeta^r)^{1/q})$$. Since $$q$$ is prime and we dealt with the relatively prime case, we need only consider the case of $$q$$ dividing $$p^f-1$$. Note that if $$q$$ divides $$p-1$$, we have that $$\mathbb{Q}_p(\zeta,(up\zeta^r)^{1/q})$$ is an abelian Galois extension of $$\mathbb{Q}_p$$, also leading to the desired extension of degree $$q$$. That leaves the case of $$q$$ not dividing $$p-1$$ but still dividing $$p^f-1$$; in these cases, the extension is not Galois. Can we say anything about intermediate extensions of degree $$q$$ in this case?

The answer is no. For example the group $$S_3\times C_3$$ occurs (as TransitiveGroup(6,5)) as the Galois group of an (automatically tame) extension of $$\mathbb{Q}_5$$ (and more generally over $$\mathbb{Q} _p$$ for all odd primes $$p\equiv 2$$ mod $$3$$, by taking the compositum of the unramified degree-$$3$$ extension with the splitting field of $$x^3-p$$), see https://www.lmfdb.org/padicField/5.6.4.2 . The point stabilizer (namely, the diagonal subgroup $$C_3$$) is contained in a maximal subgroup of index 2, but not one of index 3, meaning that there is a tame degree 6 extension not containing a degree 3 subextension.
• @RalphMorrison For this, there should be no counterexamples. Your considerations already show that there is always a subfield $F$ of some prime degree $\ell$, and inductively, between $K$ and $F$ is some intermediate field of degree $q$ (the smallest prime dividing $n/\ell$). It then suffices to solve the case $[K:\mathbb{Q}_p] = q\ell$, with $q<\ell$ both prime. As you deduce, one can reduce to the case $p^\ell \equiv 1$ mod $q$ and $p\ne 1$ mod $q$. But this means $ord_q(p) = \ell > q$, a contradiction. Jan 25 at 2:35