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I just finished the treatment of quadratic fields and cyclotomic fields in Marcus' Number Fields and I decided to approach biquadratic fields as a fun exercise. From the Ram-Rel identity we know that any decomposition of a prime $p$ of $\mathbb{Z}$ in $K = \mathbb{Q}[\sqrt{m}, \sqrt{n}]$ ($m,n$ squarefree) will decompose in the following fashion (noting $\mathcal{O}_K$ for the ring of integers of $K$ and the fact that $[K:\mathbb{Q}] = 4$):

$$ p\mathcal{O}_K = \begin{cases} P_1P_2P_3P_4 &f(P_i \mid p) = 1\\ P_1^2P_2P_3 &f(P_i \mid p) = 1\\ P_1^3P_2 &f(P_i \mid p) = 1\\ P_1^4 &f(P_1 \mid p) = 1\\ P_1P_2P_3 & f(P_1 \mid p) = 2\\ P_1P_2 & f(P_1 \mid p) = f(P_2 \mid p) = 2\\ P_1P_2 & f(P_1 \mid p) = 3, f(P_2 \mid p) = 1\\ P_1^2 &f(P_1 \mid p) = 2\\ P_1^2P_2 &f(P_2 \mid p) = 2\\ P_1 &f(P_1 \mid p) = 4\\ \end{cases} $$

Please let me know if I've missed any cases. Now, on to my question:

Is there of prime $p$ of $\mathbb{Z}$ that decomposes for every possible case above? In other word, is there a case above that never happens and can be discarded?

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    $\begingroup$ $K/\mathbb Q$ is Galois. You know more than just $\sum e_if_i=n$ in this case (see Theorem 23 and its corollary in Markus' book). $\endgroup$ Commented yesterday
  • $\begingroup$ Oh that's right, totally forgot about that! $\endgroup$ Commented yesterday
  • $\begingroup$ You may also be interested in Chebotarev density theorem. $\endgroup$ Commented 21 hours ago
  • $\begingroup$ @dummy I see, so the general problem posed in the question is still unsolved. I appreciate the reference! $\endgroup$ Commented 19 hours ago
  • $\begingroup$ I'm not too sure what reference you were using to get the "question still unsolved" conclusion. I personally don't know how to answer the question in its full generality, but let me write a remark on how you can think about the question in general which may help you rule out some cases. $\endgroup$ Commented 7 hours ago

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Here is a general approach to the problem. For a given Galois extension $K$ over $\mathbb{Q}$ of degree $n$,

  • You know that $efg = n$ (not just $\sum_i e_if_i = n$), so a lot of the possibilities are already ruled out.
  • All but finitely many primes satisfy $e = 1$. The primes that ramifies would divide the discriminant $disc(K/\mathbb{Q})$. For a given extension, you can compute the discriminant, then look at the prime factors of the discriminant, and factorize those primes in $K$. This would recover the cases where $e > 1$.
  • For the remaining cases where $e = 1$, primes involved are unramified, and $fg = n$.
    • The $f$ here can be seen as the order of Frobenius (which generates a cyclic subgroup of the Galois group $G$), and thus have group theoretic constraints. For example, if the Galois group $$G \cong \mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$$ then no element can have order 4, hence $f$ can never be 4.
    • For $f$ that doesn't contradict group-theoretic constraints (i.e. there actually exists an element in $G$ with order $f$), Chebotarev density theorem tells you that there are infinitely many primes $p$ with $f_p = f$ (in fact, you can even compute the asymptotic density).
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  • $\begingroup$ Thank you! That does help a lot. I guess it becomes way less convenient once we start looking at extension of $\mathbb{Q}$ that are not normal $\endgroup$ Commented 4 hours ago
  • $\begingroup$ The general strategy in this answer would still work in non-Galois case. See e.g. mathoverflow.net/questions/393531/… $\endgroup$ Commented 3 hours ago

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