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$\newcommand{\Q}{{\mathbb Q}} $Let $f\in \Q[x]$ be a polynomial, and let $L/\Q$ be the finite Galois extension obtaining by adjoining to $\Q$ all roots of $f$. Magma knows how to compute $\Gamma:={\rm Gal}(L/\Q)$.

I am interested in algorithms and programs that can do the following:

(1) Find a finite set $S\ni\infty$ of places of $\Q$ such that the extension $L/\Q$ is unramified for all primes $p\in\overline S:=V(\Q)\smallsetminus S$ (where $V(\Q)$ denotes the set of all places of $\Q$).

(2) For each cyclic subgroup $\Delta\subseteq G$, somehow determine all primes $p\in\overline S$ with decomposition group $D_p\subseteq \Gamma$ conjugate to $\Delta$.

Example. Let $f(x)=x^2+1$. Then $L=\Q(\sqrt{-1})$ is unramified outside $\infty$ and $2$, and by Fermat's theorem, for $\Delta=\{1\}$ we obtain $p\equiv 1\pmod 4$, while for $\Delta=\Gamma={\rm Gal}(L/\Q)$ we have $p\equiv 3\pmod 4$.

EDIT:

Motivation. I want to describe using computer the Galois cohomology set $H^1(\Q,G)$ where $G$ is a connected reductive group over $\Q$. For this it more or less suffices to compute $H^1(\Q_v,G)$ for all places $v$ of $\Q$, and the Tate-Shafarevich kernel $Ш^1(\Q,G)$. There is a program computing $H^1({\mathbb R},G)$ in my paper with Willem A. de Graaf. The other guys can be computed from the decompositions groups $D_p$ for all primes p. This motivates my question.

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    $\begingroup$ How would you like those sets of primes to be described? If the extension is not abelian they are not going to be given by congruences. $\endgroup$
    – Aurel
    Commented Jun 2 at 20:26
  • $\begingroup$ @Aurel: That's the point! I don't know how to describe those sets of primes. Is there any reasonable way to describe them? $\endgroup$
    – Mikhail Borovoi
    Commented Jun 3 at 8:16
  • $\begingroup$ @Aurel: See my edit (Motivation). $\endgroup$
    – Mikhail Borovoi
    Commented Jun 3 at 8:54
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    $\begingroup$ It sounds like you wish for an algorithm which would also calculate the Tate-Shafarevich group in case $G$ is an elliptic curve, that is known to be very hard (assuming the $G$ in the motivation is different than before). But (1) is clearly ok by taking away prime divisors of the discriminant of a version of $f$ in $\mathbb{Z}[x]$. (2) usually not, but sometimes. E.g if $K = \mathbb{Q}(E[\ell])$ for an elliptic curve $E$ and a prime $\ell$, then the Frobenius at a prime $p$ can be calculated from information of the reduction of $E$ modulo $p$. $\endgroup$
    – Chris Wuthrich
    Commented Jun 3 at 15:35
  • $\begingroup$ @Aurel: I was not sufficiently clear! In my Motivation, the group $G$ is a connected reductive group. $\endgroup$
    – Mikhail Borovoi
    Commented Jun 3 at 16:27

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