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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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    Have you tried using the hint? What is it suggesting?2011-05-21
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    The answer below is 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, the density of (totally) split primes is $1/[K : \Bbb Q]$. Thus, it forces $K = \Bbb Q$.2018-11-22
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    Perhaps, @Watson, you should post an answer.2018-11-22
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    Dear @MohammadAktar, you may notice that $\Bbb Q(\sqrt 13, \sqrt 17)$ is not isomorphic (as ring) to $\Bbb Q[X] / ((x^2-13)(x^2-17))$, nor to $\Bbb Q[X] / ((x^2-13)(x^2-17)(x^2-221))$ because those are not even _domains_.2018-11-22
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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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