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I start my question with an example. Suppose $F/K$ be the function field generated by $x^n - yx^{n-1} - 1 = 0$. It is not a cyclic over K(y), but if I set $t = yx^{n-1}$ then we have $K(x,t) \subset K(x,y)$ and because $y = tx^{p-1}$ we have $K(x,y) = K(x,t)$. So if $K$ has $n$-th root of unity then $F$ is a Kummer extension over $K(t)$.

So, my question is what are the conditions that given $F/K$ defined by equation f(x,y) = 0, of genus $\ge 2$, there exits $t \in F$ such that $F/K(t)$ is Kummer (of degree $n$ for some $n$ such that $(n,\mathrm{char}(K))=1$ ), or is there an algorithm that can answer this question.

I know already one obvious necessary condition that $\mathrm{Aut}(F/K)$ should have a cyclic subgroup of order $n$ and that the co-factor subgroup should be a finite subgroup of $PGL(2, K)$, but is there something more that I should look for?

I think the question (as Felipe said) boils down to computing the genus of $F^{\mathcal{C}}$ for cyclic $\mathcal{C}<\mathrm{Aut}(F/K)$, but is there a way to make sure that the genus of $F^{\mathcal{C}}$ is zero while avoiding the computation of the different (using only the group structure)?

For example using Riemann-Hurwitz and choosing a large $n$ (say $n > \mathrm{Genus}(F))$ we can make sure that the $Genus(F^{\mathcal{C}}) \le 2$. But how can I decide between the 3 possibilities of $0, 1, 2$?

I mean if we are going to compute the different, then my answer will be "try to see if you can build a Kummer extension then your function field" is Kummer"! I would like to look at the group structure and be able to say something about this question (without carrying out a ramification inspection).

In special case, it's to see that which function with even order automorphism group is hyper elliptic. One necessary condition in that case is that the binary subgroup be in the centre (I don't know if it's sufficient). So, I would like to know if there's a similar conditions for Kummer extensions to decide about this problem without referring to the ramification structure.

Thank you very much indeed!

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If you have the automorphism group of $F/K$, it should be easy. As you already know, this automorphism group needs to have a cyclic subgroup of order $n$, say $G$. If the fixed field $F^G$ (which is computable) has genus zero and a place of degree $1$, then it is rational. Finally, $F/F^G$ is Kummer if $K$ contains an $n$-root of unity. These are your necessary and sufficient conditions, which are computable.

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  • $\begingroup$ How is genus of $F^G$ computable if all I have is the automorphism group? $\endgroup$
    – Syed
    Commented Apr 20, 2011 at 19:32
  • $\begingroup$ I mean I need Ramification structure, is it possible to just using $Aut(F/K)$ and $G$ we can compute the genus of $F^G$? $\endgroup$
    – Syed
    Commented Apr 20, 2011 at 19:34
  • $\begingroup$ I don't think you'll be able to do it without looking at ramification. $\endgroup$
    – Felipe Voloch
    Commented Apr 21, 2011 at 0:39
  • $\begingroup$ @Felipe I read in this paper: arxiv.org/abs/math/0205314 at top of page 2: A curve with large automorphism group always has orbit genus 0. Large means |Aut(C)| > |4(g-1)|. $\endgroup$
    – Syed
    Commented May 26, 2011 at 18:27

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