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## Decomposition of primes in Galois closures of number fields

Let $L/K$ be an extension of number fields, and $M/K$ the Galois closure of $L/K$ (everything happens inside a suitably large characteristic zero field $\Omega$). Let $p$ be a discrete prime of $K$.

Given the group $G={\rm Gal}(M/K)$, its subgroup $H={\rm Gal}(M/L)$, and the splitting of $p$ in $L/K$ can one find the splitting of $p$ in $M/K$? That is, can one find the ramification index $e$ and inertial degree $f$ of a prime $P$ of $M$ lying above $p$?

Turning the above question into a group-theoretic one, I got the following: let $D$ be a finite group, and $X$ a finite set on which $D$ acts faithfully on the right. Can we obtain the order $d$ of $D$ knowing the sizes of all the $D$-orbits in $X$?

(To switch to this question from the original one, look at the natural right action of a decomposition group (resp. an inertia group) $D$ at $P$ on the coset space $X=H\backslash G$. The fact that $M$ is the Galois closure of $L$ ensures that $D$ acts on $X$ faithfully.)

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Changed answer: The answer is no: Let $L$ be a root field of $X^3-2$ over $\mathbb Q$, and $M$ the Galois closure of $L/\mathbb Q$. Then the primes $2$ and $3$ are both totally ramified in $L$, yet in both cases there is only one prime $P$ above $p=2$ or $p=3$, with $e(P)=3$ if $p=2$, and $e(P)=6$ if $p=3$.

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Thanks for your answer. I think I see the way you want to argue. One comment: knowing how $p$ decomposes in $L/K$ is equivalent to knowing the sizes of both the $D$ and the $I$-orbits in $H\backslash G$. I think those for $I$ would be different for $P$ and $P′$ in your counterexample. Right? – Tommaso Centeleghe Feb 13 at 12:51
Well, if you allow the knowledge of the $I$-orbits too, then of course you get the ramification index as the least common multiple of these orbit lengths. Probably you get the size of $D/I$ too. – Peter Mueller Feb 13 at 12:57
I definitely allow the knowledge of the $I$-orbits, too. Since I assume that the decomposition of $p$ in $L/K$ is given, i.e., we are given the $e_i$'s and $f_i$'s. Now, are you saying that the ramification index of $p$ in $M/K$ is the least common multiple of the $e_i$'s? – Tommaso Centeleghe Feb 13 at 13:02
@Tommaso, yes that's correct, your group theory question has a positive answer in the cyclic case, and therefore the ramification index in $M/K$ is the lcm of the $e_i$'s if every $e_i$ is coprime to $p$, and also the inertial degree is the lcm of the $f_i$'s if every $e_i=1$. – Michael Zieve Feb 13 at 14:49
@Tommaso: OK, my fault, I missed that you mentioned inertia groups too. I changed the answer accordingly. – Peter Mueller Feb 13 at 16:53

With regard to the group-theoretic question, the answer is no: for example, there are plenty of faithful transitive group actions of different groups on the same set (i.e. only one orbit, but it doesn't determine the size of the group). For instance, take $S_n$ and $A_n$ acting on ${1,2,\dots, n}$, as soon as $n \geq 3$.

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 Thanks! I was clearly asking too much. – Tommaso Centeleghe Feb 13 at 12:45 I guess for cyclic groups the property above holds. Do you know, more generally, to what kind of groups it applies? – Tommaso Centeleghe Feb 13 at 13:27

The answer is no. Let $K=\mathbb{Q}$ (it does not really matter) and let $M/\mathbb{Q}$ be a degree $6$ extension with Galois group $S_3$; finally, pick for $L$ any of the three cubic subfields of $M$.

By Chebotarev, there are infinitely many primes in $M$ splitting completely in $M/\mathbb{Q}$ and there are infinitely many unramified primes in $M$ whose decomposition subgroup is $\mathrm{Gal}(M/L)$. Then you cannot distinguish them if you only live in $L$ but their splitting behaviors in $M$ are different.

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 Wait: in your example primes $p$ of $Q$ lying below a prime $P$ of $M$ whose associated decomposition group $D_P$ is $Gal(M/L)$ do not split completely in $L$, I think. – Tommaso Centeleghe Feb 13 at 14:53 @Tommaso, I agree. Although $Q$ has $f=2$ in the extension $M/L$, the two other primes in the $G$-orbit of $Q$ have $f=1$ in the extension $M/L$. The prime of $L$ which lies under these two primes will have $f=2$ in $L/K$. – Michael Zieve Feb 13 at 15:01 You are right. I thought you only wanted to control the behaviour of the decomposition. I agree both with Tommaso and Michael. – Filippo Alberto Edoardo Feb 13 at 16:14

Peter's answer made me want to modify the original question by adding the residue characteristic of $p$ to the list of numerical data in the question's hypotheses. Here is an alternative example where the residue field of $p$ is fixed.

Let $f(x)=x^4-4x^2+2$ (so that $f(x+1/x)=x^4+1/x^4$). Then, if $c$ is $4$, $8$, or $10$, adjoining a root of $f(x)-c$ to $\mathbb{Q}$ yields a degree-$4$ extension $L/\mathbb{Q}$ in which the rational prime $p=2$ is totally ramified. Moreover, the Galois closure $M/\mathbb{Q}$ of $L/\mathbb{Q}$ is a degree-$8$ extension whose Galois group is dihedral. Let $q$ be the unique prime of $L$ lying over $2$. If $c=4$ then $q$ is inert in $M/L$; if $c=8$ then $q$ splits completely in $M/L$; and if $c=10$ then $q$ is totally ramified in $M/L$.

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 Nice example, thanks! I guess the question remains of whether the original question has a positive answer in the case where p is known to be tamely ramified. As we already decided, since inertia groups are cyclic in this case, the ramification of p in M/K is lcm of the ramification indeces in L/K. But what about f? – Tommaso Centeleghe Feb 15 at 13:29