Let $C$ be a curve and $K(C)$ be its function field of genus 2, where $K$ = $\mathbb{C}$.
The number of essential elliptic subfields of $K(C)$ is 0 or 2 or $\infty$.
Edit: I am looking for a proof. Thanks!
Let $C$ be a curve and $K(C)$ be its function field of genus 2, where $K$ = $\mathbb{C}$. The number of essential elliptic subfields of $K(C)$ is 0 or 2 or $\infty$. Edit: I am looking for a proof. Thanks! 


Elliptic subfields of $K(C)$ correspond to finite morphisms from $C$ to an elliptic curve, which in turn correspond to elliptic factors of the Jacobian of $C$. Thus you get $0, 2, \infty$ essential elliptic subfields according to the decomposition of $\mathrm{Jac}(C)$ : it can be simple or isogenous to a product of elliptic curves $E \times E'$. If $E'=E$ you get infinitely many elliptic factors by embedding $E$ into $E \times E$ with maps of the form $P \mapsto (mP,nP)$. EDIT : two morphisms $\varphi_1,\varphi_2 : C \to E$ give rise to the same elliptic subfield $K(E)$ inside $K(C)$ if and only if there is an automorphism $\psi : E \to E$ making the obvious diagram commutative. 

