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Reading the proof of the main theorem of complex multiplication for elliptic curves over number fields in J. Silverman's book "Advanced topics of elliptic curves" I got stuck at a passage which looks quite innocuous but which I can not provide a reasonable proof about. On page 126, proving Proposition 4.2 the following statement is given. Let $p$ be a prime ideal in $\mathbb{Z}$ which splits completely in the quadratic field extension $K$, so that $pO_{K}=\mathfrak{p}\mathfrak{q}$. Then it is always possible to find an ideal $\mathfrak{a}$ in $O_{K}$, coprime with $pO_{K}$ and such that $\mathfrak{a}\mathfrak{p}$ is principal. I do not see any immediate reason why this should be true? I have found a paper by K. Conrad stating that for any $\mathfrak{a}$ in $O_{K}$ then if $\overline{\mathfrak{a}}$ is the conjugate of $\mathfrak{a}$ (the ideal of all conjugates of the elements of $\mathfrak{a}$), we have that $\mathfrak{a}\overline{\mathfrak{a}}$ is principal. This seems weaker than the statement in J. Silverman's book and still very complicated to prove. Is there any simple proof I am missing?

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    $\begingroup$ Are you sure you've got the question right? $pO_K$ itself is principal, so you could take $\mathfrak a=O_K$. $\endgroup$
    – Wojowu
    Commented Jun 26, 2020 at 18:16
  • $\begingroup$ Thank you I was not meaning $pO_{K}$ but $\mathfrak{p}$ above $p$, I have edited the question now $\endgroup$
    – Hair80
    Commented Jun 26, 2020 at 18:23
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    $\begingroup$ 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 $\mathfrak a$. $\endgroup$
    – Joe Silverman
    Commented Jun 26, 2020 at 18:40
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    $\begingroup$ @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 $\mathfrak q$. $\endgroup$
    – Joe Silverman
    Commented Jun 26, 2020 at 20:03
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    $\begingroup$ 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 $\mathfrak q$. $\endgroup$
    – Wojowu
    Commented Jun 26, 2020 at 22:11

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