# Elliptic subfields of a function field

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!

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What's an "essentially elliptic subfield"? – Angelo May 9 '12 at 11:55
Essential subfield: A subfield of +ve genus and also "maximal" in the sense that it is not contained in any other subfield of same genus. Elliptic subfield: Genus 1 subfield of K(C). – Srilakshmi May 9 '12 at 12:36
You could take a look at Ernst Kani, "Elliptic curves on abelian surfaces". – Dan Petersen May 10 '12 at 5:13

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.
Dear Srilakshmi, every abelian variety is isogenous to a product of simple abelian varieties (Poincaré's complete reducibility theorem). Since the Jacobian of $C$ is a $2$-dimensional abelian variety, it is either simple or isogenous to a product of two elliptic curves. – François Brunault May 9 '12 at 13:17