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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
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You could take a look at Ernst Kani, "Elliptic curves on abelian surfaces". –  Dan Petersen May 10 '12 at 5:13

1 Answer 1

up vote 5 down vote accepted

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.

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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
    
Dear Francois, Thanks for your comments. –  Srilakshmi May 10 '12 at 5:58
    
What is special about genus 2 here? Can we generalize for higher genus too (looking at the jacobian decomposition of Jac(C)). –  Srilakshmi May 23 '12 at 6:21
    
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. –  François Brunault May 25 '12 at 23:11
    
Dear Francois, Thanks for your comment. I also thought that the statement can be generalized. –  Srilakshmi Jun 8 '12 at 8:43

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