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I read following paragraph from:

G. Tamme, Teilkörper höheren Geschlechts eines algebraischen Funktionenkörpers, Arch. Math. 23 (1972), 257--259

Here $C$ is a curve of genus $\ge 2$ and $J$ is the Jacobian of $C$ and $K$ is the function field of $C$:

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 $\mu = 0$.

(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 $\mu = 0$)

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.

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?

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 $K^{\mu}$

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.

A non-trivial example would be $\mu := \tau^\ast \circ \tau_\ast$ such that $\tau^*: k(C') \to k(C)$

such that

$k := \mathbb{F}_5$

$C' := v^2 - u^5 + u + 1$

$C := y^4 + (2x^5 + 2x^2 + 4x+4)y^2+x^{10}+3x^7+x^4$

$\tau(u) = 2*x/(x + 3)$

$\tau(v) = 1/(x^8 + 4*x^7 + 2*x^6 + x^5 + x^4 + 3*x^3 + 3*x^2)*y^3 + $

$(3*x^3 + 2*x^2 + 4)/(x^6 + 3*x^5 + 3*x^4 + 3*x^2)*y$

Thank you very much indeed.

share|improve this question
It depends on what you mean by compute. If $P$ is a generic point on $C$, then $\mu(\phi(P))$ is a generic point of $C^{\mu}$ and $K^{\mu}$ is its residue field. So if you have an expression for $\mu$ you might have success. I do not understand your example of $\mu$. –  Felipe Voloch Jul 19 '11 at 19:35
@Felipe, I made the example explicit. Could you please show me how can I use the generic point technique to compute $K^{\mu}$? Thanx. (My other problem is that I'm not sure if I know the definition of the function field "of" an Ab variety, my guess is the function field whose Jacobian is that variety :? ). –  Syed Jul 19 '11 at 21:43
You haven't told me how your $\mu$ is supposed to act on divisors. You are giving me maps of fields. I still don't know what you mean, but if you mean something that will factor through the Jacobian of $\mathbb{P}^1$, the result will be $0$. Your guess for the function field of an abelian variety doesn't make sense. Any variety has its function field (e.g. the field of fractions of the ring of regular functions of an affine dense subset) which is a function field of transcendence equal to the dimension of the variety, so it's not a function field in one variable usually. –  Felipe Voloch Jul 19 '11 at 22:50
@Felipe, As much as I understood from Silverman, if you have maps between function fields, then we have map between curves. This induces maps $\phi_*$ and $\phi^*$ on the divisors. –  Syed Jul 25 '11 at 21:22
@Felipe, the function field to be computed is the function field of $\mu \circ \phi(C)$ which is an image of a curve, hence is a curve and has function field of transcendental degree 1. My guess was that $Jac(\mu \circ \phi(C)$ is $\mu(Jac(C))$ and hence $K^\mu$ would the function field I mentioned in the comment for $\mu(Jac(C))$. I changed the title to be more appropriately represent my question. –  Syed Jul 25 '11 at 21:49

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