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Jul 1, 2022 at 5:01 comment added Xiao Xiao In that case, we have $K = L = \mathbb{Q}(i) = \mathbb{Q}(\zeta_4)$ so $F = \mathbb{Q}(i)$ and $\text{Gal}(\mathbb{Q}(\zeta_4)/F)$ consists of the identity element of $(\mathbb{Z}/4\mathbb{Z})^{\times} = \{[1],[3]\}$. The identity element is $[1]$. So a congruence class $[a] \in (\mathbb{Z}/4\mathbb{Z})^{\times}$ contains infinitely many primes $p$ that split completely in $\mathbb{Q}(i)$ if and only if $[a]=[1]$. Did I get this?
Jul 1, 2022 at 4:50 comment added KConrad Yes. Do you see how it is consistent with the counterexample Stanley wrote about in his comment?
Jul 1, 2022 at 4:37 comment added Xiao Xiao Just to be sure that I understand the Remark in your answer. A congruence class $[a] \in (\mathbb{Z}/d\mathbb{Z})^{\times}$ contains infinitely many primes $p$ that split completely in $K$ if and only if $[a] \in \text{Gal}(\mathbb{Q}(\zeta_d)/F) \subset \text{Gal}(\mathbb{Q}(\zeta_d)/\mathbb{Q})\cong (\mathbb{Z}/d\mathbb{Z})^{\times}$ where $F = K \cap \mathbb{Q}(\zeta_d)$?
Jun 30, 2022 at 20:59 comment added KConrad Sure, but it is easier to see the ideas at work in (1) and (2) first. Also, (1) and (2) are just what I wrote up first as a reply to the question you asked. Only later did I formulate (3) to unify (1) and (2) into a single result.
Jun 30, 2022 at 18:39 vote accept Xiao Xiao
Jun 30, 2022 at 18:39 comment added Xiao Xiao Thank you very much for the thorough response! If I understand your argument correctly, both Part (1) and (2) are corollaries of Part (3)? I do appreciate the exposition, which makes it easier to understand the obstruction.
Jun 30, 2022 at 15:17 history edited KConrad CC BY-SA 4.0
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Jun 30, 2022 at 4:25 history answered KConrad CC BY-SA 4.0