Again let \(X\) be an elliptic curve over \(k\) of characteristic \(p\), and suppose \(X\) is defined over the field \(\FF_q\) of \(q=p^r\) elements, i.e., \(X \subseteq \PP^2\) can be defined by an equation with coefficients in \(\FF_q\). Assume also that \(X\) has a rational point over \(\FF_q\). Let \(F': X_q \to X\) be the \(k\)-linear Frobenius with respect to \(q\).
Show that \(X_q \cong X\) as schemes over \(k\), and that under this identification, \(F': X \to X\) is the map obtained by the \(q\)-th power map on the coordinates of points of \(X\), embedded in \(\PP^2\).
Show that \(1_X-F'\) is a separable morphism and its kernel is just the set \(X(\FF_q)\) of points of \(X\) with coordinates in \(\FF_q\).
Using (Ex. 4.7), show that \(F'+\hat{F}'=a_X\) for some integer \(a\), and that \(N=q-a+1\), where \(N=\size X(\FF_q)\).
Use the fact that \(\deg(m+n F')>0\) for all \(m, n \in \ZZ\) to show that \(\abs{a} \leq 2 \sqrt{q}\). This is Hasse’s proof of the analogue of the Riemann hypothesis for elliptic curves (App. C, Ex. 5.6).
Now assume \(q=p\), and show that the Hasse invariant of \(X\) is 0 if and only if \(a \equiv 0 \pmod p\). Conclude for \(p \geq 5\) that \(X\) has Hasse invariant 0 if and only if \(N=p+1\).
Solution
Let \(\pi=F'\) denote the \(q\)-power Frobenius after the identification in part (a), and write \(O\) for the given \(\FF_q\)-rational origin.
1. The Frobenius twist \(X_q\) is isomorphic to \(X\) over \(k\), and under this identification\[ \boxed{\pi(x:y:z)=(x^q:y^q:z^q).} \]
Choose a homogeneous equation \[ G(x,y,z)=0 \] for \(X\) whose coefficients lie in \(\FF_q\). The \(q\)-Frobenius twist is obtained by applying the \(q\)-power automorphism of \(k\) to the coefficients. Every coefficient of \(G\) is fixed, so the twisted equation is again \(G=0\). Thus \(X_q\cong X\) over \(k\).
The \(k\)-linear Frobenius is the coordinate \(q\)-power map. Since \[ G(x^q,y^q,z^q)=G(x,y,z)^q, \] it preserves \(X\) and has the displayed form. This proves part (a).
2. The endomorphism\[ 1_X-\pi:X\longrightarrow X \]is separable.
The differential of the \(q\)-power Frobenius is zero. Hence \[ d(1_X-\pi)_O = d(1_X)_O-d\pi_O = \id_{T_OX}, \] which is nonzero. A homomorphism of elliptic curves is separable exactly when its differential at the origin is nonzero, so \(1_X-\pi\) is separable.
3. Its kernel is exactly\[ \boxed{\ker(1_X-\pi)=X(\FF_q).} \]
A geometric point \(P=(x:y:z)\) lies in the kernel exactly when \[ \pi(P)=P. \] By step 1 this is equivalent to \(P\) being fixed by the \(q\)-power Frobenius, hence to \(P\) being defined over \(\FF_q\). This proves part (b).
4. If \(N=\#X(\FF_q)\), then\[ \deg(1_X-\pi)=N. \]
By step 2 the endomorphism is separable, so its degree equals the number of geometric kernel points. Step 3 identifies these with \(X(\FF_q)\).
