I.5.10bc

8/22/2021

Ah, hi reader. Sorry about that. I was just doing weed with a Silicon Valley resident. "Yuck! I'd rather watch ゲロ-themed JAV." Hah! You go do that! We'll stick to Pornhub, sicko.

After studying the cryptic engravings in my notebook, I've concluded that the month-ago-me could not finish part (a), so I shall skip that. Funny point: I do remember that when I was working on this exercise, I wasted a lot of time because I misread the exercise and assumed that

 TP(X) = 𝔪∕𝔪2

WROOOOOOOOOOOOOOOOOOOOOOOONG. It's actually

 TP(X) = (𝔪∕𝔪2)*

It's the dual vector space of 𝔪𝔪2.

### B

Letting Q = ϕ(P), and 𝔪be the maximal ideal of Y , We already know that there is a corresponding k-algebra map (the natural "pullback")

 ϕ# : Q,Y → P,X

(Ahhh, the great staple of this blog: Black sheaf symbols). And since these are local rings, it has to map 𝔪to 𝔪. Hence we can actually consider the restriction

 ϕ# : 𝔪′ → 𝔪

Now suppose h 𝔪2, so that h = g2 for some g 𝔪. Then

 ϕ#(h) = ϕ#(g2) = ϕ#(g)ϕ#(g) ∈ 𝔪2

hence we can induce a well-defined map

 ϕ#* : 𝔪′∕𝔪′2 → 𝔪∕𝔪2

with k-linearity inherited from ϕ#. Now, you can see why I was very confused by not considering the dual (because it's a map from a vector space formed from something associated with Y to something associated with X, rather than vice versa). Well, this map induces the map of the dual spaces (X to Y ). Dun.

### C

Let

 X = Z(x - y2) (the parabola) Y = Z(x) (the x-axis)

so that the projection is

 ϕ : X → Y (y2,y) → (y2, 0)

or, you can just think of it as

 (y2,y) → y2 = x

So you can just think of Y = A1, the affine line. Look, I made a cute picture of it
Now we consider

 TO(ϕ) : TO(X) → TO(Y )

Reminder: The pullback that induces this is

 δ : 𝔪′∕𝔪′2 → 𝔪∕𝔪2f f ∘ ϕ

Here's the key point: Y is actually just the affine line, and its maximal ideal is therefore just (x), so the "f" here is necessarily in the form x g(x) for some g k[x].

 x ⋅ g (x ⋅ g) ∘ ϕ

Let's try plugging in a point into the RHS:

 ((x ⋅ g) ∘ ϕ)(y2,y) (holy shit this notation is getting cluttered) = (x ⋅ g)(y2) = y2 ⋅ g(y2) = 0

The last equality follows from the fact that the maximal ideal 𝔪 is (x,y), so y 𝔪, which means y2 𝔪2, which is hence 0 in 𝔪𝔪2. So δ is the 0-map, and hence so is its dual (i.e. TO(ϕ)). Done.