We say two elliptic curves \(X, X'\) are isogenous if there is a finite morphism \(f: X \to X'\).
Show that isogeny is an equivalence relation.
For any elliptic curve \(X\), show that the set of elliptic curves \(X'\) isogenous to \(X\), up to isomorphism, is countable. Hint: \(X'\) is uniquely determined by \(X\) and \(\ker f\).
Solution
1. Isogeny is reflexive.
For every elliptic curve \(X\), the identity \[ \id_X:X\longrightarrow X \] is finite. Hence \(X\) is isogenous to itself.
2. Isogeny is symmetric.
Suppose that \[ f:X\longrightarrow X' \] is finite. It is nonconstant, so \(\deg f>0\). By Exercise IV.4.7(a),(f), the dual morphism \[ \hat f:X'\longrightarrow X \] has \[ \deg\hat f=\deg f>0. \] A nonconstant morphism of projective nonsingular curves is finite. Thus \(\hat f\) is a finite morphism from \(X'\) to \(X\).
3. Isogeny is transitive.
If \[ X\xrightarrow{f}X'\xrightarrow{g}X'' \] are finite morphisms, then \(g\circ f\) is finite. Hence an elliptic curve isogenous to one isogenous to \(X\) is itself isogenous to \(X\). Together with steps <1>1–<1>2, this proves part (a).
4. For counting targets, every finite morphism \(f:X\to X'\) may bereplaced by an isogeny preserving the origins without changing \(X'\).
Let \(O\in X\) and \(O'\in X'\) be the chosen origins. Translation on \(X'\) is an automorphism. Therefore \[ g(P)=f(P)-f(O) \] is again finite and has the same target curve. It sends \(O\) to \(O'\), hence is a homomorphism of elliptic curves.
5. A separable isogeny\[ h:E\longrightarrow E' \]is determined, up to isomorphism of its target, by the finite subgroup\[ G=\ker h\subseteq E(k). \]
For each \(T\in G\), translation \(\tau_T\) is an automorphism of \(E\) over \(E'\). Since \(h\) is separable, \[ [k(E):h^*k(E')]=\deg h=\#G. \] The translations by \(G\) already give \(\#G\) distinct automorphisms of this function-field extension. Hence \[ h^*k(E')=k(E)^G. \] The right-hand side depends only on \(E\) and \(G\). A nonsingular projective curve is determined up to isomorphism by its function field, so the target \(E'\) is determined up to isomorphism by \(E\) and \(G\).
6. For a fixed elliptic curve \(E\), there are only countably many pairs\((r,G)\) in which \(r\ge0\) and \(G\) is a finite subgroup of\(E^{(p^r)}(k)\).
Fix \(r\). Every finite subgroup \(G\subseteq E^{(p^r)}(k)\) has finite exponent, say \(n\), and therefore \[ G\subseteq E^{(p^r)}[n](k). \] The \(n\)-torsion set is finite. Hence it has only finitely many subgroups. Taking the union over \(n\ge1\) shows that \(E^{(p^r)}(k)\) has only countably many finite subgroups. Taking the further union over \(r\ge0\) remains countable.
In characteristic zero, only the case \(r=0\) is needed.
7. Up to isomorphism, only countably many elliptic curves are isogenousto \(X\).
By step 4, consider only isogenies \(g:X\to X'\) preserving origins. In characteristic zero, \(g\) is separable, so step 5 says that \(X'\) is determined by the finite subgroup \(\ker g\subseteq X(k)\). Step 6 gives only countably many possibilities.
6. This proves part (b).
8. Q.E.D.
Steps <1>1–<1>3 prove that isogeny is an equivalence relation, and steps <1>4–<1>7 prove that each isogeny class contains only countably many isomorphism classes of elliptic curves.