# How can I prove this claim about splitting of prime ideals in real cyclotomic fields?

Let $$L_k = \mathbb{Q}(\zeta_{2^k} + \zeta_{2^k}^{-1})$$ be the maximal real subfield of the cyclotomic field of conductor $$2^k, k \ge 2$$ and $$f_k(x)$$ be the minimal polynomial of $$\zeta_{2^k} + \zeta_{2^k}^{-1}$$.

Define $$L = L_{k+1}, K = L_{k}$$ so $$L/K$$ has degree 2. Assume a prime ideal $$\mathfrak{p} \subset \mathcal{O}_K$$ above $$p$$ splits in $$\mathcal{O}_L$$ as $$\mathfrak{p}\mathcal{O}_L = \mathfrak{p}_1 \mathfrak{p}_2$$. Then I suspect that $$p\mathcal{O}_L$$ totally splits (in which case $$p\mathcal{O}_K$$ totally splits as well). How can I prove this?

I believe this is equivalent to showing that if $$f_{k+1}(x)$$ has 2 roots over $$\mathbb{F}_p$$ then it factors completely over $$\mathbb{F}_p$$. I'm not sure how that can be shown. Are there any general approaches to take here?

Use that $$\text{Gal}(L_k/\mathbb{Q})$$ is cyclic and look at the fixed field of the decomposition group.
• 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 23:43
• 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 17, 2021 at 9:42