When is norm surjective mod an ideal for global fields?

Question

Given $K/k$ Abelian, for which ideals of $\frak p\unlhd\cal O_k$ will we have $N^K_k:\cal O_K\rightarrow\cal O_k/\frak p$ surjective? $k$ is an algebraic number field.

In his article "On the norm groups of global fields", Stern wrote that the norm function is never surjective for Abelian local extensions of global fields. Can surjectivity still be salvaged if we are looking mod an ideal? In particular, if the ideal is either small enough, or has certain properties (like not being ramified), then this should be possible: especially since the latter group is finite.

I am getting a feeling this is either completely obvious, or else there is some higher-level terminology in which my question should be rephrased to get an answer. Thanks!

Answer 1

Possibly partial answer; at least someone please look over and see if I am missing something:

If $\frak p = \frak p_1^{e_1}\cdots\frak p_r^{e_r}$, then by CRT, surjectivity of $\cal O_K\rightarrow\cal O_k/\frak p$ is equivalent to surjectivity of each of the $\cal O_K\rightarrow\cal O_k/\frak p_i^{e_i}$. So it suffices to show this for $\frak p$ prime power.

Now if $\frak p$ lifts to $\frak (P_1\cdots\frak P_r)^e$ (ramification degree is the same because the extension is Galois), then $N(\frak P_i) = \frak p$ is in the kernel of the above map; so it suffices to assume that $N:\cal O_K/\frak P_1\cdots\frak P_r\rightarrow\cal O_k/\frak p$, and both groups are finite. In such a case, it is a standard argument that the induced norm map ($\frak P_i\mapsto\frak P_1\cdots\frak P_r$) will be surjective. However, if $\frak p$ is ramified, then the $N$ map sends $\frak P_i$ to $(\frak P_1\cdots\frak P_r)^e$, and so it equals the induced norm map ^e. In this case, if ($e, |\cal O_k/\frak p|$) $\neq$ 1 [which I believe happens every time the extension is ramified?], then we only get the set of $e$-th powers in $\cal O_k/\frak p$, and so the map is not surjective.

Tracing backwards, given an ideal $\frak p$ we can figure out whether or not the norm map mod that ideal is surjective.

Answer 2

Denote by $N$ the norm in the finite extension of number fields $K/k$. Because the norm of an integral element is an integral element, $N$ restricts to a map $O_K$ -->$O_k$, still denoted by $N$. If $Q$ is a prime ideal of $O_K$ lying above a prime ideal of $ P$ of $O_k$, $N ( Q) = P^f$ (where $f$ is the inertia degree) is contained in $ P$, hence induces a map $O_K / Q $--> $O_k / P $. But these residue fields are finite , hence the norm is surjective. Finally, your original map $O_K$-->$O_k / P $ is surjective because it factors through $O_K / Q $-->$O_k / P $. Note that we don’t need $K/k$ to be abelian, not even Galois.

If $K / k$ is an abelian extension of local fields, then local CFT says that the norm index $(k^* :N(K^*))$ equals the degree $[K : k]$. For an abelian extension of global fields, see e.g. the question « Index of norm group of a global field », posted by Jack Yoon, May 9'16.