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I have an algebraic function field $\mathbb{Q}(x,y)$, where $y$ satisfies $$ (y^2-1)^2 = x^2(1+x^2), $$ and I need to find a rational function that has a first order root at $x=0,y=1$ a first order pole at $x=0,y=-1$, and no other roots or poles.

Is this possible? How do I go about finding such a function or proving that it cannot exist? Is it an issue that $y+1$ has a branch point of order $\frac12$ at $z=\pm i$? I need to automate this process, so is there any standard algorithm for this that I need to look at?

This might be easy (I don't know), but I asked a related question on MSE and got no responses at all, so I want to ask here this time.

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    $\begingroup$ Your points are singular points on the curve so talking about order of zeros and poles of functions at these points is not well-defined. Having said that, Magma can do these kind of computations. $\endgroup$ Oct 11, 2013 at 15:51
  • $\begingroup$ @FelipeVoloch Hmm. How do I modify the question so that it makes sense? I don't have access to Magma. I have a certain differential $f\,dx$ with known poles and their residues, and I need to find $v$ so that $f\,dx = c\,d(\log v)$, hence the question. $\endgroup$
    – Kirill
    Oct 11, 2013 at 16:13
  • $\begingroup$ You need to find a non-singular model of the curve. $\endgroup$ Oct 11, 2013 at 17:21
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    $\begingroup$ Since your function field has positive genus, there is no rational function having only a single root if that root has degree $1$ and order $1$ (because such a function would be an isomorphism between your field and $\mathbf{Q}(z)$). But there are two places $P,Q$ of the field which contain both $x$ and $y-1$, so you need to be careful when talking about the order of the root at $x=0,y=1$. In order to have a chance of the function existing, you must mean that it has first order roots at both $P$ and $Q$, and likewise for its poles. Magma says that no such function exists. $\endgroup$ Oct 13, 2013 at 11:48

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