Let $K=\mathbb{Q}(\sqrt{D})$ be an imaginary quadratic field of discriminant $D<0$. I want to know the image of the norm map $$ N^K_{\mathbb{Q}}:\mathcal{O}_K\to\mathbb{Z} $$ and the values of $N^K_{\mathbb{Q}}(\mathcal{O}_K)$ modulo $D$, as explicitly as possible. For example, if $D=-4$ then $N^K_{\mathbb{Q}}(\mathcal{O}_K)$ are those non-negative integers $n$ such that if $p^k||n$ and $p\equiv3(4)$, then $k$ is even, and the values of the norm modulo $4$ are $\{0,1,2\}$.

Are there similar formulations (congruence conditions, etc.) for all values of $D<0$? I'm especially interested in the values of the norm modulo $D$ as this condition occurs in something I'm working on.

  • 1
    $\begingroup$ Undoubtedly you know that the image $N^K_{\Bbb{Q}}(K^*)$ is generated by the rational primes contained in it (and the squares of the others) iff $h=1$.The split (and ramified) primes occur as simple factors, but not independently from each other when $h>1$. IIRC the case $D=-20$ is already illuminating. $N(1+\sqrt{-5})=6$, but neither $2$ nor $3$ are in the image. Similarly $N(1+2\sqrt{-5})=3\cdot7$ and $N(3+\sqrt{-5})=2\cdot7$, so any integer $2^a3^b7^c$ with $a+b+c\equiv0\pmod2$ will appear. The primes splitting to principal prime ideals OTOH appear alone: $N(6+\sqrt{-5})=41$. $\endgroup$ – Jyrki Lahtonen Jun 3 '16 at 22:48
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    $\begingroup$ If we bring the problem back to representing rational primes as values of norm maps, not only from rings of integers, but from orders in imaginary quadratic fields, say p = x^2 + ny^2, then the complete answer is known, see e.g. D. Cox's wonderful book bearing approximatively the same title, 2nd edition. I myself only discovered it recently, thanks to an answer posted by Will Jagy. The enlightening introduction points out in particular the necessity to appeal to CFT and the theory of complex multiplication. $\endgroup$ – nguyen quang do Jun 6 '16 at 6:24

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