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.

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.
