9
$\begingroup$

Let $E$ be a supersingular elliptic curve over a finite field of characteristic $p$, and $\mathbb{F}_q\supset \mathbb{F}_{p^2}$ be a finite field large enough such that all (absolute) endomorphisms of $E$ is defined over $\mathbb{F}_q$. We write $G$ for the absolute Galois group of $\mathbb{F}_q$. It is well known that the Frobenius automorphism $\varphi$ is a (topological) generator of $G$. Let us fix a prime $\ell\neq p$. By the Tate Conjecture, $$\mathrm{End}(E)\otimes\mathbb{Q}_\ell=\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))^G=\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))^\varphi. $$

Since $E$ is supersingular, $\mathrm{End}(E)$ is an order in a quaternion algebra. In particular, $\mathrm{rank}_\mathbb{Z}(\mathrm{End}(E))=4$. So $$ \mathrm{End}(E)\otimes\mathbb{Q}_\ell=\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))$$ It follows that $\varphi$ is in the center of $\mathrm{End}_{\mathbb{Q}_\ell}(V_\ell(E))$. In other words, $\varphi$ is a scalar. This clearly leads to a contradiction (say, with the Riemann Hypothesis). Where did the argument go wrong?

Improve this question
$\endgroup$
3
  • 17
    $\begingroup$ This question is a fabulous instance of the phenomenon that surely many people have observed, many times over: the mistake in the argument is at the unique point where the author writes the word "clearly". $\endgroup$
    – Kevin Buzzard
    Commented Jan 14, 2011 at 11:58
  • 7
    $\begingroup$ I think it's a well-known entry in the mathematical thesaurus that "clearly" is a synonym for "I should probably check that." $\endgroup$
    – JSE
    Commented Jan 14, 2011 at 14:24
  • $\begingroup$ Let $E/\mathbb F_p$ be an elliptic curve, with $p\ge5.$ Then $E$ is super-singular if and only if the Frob. trace $a_p$ on the Tate module is 0 (see Silverman I, an exercise in the finite field chapter). If $\alpha,\bar{\alpha}$ are the Frob. eigenvalues, then $\alpha+\bar{\alpha}=a_p=0\Rightarrow\alpha$ is purely imaginary. Hence the Frob. eigenvalues of $E\otimes\mathbb F_{p^2}$ are equal and are real (as Emerton pointed out in his answer). I learned this from B. Poonen. $\endgroup$
    – shenghao
    Commented May 20, 2011 at 12:24

1 Answer 1

Reset to default
11
$\begingroup$

There is nothing wrong. A typical case will be $\mathbb F_q = \mathbb F_{p^2}$ and $\varphi = -p$. Since $q^{1/2} = (p^2)^{1/2} = p$, this is consistent with RH.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you. For some reason I always thought for the elliptic curves, the eigenvalues of the Frobenius is in $\mathbb{C}-\mathbb{R}$. I guess this is an instance where it is not. $\endgroup$
    – Jiangwei Xue
    Commented Jan 14, 2011 at 11:37
  • 2
    $\begingroup$ Jiangwei, I recommend to you in the strongest terms that you spend a few days doing explicit computations with the chord-and-tangent algorithm for adding points on an elliptic curve. A specific supersingular curve, and a specific $p$. And then a specific ordinary one over the same base. $\endgroup$
    – Lubin
    Commented Jan 14, 2011 at 19:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .