Let $F$ be a function field of "transcendental degree one" over its full constant field $K$. Let $x \in F \backslash K$. We know the divisor of $(x) = (x) - (1/x)$ in $K(x)$. Could you please give me an algorithm to compute the places over two above places in $F$ and the ramification degrees.

If this setting is too abstract, what if we have $F$ is the field of fraction of $K(x)[y]/f(x,y)$ where $K$ is a finite field, could you show me any algorithm to find places over zero place and infinite place of $x$.

As KConrad suggested, I'm telling you a little about how I got involved with this problem.

Once upon a time when I was a bit younger (and a bit more stupid but not much less than now) I dared to ask Noam Elkies that how I can represent a curve with an equations of different degree than the one I'm given. For example an elliptic curve of degree 5 (you see, it's not only your time that I waste, so don't take it personal). He wrote me something that time I didn't quite understand at the time but today I went back to the email and fortunately I understood almost all of it:

*start from your sample curve
y^2 + xy + x^3 + 1 = 0 over Z/2Z*

and choose any function of degree 5, say z = x*y. Then eliminate y from the equations by computing the resultant with respect to y of y^2 + xy + x^3 + 1 with the equation
satisfied by x,y,z, which is here z - x*y. This gives z^2 + x*z = x^2 + x^5 with x,z functions of degree 2 and 5 on the curve.*

*Sincerely,
--Noam D. Elkies*

The only point which wasn't clear for me was "function of degree 5, say z = x*y". So I assumed it means that the degree of the zero divisor or the pole divisor should be 5. Although I checked it with Magma and it was the case, but I felt the need to compute the divisor for function $z$ in $K(x,y)$ myself. So I tried to compute the divisor of $x$ as the first step. Using the "Extensions = Ramified covers" rule of thumb, and looking at $x$ (the coordinate function) as the covering map to $\mathbb{P}^1$, I said that $(x)$ (the function) correspond to point $x - 0$ in $\mathbb{P}^1$ scheme so I put zero instead of $x$ in my equation and I get my two ramified points $y^2 = 1$. But for the places of over place at infinity downstairs $(1/x)$, I couldn't go that far. I changed the variable $1/\theta = x$ and put zero in $\theta$, I'll get 1=0, unless I replace $y$ with something like $\omega/\theta^2$ as well (which I don't see why) to see my ramification at infinity.

Now my question unfolded is: 1. Do you think what I'm doing makes sense and why it doesn't work for the infinite place. 2. Is there an algebraic/arithmetic way to do what I did instead of the geometric approach of covering space that I used, which I suppose would be more algorithmic friendly.

Sorry I think I gave too much of background.