Let \(X\) be an elliptic curve over a field \(k\) of characteristic \(p>0\), and let \(R=\operatorname{End}(X, P_0)\) be its ring of endomorphisms.
Let \(X_p\) be the curve over \(k\) defined by changing the \(k\)-structure of \(X\) (2.4.1). Show that \(j(X_p)=j(X)^{1/p}\). Thus \(X \cong X_p\) over \(k\) if and only if \(j \in \FF_p\).
Show that \(p_X\) in \(R\) factors into a product \(\pi \hat{\pi}\) of two elements of degree \(p\) if and only if \(X \cong X_p\). In this case, the Hasse invariant of \(X\) is 0 if and only if \(\pi\) and \(\hat{\pi}\) are associates in \(R\) (i.e., differ by a unit). (Use (2.5).)
If \(\Hasse(X)=0\) show in any case \(j \in \FF_{p^2}\).
For any \(f \in R\), there is an induced map \(f^*: H^1(\OO_X) \to H^1(\OO_X)\). This must be multiplication by an element \(\lambda_f \in k\). So we obtain a ring homomorphism \(\varphi: R \to k\) by sending \(f\) to \(\lambda_f\). Show that any \(f \in R\) commutes with the (nonlinear) Frobenius morphism \(F: X \to X\), and conclude that if \(\Hasse(X) \neq 0\), then the image of \(\varphi\) is \(\FF_p\). Therefore, \(R\) contains a prime ideal \(\mfp\) with \(R/\mfp \cong \FF_p\).
Solution
We use the standing convention of Chapter IV that \(k\) is algebraically closed. Let \[ F':X_p\longrightarrow X \] be Hartshorne’s \(k\)-linear Frobenius, and put \[ V=\widehat{F'}:X\longrightarrow X_p. \] By Exercise IV.4.15, \[ F'\circ V=[p]_X, \qquad \deg F'=\deg V=p. \]
1. One has\[ \boxed{j(X_p)=j(X)^{1/p}.} \]
Choose a Weierstrass equation for \(X\). The twist \(X_p\) of IV.2.4.1 is obtained by applying the inverse Frobenius automorphism of the algebraically closed field \(k\) to its coefficients, that is, by replacing each coefficient \(a\) by its unique \(p\)th root \(a^{1/p}\).
The \(j\)-invariant is a rational expression in the Weierstrass coefficients with coefficients in the prime field. Applying inverse Frobenius to all coefficients therefore applies inverse Frobenius to \(j\) itself. Hence \[ j(X_p)=j(X)^{1/p}. \]
2. The curves \(X\) and \(X_p\) are isomorphic over \(k\) if and only if\[ \boxed{j(X)\in\FF_p.} \]
Over the algebraically closed field \(k\), elliptic curves are isomorphic if and only if they have the same \(j\)-invariant. By step 1, \[ X\cong X_p \iff j(X)=j(X)^{1/p} \iff j(X)^p=j(X). \] The roots in \(k\) of \(t^p-t\) are exactly the elements of \(\FF_p\). This proves part (a).
3. If \(X\cong X_p\), then \([p]_X\) factors in \(R\) as\[ \boxed{[p]_X=\pi\circ\widehat\pi} \]with\[ \deg\pi=\deg\widehat\pi=p. \]
Choose an origin-preserving isomorphism \[ \alpha:X\overset\sim\longrightarrow X_p \] which exists because any isomorphism of the underlying genus-one curves can be followed by a translation on \(X_p\) to send the chosen origin to the chosen origin. Define \[ \pi=F'\circ\alpha:X\longrightarrow X. \] Then \(\pi\in R\) and \(\deg\pi=p\). By contravariance of the dual and Exercise IV.4.7, \[ \widehat\pi = \widehat\alpha\circ V. \] The dual of an isomorphism is its inverse, so \[ \widehat\alpha=\alpha^{-1}. \] Consequently \[ \pi\circ\widehat\pi = F'\circ\alpha\circ\alpha^{-1}\circ V = F'\circ V = [p]_X. \] Exercise IV.4.7(f) gives \[ \deg\widehat\pi=\deg\pi=p. \]
4. Conversely, if\[ [p]_X=\pi\circ\widehat\pi \]with \(\pi,\widehat\pi\in R\) both of degree \(p\), then\[ \boxed{X\cong X_p.} \]
The differential of \([p]_X\) is multiplication by \(p\), hence is zero in characteristic \(p\). Thus \([p]_X\) is inseparable. If both \(\pi\) and \(\widehat\pi\) were separable, their composition would be separable. Therefore at least one of them is inseparable.
An inseparable morphism of degree \(p\) between nonsingular curves is purely inseparable of degree \(p\). Proposition IV.2.5 says that any such morphism \[ f:X\longrightarrow X \] is the \(k\)-linear Frobenius \[ X_p\longrightarrow X \] after an isomorphism of its source with \(X_p\). Hence \(X\cong X_p\). Together with step 3 this proves the first assertion of part (b).
5. Assume \(X\cong X_p\) and choose \(\pi\) as in step
3. Then\[ \boxed{ \Hasse(X)=0 \iff \pi\text{ and }\widehat\pi\text{ are associates in }R. } \]
The morphism \[ \pi=F'\circ\alpha \] is purely inseparable of degree \(p\). Moreover step 3 gives \[ \widehat\pi=\alpha^{-1}\circ V. \] Thus \(\widehat\pi\) is inseparable exactly when \(V\) is inseparable. By Exercise IV.4.15, this is equivalent to \(\Hasse(X)=0\).
