Principal divisor of degree $\neq 0$

Question

Consider the curve $$C:x+y+y^3+x^4=0\subset\mathbb{A}_k^2.$$ I want to calculate the principal divisor of $f:x+y\in k(C)$.

Since $f$ has no poles in $C$, we just look at point $(a,b)\in C$ with $a+b=0$. That is $(0,0)$ and $(1,-1)$.

We have $v_{(0,0)}(f)=1=v_{(1,-1)}(f)$. So the principal divisor is $$\operatorname{div} (f)=1\cdot(0,0)+1\cdot (1,-1).$$

Is this right so far?

Now consider the homogenization $\widetilde{C}$ of $C$, that is $$\widetilde{C}:x_0^3x_1+x_0^3x_2+x_0x_2^2+x_1^4=0\subset\mathbb{P}_k^2.$$ The points with $f(a:b:c)=b+c=0$ on $\widetilde{C}$ are $(1:0:0)$ and $(1:1:-1)$. So the divisor of $f$ on $\widetilde{C}$ is $$\operatorname{div} (f)=1\cdot(1:0:0)+1\cdot (1:1:-1).$$

Now my question:

I have learned that the degree of a principal divisor on a projective curve is equal to $0$. But here we have $\deg \operatorname{div}(f)=2\neq 0$. So where is the problem?

Answer 1

Look at the "point at infinity" on your projective curve, namely, $[0:0:1]$.

To deal with the point at infinity, write $f = x_1/x_0 + x_2/x_0$ so that $f$ is well-defined on the entire projective curve. Go to the affine chart $[x_0 : x_1: x_2] = [u: v: 1]$, where the equation for the curve becomes $$ u^3 v + u^3 + u + v^4 = 0$$ and $f = 1/u + v/u$.

The point at infinity is has coordinates $(u,v) = (0,0)$. Notice that $v$ is a local paramater here and $u = -v^4/(1+u^2+u^2v)$ has valuation $4$ in the local ring, so $v_{[0:0:1]}(f) = -4$.

Also, $v_{(0,0)}(f)$ is not $1$. $y$ is a local parameter at $(0,0)$ and $f = y^3 + x^4 = y^3 [1+ y(1+y^2)^4/(1+x^3)^4]$ in the local ring at $(0,0)$, so $v_{(0,0)}(f) = 3$.

Now the sums work.

Answer 2

The local description of $f=x+y$ corresponds to $\frac{X+Y}{Z}$ in homogeneous coordinates. So you're going to pick up a pole at $[0:1:0]$.