# Roots of unity in fixed field of decomposition group.

$\zeta_q\in L^{G({\frak P})}\Leftrightarrow$ if $\sigma\in$ $G(\frak p$) then $\sigma(\zeta_q)=\zeta_q\Leftrightarrow$ if $\sigma$ fixes $\frak P$ then $\sigma$ fixes $\zeta_q$. What's next?

Suppose $L/k$ is a Galois extension with Galois group $G$. For $\frak P$ an ideal of $\cal O$$_L, the decomposition group G(\frak P) is the set {\sigma\in G|\sigma(\frak P)=\frak P}. I am looking for roots of unity that appear in L^{G({\frak P})}. Let \zeta_q\in L denote the q-th primitive root of unity (p either prime, or the power of a prime, not necessarily the one that lies over \frak P). Then the above chain of equivalences is true. What's next? I know that does not imply \zeta_q\in\frak P. Does that tell us anything about \cal O$$_L/\frak P$? I am having trouble making sense of it. One-way implications are also sufficient.

• I'm slightly confused by your notation. Is $p$ related to $\mathfrak P$ in any way? – Mathmo123 Apr 1 '16 at 17:56
• No, it is not. I'll change it now, thanks. – Alex Apr 1 '16 at 17:58
• Do you already know that $\zeta_q\in L$ and you want to check that it is in $L^G$? – Mathmo123 Apr 1 '16 at 18:09
• That is correct, thanks. – Alex Apr 1 '16 at 18:12
• What kind of condition are you after. For example, these conditions are equivalent to $\mathfrak P\cap k$ not splitting at all in $k(\zeta_q)$ - i.e. there is exactly one prime in $k(\zeta_q)$ above $\mathfrak P\cap k$. – Mathmo123 Apr 1 '16 at 18:18