2
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Use that $\text{Gal}(L_k/\mathbb{Q})$ is cyclic and look at the fixed field of the decomposition group.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – user106850
    Commented Jun 16, 2021 at 23:43
  • $\begingroup$ 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. $\endgroup$
    – P. Koymans
    Commented Jun 17, 2021 at 9:42
  • $\begingroup$ I see now, thanks! $\endgroup$
    – user106850
    Commented Jun 17, 2021 at 20:54

You must log in to answer this question.