This is from Silverman's 'the arithmetic of elliptic curves', exercise 5.5.


Let $K$ be an imaginary quadratic field, and let $R\_1...R\_n$ be orders in $K$.
I would like to prove that there are more than 2 prime numbers $p$ such that $pR\_i$ is a prime ideal of $pR\_i$ for all $i$.


By using this, we can characterize supersingular elliptic curve as $End(E)$ is an order in a quaternion algebra.
Any reference is also appreciated. Thank you in advance.



Common prime of the finite number of order of imaginary quadratic field
If $R$ is a (nonmaximal) order in $O\_K$, then it has finite index in $O\_K$. If $p$ is a prime not dividing this index indices, the obvious map $R\to O\_K/pO\_K$ is surjective and its kernel is $pR$, so $R/pR\cong O\_K/pO\_K$. Hence if $p$ stays inert in $O\_K$, $pR$ will be prime in $R$.


Therefore, given a finite number of orders $R\_i$, since infinitely many rational primes stay inert in $K$, we can pick more than two such primes larger than all the indices $[O\_K:R\_i]$ and they will satisfy the required conditions.




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Proof that $R\to O\_K/pO\_K$ is surjective: let $[O\_K:R]=n$ be coprime to $p$. Let $m$ be modular inverse of $n$ modulo $p$. For any $a\in O\_K$ we have $na\in R$ and hence $mna\in R$ and $mna\equiv a\pmod{pO\_K}$.


Proof that the kernel is $pR$: clearly $pR$ is contained in the kernel. Let $a\in R\cap pO\_K$. Write $a=pb$ with $b\in O\_K$. Since the group quotient $O\_K/R$ has order $n$ coprime to $p$, it has no $p$-torsion, so $pb\in R$ implies $b\in R$, so $a=pb\in pR$.


