Class field theory and writing down explicit fields I'm taking a class in CFT and I'm trying to figure out what the theorems say and what they can be used for to get a "feel" for them. More explicitly, say I take $\mathbb{Q}_p$, so we have the local Artin homomorphism:
$\theta:\mathbb{Q}_p^\times \to \textrm{Gal}(\mathbb{Q}_p^{ab}/\mathbb{Q}_p)$.
This map is only an isomorphism if the take the profinite completion of $\mathbb{Q}_p^\times$ on the left, but otherwise we have a bijection between the finite-index open subgroups of $\mathbb{Q}_p^\times$ and abelian extensions of $\mathbb{Q}_p$.
Let's say we take the subgroup $\mathbb{Q}_p^{\times 2}$ of $\mathbb{Q}_p^\times$ and for simplicity let's assume that $p\neq 2$. Can we actually somehow explicitly write down the abelian extension that this subgroup corresponds to? None of the theorems look constructive, but because they are considered so useful, I guess there has to be ways of writing down the corresponding abelian extension explicitly through generators?
I still need to look into profinite groups, but my understanding is that for the previous example the order of the abelian extension corresponding to $\mathbb{Q}_p^{\times 2}$ has order $(\mathbb{Q}_p^\times:\mathbb{Q}_p^{\times 2})$.
If this is possible, does anyone know of a source where I could find examples of this?
 A: As jspecter's comment says, the general case of explicit class field theory over local fields is given by Lubin-Tate theory.  But in your particular case of $\mathbb{Q}_p^{\times 2}$ in $\mathbb{Q}_p^{\times}$, we can work it out by hand.  For simplicity, I will assume that $p$ is odd.
The squares in $\mathbb{Q}_p^{\times}$ are elements of the form $p^{2n} u$ where $u$ is an element of $\mathbb{Z}_p$ which congruent to a non-zero square modulo $p$.  As a result,  $(\mathbb{Q}_p^{\times} : \mathbb{Q}_p^{\times 2}) \cong C_2 \times C_2$ (the direct product of two cyclic groups of order 2).  By class field theory, the corresponding abelian extension of $\mathbb{Q}_p$ must have Galois group $C_2 \times C_2$.  What could it be?
Well, there's really only one way to get elements of order 2 in a Galois group: adjoin square roots, and there aren't too many square roots we can adjoin that aren't already in $\mathbb{Q}_p$.  In fact, a set of coset representatives for $\mathbb{Q}_p^{\times 2}$ in $\mathbb{Q}_p^{\times}$ is given by $\{1,s,p,sp\}$ where $s$ is any element of $\mathbb{Z}_p$ which is a non-square modulo $p$.  So we should try the field $K = \mathbb{Q}_p(\sqrt{s}, \sqrt{p})$, which does have the right Galois group $C_2 \times C_2$.
Is $K$ the right field?  Yes, it is, because, as you can easily check, $\mathbb{Q}_p^{\times 2}$ is the unique subgroup of $\mathbb{Q}_p^{\times}$ whose quotient group is isomorphic to $C_2 \times C_2$.  Therefore, by class field theory, there is a unique abelian extension with Galois group $C_2 \times C_2$, so $K$ is the field we are looking for.
Some other cases you can think about: 


*

*There are 3 subgroups of index 2 between $\mathbb{Q}_p^{\times 2}$ and $\mathbb{Q}_p^{\times}$ ($p$ odd).  What are the extensions corresponding to each of them?

*What happens when $p=2$ ?  What is the extension corresponding to $\mathbb{Q}_2^{\times 2}$?  What about all the subgroups in between $\mathbb{Q}_2^{\times 2}$ and $\mathbb{Q}_2^{\times}$?
