Group decomposition extension of number fields (and question about normal closure)

Question

Again, I have a question ! Actually, everything begin with this exercise :

Let $L/K$ be a Galois extension of number fields of Galois group $G$, and let $H$ be a subgroup. Let $E = L^H$ the fixed fields induced by $H$. Let $v$ be a place of $K$, and $G^v$ be a decomposition group. Show that $v$ splits completely in $E$ if and only if $H$ contains all the conjugacy of $G^v$ (Cassels Froehlich, exercise 6.2).

So, as we know that the conjugacy of $G^v$ are the $G_w$ for an other $w/v$, we have to show that $G_w \subset H$. This is what I want to do :

Necessary condition : Let's suppose $v$ splits completely in $E$. Let $\sigma \in D_w$, for a certain $w/v$ on $L$, $w \cap E = w_{E}$ divising $v$ on $E$. Then (*) : $\sigma_{| E} \in D_{w_E} = {Id_E}$ as $v$ splits completely in $E$. So, $\sigma \in Gal(L/E)=H$.

Sufficient condition : Suppose that $H$ contains all the $G_w$ for $w / v$ in $L$. Let $w_E$ a place which divide $v$ in $E$. Then, let $w$ be a place of $L$ dividing $w_E$, and so dividing $v$. We have $D_w \subset H$. Let $s \in D_{w_E}$. We can find $\sigma \in Gal(L/K)$ such that : $s = \sigma_E$. We have $\sigma(w) = w'$. As $w$ and $w'$ divides $w_E$, there is a $g \in Gal(L/E)$ such that $f(w') = w$ by transitivity of the action of $Gal(L/E)$. So : $f \circ \sigma \in D_w \subset H$. But as $f \in H$, $\sigma$ is in $H$ aswell. Then, $s= \sigma_{|E} = Id_E$, and then $D_{w_E} = {Id_E}$.

But I have a problem. I used during all the proof that $v$ splits completely in $E$ if and only if $D_{w_E} = \{Id_E\}$. The problem is that in my definition of $D_{w_E}$, we have $E/K$ a Galois extension ($D_{w_E} \subset Gal(E/K)$), which could be unverified here.

1) So, is there a definition of group decomposition of an non Galois extension ?

2) Trying to fix my problem, I wanted to go in an normal closure $\hat{E}$ of $E$ on $K$. But, if $v$ is ramified/unramified/inert/splits completely/etc... in $\hat{E}$, is it the same over $E$ ? A good solution would be that the integer rings of $\hat{E}$ and $E$ on $K$ are the same. But is it true ? (I didn't succeed to prove it) At least, we would like $Spec(\hat{E}) = Spec(E)$ (it has to be understood as the Spec of the integer rings), but again, is it really the case ?

Thank you !

Answer 1

$L/K$ is Galois, $E=L^H$, let $p$ be a prime of $O_K$, it factorizes as $$pO_E=\prod P^{e(P|p)},\qquad f(P|p)=[O_E/P : O_K/p]=|Gal((O_E/P)/(O_K/p))|$$ $$\sum e(P|p)f(P|p)=[E:K]$$ $p$ splits completely in $O_E$ iff there are $[E:K]$ primes in the product iff all the $e(P|p)=1,f(P|p)=1$.

• Fix one of those primes $P$ of $O_E$, as well as a prime $Q$ of $O_L$ above $P$.

  $D(Q|p)$ is the subgroup of $Gal(L/K)$ sending $Q$ to itself and $I(Q|p)$ is the subgroup of those acting trivially on the residue field.

  The multiplicity of $Q$ in the prime ideal factorization of $pO_L$ is $e(Q|p)=|I(Q|p)|$, the multiplicity of $Q $ in $P O_L$ is $e(Q|P)=|I(Q|P)|=|I(Q|p)\cap H|$.

  Note that $e(Q|p)=e(P|p)e(Q|P)$.

  Thus, $e(P|p)=1$ means $e(Q|p)=e(Q|P)$ ie. $I(Q|P)=I(Q|p)$ and $I(Q|p)\subset H$.

• If it is the case, since $Gal((O_L/Q)/(O_K/p))=D(Q|p)/I(Q|p)$ then $$Gal((O_L/Q)/(O_E/P))=D(Q|P)/I(Q|P)= (D(Q|p)\cap H)/I(Q|p)$$

• That $f(P|p)=1$ means $f(Q|P)=f(Q|p)$ so that $$|Gal((O_L/Q)/(O_E/P))|=|Gal((O_L/Q)/(O_K/p))|$$ that is $D(Q|p)/I(Q|p)=(D(Q|p)\cap H)/I(Q|p)$

And hence $D(Q|p)\subset H$.

Since $L/K$ is Galois, the other primes above $p$ are of the form $\sigma(Q)$,

We get that for all prime $P'$ of $O_E$ above $p$, $\ e(P'|p)=1,f(P'|p)=1$

iff for all $\sigma\in Gal(L/K)$, $D(\sigma(Q)|p)=\sigma D(Q|p)\sigma^{-1}\subset H$