This question was previously asked on Math.Stackexchange.

It is about a proposition from R. Schoof's article Nonsingular Plane Cubic Curves over Finite Fields:

Proposition 3.7: Let $E$ be an elliptic curve over the finite field ${\mathbb F}_q$ of characteristic $p$ and $n\in {\mathbb N}$ such that $p\nmid n$. Further, let $t$ be the trace of the Frobenius $\phi$, that is, $t := q + 1 - |E({\mathbb F}_q)|$. Then the following are equivalent:

  1. $E(\overline{{\mathbb F}_q})[n]\subset E({\mathbb F}_q)$, i.e. every $n$-torsion point is defined over ${\mathbb F}_q$.
  2. $n | q-1$, $n^2 | q + 1 - t$, and either $\phi\in{\mathbb Z}$ or ${\mathcal O}\left(\frac{t^2-4q}{n^2}\right)\subset\text{End}_{{\mathbb F}_q}(E)$.

Here ${\mathcal O}(\Delta)$ is introduced earlier as the complex quadratic order with discriminant $\Delta$.

I'm having trouble understanding the precise meaning of the second statement if $E$ is not ordinary. First, under the assumption that $ n | q - 1$ and $n^2 | q + 1 - t$, one has $n^2 | t^2 - 4q$, so ${\mathcal O}\left(\frac{t^2-4q}{n^2}\right)$ is defined. But:

Q: Is ${\mathcal O}\left(\frac{t^2-4q}{n^2}\right)\subset\text{End}_{{\mathbb F}_q}(E)$ really only supposed to mean that there's some embedding, even if $\text{End}_{{\mathbb F}_q}(E)$ is not assumed to be commutative?

As far as I understand, for a commutative domain $R$ the image of an embedding ${\mathcal O}\left(\Delta\right)\to R$ is unique if it exists, namely the subring of $R$ spanned by all solutions of quadratic equations with discriminant $\Delta$ - and in fact, a single such would suffice. However, over the non-commutative domain $\text{End}_{{\mathbb F}_q}(E)$ this canonicity seems to fail - e.g. in the ring of integral quaternions ${\mathbb Z}[i,j,k]$ we have $X^2 + 1 = (X-i)(X+i) = (X-j)(X+j) = (X-k)(X+k)$ corresponding to different embeddings ${\mathcal O}(-4)={\mathbb Z}[i]\hookrightarrow {\mathbb Z}[i,j,k]$. Given that such quaternion algebras actually arise as endomorphism algebras of supersingular elliptic curves, this example doesn't seem to be too far-fetched. In fact, this is the case I am most interested in.

Q: Considering that the proposition is specifically about describing $\frac{\phi-1}{n}$ as opposed to any quadratic element of $\text{End}_{{\mathbb F}_q}(E)$ of discriminant $\frac{t^2-4q}{n^2}$, perhaps one should restrict the discussion to the intersection of $\text{End}_{{\mathbb F}_q}(E)$ with the ${\mathbb Q}$-subalgebra of $\text{End}_{\overline{{\mathbb F}_q}}(E)\otimes_{\mathbb Z}{\mathbb Q}$ generated by $\phi$?

Looking at the proof doesn't help me, unfortunately, because it also uses notation I'm lacking a precise definition for.

I'd be happy if somebody could clarify. Thanks alot!


We have $\phi \in \mathbb Z$ if and only if $\operatorname{End}_{\mathbb F_q}(E)$ is noncommutative.

So it is irrelevant how the last clause of (2) treats the noncommutative case as it is already handled in the first clause of (2).

The "if" direction, which is the really relevant direction here, is easy - because Frobenius commutes with the endomorphism algebra on the $\ell$-adic Tate module, it must act by scalar multiplication on the $\ell$-adic Tate module, hence by scalar multiplication. The "only if" direction follows for instance from the Tate conjecture that the endomorphism algebra is the centralizer of Frobenius, although it's probably possible to give an easier proof in this case.

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  • $\begingroup$ Thanks for your answer! I have trouble understanding how $\phi\in{\mathbb Z}$ can be true for non-commutative $\text{End}_{{\mathbb F}_q}(E)$ in general - for example, if $\text{tr}(\phi)=0$ then we have $\phi^2=-q$, don't we? $\endgroup$ – Hanno Becker Nov 28 '17 at 8:42
  • $\begingroup$ @HannoBecker Yes, and then $\operatorname{End}_{\mathbb F_q}(E)$ is $\mathbb Z[\sqrt{-q}]$, or an order containing it, but $\operatorname{End}_{\overline{\mathbb F}_q}(E)$ is larger. $\endgroup$ – Will Sawin Nov 28 '17 at 9:03

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