I.7.4

9/25/2021

Anyway, this exercise is slightly confusing the way it's phrased, but it's pretty simple once you decode it. Let's start by following the hint. btw: I should have mentioned this last time, but the best way to think of (P2)* is as a subset of P2 (with the Zariski topology, which Hartshorne should have mentioned last time, the fucker). But sometimes we take points in (P2)* and identify them with lines in P2. So points in (P2)* "stand for" lines in P2. So what the hint is telling us to do is find all the points that stand for tangent lines (or pass through singular points). Letting Y = Z(f), let's set

 gx = - x gy = - y gz = - z

Then the tangent lines (the points that stand for tangent lines) are given by the closed set Z1 = Z(gx,gy,gz)

Okay so having established that, we'd like to show that the points in (P2)* whose representative lines intersect S = SingY makes a closed set. Well, I know that S is closed (it's just the zero set of the partials of f), but those are the points in Y that are singular, not the lines intersecting Y .................... Fuck this. I can't figure it out. I'm sorry, I just fucking can't. I'm calling the god damn motherfucking set Z2 and going with it. Fuck this fucking god damn shit. uck this I spent so much time and I can't figure it out fucking worthless fuck.

So, I'm just going to set K = Z1 Z2 which is "obviously" a proper closed subset. Now, the implication of the hint is that the set U must be the complement of K in P2. And this makes sense. Taking a point L U, think of the line L (ABUSE OF NOTATION) it represents in P2, then look at the MAIN RESULT of this section, Bezout's theorem:

Letting Z = L for this theorem, which has degree 1, we have

 ∑ i(Y,L; Pj) = d

Now, without going into too much abstract algebra, it's intuitively clear that every point of intersection of Y with L has multiplicity exactly 1, because these are "ordinary" intersections (intersections that aren't tangent, which would have multiplicity greater than 1 according to last time, and intersections that don't cross singular points, which thinking about it visually would also possibly have multiplicity greater than 1). So there must be exactly d terms in the summation, i.e. there are exactly d points of intersection. Done.

That was a lackluster explanation, lol. Fucking pathetic. SEEYA NEXT TIME.