What is the best way to find an actual divisor of an affine curve? I.E. if I am interested in finding a canonical divisor of a curve in two variables, is there a general way to go about it? Do I need to consider a projection on the x-axis?

I should clarify. I'm assuming the field is characteristic 0, and the curve is affine of the form f(x,y)=0. I computed the closure in P2, it was smooth, and am now trying to compute a canonical divisor on this curve. Thanks for the comments am reading up on it now.


4 Answers 4


As others have pointed out, there are several different ways to explicitly write down a divisor. So it would be helpful to know what kind of answer you're looking for.

Anyhow, here's one answer. For a curve, the canonical divisor is the same as the sheaf of differential 1-forms. Let's assume that your curve $C$ in the affine plane is cut out by the equation $f(x,y)=0$. Theorem II 8.17 in Hartshorne's Algebraic Geometry book yields an exact sequence for computing $\Omega^1_{C/k}$. Namely, we have an exact sequence:

$ I/I^2 \to \Omega^1_{\mathbb A^2_k}\otimes \mathcal O_C\to \Omega^1_{C/k}\to 0 $

where $I$ is the defining ideal of the curve $C$. Since $I$ is generated by $f$, and since $\Omega^1_{\mathbb A^2_k}\otimes \mathcal O_C$ is the free $\mathcal O_C$ module with generators $dx, dy$, it follows that $\Omega^1_{C/k}$ is generated by $dx$ and $dy$ modulo the relation $df$. This provides an explicit presentation for the canonical divisor as a module. When $C$ is smooth, this will be a locally free rank 1 $\mathcal O_C$-module.


Hey Elijah, the answer to your question is quite simple, elementary and explicit! You don't need to read up on anything fancy. Here it goes:

Fact: The canonical divisor of a smooth affine hypersurface is zero! In particular the canonical divisor of your curve $f(x,y)$ is $0$ since as you have mentioned the curve has a smooth projectivisation (so the curve itself must be smooth).

Proof: Let $X\subset\mathbb{A}^2$ be the affine curve defined by $f(x,y)$ which we are assuming to be smooth. Define the open sets $U_1,U_2$ in the plane by $\frac{df}{dx}\neq{0}$ and $\frac{df}{dy}\neq{0}$. Then $y$ and $x$ are local parameters in $U_1$ and $U_2$ respectively and the forms $dy$ and $dx$ are the basis of $\Omega^{1}[U_1]$ over $k[U_1]$ (respectively $\Omega^{1}[U_2]$ over $k[U_2]$). However, let us choose more convenient basis like $\omega_1=-\frac{dy}{df/dx}$ and $\omega_2=\frac{dx}{df/dy}$ on $U_1$ and $U_2$ respectively. This is permissible since the denominators don't vanish on the respective open sets. Now note that on $U_1\cap{U_2}$ both the forms are equal since $\frac{df}{dx}dx+\frac{df}{dy}dy=0$, therefore they patch to give a form $\omega$ that is regular and everywhere nonzero on $U$, so that $div\ \omega=0$ in $U$. In other words the canonical divisor is zero.

Note: This works analogously for any smooth affine hypersurface.

P.S.: Quoting an exact sequence is not a substitute for making even one small and simple calculation. Hope this motivates you for more algebraic geometry!

  • 1
    $\begingroup$ I am confused, take the line $y = 0$ ( x-axis) in $\mathbb{A}^{2}$. The differential form $x\;dx$ will have a divisor $[0]$? $\endgroup$
    – isildur
    Commented Jun 20, 2011 at 5:33
  • $\begingroup$ Dear @isildur: you are absolutely right, the formulation of the answer is a bit sloppy. The correct formulation is that the canonical divisor class of a smooth hypersurface in affine space $\mathbb A^n$ is trivial. In your case the divisor of your differential form $xdx$ is the divisor $D=1.[0]$, which is non-zero but is linearly equivalent to zero since $D= \operatorname {div}(x)$. Hence its class $D]$ is zero in the divisor class group of your curve $y=0$. $\endgroup$ Commented Mar 15, 2019 at 10:14

Do you mean actual divisor corresponding to some rational function? Then look at the zeros and poles (ie. factor the numerator and denominator, this part might need some commutative algebra in general). Otherwise, as Kevin said divisors are just formal sums of points. They are book-keeping devices that are boring without object that they book-keep for (if this made sense).

I suggest you look at projective curves, where all the nice properties are more obvious (for compactness reasons). For canonical divisors explicitly factor a differential (again the class is canonical, there are many divisors depending on the choice of a differential). The only difference is that now you need the differential part (dx) and it will change your function as you go from chart to chart. Otherwise computing its divisor is basically the same.

This is the explicit approach taken in Miranda's Algebraic Curves and Riemann Surfaces. If you have good understanding (i don't) of manifolds and bundles you can take the tangent bundle approach.


Not sure if I understand your question, but here are some generalities.

There is a correspondence between line bundles and divisors (you can find this in any algebraic geometry book). There is a canonical line bundle on any curve -- in the smooth case this is the cotangent bundle, or the sheaf of differentials, or whatever you prefer to call it. In the singular case there is still the dualizing sheaf. Then you get a canonical divisor, namely the one which corresponds to this canonical line bundle.

It's easy to find "an actual divisor" of a curve. Divisors on a curve are just formal combinations of points. So find some points of your curve, and voila.

Edit: Let me now elucidate a bit. Dan's answer provides a way to identify the canonical line bundle in your situation of interest. The next step is to identify the corresponding divisor (or more precisely, as Rado points out, the class of the divisor). You can do this by writing down any nonzero (meromorphic) differential form, i.e. any nonzero (meromorphic) section of the canonical line bundle, and finding its zeroes and poles (counting multiplicities); then the canonical divisor will be the formal sum of the zeroes minus the formal sum of the poles (counting multiplicities).


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