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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1$\begingroup$ What's an "essentially elliptic subfield"? $\endgroup$– AngeloCommented May 9, 2012 at 11:55
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$\begingroup$ 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). $\endgroup$– SrilakshmiCommented May 9, 2012 at 12:36
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1$\begingroup$ You could take a look at Ernst Kani, "Elliptic curves on abelian surfaces". $\endgroup$– Dan PetersenCommented May 10, 2012 at 5:13
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1 Answer
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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.
answered May 9, 2012 at 12:54
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$\begingroup$ 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. $\endgroup$ Commented May 9, 2012 at 13:17
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$\begingroup$ Dear Francois, Thanks for your comments. $\endgroup$ Commented May 10, 2012 at 5:58
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$\begingroup$ What is special about genus 2 here? Can we generalize for higher genus too (looking at the jacobian decomposition of Jac(C)). $\endgroup$ Commented May 23, 2012 at 6:21
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$\begingroup$ Dear Srilakshmi, there is nothing special about the case of genus 2 : everything can indeed be read from the decomposition of Jac(C) into simple factors, more precisely the answer will depend on the number of elliptic factors and whether or not there is a repeated elliptic factor. $\endgroup$ Commented May 25, 2012 at 23:11
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$\begingroup$ Dear Francois, Thanks for your comment. I also thought that the statement can be generalized. $\endgroup$ Commented Jun 8, 2012 at 8:43