# Are the abelian absolute Galois groups of these local fields isomorphic?

For a field $F$ we denote by $F^{\mathrm{ab}}$ the compositum of all finite Galois abelian extensions of $F$.

Is $\mathrm{Gal}(\mathbb{Q}_2(\sqrt{3})^{\mathrm{ab}}/\mathbb{Q}_2(\sqrt{3})) \cong \mathrm{Gal}(\mathbb{Q}_2(\sqrt{48})^{\mathrm{ab}}/\mathbb{Q}_2(\sqrt{48})) ?$

Equivalently,

Is every finite abelian group which is a Galois group of some finite Galois extension of $\mathbb{Q}_2(\sqrt{3})$, is also a Galois group of some finite Galois extension of $\mathbb{Q}_2(\sqrt{48})$ (and vice versa) ?

• I don't see that two statements are equivalent. If $G_1$ and $G_2$ are the two Galois groups in the first statement, then the second statement merely says that for any open subgroup $H_1\leq G_1$ there is an open subgroup $H_2\leq G_2$ such that $G_1/H_1\cong G_2/H_2$ and vice versa. For example, the latter property is true if $G_1$ and $G_2$ are factors of each other, but I don't see how this implies that $G_1$ and $G_2$ are isomorphic. Mar 10, 2016 at 16:22
• @GHfromMO It is well known that these absolute (abelian) Galois groups are finitely generated (as profinite groups). Finitely generated profinite groups are isomorphic if and only if they have the same continuous finite images (you can see this in the book of Ribes-Zalesskii on Profinite groups). Mar 10, 2016 at 17:22

Let $K_1$ and $K_2$ be the fields in question. By local class field theory you're asking if $\mathcal{O}_{K_1}^\times\times\widehat{\mathbb{Z}}$ and $\mathcal{O}_{K_2}^\times\times\widehat{\mathbb{Z}}$ are isomorphic (the usual exact sequence splits). Because everything is finitely-generated an equivalent question is whether $\mathcal{O}_{K_1}^\times$ and $\mathcal{O}_{K_2}^\times$ are isomorphic. Now for $K=K_1$ or $K_2$ we have $\mathcal{O}_K^\times$ is a finitely-generated group isomorphic in this case to the product of a finite group of odd order (which you can compute by computing the odd order torsion, which is the same as the torsion in the residue field by a Hensel argument) and the Sylow 2-subgroup of $\mathcal{O}_K^\times$, which is the kernel $1+m_K$ of the reduction map onto the residue field. Finally $1+m_K$ is a finitely-generated $\mathbb{Z}_2$-module so is classified by its torsion subgroup and rank.

For $K=K_1$, $K_2$ both degrees are 8 ($x^8-3$ and $x^8-48$ are both irreducible over $\mathbb{Q}_2$ according to pari) so all you have to do is to check that the torsion in the unit groups are the same and this looks like a much easier question. However I don't know how to do this offhand without more thought. This question should be easy with a computer though -- compute the residue fields and then try and fathom out for which $n$ you have a $2^n$th root of unity in each $K_i$.

• It’s clear that the only torsion in $\Bbb Q_2(3^{1/8})$ is $\{\pm1\}$, ’cause if $i$ were there, $\sqrt{-3}$ would be there too, inducing an unramified quadratic subfield. But $\Bbb Q_2(3^{1/8})$ is certainly totally ramified of degree $8$, the Eisenstein polynomial being $(X+1)^8-3$. I have no insight about $\Bbb Q_2(48^{1/8})$, though. It would help to know a prime element of that field. Mar 12, 2016 at 4:41
• But you're right, I should edit.
– znt
Mar 12, 2016 at 14:18
• It seems to turn out that $\Bbb Q_2(3^{1/8},\sqrt2\,)$, which contains both fields, is totally ramified (of degree $16$). The argument that I gave above therefore applies to both fields: neither has any more $2$-power torsion in the principal units than $\{\pm1\}$, and the residue field is $\Bbb F_2$ for both as well. My argument involves Higher Ramification, and right now I’m so sick of ramified extensions of $\Bbb Q_2$ that it’ll have to wait till tonight for me to write the whole mess up. Mar 12, 2016 at 18:26
• pari-gp says [type "idealfactor(nfinit(x^8-48),2)" into it ] that Q_2(48^{1/8}) is also totally ramified of degree 8.
– znt
Mar 12, 2016 at 19:43
• I think it must be positive. There is no odd order torsion, and for 2-power order we see that both fields contain $\pm1$ and if either one contained $i$ then it would contain $1+i$ which is an $8$th root of 16, meaning that the fields themselves would be isomorphic (this is Lubin's observation). Hence either the fields aren't isomorphic and the only torsion in their units is $\pm1$ or they're isomorphic. I think we're done.
– znt
Mar 12, 2016 at 21:16

Although @znt has gotten the answer through pari, I think it may be instructive to outline my argument.

