This is an exercise I’ve never done, but it should be a lot of fun. What is the general Eisenstein polynomial in this case? it’ll be
$$
X^3 + 2aX^2+2bX+2(1+2c)\,,
$$
where $a$, $b$, and $c$ can be any $2$-adic integers. Notice that the constant term has to be indivisible by any higher power of $2$, so of form $2$ times a unit, and the units of $\Bbb Z_2$ are exactly the things of form $1+2c$. So your parameter space is $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$, pleasingly compact, and a general result of Krasner says that if you jiggle the coefficients a little, the extension doesn’t change. You might be able to use all this to construct your (finitely many) fields.
Not much of an answer, I know, but it was too long for a comment. It’s a nice question, though, and I think I’m going to worry it over a little.
EDIT — Expansion:
I told no lies above, but that’s not the way to look at this problem. As I reached the solution, I realized that there are really two questions here. Consider the simplest case, which you mentioned, the Eisenstein polynomial $X^3-2$. If you think of it abstractly, there’s only the one extension of $\Bbb Q_2$ here, but if you think of the subfields of some algebraically closed containing field, there are three fields, generated by $\lambda$, $\omega\lambda$, and $\omega^2\lambda$, where $\lambda$ is a chosen cube root of $2$ and $\omega$ is a primitive cube root of unity.
As usual, if you take the displayed cubic above and make a substitution $X'=X-2a/3$, you’ll get a new Eisenstein polynomial, but without a quadratic term. Now, if you calculate the discriminant of $X^3+2bX+2(1+2c)$, you’ll get $\Delta=-32b^3-27(1+4c+4c^2)$; and since $c$ and $c^2$ have same parity, we get $\Delta\equiv-3\pmod8$, definitely not a square, indeed $\sqrt\Delta\in\Bbb Q(\omega)$, hardly a surprise, I suppose. And the splitting field of our polynomial will be a cubic extension of $k=\Bbb Q_2(\omega)$, all of which we know. We need only calculate the group $k^*/(k^*)^3$, and its cyclic subgroups (of order $3$) will tell us the cubic extensions of $k$. That’s Kummer Theory, as I’m sure you know.
Let’s call $\Bbb Z_2[\omega]=\mathfrak o$, that’s the ring of integers of $k$.
To know $k^*/(k^*)^3$ we have to look at the groups $1+2\mathfrak o\subset \mathfrak o^*\subset k^*$. Now the principal units $1+2\mathfrak o$ are uniquely $3$-divisible, so no contribution to $k^*/(k^*)^3$; the next layer, $\mathfrak o^*/(1+2\mathfrak o)$ is cyclic of order $3$, generated by $\omega$, and $k^*/\mathfrak o^*$ is infinite cyclic, that’s the value group. So $k^*/(k^*)^3$ is of dimension two as an $\Bbb F_3$-vector space, and has only four one-dimensional subspaces. One is spanned by $\omega$, and its cube roots generate an unramified extension, so is not of interest to us. The other three are spanned by $2$, $2\omega$, and $2\omega^2$. ( ! )
And that’s it. Contrary to my expectation and perhaps yours, the only cubic ramified extensions of $\Bbb Q_2$ within an algebraic closure are the three I mentioned in the first paragraph of this Edit.
If a primitive root of unity $\zeta_n\in F$ and $a^n\in F$ then $F(a)/F$ is separable because $a$ is a root of the separable polynomial $x^n-a^n$, which splits completely in $F(a)$ so $F(a)/F$ is Galois,
It is cyclic because its automorphisms are of the form $\sigma : a\to \zeta_n^{\phi(\sigma)} a$ making $Gal(F(a)/F)$ a subgroup of $\Bbb{Z}/n\Bbb{Z}$.
What Nico said is the converse: every degree $d|n$ cyclic extension of $F$ is of the form $F(c)/F$ with $c^n\in F$, one extension per cyclic subgroup of $F^\times/F^{\times n}$.
Best Answer
Let $K$ be a local field with residue field $k$ and uniformiser $\pi$. Let $F$ be a finite extension with residue field $k_F$ and uniformiser $\pi_F$. An extension $F/K$ can take one of two flavours:
If $\pi\mathcal O_F = (\pi_F)^e$, we say the extension is ramified. It is totally ramified if $e = [F:K]$.
If $\pi\mathcal O_F = (\pi_F)$ is still prime, we say the extension is unramified. In this case, we can choose $\pi_F = \pi$.
In the ramified case, the extension is obtained by adding a new uniformiser of smaller valuation: if we normalise the valuation on $F$ so that $v(\pi_F) = 1$, then $v(\pi) = e$. In the totally ramified case, the residue field $k_F = k$.
In the unramified case, the uniformiser does not change. Instead, the extension is happening on the residue field: $[k_F: k] = [F:K]$.
Now, suppose that $F = K(\alpha)$ where $\alpha$ is a root of the irreducible polynomial $f$. We can always assume that $f$ is monic with integer coefficients. In this case, $\alpha$ is an algebraic integer. Let $\overline \alpha$ be its image in $k_F$. Then $\overline \alpha$ is a root of $f\pmod \pi$. In particular, if $F/K$ is totally ramified, then $k(\overline \alpha) = k$. It follows that $f\pmod \pi$ is completely reducible.
The converse is false in general, as $f$ could be a polynomial like $X^2 + p^2$, and $F(\sqrt{-p^2}) = F(\sqrt{-1})$ is unramified if the residue characteristic is not $2$. The problem here is that the map $\mathcal O_K[\sqrt{-p^2}]\to k_F$ is not surjective: its image is exactly $k$! However, if $\mathcal O_F = \mathcal O_K[\alpha]$, then the map $\mathcal O_K[\alpha]\to k_F$ is surjective. In particular, $k_F = k(\overline \alpha)$. So if $f\pmod \pi$ is completely reducible, then $k_F = k$, so $F$ is totally ramified.