Let $\mathcal O$ be an order in an imaginary quadratic field $K$.

  1. Does there exists an element $\lambda\in \mathcal O$ such that the norm $N(\lambda)$ is not a square?

  2. Does there exists an element $\lambda\in \mathcal O$ such that the norm $N(\lambda)$ is squarefree and not equal to $1$?

  3. Is there an elementary solution for 1. and 2.?

Note that if there was a simple proof of the first statement, then we could perhaps simplify the proof of the integrality of the $j$-invariant at $CM$ points by avoiding reduction to the case of the maximal order.

  • 2
    $\begingroup$ Part 1 is easy; e.g. if the order contains $\sqrt{-D}$ then $n+\sqrt{-D}$ has norm $n^2+D$, which is not a square once $2n+1>D$. Part 2 is too easy as stated because you can take $\lambda = 1$, so I guess you meant to impose some additional constraint. $\endgroup$ – Noam D. Elkies May 29 '19 at 21:06
  • $\begingroup$ @NoamD.Elkies, Does every order contain a number of the form $\sqrt{-D}$? $\endgroup$ – Shimrod May 29 '19 at 21:14
  • 2
    $\begingroup$ Yes; for instance, the full ring of integers $O_K$ contains some $\sqrt{-D_0}$, and $\cal O$ is contained in $O_K$ with finite index, say index $m$, so $\cal O$ contains $m \sqrt{-D_0}$, and we may take $D = m^2 D_0$. $\endgroup$ – Noam D. Elkies May 29 '19 at 22:27

Yes to both questions.

Let $C$ be the class group of the order $\mathcal{O}$. Then there exists infinitely many prime ideals of degree $1$ in $K$, invertible in $\mathcal{O}$, that represent the trivial class in $C$. If $\mathfrak{p}$ is such a prime ideal, then its norm is a prime integer $p$, and there exists $\lambda\in\mathcal{O}$ such that $\lambda\mathcal{O}=\mathfrak{p}$: in particular $N(\lambda)=p$.

  • $\begingroup$ Is there a more elementary proof? $\endgroup$ – Shimrod May 29 '19 at 20:26

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