It all depends on the transition function of Higher Ramification Theory: for a finite extension $K\supset k$ of local fields, $\varphi^K_k$ is a function from $\Bbb R^{\ge0}$ to itself, polygonal and concave, in which, if the (finitely many) vertices are $\{(x_i,y_i)\}$, each $x_i$ is a lower number of a ramification break, and $y_i$ is the corresponding upper number. Perhaps the most useful property of the transition function is that it is functorial: if $L\supset K\supset k$, then $\varphi^L_k=\varphi^K_k\circ\varphi^L_K$.

Although you’ll see below that there are other methods of calculating $\varphi^K_k$, here’s a relatively painless way for totally ramified extensions, if you know a prime element $\pi$ of the integers of the big field. Let $F(X)\in k[X]$ be the minimal $k$-polynomial for $\pi$, and draw the Newton copolygon of $F(X+\pi)$.

The copolygon of $g(X)=\sum_na_nx^n$ is the intersection of all the lower halfplanes $\eta\le n\xi+v(a_n)$, where you may take $v$ to be the (additive) valuation for which $v(k^\times)=\Bbb Z$. One sees that the segments of the copolygon are in one-to-one correspondence with the vertices of the Newton polygon: there’s a segment along the line $\eta=n\xi+v(a_n)$ if and only if $(n,v(a_n))$ is a vertex of the polygon.

And you get the transition function from this process by stretching the boundary of the copolygon horizontally by a factor of $e^K_k$, the ramification index (equal to the degree since our extension is presumed totally ramified). That is, every point $(\xi,\eta)$ on the boundary of the copolygon gets moved to the point $(e^K_k\xi,\eta)$ on the graph of the transition function. The function achieved in this way is shifted one unit up and one to the right from that defined in most standard texts, like Corps Locaux of Serre. In all cases, the infinite segment in the graph has slope $1/e^K_k$, and in the case of wildly ramified extensions, all slopes are powers of $1/p$.

Now calculate the transition functions of $\Bbb Q_2(3^{1/8})$, $\Bbb Q_2(\sqrt3\,)$, and $\Bbb Q_2(\sqrt2\,)$, all as extensions of $\Bbb Q_2$. I won’t go through the details, except to say that $(X+1)^8-3$ and $(X+1)^2-3$ are the Eisenstein polynomials for the first two of the fields above. You find that the transition functions are: $\varphi^{\Bbb Q_2(3^{1/8})}_{\Bbb Q_2}$ has vertices $(2,2)$, $(4,3)$, and $(8,4)$; $\varphi^{\Bbb Q_2(\sqrt3\,)}_{\Bbb Q_2}$ has the single vertex at $(2,2)$, and $\varphi^{\Bbb Q_2(\sqrt3\,)}_{\Bbb Q_2}$ has the single vertex at $(3,3)$.

Even though the transition function of $\Bbb Q_2(\sqrt2,\sqrt3\,)$ over $\Bbb Q_2$ can be calculated directly knowing that a prime element is $\pi=1-(\sqrt3-1)\big/\sqrt2$, you can also get it from the transition functions of the subfields, using functoriality. The upper numbers ($\eta$-values) of the subfields must appear among the upper numbers of the whole extension. Since these numbers are $2$ and $3$, $\varphi^{\Bbb Q_2(\sqrt2,\sqrt3\,)}_{\Bbb Q_2}$ must have the vertices $(2,2)$ and $(4,3)$.

All the above is straightforward and easy. But I wanted the transition function of the extension $\Bbb Q_2(\sqrt2,3^{1/4})$ over $\Bbb Q_2(\sqrt2,\sqrt3\,)$. For that I needed a prime element, and I found $$\beta=\frac{\frac{3^{1/4}-1}{\pi}-1}{\pi^2}-1\,,$$ for which the minimal polynomial over $\Bbb Q_2(\sqrt2,\sqrt3\,)$ is of form $X^2+u_1\pi X+u_2\pi$, for units $u_i$. The upshot is that the transition function has the single vertex $(2,2)$. Compose this function first, then the transition function $\varphi^{\Bbb Q(\sqrt2,\sqrt3\,)}_{\Bbb Q_2}$, and you see that the transition function for the degree-eight extension $\Bbb Q_2(\sqrt2,3^{1/4})$ over $\Bbb Q_2$ has two vertices $(2,2)$ and $(6,3)$. The slope ratio at the lefthand vertex is $4$ rather than $2$.

And that’s enough to get our conclusion that $\Bbb Q_2(\sqrt2,3^{1/8})$ is totally ramified of degree $16$ over $\Bbb Q_2$, since its transition function must have a vertex with height $4$, coming from the subfield $\Bbb Q_2(3^{1/8})$. So the full transition function has vertices $(2,2)$, $(6,3)$, and $(14,4)$.