$\zeta_q\in L^{G({\frak P})}\Leftrightarrow$ 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.

  • $\begingroup$ I'm slightly confused by your notation. Is $p$ related to $\mathfrak P$ in any way? $\endgroup$ – Mathmo123 Apr 1 '16 at 17:56
  • $\begingroup$ No, it is not. I'll change it now, thanks. $\endgroup$ – Alex Apr 1 '16 at 17:58
  • $\begingroup$ Do you already know that $\zeta_q\in L$ and you want to check that it is in $L^G$? $\endgroup$ – Mathmo123 Apr 1 '16 at 18:09
  • $\begingroup$ That is correct, thanks. $\endgroup$ – Alex Apr 1 '16 at 18:12
  • $\begingroup$ 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 $\mathfrak P\cap k$. $\endgroup$ – Mathmo123 Apr 1 '16 at 18:18

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