I.2.16

3/10/2021

Oh, twisted cubic, you appear to be too useful a source of examples in the world of algebraic geometry to
dismiss. Alas, I must dance with you if I want to do algebraic geometry. One day, I'll look back at
2.12d and figure out what went so horribly wrong there. I'm not ready yet. But maybe one day...

By the way, this is the second last exercise of this section. Whew. Almost one and a half months since I started this
section. What a slowpoke I am, huh? I-I just like taking my time, that's all.

Both parts encompass easy to make misconceptions, things that I would be tempted to conclude under
tense circumstances, and provide elegant counterexamples to those. Okay, I'm interested. Let's go.

PART (a):

Now, going off last time, I noted that for the purposes of the surface Q, we're using the coordinate ring
k[w,x,y,z], whereas the Wikipedia page (as well as my 2.12d work) for the twisted cubic curve uses the
coordinate ring k[x,y,z,w]. Thus, I'll have to translate the Wikipedia page's generators F_{0},F_{1},F_{2}, into our new
coordinates. Here's the version with new coordinates:

F_{0} | = wy - x^{2} | ||

F_{1} | = xz - y^{2} | ||

F_{2} | = wz - xy |

This exercise is asking us to compare the twisted cubic curve T = Z(F

Now give me a point P = (W,X,Y,Z) ∈ Q

F_{0}(P) | = 0 | (1) |

F_{2}(P) | = 0 | (2) |

Now, if F

Specifically,

F_{1}(P) | ≠0 | |||||

F_{1}(P) | = a | (For some nonzero a) | ||||

XZ - Y ^{2} | = a | |||||

XZ | = Y ^{2} + a | |||||

WXZ | = WY ^{2} + Wa | (yep, I multiplied by W on both sides on the basis of "All the variables except W are here, so I guess I'll just add it to the party". Sometimes stupid intuition can take you places) | ||||

(WZ)X | = (WY )Y + Wa | |||||

(XY )X | = (X^{2})Y + Wa | |||||

X^{2}Y | = X^{2}Y + Wa | |||||

Wa | = 0 | |||||

W | = 0 | (Since k is an integral domain and a≠0 by assumption) |

So W = 0. But if that's true, then (1) tells us that X = 0

So W = 0,X = 0. Sound familiar? It's a TYPE K line ! (where m = 0). So our L is actually L = Z(w,x) And hence, we've shown that Q

PART (b):

Repasting the exercise for reference:

Okay, C = Z(x

So if I'm given a P = (X,Y,Z) ∈ C ∩L (note we're in P

Now, I(P) = (x,y). And I(C) + I(L) = (x

BTW: A little errata for ya: Back in 2.15b , I said this:

Rank 2 is the 2 dimensional case (the lines coincide), rank 3 is the 1D case (singleton), rank 4 is the 0D case (empty set).

I am one off on every count. Rank 2 is 1D, rank 3 is 0D, and rank 4 is an empty set, which arguably you can't really give a "dimension" to. Doesn't affect the rest of the proof, afaik.