5. There is an integer\[ a=q+1-N \]such that\[ \boxed{\pi+\widehat\pi=[a]_X.} \]
By additivity of dualization from Exercise IV.4.7(d),(e), \[ \widehat{(1_X-\pi)}=1_X-\widehat\pi. \] Exercise IV.4.7(c) applied first to \(1_X-\pi\) and then to \(\pi\) gives \[ \widehat{(1_X-\pi)}\circ(1_X-\pi)=[N]_X \] and \[ \widehat\pi\circ\pi=[q]_X. \] Therefore \[ \begin{aligned} [N]_X &=(1_X-\widehat\pi)\circ(1_X-\pi)\\ &=[q+1]_X-(\pi+\widehat\pi). \end{aligned} \] Hence \[ \pi+\widehat\pi=[q+1-N]_X. \] Taking \(a=q+1-N\) proves part (c), including \(N=q-a+1\).
6. For all \(m,n\in\ZZ\),\[ \boxed{ \deg([m]_X+[n]_X\circ\pi) = m^2+amn+qn^2. } \]
Put \[ \phi=[m]_X+[n]_X\circ\pi. \] By Exercise IV.4.7(d),(e), \[ \widehat\phi=[m]_X+[n]_X\circ\widehat\pi. \] Using step 5 and \(\widehat\pi\circ\pi=[q]_X\), we obtain \[ \begin{aligned} \widehat\phi\circ\phi &=[m^2]_X+[mn]_X\circ(\pi+\widehat\pi) +[n^2]_X\circ\widehat\pi\circ\pi\\ &=[m^2+amn+qn^2]_X. \end{aligned} \] Exercise IV.4.7(c) also gives \[ \widehat\phi\circ\phi=[\deg\phi]_X. \] Since \(\ZZ\to\Endo(X,O)\) is injective, the two integers are equal.
7. One has Hasse’s bound\[ \boxed{\abs a\le2\sqrt q.} \]
Step 6 gives \[ Q(m,n)=m^2+amn+qn^2\ge0 \] for all \(m,n\in\ZZ\), because \(Q(m,n)\) is a degree. The strict positivity in the exercise applies when the corresponding endomorphism is nonzero; nonnegativity is sufficient for the argument.
For \(n\ne0\), division by \(n^2\) gives \[ \left(\frac mn\right)^2+a\frac mn+q\ge0. \] The rational numbers are dense in \(\RR\), so continuity implies \[ x^2+ax+q\ge0 \] for every real \(x\). Hence the discriminant is nonpositive: \[ a^2-4q\le0. \] Thus \(\abs a\le2\sqrt q\), proving part (d).
8. Now assume \(q=p\).The Hasse invariant of \(X\) is zero if and only if\[ \boxed{a\equiv0\pmod p.} \]
Let \[ V=\widehat\pi. \] Step 5 gives \[ V=[a]_X-\pi. \] The differential of \(\pi\) is zero, while the differential of multiplication by \(a\) on the one-dimensional tangent space at the origin is multiplication by the image of \(a\) in \(k\). Hence \[ dV_O=a\cdot\id_{T_OX}. \] By Exercise IV.4.15, the Hasse invariant is zero exactly when the dual Frobenius \(V\) is inseparable. An isogeny of elliptic curves is inseparable exactly when its differential at the origin vanishes. Therefore \[ \operatorname{Hasse}(X)=0 \iff dV_O=0 \iff a\equiv0\pmod p. \]
9. If \(p\ge5\), then the Hasse invariant of \(X\) is zero if and only if\[ \boxed{N=p+1.} \]
When \(q=p\), step 7 gives \[ \abs a\le2\sqrt p. \] For \(p\ge5\), \[ 2\sqrt p<p. \] Thus an integer \(a\) in this range is divisible by \(p\) exactly when \(a=0\). By step 8, \[ \operatorname{Hasse}(X)=0 \iff a=0. \] Finally step 5 gives \[ N=p-a+1, \] so \(a=0\) is equivalent to \(N=p+1\). This proves part (e).
10. Q.E.D.
Step 1 proves part (a), steps <1>2–<1>3 prove part (b), steps <1>4–<1>5 prove part (c), steps <1>6–<1>7 prove part (d), and steps <1>8–<1>9 prove part (e).