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Let $~E:y^2=x^3+x~$ be an elliptic curve over finite field $\mathbb{F}_{5},$ I compute the trace of Frobenius is $2$($E/\mathbb{F}_{5}$ obvious is ordinary).

(By the theory of CM, I know (when $E$ is an ordinary elliptic curve over finite field $\mathbb{F}_{q}$):$$\mathbb{Z}[\pi_{E}]\subseteq End(E)\subseteq \mathcal{O}_{K}$$ )

in the above example I got $\mathcal{O}_{K}=\mathbb{Z}[i]$, and $\mathbb{Z}[\pi_{E}]=\mathbb{Z}+2\mathcal{O}_{K}.$

Therefore $End(E)$ have two possible result: $\mathcal{O}_{K}$ or $\mathbb{Z}[\pi_{E}]$.

I want to know which is the true $End(E)$.

Actually, there is a bound for the conductor $[\mathcal{O}_{K}:\mathbb{Z}[\pi_{E}]].$ I'am not sure whether or not there exists any better result for deciding the conductor.

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In your case, $End(E)=\mathcal O_K$, because $E$ has an automorphism of order $4$, namely $x\mapsto -x$ and $y\mapsto -2y$. This endomorphism cannot belong to $\mathbb Z[\pi_E]$, because you can check that there are no elements of $\mathbb Z[\pi_E]$ whose square is $-1$.

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  • $\begingroup$ Thanks for your cute answer $\endgroup$
    – sherry
    May 23 '14 at 10:47
  • $\begingroup$ @FerraI'd love to share this general method with you. $\endgroup$
    – sherry
    May 23 '14 at 13:26
  • $\begingroup$ Can I ask how do you determine the field $K$ here? $\endgroup$ Apr 20 '15 at 16:53
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    $\begingroup$ ^^Never mind, I just found the formula $K=\mathbb{Q}(\sqrt{2^2-4\cdot 5})=\mathbb{Q}(i)$ where 2 is the Frobenius trace and 5 is the characteristic of the field $\endgroup$ Apr 20 '15 at 20:30

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