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Let $\mathbb{F}_q$ be a finite field of characteristic $p$ with $q=p^a$ elements. Let $E$ be an ordinary elliptic curve defined over $\mathbb{F}_q$. We know that $\text{End}(E)\otimes\mathbb{Q}$ is isomorphic to a quadratic imaginary field $K\subset\mathbb{C}$ and that $\text{End}(E)\cong\mathcal{O}$ is an order in $K$.

Let $L$ be the ring class field of the order $\mathcal{O}$. Then we can choose elliptic curves $E_1,\ldots,E_h$ defined over $L$ such that $\text{End}(E_i)=\mathcal{O}$ and that $E_i\ncong E_j$ for $i\ne j$, where $h$ is the class number of $\mathcal{O}$. Let $\mathfrak{P}$ be a prime of $L$ lying over $p$. Then $E_i$ has potential good reduction over the completion $L_{\mathfrak{P}}$ of $L$ at $\mathfrak{P}$, hence there is a finite extension $L'/L$ so that $E_i$ has good reducion over $L'L_{\mathfrak{P}}$. Extending $L'$, we can assume that $E_i$ has good reduction at the prime $\mathfrak{P}'$ of $L'$ for $i=1,\ldots,h$.

Let $\tilde{E_i}$ be the reduction of $E_i$ modulo $\mathfrak{P}'$. Since the degree of residue field extension $[\mathcal{O}_L/\mathfrak{P}:\mathcal{O}_K/\mathfrak{P}\cap\mathcal{O}_K]$ is equal to the order of $\mathfrak{P}\cap\mathcal{O}$, the j-invariants $j(\tilde{E_i})\in\mathcal{O}_L/\mathfrak{P}\subseteq\mathbb{F}_q$.

Now we have the following:
1) $j(\tilde{E_i})\in\mathbb{F}_q$
2) $j(\tilde{E_i})\ne j(\tilde{E_j})$ for $i\ne j$
3) $\text{End}(\tilde{E_i})=\mathcal{O}$
The $\tilde{E_i}$ may not be defined over $\mathbb{F}_q$, but we have elliptic curves $E_i'/\mathbb{F}_q$ such that $\tilde{E_i}\cong E_i'$ over the algebraic closure of $\mathbb{F}_q$.

Can we choose $E_i'$ such that $\#E_i'(\mathbb{F}_q)=\#E(\mathbb{F}_q)$? If yes, how can we achieve this? I just want to know that whether the two sets $$\{j(E'):E'/\mathbb{F}_q,\,\text{End}(E ')=\mathcal{O}\}$$

and

$$\{j(E'):E'/\mathbb{F}_q,\,\text{End}(E ')=\mathcal{O},\,\#E'(\mathbb{F}_q)=\#E(\mathbb{F}_q)\}$$ (The cardinality of the first set is the class number of $\mathcal{O}$.)

Thanks!

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  • $\begingroup$ The questions in the title and body of the post are not the same. There are pairs of elliptic curves over $\mathbb{F}_q$ that have the same $j$-invariant but are not isogenous over $\mathbb{F}_q$. $\endgroup$
    – dgulotta
    Apr 16, 2017 at 16:28
  • $\begingroup$ Thanks! Let $\{j(E'):E'/\mathbb{F}_q, \,\text{End}(E')=\mathcal{O}\}=\{j_1,j_2,\ldots,j_h\}$. My question is can we choose $E_i'$ such that $j(E_i')=j_i$ and that $\#E_1'(\mathbb{F}_q)=\#E_2'(\mathbb{F}_q)=\cdots=\#E_h'(\mathbb{F}_q)$. $\endgroup$
    – tiansong
    Apr 16, 2017 at 23:38

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