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](https://mathoverflow.net/users/3272/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.



Computing places over x in F/K(x)
For $x\in F\setminus K$, the degree of $x$ is the degree of the field extension $F/K(x)$. For example, in the $F$ corresponding to your curve, the degree of $x$ is $2$, since the extension $F/K(x)$ is the simple extension corresponding to $y^2 + xy + x^3 + 1 = 0$. Similarly, the degree of $y$ is $3$, since $F/K(y)$ is the simple extension corresponding to $x^3 + yx + y^2 + 1 = 0$.


The degree of $x$ is also the degree of the positive (or negative) part of the divisor of $x$. Thus if $x$ and $y$ are such that the divisor of zeros of each is disjoint from the divisor of poles of the other, then the degree of $xy$ is the sum of the degrees of $x$ and of $y$. These last conditions can be checked as follows: if $y$ is integral over $k[x]$ (i.e. if it satisfies a monic polynomial with coefficients in $k[x]$), then $y$ is finite wherever $x$ is finite. In particular, $y$ never has a pole where $x$ vanishes. Since for your $x$ and $y$, you have $y$ integral over $k[x]$ and $x$ integral over $k[y]$, you know without further calculation that the degree of $xy$ is $5$. Perhaps Elkies had something like this in mind when he wrote you.


Going back to the general case, if you want to compute the actual places where an $x\in F\setminus K$ vanishes (and the vanishing multiplicities), let me recommend
<http://www.cse.chalmers.se/~coquand/place.pdf>, which gives an algorithm. The essential step of the calculation is this: take $K$ to be algebraically closed, and suppose your field $F$ is presented as the fraction field of $K[x,y]/f(x,y)$. Suppose furthermore that you have a solution $(a,b)$ of $f(x,y) = 0$. You must then find the places of $F$ centered at $(a,b)$. If $(a,b)$ is a non-singular point of the affine model, then there is a single place, but generally one needs to resolve the singularity.


For your example, here is how to do the calculations by hand: let's compute the places where $1/x$ vanishes. Let $u = 1/x$ so that $F/K(u)$ is the extension corresponding to $u^3y^2 + u^2y+1 +u^3 = 0$. As you point out, setting $u=0$ gives no solutions. That should not worry us, since $y$ is not integral over $K[u]$, and so we have no reason to expect that $y$ should be finite where $u$ is finite.


Let's try again with $v = 1/y$. We then have $(1+u^3)v^2 + u^2v + u^3 = 0$, which is not monic in $v$, but for which the leading coefficient is invertible when $u$ vanishes. Thus $v$ will be finite when $u$ vanishes. Setting $u=0$, we find $v=0$. We still have a problem here, since
$$
K[u,v]/((1+u^3)v^2 + u^2v + u^3)
$$
is not non-singular at the prime ideal $(u,v)$, as there is no linear term in the defining polynomial, and so it is not clear how many places of $F$ are centered at this prime.


Finally, let's try $w = v/u = x/y$. We then have $(1+u^3)w^2 + uw + u = 0$. As before, $w$ will be finite when $u$ vanishes. We set $u=0$ and find $w=0$, but now we get a place of $F$, since
$$
K[u,w]/((1+u^3)w^2 + uw + u)
$$
is non-singular at the prime ideal $(u,w)$. Furthermore, $w$ is a local uniformizer, since $u = w((1+u^3)w + u)$ implies that $u$ is divisible by $w$. Pulling one more factor of $w$ out on the right, we see $u$ is divisible by $w^2$. Finally, since $u = w^2(\mathrm{unit} + \mathrm{multiple of }w)$, we find that $u$ vanishes to order exactly $2$ at this place. Since $u = 1/x$, we find that $x$ has a pole only at the place of $F$ corresponding to $(1/x,x/y)$ and that the pole order is $2$.


