I.2.9b

2/13/2021

Hey, who remembers the twisted cubic curve?

Guess what? I(Y ) = (y - x^{2},z - x^{3}) = (y - x^{2},z - xy)! I've already done that part back in
1.2. YAAAAAYYYY. Now I have to figure out I(Y ), and here is where I got stuck for 3 fucking
days.

Read at your own risk. What follows is amateur math hell.

Okay: midway, in the midst of trying way too overcomplicated stuff and going insane (I was originally trying to compute it
using the I(Y ) = Sβ(I(Y )), which you'll have to forgive me for since part fucking a was all about that equation),
I had to look up the Wikipedia page on it to unstuck myself. And according to the wikipedia page, generators
are:

F_{0} | = XZ - Y ^{2} | ||

F_{1} | = Y W - Z^{2} | ||

F_{2} | = XW - Y Z |

And there we are. That's the answer. YAAAAAAAAAAAAAY ONTO THE NEXT POST just kidding obviously I have to show why these are the generators blah blah fuck.

Following the exposition of this exercise, I can start by finding a set isomorphic to Y in projective space. I.e. Let me construct some T ⊂ S

| (1) |

is a homeomorphism.

(BTW: to match the Wikipedia article, I'm going to let the coordinate ring of P^{3} be S = k[X,Y,Z,W], and, to
match my old work, I'll let the coordinate ring of A^{3} be k[x,y,z]).

So given (a,b,c,d) ∈ Z(T) ∩ U_{0}, I need

ϕ_{0}(a,b,c,d) | ∈ Y | |||||

ϕ_{0}(a,b,c,d) | = (t,t^{2},t^{3}) | (Definition of affine twisted cubic) | ||||

(b∕a,c∕a,d∕a) | = (t,t^{2},t^{3}) | (Definition of ϕ_{0}) | ||||

Hence we must have

(b∕a)^{2} | = c∕a | ||

(b∕a)(c∕a) | = d∕a |

(like back in 1.2, the second equation could be replaced by (b∕a)

Some basic algebra yields:

b^{2} | = ac | ||

bc | = ad |

i.e. (a,b,c,d) must satisfy the homogenous polynomials:

F_{0} | = XZ - Y ^{2} | ||||

F_{2} | = XW - Y Z |

And thus, if you set T = F

"Hmmm, aren't we missing something? Where's F

Y | = Z(T) ∩ U_{0} | ||

Y | = Z(T) ∩ U_{0} |

But careful not to repeat my own errors, I know that it's NOT necessarily the case that Y = Z(T) ∩ U

In fact, I know for sure that Y ≠Z(T) ∩ U

So where does the third generator, F

bd - c^{2} | = 0 | (2) |

⇐⇒bd | = c^{2} | (3) |

⇐⇒(b∕a)(d∕a) | = (c∕a)^{2} | (4) |

Which represents the relationship xz = y

Now, I tried to get it "from scratch". Let me take you part of the way there. Let Y = Z(T′). Recall, I know that Z(T′) ⊂ Z(T). So if there's any element P in Z(T′) = Y = Z(T′) ∩ U

b^{2} | = ac | ||

bc | = ad |

Now combining this with a = 0, it's clear that we must also have b = 0. Which means P = (0, 0,c,d).

That is as far as I can get. I tried using the definition of closure (that every neighborhood containing P has a nonempty intersection with Z(T′) ∩ U

Fine, let's add this fucking F

So Y = Z(R) = Z(F

I(Y ) | = I(Z(F_{0},F_{1},F_{2})) | ||

= | |||

= (F_{0},F_{1},F_{2}) |

The last equality follows from LOL I'M TOO LAZY TO VERIFY IT. I expect that the work to show that (F

*Gasp* *Gasp*. Okay. Last thing the exercise asks us to show is essentially that (F

AND WE ARE FUCKING DONE.

THREE DAYS OF GETTING PWNED.

AND I STILL DIDN'T REALLY COME OUT ON TOP

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA.

Dude. I'm fucking melting. I'm fucking so exhausted. I still don't know how you're supposed to just fucking get the F