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Jun 17, 2021 at 20:54 comment added user106850 I see now, thanks!
Jun 17, 2021 at 20:53 vote accept CommunityBot
Jun 17, 2021 at 9:42 comment added P. Koymans Look at the decomposition group of the rational prime $p$ in $\text{Gal}(L_k/\mathbb{Q})$ not in $\text{Gal}(L_k/L_{k - 1})$. If the fixed field of the decomposition group is $E$, then there is no more splitting in the extension $L_k/E$. But we know that there is splitting in $L_k/L_{k - 1}$ by assumption, hence $E = L_k$ and the decomposition group is trivial, so $p$ splits completely.
Jun 16, 2021 at 23:43 comment added user106850 Can you elaborate on this? If $p\mathcal{O}_K$ totally splits then I think we should have trivial decomposition group $D_\mathfrak{p}$ with fixed field = $K$, but I don't see how to use this.
Jun 16, 2021 at 23:19 history answered P. Koymans CC BY-SA 4.0