Suppose first that \(\Hasse(X)=0\). Then both \(\pi\) and \(\widehat\pi\) are purely inseparable of degree \(p\). Proposition IV.2.5 therefore gives isomorphisms \[ \alpha,\beta:X\overset\sim\longrightarrow X_p \] such that \[ \pi=F'\circ\alpha, \qquad \widehat\pi=F'\circ\beta. \] These isomorphisms preserve the origins: both endomorphisms and \(F'\) send origin to origin, while \(F'\) is radicial and hence has only one point over the origin. Thus \[ \alpha(P_0)=\beta(P_0)=P_{0,p}. \] Put \[ u=\beta^{-1}\circ\alpha\in R. \] Then \(u\) is a unit and \[ \pi =F'\circ\beta\circ u =\widehat\pi\circ u. \] Hence \(\pi\) and \(\widehat\pi\) are associates.
Conversely, suppose that they are associates. Composition with a unit preserves separability. Since \(\pi\) is purely inseparable, \(\widehat\pi\) is inseparable as well. Hence \[ V=\alpha\circ\widehat\pi \] is inseparable, and Exercise IV.4.15 gives \(\Hasse(X)=0\). This completes part (b).
6. If \(\Hasse(X)=0\), then\[ \boxed{j(X)\in\FF_{p^2}.} \]
By Exercise IV.4.15, \(\Hasse(X)=0\) means that \[ V:X\longrightarrow X_p \] is inseparable. Since \(\deg V=p\), it is purely inseparable of degree \(p\). Apply Proposition IV.2.5 to \(V\). Its target is \(X_p\), so its source is isomorphic to the Frobenius twist of \(X_p\): \[ X\cong (X_p)_p=X_{p^2}. \] Applying step 1 twice gives \[ j(X_{p^2})=j(X)^{1/p^2}. \] Isomorphic elliptic curves have the same \(j\)-invariant, hence \[ j(X)=j(X)^{1/p^2}. \] Equivalently, \[ j(X)^{p^2}=j(X), \] whose solutions in \(k\) are exactly \(\FF_{p^2}\). This proves part (c).
7. Every \(f\in R\) commutes with the nonlinear Frobenius\[ F:X\longrightarrow X. \]
The nonlinear Frobenius here is the absolute Frobenius: on the structure sheaf it sends a local function \(a\) to \(a^p\). For every morphism of schemes \(f:X\to X\) in characteristic \(p\), \[ f^\sharp(a^p)=f^\sharp(a)^p. \] Therefore the square \[ \begin{CD} X @>{F}>> X\\ @V{f}VV @VV{f}V\\ X @>{F}>> X \end{CD} \] commutes, so \[ f\circ F=F\circ f. \]
8. If \(\Hasse(X)\ne0\), then for every \(f\in R\) the scalar\(\lambda_f\) defined by\[ f^*:H^1(X,\OO_X)\longrightarrow H^1(X,\OO_X) \]satisfies\[ \boxed{\lambda_f\in\FF_p.} \]
The vector space \(H^1(X,\OO_X)\) is one-dimensional over \(k\), while the absolute Frobenius induces a \(p\)-semilinear map \[ F^*:H^1(X,\OO_X)\longrightarrow H^1(X,\OO_X), \qquad F^*(a\xi)=a^pF^*(\xi). \] The condition \(\Hasse(X)\ne0\) says that \(F^*\) is nonzero. Choose \(\xi\) with \[ F^*(\xi)\ne0. \]
Step 7 and contravariance of pullback give \[ F^*\circ f^*=f^*\circ F^*. \] Since \(f^*\) is multiplication by \(\lambda_f\), \[ \lambda_f^pF^*(\xi) = F^*(\lambda_f\xi) = F^*f^*(\xi) = f^*F^*(\xi) = \lambda_fF^*(\xi). \] Because \(F^*(\xi)\ne0\), \[ \lambda_f^p=\lambda_f. \] Thus \(\lambda_f\) is a root of \(t^p-t\), so \(\lambda_f\in\FF_p\).
9. If \(\Hasse(X)\ne0\), the image of\[ \varphi:R\longrightarrow k, \qquad f\longmapsto\lambda_f, \]is\[ \boxed{\operatorname{im}\varphi=\FF_p.} \]
Step 8 gives \[ \operatorname{im}\varphi\subseteq\FF_p. \] For every integer \(m\), the endomorphism \([m]_X\) belongs to \(R\). Under the identification \[ T_0\Pic^0(X)\cong H^1(X,\OO_X), \] the pullback \([m]_X^*\) is the tangent map of the dual endomorphism \(\widehat{[m]_X}\). Exercise IV.4.7(e) gives \[ \widehat{[m]_X}=[m]_X, \] whose tangent map is multiplication by \(m\). Therefore \[ \varphi([m]_X)=m\pmod p. \] As \(m\) varies, these elements exhaust \(\FF_p\). Hence the inclusion is an equality.
10. If \(\Hasse(X)\ne0\), the ideal\[ \mfp=\ker\varphi \]is prime and\[ \boxed{R/\mfp\cong\FF_p.} \]
By step 9, \(\varphi\) is a surjective ring homomorphism onto the field \(\FF_p\). The first isomorphism theorem gives \[ R/\ker\varphi\cong\FF_p. \] Moreover, if \(ab\in\ker\varphi\), then \[ 0=\varphi(ab)=\varphi(a)\varphi(b) \] in the field \(\FF_p\), so \(\varphi(a)=0\) or \(\varphi(b)=0\). Hence \(\ker\varphi\) is prime. This proves part (d).
11. Q.E.D.
Steps <1>1–<1>2 prove part (a), steps <1>3–<1>5 prove part (b), step 6 proves part (c), and steps <1>7–<1>10 prove part (d).