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Dear all, I would be grateful if someone could provide a solution to the following problem (using decomposition and inertia groups):

Find a finite extension of $\mathbb{Q}$ in which all primes split.

[Hint: Use the fact that a prime splits if and only if its decomposition group is not the full Galois group (and that the decomposition group is cyclic for all unramified primes)]

Many thanks, Mohammad.

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    Typically, a monic irreducible polynomial $f \in \Bbb Z[X]$ which has root mod $p$ (i.e. $p$ splits in the splitting field of $f$) for every $p$ must be linear (consequence of Cebotarev). Irreducibility is necessary, as $(x^2-2)(x^2-3)(x^2-6)$ shows.2018-12-02

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It may be possible to do this without using the hint. The polynomial $(x^2-13)(x^2-17)(x^2-221)$ has a root mod $p$ for all $p$; does that imply that every $p$ splits in ${\bf Q}(\sqrt{13},\sqrt{17})$? Or do I have my wires crossed?

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    Your answer is indeed wrong. 1) To check that $p$ splits in a number field $K = \Bbb Q[X] / (f(X))$ (where $f$ is irreducible over $\Bbb Q$), we have to look at the decomposition of the polynomial $f$ modulo $p$. Your polynomial is _not_ the polynomial $f$ that is needed for this. 2) Let $K/\Bbb Q$ be a finite extension such that every rational prime splits totally. From [this](https://math.stackexchange.com/questions/189986/), we may assume that $K/\Bbb Q$ is Galois. Then, by Cebotarev, it follows that the density of (totally) split primes is $1/[K : \Bbb Q]$. Thus, it forces $K = \Bbb Q$.2018-11-22