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I know that if $K/\mathbb Q$ is a finite Galois extension (i.e. a Galois number field), then for any prime $(p)\subseteq \mathbb Z$, the Galois group $G=\operatorname{Gal}(K/\mathbb Q)$ acts transitively on the finite set of primes of $\mathcal O_K$ lying over $p$. I was wondering if something similar is true for the infinite case $K=\overline{\mathbb{Q}}$, but the proof I know for the finite case does not generalize to the infinite case at all, because it asks us to use the CRT on the set of primes (whose finiteness is established by the unique factorization in the Dedekind domain $\mathcal O_K$). The Going-Up Lemma tells us that if $(p)\subseteq \mathbb Z$ is a prime, then there is at least one prime $\mathfrak p\subseteq \overline{\mathbb{Z}}$ lying over it. However, since $\overline{\mathbb{Z}}$ is no longer a Dedekind domain, this proof does not apply. Do we know if there are only finitely many in this case too, or can there be infinitely many such primes? Do we know if the absolute Galois group $G_\mathbb{Q}$ acts transitively on this set of primes too?

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    $\begingroup$ Not only there can be, there necessarily will be. In fact I believe the set of primes lying over any prine $p\in\mathbb Z$ will be uncountable, and I think $G_{\mathbb Q}$ does act transitively on it. To prove it, note that for any two such primes $\mathbb p_1,\mathbb p_2$, the set of automorphisms taking $K\cap\mathbb p_1$ to $K\cap\mathbb p_2$ form an inverse system as we go over all Galois extensions $K/\mathbb Q$, and the inverse limit is hence nonempty. $\endgroup$ – Wojowu Jul 7 at 13:35
  • $\begingroup$ @Wojowu, can you point me to a source for your first assertion? $\endgroup$ – Gaurav Goel Jul 7 at 13:49
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There are infinitely many prime ideals $\mathfrak p$ in $\overline{\mathbf Z}$ that lie over $p$ since you can find an arbitrarily large (finite) number of prime ideals lying over $p$ in suitable number fields. I'll show how to do this using composites of quadratic fields. For each $n \geq 1$ there are $n$ primes $q_1, \ldots, q_n$ such that $p$ splits completely in $\mathbf Q(\sqrt{q_i})$ by using $q_i \equiv 1 \bmod 4p$ for all $i$. When a prime splits completely in number fields $K_1, \ldots, K_n$, it splits completely in the composite field $K_1\cdots K_n$, and the composite of the fields $\mathbf Q(\sqrt{q_i})$ has degree $2^n$ over $\mathbf Q$ because those quadratic fields are linearly disjoint, either since $q_1, \ldots, q_n$ are multiplicatively independent mod squares (Kummer theory proof), or since the discriminants of those quadratic fields are $q_1, \ldots, q_n$ and these are pairwise relatively prime (not Kummer theory proof). Thus $p$ has $2^n$ distinct primes lying over it in the number field $\mathbf Q(\sqrt{q_1}, \ldots, \sqrt{q_n})$. For each prime over $p$ in $\mathbf Q(\sqrt{q_1},\ldots,\sqrt{q_n})$, pick a prime ideal lying over it in $\overline{\mathbf Z}$ and that gives you $2^n$ different primes over $p$ in $\overline{\mathbf Z}$.

The transitivity of the action of $G_\mathbf Q$ on the primes over $p$ in $\overline{\mathbf Z}$ is proved in the appendix on infinite Galois theory in Washington's book on cyclotomic fields. The proof uses transitivity of the Galois action on number fields plus compactness of $G_\mathbf Q$, as mentioned in Wojowu's comment above.

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  • $\begingroup$ Thank you for the excellent answer! That was very helpful! $\endgroup$ – Gaurav Goel Jul 7 at 15:58

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