Suppose that $L/k$ is a Galois extension of number fields and that $G$ is the corresponding Galois group. Further, for $\frak p$ a prime ideal of $\cal O$$_L$, let $K=L^{G(\frak p)}$, where [$L$ : $K$] = $p$ is prime. Suppose $m$ is the highest integer such that $\mu_{p^m}\subseteq L$, and $\mu_{p^n}\subseteq K$ with $n\geq m-1$ [$\mu_{p^m}$=the group of $p^m$-th roots of unity]. We know that $N_{L/K}(\mu_{p^m})\subseteq\mu_{p^n}$. How do we determine its exact image?

While I only need the more specific case in the paragraph above, I'm pretty sure I read about the (more general) question in the title in some well-known book; and surprisingly, I can't find any references for this on the internet.


The answer should be: if $n = m$, then the image is the roots of unity of order $p^{n-1}$. If $n = m - 1$ then the norm map should be surjective on the roots of unity.

In the first case, since all roots of unity $\zeta$ are already contained in $K$ and $[L:K] = p$, then $N_{L/K}(\zeta) = \zeta^p$, and $\zeta \mapsto \zeta^p$ is surjective onto the index-$p$ subgroup of $p^{n-1}$-order roots of unity.

In the second case, fix $\zeta \in K$, and there is a $\zeta_0$ in $L$ with $\zeta_0^p = \zeta$. Then the minimal polynomial of $\zeta_0$ is $x^p - \zeta = 0$, so the norm of $\zeta_0$ is $\zeta$ (unless $p = 2$, in which case it is $-\zeta$, but we still get surjectivity).

  • $\begingroup$ There is a little more subtlety when $p=2$, take $K=\mathbb Q$, $L= \mathbb Q(i)$, $m=1$, $n=2$, the norm map is not surjective. This can only problems for $n=2$ though as if $n=3$ or higher, $-\zeta$ has the same order as $\zeta$. $\endgroup$ – Will Sawin Mar 28 '16 at 21:48

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