Let $K$ be a number field with ring of integers $O_K$ is not PID. Can we estimate the cardinality of the following sets $$\mathcal{A}= \{\mathcal{P}\subset O_K \ |\ Nm(\mathcal{P})\leq x, \mathcal{P}\ \text{ is not principal}\},$$ $$\mathcal{B}= \{\mathcal{P}\subset O_K \ |\ Nm(\mathcal{P})\leq x, \mathcal{P}\ \text{ is principal}\}.$$ Any hint or comments are welcome. Thanks.
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1$\begingroup$ The title says "non prime" but body asks for principal and nonprincipal ideals. You probably want to edit the title. Is $\mathcal P$ supposed to be prime? It is not explicitly said. $\endgroup$– WojowuCommented Oct 29, 2020 at 11:03
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1$\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$– GH from MOCommented Oct 29, 2020 at 12:56
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I assume that, in your question, $\mathcal{P}$ means a prime ideal of $\mathcal{O}_K$.
It follows from the non-vanishing of Hecke $L$-functions $L(s,\chi)$ at $s=1$ (where $\chi$ is an unramified Hecke character of $K$) that $\#\mathcal{B}$ is asymptotically $\mathrm{li}(x)/h(K)$, and $\#\mathcal{A}$ is asymptotically $(1-1/h(K))\mathrm{li(x)}$. More generally, the prime ideals of $\mathcal{O}_K$ are equidistributed in the ideal class group of $K$. The proof is the same as the proof of Dirichlet's theorem on arithmetic progressions. (Here of course $\mathrm{li}(x)$ can be replaced by $x/\ln(x)$, but $\mathrm{li}(x)$ is a better approximation.)
answered Oct 29, 2020 at 5:23
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$\begingroup$ Thanks, GH. Could you provide some references? $\endgroup$ Commented Oct 29, 2020 at 6:04
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$\begingroup$ @SUNILPAUPULATI: See Section VII.13 in Neukirch: Algebraische Zahlentheorie (or its English translation). $\endgroup$ Commented Oct 29, 2020 at 8:42
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2$\begingroup$ It may be useful to add that this is a special case of the Chebotarev density theorem, applied in the case of the Hilbert class field of $K$. $\endgroup$ Commented Oct 29, 2020 at 8:48
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$\begingroup$ @DanielLoughran: I agree, but the proof of the Chebotarev density theorem is more complicated. I mean, there is more algebraic input. $\endgroup$ Commented Oct 29, 2020 at 9:00