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Let $E$ be an elliptic curve over $\mathbb{F}_q$. I want to show $E(\mathbb{F}_q) \cong (\mathbb{Z}/m_1\mathbb{Z}) \times (\mathbb{Z}/m_2\mathbb{Z})$ where $m_1,m_2 \in \mathbb{Z}$ are such that

  • $m_1 \mid m_2$;
  • $m_1$ is the largest integer such that $F-1$ (with $F$ the $q$-th power Frobenius) is a multiple of $[m_1]$ in $\mathrm{End}(E)$.

I know numbers $m_1,m_2$ satisfying these two points exist as follows. We have some $m\in \mathbb{Z}$ such that $F-1 = \varphi \circ [m]$ for some $\varphi \in \mathrm{End}(E)$, for example $m=1, \varphi = F-1$, and the set of all such $m$ must be bounded, which can be seen from the degree of $F-1$.

Now take $m_1$ maximal amongst these $m$, and $\varphi \in \mathrm{End}(E)$ such that $F-1 = \varphi \circ [m_1]$. If we write $k$ for the number of elements in $E(\mathbb{F}_q)$ then we have that the degree of $F-1$ is $k$, for $F-1$ is separable and has kernel equal to $E(\mathbb{F}_q)$. So $\varphi \circ [m_1]$ is of degree $k = m_1^2 \cdot \deg(\varphi)$. Therefore, if we take $m_2 := m_1 \cdot \deg(\varphi)$, then $k= m_1 \cdot m_2$ and $m_1 \mid m_2$.

How can we now see that $E(\mathbb{F}_q) \cong (\mathbb{Z}/m_1\mathbb{Z}) \times (\mathbb{Z}/m_2\mathbb{Z})$ holds? I know $E(\mathbb{F}_q) = \ker(F-1) = \ker(\varphi \circ [m_1])$, but the problem seems to be we know little about $\varphi$.

Maanroof
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1 Answers1

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If $E$ is an elliptic curve over $\mathbb{F}_q$, then the Weil pairing $E(\mathbb{F}_q)\times E(\mathbb{F}_q)\rightarrow \mathbb{F}_q^*$ shows that there exist positive integers $m_1,m_2$ such that $$ E(\mathbb{F}_q)\cong \mathbb{Z}/m_1 \mathbb{Z} \times \mathbb{Z}/m_2 \mathbb{Z}, $$ with $m_1\mid gcd(m_2,q-1)$, see Chapter III, Corollary $8.1.1$ in Silverman's book "The Arithmetic of Elliptic Curves". So the subjectivity of the Weil pairing should help.

Dietrich Burde
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  • Thanks for your answer. Unfortunately, I don't really see how Corollary 8.1.1 relates to the question (I'm looking at it in the 2nd edition). Furthermore, the question should presumably be solvable without using the Weil paring, since we haven't covered this in class. – Maanroof Jan 23 '16 at 19:29
  • Silverman says: " The basic properties of the Weil pairing imply its subjectivity, as we now show", in Corollary $8.1.1$. I think, that Silverman does similar things as above before in chapter III; but there are other references with a proof, of course, if you don't want to use the Weil pairing. – Dietrich Burde Jan 23 '16 at 19:43