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.
Your two polynomials are both biquadratic, so you can determine their Galois groups simply by the classification of biquadratic extensions, without any local field theory:
For $h = (x^2 - 1)^2 + 3$, we have $(a,b,c)=(1,-3,4)$. Here $b$ is not a square in $\Bbb Q_2$, and $c$ is a square, and $2(a \pm \sqrt c) = \{6, -2\}$ are not squares in $\Bbb Q_2$, so the Galois group is $C_2 \times C_2$, and the splitting field is $\Bbb Q_2(\sqrt{-2}, \sqrt{-3})$.
For $h_2 = (x^2 + \frac32)^2 - \frac54$, we have $(a,b,c) = (-\frac32, \frac54, 1)$. Again $b$ is not a square, but $c$ is a square, and $2(a\pm\sqrt c) = \{-1,-5\}$ are not squares, so the splitting field is $\Bbb Q_2(\sqrt5, \sqrt{-1})$, and the Galois group is $C_2 \times C_2$.
Since these fields are made from quadratic extensions of $\Bbb Q_2$, we use Kummer theory to study them.
By Kummer theory, quadratic extensions of $\Bbb Q_2$ corresponds to non-trivial elements of $\Bbb Q_2^\times / \Bbb Q_2^{\times 2}$, which are represented by $\{2, 3, 5, 6, 7, 10, 14\}$. Here the number $d$ corresponds to the quadratic extension $\Bbb Q_2(\sqrt d)$.
The unique unramified extension among them is $\Bbb Q_2(\mu_3) = \Bbb Q_2(\sqrt{-3}) = \Bbb Q_2(\sqrt{5})$.
Therefore, $e_h = f_h = e_{h_2} = f_{h_2} = 2$.
Appendix: Classification of biquadratic polynomials
Let $K$ be a field of characteristic $\ne 2$, and let $L$ be the splitting field of $(x^2 - a)^2 - b$ with Galois group $G$. Also, let $c = a^2 - b$. Assume that $b$ is not a square in $K$.
- If $bc$ and $c$ are not squares, then $G = D_8$.
- If $bc$ is a square (then $c$ is not), then $G = C_4$ and $L = K(\sqrt{a+\sqrt b})$.
- If $c$ is a square (then $bc$ is not), and one of $2(a \pm \sqrt c)$ is square, then $G = C_2$ with $L = K(\sqrt b)$.
- If $c$ is a square (then $bc$ is not), and none of $2(a \pm \sqrt c)$ is square, then $G = C_2 \times C_2$ with $L = K(\sqrt b, \sqrt{2(a + \sqrt c})$.
Appendix: a trick
For $h$, let $y = -2x^2$, so $y^2 + y + 1 = 0$, so $y$ is a primitive cube root of unity $\omega$. This is the unique unramified quadratic extension of $\Bbb Q_2$. Now $\omega = (\omega^2)^2$ is a square, but $-2$ is not, so it remains to adjoin the square root of $-2$, which gives a ramified quadratic extension.
Therefore $e_h = f_h = 2$.
Best Answer
Since @PaulGarrett has given me the prompt, I’ll say something about higher ramification. This is a way of examining how close to each other the various roots $\lambda,\lambda'$ of the defining equation $f=0$ may be, when you choose $f$ so that one root $\lambda$ generates the ring of integers of the extension field $K$, $\mathcal O_K=\Bbb Z_p[\lambda]$. In the case of an unramified extension, as I recall, the differences between two roots will all be units. It’s the totally ramified extensions where higher ramification theory comes into its own.
Then let $\lambda$ be a uniformizing element of $\mathcal O_K$, $K$ totally ramified over $k$, with $v_k(\lambda)=1/e$ where $e$ is the ramification index (and the degree of the extension). We take $v_k$ normalized so that $v_k(k^\times)=\Bbb Z$. And let $f(x)$ be the minimal polynomial for $\lambda$, necessarily an Eisenstein polynomial over $\mathcal O_k$. Then put $g(x)=f(x+\lambda)$, whose roots are the quantities $\lambda'-\lambda$. We want a good way of measuring the size(s) of the roots of $g$.
Well, you measure the size of the roots of a polynomial with the Newton Polygon, but if you encode the same information in a kind of dual convex body that’s sometimes called the Newton Copolygon, you get something a little more useful. You can stretch this out horizontally by a factor of $e$ and consider the boundary to be the graph of a function, and thereby get the Hassse-Herbrand Transition Function $\psi^K_k$, which also has the wonderful property of being functorial: if $L\supset K\supset k$ is a chain of totally ramified extensions, then $\psi^L_k=\psi^K_k\circ\psi^L_K$.
Note that there are no Galois groups in this story. We can apply it to the two extensions mentioned at the beginning, $\Bbb Q_3(\zeta_9)$ and $\Bbb Q_3((-3)^{1/6})$. The two uniformizers are, respectively, $\tau=\zeta_9-1$ and $\pi=(-3)^{1/6}$. The minimal polynomial for $\zeta_9$ is $\frac{x^9-1}{x^3-1}$, so the minimal polynomial for $\tau$ is $f_1(x)=\frac{(x+1)^9-1}{(x+1)^3-1}$. The numerator and denominator have no constant, while the first-degree coefficient upstairs is $9$, downstairs is $3$, thus the quotient, known to be a polynomial, has constant coefficient $9/3=3$. Looking at the quotient modulo $3$, you see that there, the quotient is $x^6$. Thus $f_1$ is Eisenstein, which we already knew, didn’t we? Forming $g_1(x)=f_1(x+\tau)$, as above, we need to look closely at $g_1$ and its Newton Polygon (and Copolygon). It’s somewhat of a pain to do the computation this way, but you find that the $v_p$-value of the first coefficient is $3/2$, the value of the third is $1/2$ and of the sixth is zero. This gives vertices of the Polygon at $(1,3/2)$, $(3,1/2)$, and $(6,0)$. The Copolygon (in the Cartesian $(\xi,\eta)$-plane) is described by the corresponding inequalities $\eta\le\xi+3/2$, $\eta\le3\xi+1/2$, and $\eta\le6\xi$. The vertices are where two half-planes’ boundaries intersect, namely at $(1/6,1)$ and $(1/2,2)$. Stretch by a factor of six to get the points $(1,1)$ and $(3,2)$ (the “breaks” in traditional terminology), exactly what you’ll find in Serre’s Corps Locaux, except that my coordinatization of the Cartesian plane is off from Serre’s by one in each direction.
If you do the corresponding computation for $\sqrt[6]{-3}$, you’ll see that it’s much less messy, giving $g_2(x)=(x+\pi)^6+3$ with the Polygon’s vertices at $(1,11/6)$, $(3,1/2)$ and $(6,0)$. The Copolygon is defined by the corresponding inequalities $\eta\le\xi+11/6$, $\eta\le3\xi+1/2$, and $\eta\le6\xi$. Here, the half-planes’ boundaries intersect at $(1/6,1)$ and $(2/3,5/2)$. Stretching by six gives the vertices of the polygonal Hasse-Herbrand function at $(1,1)$ and $(4,5/2)$. Here, the “breaks” are not at integer points, which is fine, because the Galois group of the normal closure of our field is not abelian.