# Just how to define the Galois team of the compositum of all square expansions of Q?

This is Problem 1.7 from Gouvea's lecture notes on deformations of Galois representations. Specifically, he asks you to show that it has several subgroups of limited index which are not shut. So below's what I've obtained until now, which might be incorrect.

I can write the compositum as F = Q [√-- 1, √ 2, √ 3, √ 5, √ 7 ...] (can I? ) and afterwards the Galois team G = Gal (F/ Q) is isomorphic to a straight item Π p (Z/ 2Z) where the item is taken control of all tops p, along with p= -1, and also the pth part is created by the conjugation σ p specified by √ p ->-- √ p.

An instance of a subgroup which isn't shut would certainly be the subgroup H containing limited items of conjugations, given that as an example, H has the series σ 2 , σ 2 σ 3 , σ 2 σ 3 σ 5 , σ 2 σ 3 σ 5 σ 7 ... which merges to the automorphism "conjugate every little thing", and also this automorphism is not had in H.

Nonetheless this subgroup is no place near being limited index-- it has the cardinality of the all-natural numbers, whereas G has the cardinality of the reals. The only limited index subgroups I can consider take are of the kind Gal (F/K) where K is a limited expansion of Q, yet certainly these are necessarily all shut. So I presume I've packed up someplace, and also I would certainly be actually happy for any kind of aid ?! In recognize this might appear a little bit "research questiony" yet it's not, it's simply something that's actually badgering me!

0
2019-05-07 00:07:43
Source Share

Let's neglect the Galois team framework : it suffices to take into consideration the boundless item $G = \mathbb{Z}/(2)^\mathbb{N}$ with its profinite geography and also show non - shut subgroups of limited index.

A component of $G$ is normally understood a part of $\mathbb{N}$ : the string $(a_i)$ represents the part $A$ such that $n \in A$ if and also just if $a_n = 1$. With this analysis, the procedure on $G$ simply comes to be the symmetrical distinction of collections.

Currently take into consideration the filter $F'$ of all cofinite parts of $\mathbb{N}$, and also allow $F$ be any kind of ultrafilter having it. I show that we can associate to $F$ a subgroup of index $2$. The $F$ will certainly be non shut given that $F$ is not a major ultrafilter (without a doubt any kind of non - major ultrafilter will certainly do).

Take into consideration the particular function of $F$ : $$f = \chi_F \colon G = \mathcal{P}(\mathbb{N}) \to \mathbb{Z}/(2)$$. I assert that this is a homomorphism, and also its bit is the wanted subgroup.

This is very easy to validate straight. Allow $A, B \in G$ and also think $f(A) = f(B) = 0$. This suggests that neither $A$ neither $B$ come from $F$, so their corresponding collections do. It adheres to that $A^c \cap B^c \in F$, to make sure that $A \cup B \notin F$. A fortiori the symmetrical distinction $A \Delta B \notin F$, whic methods $f(A \Delta B) = 0$. The various other instances are comparable.

0
2019-05-08 21:45:50
Source

The limited index subgroups that you claim that you can consider are specifically the shut (equivalently open) limited index subgroups of $Gal(F/\mathbb{Q})$ relative to the Krull geography on the team. These are actually the only pertinent ones in regards to Galois theory. Yet the team has non - shut subgroups also, yet their presence calls for some non - positive concept such as Zorn's. lemma. These subgroups will not be open or enclosed the Krull geography.

0
2019-05-08 21:43:49
Source

One means to consider the scenario is that the countable item of $\mathbf Z/2\mathbf Z$s that you are taking into consideration is a vector room $V$ over $\mathbf F_2$ that is vast, and also hence has an (a lot more) vast twin room. If $f:V \to \mathbf F_2$ is a non - absolutely no component of the twin room, its bit is an index 2 subgroup of $V$.

Currently the open subgroups that you find out about represent (i.e. are the bits of) the estimates onto the numerous variables in the item, and also these are countable in number. So there is a substantial host of various other index 2 subgroups, representing all the various other non - absolutely no components $f$ of the twin room.

As others have actually kept in mind, you will not have the ability to write these down clearly however.

0
2019-05-08 21:08:02
Source

Here is a very easy factor which is implied in the various other solutions, yet which becomes part of the inquiry (as I read it).

Place $(p_1,p_2,p_3,p_4,\dots):=(-1,2,3,5,\dots)$, and also specify the areas $K_0,K_1,\dots$ by $K_0:=\mathbb Q$ and also $K_n:=K_{n-1}(\sqrt{p_n})$ for $n\ge 1$. [It's clear that any kind of square area is had right into some $K_n$. ] To see to it that the Galois team concerned is the shown one, we require to examine that $\sqrt{p_n}$ is not in $K_{n-1}$. To do that we confirm the adhering to evidently more powerful case.

Allow $n$ be a favorable integer and also $a$ the item of the $p_k^{e(k)}$, where $k$ runs over the integers $\ge n$ and also $e(k)$ is an integer equivalent to $0$ for mostly all $k$. Think $\sqrt a\in K_{n-1}$. After that all the $e(k)$ are also.

This is conveniently confirmed by induction.

EDIT. Points can be defined this way : The noticeable morphism from $\mathbb Q^{\times}/\mathbb Q^{\times 2}$ right into our Galois team generates an isomorphism in between the profinite conclusions. [It's wonderful to get a Galois team by using a straightforward building and construction to the base area. ]

0
2019-05-08 20:58:59
Source