Let \(X\) be an elliptic curve over a field \(k\) of characteristic \(p\). Let \(F': X_p \to X\) be the \(k\)-linear Frobenius morphism (2.4.1). Use (4.10.7) to show that the dual morphism \(\hat{F}': X \to X_p\) is separable if and only if the Hasse invariant of \(X\) is 1.
Now use (Ex. 4.7) to show that if the Hasse invariant is 1, then the subgroup of points of order \(p\) on \(X\) is isomorphic to \(\ZZ/p\); if the Hasse invariant is 0, it is 0.
Solution
Let \[ F':X_p\longrightarrow X \] be the \(k\)-linear Frobenius morphism and let \[ V=\widehat{F'}:X\longrightarrow X_p \] be its dual.
1. Under the identifications of an elliptic curve with its Jacobian, themap \(V\) is the morphism of Jacobians induced by pullback along \(F'\).
Exercise IV.4.7 defines the dual of a morphism \[ f:Y\longrightarrow Z \] by the pullback homomorphism \[ f^*:\Pic^0(Z)\longrightarrow\Pic^0(Y), \] after identifying each elliptic curve with its degree-zero Picard variety. Applying this to \(F'\) gives exactly the assertion.
2. The tangent map of \(V\) at the origin identifies with\[ (F')^*: H^1(X,\OO_X) \longrightarrow H^1(X_p,\OO_{X_p}). \]
Remark IV.4.10.7 identifies the Zariski tangent space at the origin of the Jacobian of a curve \(Y\) with \[ T_0\Pic^0(Y)\cong H^1(Y,\OO_Y). \] This identification is functorial for pullback of line bundles. By step 1, \(V\) is precisely the pullback map on the two Jacobians, so its differential at the origin is \((F')^*\) on \(H^1(\OO)\).
3. The dual Frobenius \(V\) is separable if and only if the Hasse invariantof \(X\) is \(1\).
The map in step 2 is the \(k\)-linear form of the Frobenius action \[ F^*:H^1(X,\OO_X)\longrightarrow H^1(X,\OO_X) \] used to define the Hasse invariant. Since \(H^1(X,\OO_X)\) is one-dimensional, the Hasse invariant is \(1\) exactly when this map is nonzero.
An isogeny between smooth curves is separable exactly when its differential is nonzero. For a group homomorphism it suffices to test the differential at the origin, because translations identify all tangent maps. Therefore, by step 2, \[ V\text{ is separable} \iff dV_0\ne0 \iff F^*\ne0 \iff \operatorname{Hasse}(X)=1. \] This proves the first assertion.
4. The Frobenius \(F'\) has degree \(p\), and Exercise IV.4.7 gives\[ \boxed{ F'\circ V=[p]_X. } \]
The \(k\)-linear Frobenius is purely inseparable of degree \(p\). Exercise IV.4.7(c), applied to \(F'\), says that composition with the dual is multiplication by the degree, hence \[ F'\circ\widehat{F'}=[p]_X. \]
5. The geometric \(p\)-torsion points of \(X\) are exactly the geometrickernel points of \(V\):\[ \boxed{X[p](k)=\ker V(k).} \]
The inclusion \[ \ker V(k)\subseteq X[p](k) \] follows immediately from step 4.
Conversely, let \(P\in X[p](k)\). Then step 4 gives \[ F'(V(P))=[p](P)=0. \] The Frobenius \(F'\) is radicial, so its geometric kernel has only the identity point. Hence \(V(P)=0\), proving the reverse inclusion.
6. If the Hasse invariant is \(1\), then\[ \boxed{X[p](k)\cong\ZZ/p\ZZ.} \]
By step 3, \(V\) is separable. Exercise IV.4.7(f) gives \[ \deg V=\deg F'=p. \] For a separable isogeny, the number of geometric kernel points equals its degree, so \[ \#\ker V(k)=p. \] By step 5 this is \(X[p](k)\). A group of prime order is cyclic, hence \[ X[p](k)\cong\ZZ/p\ZZ. \]
7. If the Hasse invariant is \(0\), then\[ \boxed{X[p](k)=0.} \]
By step 3, \(V\) is inseparable. Since \(\deg V=p\) is prime, its inseparable degree is \(p\) and its separable degree is \(1\). Thus its geometric kernel has only the identity point. Step 5 now gives \[ X[p](k)=\ker V(k)=\{0\}. \]
8. Q.E.D.
Step 3 proves the separability criterion, while steps <1>6–<1>7 give the two possible groups of \(p\)-torsion points.