Elliptic subfields of a function field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:58:18Z http://mathoverflow.net/feeds/question/96426 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field Elliptic subfields of a function field Srilakshmi 2012-05-09T10:37:59Z 2012-05-10T04:57:48Z <p>Let $C$ be a curve and $K(C)$ be its function field of genus 2, where $K$ = $\mathbb{C}$. </p> <p>The number of essential elliptic subfields of $K(C)$ is 0 or 2 or $\infty$.</p> <p>Edit: I am looking for a proof. Thanks!</p> http://mathoverflow.net/questions/96426/elliptic-subfields-of-a-function-field/96444#96444 Answer by François Brunault for Elliptic subfields of a function field François Brunault 2012-05-09T12:54:37Z 2012-05-09T13:00:57Z <p>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 $P \mapsto (mP,nP)$.</p> <p>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.</p>