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Jun 27, 2020 at 13:10 history edited GH from MO
edited tags
Jun 26, 2020 at 22:51 history edited YCor CC BY-SA 4.0
made title more neutral, formatting
Jun 26, 2020 at 22:11 comment added Wojowu Let $a\in O_K$ be such that $a\in\mathfrak p,a\not\in\mathfrak p^2,a\not\in\mathfrak q$ (CRT). Then $(a)\subseteq\mathfrak p$, so $(a)=\mathfrak a\mathfrak p$ for some $\mathfrak a$. The choice of $a$ guarantees $\mathfrak a$ is not divisible by $\mathfrak p$ or $\mathfrak q$.
Jun 26, 2020 at 20:03 comment added Joe Silverman @NoamD.Elkies I agree, there probably is an easier way. But for a book at this level, I think it's fair to use results from a first course in algbraic number theory. Also, what I wrote isn't quite right, one should take a prime ideal in the ideal class of $\mathfrak p^{-1}$ other than $\mathfrak p$ and $\mathfrak q$.
Jun 26, 2020 at 19:09 comment added Noam D. Elkies It looks even easier because there was no requirement that 𝔞 be prime.
Jun 26, 2020 at 18:40 comment added Joe Silverman Quickest, although maybe overkill, consider the ideal class of $\mathfrak p^{-1}$, Every ideal class contains infinitely many prime ideals, take any of them (other than $\mathfrak p$ itself if $\mathfrak p^2$ is principal) for $\mathfrak a$.
Jun 26, 2020 at 18:23 comment added Hair80 Thank you I was not meaning $pO_{K}$ but $\mathfrak{p}$ above $p$, I have edited the question now
Jun 26, 2020 at 18:23 history edited Hair80 CC BY-SA 4.0
reference to wrong ideal corrected
Jun 26, 2020 at 18:16 comment added Wojowu Are you sure you've got the question right? $pO_K$ itself is principal, so you could take $\mathfrak a=O_K$.
Jun 26, 2020 at 18:03 history asked Hair80 CC BY-SA 4.0