I am finding the divisor of f=(x1x0)−1 on C, where C=V(x12+x22−x02)⊂PP2. 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=x2=0. Then I should look at ord(f) on this U. Then I get confused, since f is a rational function, how can f belong to k[U]? Who knows?

Question

I am finding the divisor of f=(x1x0)−1 on C, where C=V(x12+x22−x02)⊂PP2. 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=x2=0. Then I should look at ord(f) on this U. Then I get confused, since f is a rational function, how can f belong to k[U]?    Who knows?

Julien Manning · Accepted Answer

First, I would take the open set where x0≠0, and dehomogenize by setting x0=1, to get x12+x22=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 (x1,x2)-plane, the poles must lie on the line x0=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 d1 and a curve of degree d2 will always have d1d2 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.

Vikki Chapman · Accepted Answer

Your curve is a circle which in the affine plane U0:x0≠0 has equation x2+y2−1=0 in the affine coordinates x=x1x0,y=x2x0.    The function f restricted to C0=C∩U0 can be written f(x,y)=x-1. It should be pretty clear that its divisor is (divf)∣C0=2⋅P where P=(1,0)=[1:1:0].    At the points at infinity of C , namely    ∞+=[0:1:i],∞≡[0:1:−i], use the coordinates u=x0x1,v=x2x1.    Then near these points C has equation 1+v2−u2=0, the point ∞+ has the expression f=1−uu.     Since u is a uniformizing parameter at ∞+ for we see that (divf)(∞+)=−1.    Similarly (divf)(∞−)=−1 and finally we get    divf=2⋅P−1⋅∞±1⋅∞−    Note that that the degree of that divisor is zero, just as it should.

I am finding the divisor of f=(\frac{x_1}{x_0})-1 on C, where C=V(x^2_1+x^2_2-x^2_0)\subset

Answered question

Answer & Explanation

New Questions in Algebra II