# How is the Frobenius morphism of a supersingular curve non-integral if its characteristic equation splits over Z?

Let's take a supersingular elliptic curve $$E / \mathbb{F}_q$$, where $$q = p^2$$ for some $$p \equiv 1 (\text{mod } 4)$$, and $$p \geq 5$$, let's say. Then $$\# E(\mathbb{F}_q) = (p - 1)^2$$, which by Hasse's theorem implies that the $$q$$-power Frobenius endomorphism $$\phi_q$$ satisfies the characteristic equation:

$$\phi_q^2 - 2p \cdot \phi_q + q = 0 = (\phi_q - p)^2$$

in $$\text{End}(E)$$.

How does this not imply that $$\phi_q$$ is identically equal to the multiplication-by-p map $$[p]$$? It's surely not, or else $$\phi_q$$ would represent an integral element of the quaternion-order $$\text{End}(E)$$.

EDITED: Looks like I flipped a sign in the trace equation (thanks @reuns). Thus requiring now $$p \equiv 1 (\text{mod } 4)$$, re: @hutner's comment, at least we have no contradiction in this case, as indeed $$E(\mathbb{F}_q)$$ is a $$p - 1$$-group.

The corresponding fact for $$p \equiv 3 (\text{mod } 4)$$ is that $$\phi_q = [-p]$$ should act as the identity on $$E(\mathbb{F}_q) \cong \left( \mathbb{Z} / (p + 1)\mathbb{Z} \right)^2$$. But this in turn is equivalent to $$[p + 1]$$ being the zero map, so we're again ok.

Thus at least no contradiction follows from @Lord Shark the Unknown's claim! Thanks everyone.

• What's wrong with $\phi$ equalling $[p]$? It's definitely in the quaternion order, and one expects $\hat\phi\circ\phi=[q]$. Nov 9, 2018 at 2:54
• Thanks for your response. But then where are the non-integral endomorphisms going to come from (there should be three which are $\mathbb{Z}$-linearly independent). Nov 9, 2018 at 3:10
• Alas, in this situation, these non-trivial endomorphisms don't come "for free". Nov 9, 2018 at 3:21
• @LordSharktheUnknown I don't think that $\phi$ could equal $[p]$. For example, this would imply that every point in the ground field was annihilated by $[p-1]$, but this contradicts the given count since there are a maximum of $(p-1)^2$ such $\mathbb{F}_q$-points. Nov 9, 2018 at 3:36
• So it would work better if $\phi^2+2p \phi + p^2 = 0$ Nov 9, 2018 at 3:43