Divisor of rational function I am finding the divisor of $f = (x_1/x_0) − 1 $ on $C$, where $C = V ( x_1^2 + x_2 ^2 − x_0^ 2 ) ⊂ \mathbb P^ 2 $. Characteristic is not 2.
I am totally new to divisors. So the plan in my mind is first find an open subset $U$ of $C$. In this case maybe it should be the complement of $X={x_2=0}$. Then I should look at $\text{ord}(f)$ on this $U$. Then I get confused, since $f$ is a rational function, how can $f$ belong to $k[U]$?
I know this is a stupid question, by the way.
 A: Here’s how I would do it, but my method and understanding are irremediably old-fashioned, to the extent that they may be of limited assistance to you.
First, I would take the open set where $x_0\ne0$, and dehomogenize by setting $x_0=1$, to get $x_1^2+x_2^2=1$, the unit circle! The function $f$ is now $x-1$, and this clearly has a double zero at $(1,0)$ in the affine plane, corresponding to the projective point $(1:1:0)$. Since $f$ certainly has no poles in the affine $(x_1,x_2)$-plane, the poles must lie on the line $x_0=0$, so we’re looking for points on our conic of form $(0:x:y)$. But there you are: they are $(0:1:i)$ and $(1:-i:0)=(0:i:1)$.
The upshot? The divisor is $2(1:1:0)-(0:1:i)-(0:i:1)$.
Let me add for an amusing point that your curve is a circle, and if you’re applying the theorem of Bézout that says that in the projective plane, a curve of degree $d_1$ and a curve of degree $d_2$ will always have $d_1d_2$ points of intersection, the four points of intersection of two circles are the two points that you know about from high-school geometry, together with the points $(x:y:z)=(1:i:0)$ and $(i:1:0)$, through which every circle passes.
A: Your curve is a circle which in the affine plane $U_0:x_0\neq0$ has equation $x^2+y^2-1=0$ in the affine coordinates $x=\frac {x_1}{x_0}, y=\frac {x_2}{x_0}$.
The function $f$ restricted to $C_0=C\cap U_0$ can be written $f(x,y)=x-1$. 
It should be pretty clear that its divisor is $(\operatorname {div}f)\vert C_0=2\cdot P$ where $P=(1,0)=[1:1:0]$.
At the points at infinity of $C$ , namely $\infty_+=[0:1:i],\infty_-=[0:1:-i]$, use the coordinates $u=\frac {x_0}{x_1}, v=\frac {x_2}{x_1}$.
Then near these points $C$ has equation $1+v^2-u^2=0$, the point $\infty_+$ has coordinates $(u=0,v=i)$ and the function $f$ has the expression $f=\frac {1-u}{u}$ .
Since $u$ is a uniformizing parameter at  $\infty_+$ for $C$ we see that $(\operatorname {div}f)(\infty_+)=-1$.
Similarly  $(\operatorname {div}f)(\infty_-)=-1$ and finally we get $$ \operatorname {div}f=2\cdot P- 1\cdot \infty_+     - 1\cdot \infty_-    $$ Note that that the degree of that divisor is zero, just as it should.
