# Norms of elements in a quadratic order - can you do it elementarily?

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.

• 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. – Noam D. Elkies May 29 '19 at 21:06
• @NoamD.Elkies, Does every order contain a number of the form $\sqrt{-D}$? – Shimrod May 29 '19 at 21:14
• 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 $D = m^2 D_0$. – Noam D. Elkies May 29 '19 at 22:27

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 $$N(\lambda)=p$$.