Computing the function field of a curve given as a subvariety of the Jacobian of its cover or merely the degree of the covering - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:13:56Z http://mathoverflow.net/feeds/question/70759 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70759/computing-the-function-field-of-a-curve-given-as-a-subvariety-of-the-jacobian-of Computing the function field of a curve given as a subvariety of the Jacobian of its cover or merely the degree of the covering Syed 2011-07-19T18:06:49Z 2011-07-25T23:16:47Z <p>I read following paragraph from: </p> <p>G. Tamme, Teilkörper höheren Geschlechts eines algebraischen Funktionenkörpers, Arch. Math. 23 (1972), 257--259</p> <p>Here $C$ is a curve of genus $\ge 2$ and $J$ is the Jacobian of $C$ and $K$ is the function field of $C$:</p> <p>Es sei $\mu$ ein beliebiger $k$-Endomorphismus von $J$. Dann hat man einen $k$-Morphismus $C \underrightarrow{\hspace{2mm}\phi\hspace{2mm}} J \underrightarrow{\hspace{2mm} \mu \hspace{2mm}} J$ ; das geometrisehe Bild unter diesem $k$-Morphismus ist eine irreduzible vollstäindige Untervarietät $C^{\mu}$ von $J$. Der $k$-Morphismus $C \to C^{\mu}$ induziert einen $k$-Isomorphismus des Funktionenkörpers von $C^{\mu}$ auf einen Teilkörper von $K$, den wir mit $K^{\mu}$ bezeichnen. Es gilt $K^{\mu} = k$ d.u.n.d., wenn $\mu = 0$.</p> <p>(Disclaimed translation: Let $\mu$ be an arbitrary $k$-endomorphism of $J$. Then one has a $k$-morphism $C \underrightarrow{\hspace{2mm}\phi\hspace{2mm}} J \underrightarrow{\hspace{2mm} \mu \hspace{2mm}} J$; The geometric image under this $k$-morphism is an irreducible complete subvariety $C^\mu$ of $J$. The $k$-morphism $C\to C^\mu$ induces a $k$-Isomorphism of the function field of $C^{\mu}$ to a subfield of $K$, which we denote by $K^\mu$. Note tha $K^\mu=k$ if and only if $\mu = 0$)</p> <p>Now my question is that if we have $\mu$ as a function that gets a divisor and gives back a divisor in $J$, how can we compute $K^{\mu}$ as a function field over $k$. "Compute" in a sense of computing the defining equation(s) of the function field.</p> <p>And if it's too much to ask, is there a way to compute the degree of $\mu \circ \phi$ map as a covering of curves?</p> <p>For example if $\mu$ is simply the pullback of the Norm of a divisor (such that Norm$:K\to k(x))$ then what would be $K^{\mu}$</p> <p>To be more explicit, for example, let $k := \mathbb{F}_5$, $C := y^2 -x^3-x-2$ and let $\tau^{\ast}: k(x) \hookrightarrow k(C)$, and $Norm:k(C)\to k(x)$. Let $\mu$ be the map induced by $\tau^{\ast} \circ Norm$. How can I compute $k(C)^{\mu}$ in this situation.</p> <p>A non-trivial example would be $\mu := \tau^\ast \circ \tau_\ast$ such that $\tau^*: k(C') \to k(C)$ </p> <p>such that </p> <p>$k := \mathbb{F}_5$</p> <p>$C' := v^2 - u^5 + u + 1$</p> <p>$C := y^4 + (2x^5 + 2x^2 + 4x+4)y^2+x^{10}+3x^7+x^4$</p> <p>$\tau(u) = 2*x/(x + 3)$</p> <p>$\tau(v) = 1/(x^8 + 4*x^7 + 2*x^6 + x^5 + x^4 + 3*x^3 + 3*x^2)*y^3 +$</p> <p>$(3*x^3 + 2*x^2 + 4)/(x^6 + 3*x^5 + 3*x^4 + 3*x^2)*y$</p> <p>Thank you very much indeed.</p>