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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.

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Is this something you know how to do for an extension of number fields already (to "find" the primes over a given prime)? What is the nature of your constant field? Is it perfect (e.g., finite or char. 0)? Some context behind why you're asking the question may be useful, in case someone has a good suggestion about the goal behind the question. –  KConrad Jul 1 '10 at 23:30
    
Thank you very much for your comment. I'll re-arrange it right away. –  Syed Jul 2 '10 at 2:26
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2 Answers

up vote 2 down vote accepted

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 $2$.

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it seems that u should take a look at stichtenoth's book "algebraic function fields and codes" where you'll find many general results and examples on computing number of places, ramification degrees and so on for finite and separable extensions of a rational function field...also, there is this software called kash (it's free and works under linux and windows) which gives you all of that for concrete examples over finite fields.

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