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Let us work over the complex numbers for simplicity. Consider a curve $C$ presented as a cyclic cover of some lower genus curve $C'$. When $C'$ has genus $0$, we can write $C$ as the normalization of the projective curve given by $$ y^n= \Pi (x- \alpha_i) $$ where $\alpha_i \in \mathbb{C}$ are the branch points.

Question: how can I decide if the curve presented this way is or not hyperelliptic?

One easy sufficient condition (for $C$ to be hyperelliptic) is: when $n$ is even and the curve defined by $$ y^{n/2} = \Pi (x- \alpha_i) $$ has genus $0$. Of course this is not necessary: the automorphism group of the curve $C$ may be bigger than $\mathbb{Z}/n\mathbb{Z}$.

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Why do you think that $y^{n/2}=\prod(x-\alpha_j)$ has genus 0??? – Serge Lvovski Feb 21 '13 at 14:19
I am saying that is a sufficient condition for $C$ to be hyperelliptic. – calc Feb 21 '13 at 14:25
Just to clarify the statement of the question: when $C'$ has genus $0$, you might need to allow some of the $\alpha_i$ to equal one another, and also it can happen that $\infty$ is a branch point of your cyclic cover of $\mathbb{P}^1$. – Michael Zieve Feb 21 '13 at 15:19
up vote 4 down vote accepted

Consider an automorphism of a hyperelliptic curve. It is sufficient to ask what the quotient by this automorphism can be. Because there is a unique hyperelliptic projection to $\mathbb P^1$, it is canonical, hence the automorphism factors through to an automorphism of $\mathbb P^1$. As a finite order automorphism of $\mathbb P^1$ it fixes two points, and if those points are moved to $0$ and $\infty$ it acts by multiplication by $\mu_n$.

Either the cyclic subgroup generated by this automorphism contains the hyperelliptic involution, or it does not. If it does, then clearly we are in your case. There are only two ramification points so the cover must have the form $y^{n/2}=\frac{(x-\alpha_1)}{(x-\alpha_2)} \prod (x-\beta_i)^{n/2}$, so the full cover has the form $y^n=\frac{(x-\alpha_1)}{(x-\alpha_2)} \prod (x-\beta_i)^{n/2}$.

In the other case, the quotient curve is hyperelliptic or elliptic, and you are taking the fiber product of the hyperelliptic cover with another cyclic cover of $\mathbb P^1$. Thus the cover must have the form $y^n=\frac{x-\alpha_1}{x-\alpha_2}$, where the meromorphic function $x$ is the hyperelliptic projection. Special attention should be devoted to the the case when $\alpha_1$ and $\alpha_2$ are ramification points of the hyperelliptic projection. This will create singularities that must be normalized to get the actual ramified cover. In particular, this can make the cover not actually ramified at those points if $n$ is even. Thus the total number of ramification points can be anything from $0$ to $4$.

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Nice answer! A minor reformulation: if the curve $y^n=f(x)$ is hyperelliptic, where $f(x)\in\mathbb{C}[x]$ is a monic polynomial in which every root has multiplicity less than $n$, then $f(x)$ is either $(x-a)^i (x-b)^{n-i}\prod (x-c)^{n/2}$ or $(x-a)^i \prod (x-c)^{n/2}$ or $g(x)^n\cdot\mu(\nu(x)^2)^j$ where $\mu(x),\nu(x)\in\mathbb{C}(x)$ have degree one. Conversely, if $f(x)$ has one of these forms and $y^n=f(x)$ is irreducible then $y^n=f(x)$ is hyperelliptic; irreducibility is equivalent to the conditions $a\ne b$ and $\gcd(i,n/2)$ in the first two cases. In the last case, ... – Michael Zieve Feb 21 '13 at 16:01
...irreducibility means that $\gcd(j,n)=1$ and in addition, if $n$ is even then $\mu(x^2)$ is not a square. – Michael Zieve Feb 21 '13 at 16:04
I do not understand your last case. You say that all roots of $f$ have multiplicity less than $n$, but then one factor of $f$ is $g^n$, whose roots will certainly have multiplicity at least $n$. Am I misunderstanding or there is a typo? – calc Feb 21 '13 at 16:37
I put in the $g^n$ because $\mu(\nu(x)^2)$ can have poles; multiplying by $g^n$ for a suitable $g$ will convert all the poles into zeroes of multiplicity between $0$ and $n-1$. – Michael Zieve Feb 21 '13 at 16:59
Is your second case equal to the first, with the second ramification pushed to infinity? – calc Feb 22 '13 at 15:28

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