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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
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. $\endgroup$
    – wyoumans
    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
    Jun 17, 2021 at 9:42
  • $\begingroup$ I see now, thanks! $\endgroup$
    – wyoumans
    Jun 17, 2021 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.