Given a field $K$ and an irreducible polynomial $f(x,y) \in K[x,y]$ that defines an algebraic curve, we can construct the maximal order $O$, consisting of all algebraic function with finite values at all finite points, and define a place as a prime ideal in $O$.

We need to distinguish between points (prime ideals in the coordinate ring $K[x,y] \bmod f$) and places (prime ideals in the maximal order $O$) in order to handle singularities, where there are multiple places at a single point.

For example, the curve $y^2 = x^3+x^2$ has a singularity at the origin; there are two places at a single point:

implicit_plot(y^2-(x^3+x^2), (-2,2), (-2,2), axes=True)

A function like $x/y$ can not be evaluated at the origin just by plugging in values for $x$ and $y$. Not only do you get zero divided by zero, but the value is indeterminate because we need to specify which branch of the curve we're on, i.e, which place. Once we've specified one of the two places/branches, we can then ask for the value of $\lim_{(x,y)\to(0,0)} x/y$.

We can do a lot of this calculation in Sage 9:

sage: K.<x> = FunctionField(QQbar); _.<Y> = K[]
sage: A.<y> = K.extension(Y^2-(x^3+x^2))
sage: OA=A.maximal_order()
sage: OA.basis()
(1, 1/x*y)
sage: OA.ideal(x,y)
Ideal (x) of Maximal order of Function field in y defined by y^2 - x^3 - x^2
sage: OA.ideal(x,y).factor()
(Ideal (1/x*y - 1) of Maximal order of Function field in y defined by y^2 - x^3 - x^2) * 
(Ideal (1/x*y + 1) of Maximal order of Function field in y defined by y^2 - x^3 - x^2)

This tells us:

  1. The basis for the maximal order $O$ as a $K[x,y]$-module is $\{1,y/x\}$, i.e, $\{1,y/x\}$ is an integral basis for this curve, and all algebraic functions with finite values at all finite places can be written as $a+b(y/x)$ where $a,b \in K[x,y]$,

  2. There's a singularity at the origin, since the ideal (in $O$) generated by $x$ and $y$ is not prime,

  3. The prime factors of the ideal correspond to the two places of the curve at the origin, and

  4. The value of $y/x$ is $1$ at the one place and $-1$ at the other.

So we can conclude that the value of $x/y$ is also $1$ at the place represented by Ideal (1/x*y - 1) and $-1$ at the place represented by Ideal (1/x*y + 1).

My question: given the irreducible polynomial $f(x,y)$ and the generators of one of these prime ideals in $O$, how to compute the value of an arbitrary algebraic function like $x/y$? The example I gave is easy, but how to do it in general?

An obvious way is to approach it as a limit, but I'm not even sure that I can guarantee that L'Hospital's rule always terminates here. Also, I'm looking for a more algebraic approach. Obviously, substituting in values for $x$ and $y$ works in some cases, but how do I deal with the cases where that produces $0/0$?


Let's work in the situation where $K$ is algebraically closed. This is the "geometric" situation.

In this situation, for any place, i.e., nonzero prime ideal $\mathfrak{p}$ of $O$, the quotient ring $O/\mathfrak{p}$ is canonically isomorphic to $K$. (I've outlined the reason why in a postscript.)

For any $f\in O$, you can ask what the image of $f$ is in the quotient ring $O/\mathfrak{p}$. This is an element of $K$, because $O/\mathfrak{p}$ is canonically identified with $K$. And the value of $f$ at the place $\mathfrak{p}$ is simply the image of $f$ in $O/\mathfrak{p}\cong K$!

In your example, with $f = y/x$, the image of $f$ in $K[y/x]/(y/x - 1)$ is $1$ because $y/x-1=0$ in that ring, therefore $f=y/x=1$ in that ring. Meanwhile, the image of $f$ in $K[y/x]/(y/x+1)$ is $-1$, because $y/x+1=0$ in that ring, therefore $f=y/x = -1$ in that ring.

Postscript: The reason why $O/\mathfrak{p}$ is canonically isomorphic to $K$.

This is because the curve is one-dimensional, so $O$ is a one-dimensional ring, so $\mathfrak{p}$ is a maximal ideal, so $O/\mathfrak{p}$ is a field; and, $O$ is finitely generated as a $K$-algebra, so $O/\mathfrak{p}$ is also finitely generated as a $K$-algebra, and Hilbert's Nullstellensatz tells us that a field that's a finitely generated algebra over an algebraically closed field $K$ is actually $K$. Thus $O/\mathfrak{p}\cong K$. The isomorphism is canonical because all the rings in sight contain $K$, so the canonical homomorphism from $O$ to $O/\mathfrak{p}$ amounts to a surjection onto the subring $K$.

  • $\begingroup$ OK, makes sense, but my question is how to actually compute that value for a generic rational function f, in particular an f that yields 0/0 if you just plug in values for x and y. $\endgroup$ Nov 7 '19 at 15:27
  • $\begingroup$ You can do it with Grobner bases. I'll add some info to the answer. $\endgroup$ Nov 8 '19 at 3:12
  • $\begingroup$ Oops, I started to make the edit but I don't have time now. Will try to circle back to this in the next few days. $\endgroup$ Nov 8 '19 at 3:48

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