It is well known that the maximal unramified extension of $\mathbb{Q}\_p$ can be extended by adding the roots of unity of order prime to $p$. Is there any explicit description of the maximal totally ramified extension of $\mathbb{Q}\_p$?



Is there any explicit description of the maximal totally ramified extension of $\mathbb{Q}_p$?
A composite of totally ramified extensions need not be totally ramified:


**Example 1.** (As per LSpice's suggestion) Consider the extensions $\mathbb Q\_p(\sqrt{p})$ and $\mathbb Q\_p(\sqrt{\varepsilon p})$, where $\varepsilon \in \mathbb Z\_p^\times$ is a nonsquare unit (corresponding, for example, to any lift of a generator of $\mathbb F\_p^\times$). Then the compositum $\mathbb Q\_p(\sqrt{p},\sqrt{\varepsilon p})$ contains the unramified extension $\mathbb Q\_p(\sqrt{\varepsilon})$.


**Example 2.** Consider the extension $\mathbb Q\_p \subseteq \mathbb Q\_p(\sqrt[n]{p})$, where $n = p^r - 1$. Then the Galois closure of this extension contains all $n$-th roots of unity. But $\mathbb Q\_p \subseteq \mathbb Q\_p(\zeta\_{p^r-1})$ is the unique degree $r$ unramified extension. Thus, we see that the Galois closure of a totally ramified extension need not even be totally ramified.


So there is no such thing as *the* maximal totally ramified extension. You could in principle still construct some maximal totally ramified extension (i.e. no further extension is totally ramified), but as far as I can tell these fields are not very explicit. By the second example above, it is not a Galois extension of $\mathbb Q\_p$.


On the other hand, maximal *abelian* totally ramified extensions are very well studied: Lubin–Tate theory gives a relatively explicit construction of an abelian field extension $K \to K^{\operatorname{LT}}$ that is linearly disjoint from $K^{\operatorname{ab}}\_{\operatorname{nr}}$ such that their compositum equals $K^{\operatorname{ab}}$. Thus, $K^{\operatorname{LT}}$ plays the role of a maximal abelian totally ramified extension in a relatively strong sense. The most explicit case is $K=\mathbb Q\_p$; in this case we have $K^{\operatorname{LT}} = \mathbb Q\_p(\zeta\_{p^\infty})$.


