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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 a any nonzero (meromorphic) differential form, i.e. a 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).

2 added 541 characters in body; added 8 characters in body; deleted 8 characters in body

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 an affine 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 a differential form, i.e. a 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).

1

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 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 bundle.

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