Finite Extension(s) of $\mathbb{Q}^{\text{ur}}_p$, the Maximal Unramified Extension of $\mathbb{Q}_p$

Question

During a discussion with a professor of mine, he happened to mention that there's a unique quadratic and cubic ramified extension of the Maximal Unramified Extension of $\mathbb{Q}^{\text{ur}}_p$ (for $p \neq 2,3$), being $\mathbb{Q}^{\text{ur}}_p[\sqrt{p}]$ and $\mathbb{Q}^{\text{ur}}_p[\sqrt[3]{p}]$, respectively. I was wondering if anyone could explain this or point me towards a book or notes where this is discussed. And also if there's a way to classify finite extensions of $\mathbb{Q}^{\text{ur}}_p$.

Answer 1

It follows from that $|.|_p$ extends uniquely to the local field $\Bbb{Q}_p^{ur}$ (its integers are a DVR because the ramification index is $=1$ in particular is finite) and that $1/2$ thus $ {1/2 \choose n}$ is the $p$-adic limit of a sequence of integers so that $(1+x)^{1/2} = \sum_{n=0}^\infty {1/2 \choose n} x^n$ converges for $|x|_p < 1$.

If $[K:\Bbb{Q}_p^{ur}]=2$ then $K/\Bbb{Q}_p^{ur}$ is totally ramified of degree $d=2$, let $O_K = \{ x \in K, |x|_p \le 1\}$ and $\pi_K$ an uniformizer, we must have $\pi_K^d = u p$ for some $u \in O_K^\times$, thus $u^{-1} \in \zeta_m (1+\pi_K O_K)$, let $x = \zeta_m^{-1} u^{-1}-1$ then $|x|_p< 1$ so that $u^{-1/2} = \zeta_m^{1/2} (1+x)^{1/2} \in O_K^\times$ which implies $\varpi_K^2 = p$ where $\varpi_K = \pi_K u^{-1/2} \in O_K$ is also an uniformizer, ie. $ \varpi_K = \pm p^{1/2}$, $K =\Bbb{Q}_p^{ur}(p^{1/2})$.

Answer 2

Your professor simplified the situation (telling no lies, I assure you) by restricting to the case of degree $2$ and $3$. The story is intimately bound up with the phenomenon of tame ramification: a totally ramified extension is tame if its degree is prime to $p$. For finite Galois extensions that are tame, say with group $\Gamma$, you get an injection of $\Gamma$ into the multiplicative group of the residue field, so that $\Gamma$ will necessarily be cyclic. My main source for this is Chapter IV of Serre’s Corps Locaux/Local Fields. It’s old, but not likely to have a better, since Serre is Serre.