Part a

As I step up to the plate to write this post, I am looking at my weeks-old notes and learning that I did not record sufficiently my solution to this exercise.

One little realization I remember making is that f is homogenous. "Ummmm, duh? It's a curve in P², of course it's homogenous." Well, reader, Ispent days stuck on this exercise because I wrote f as the sum of its homogenous components, all like

f

= f2 + f3 + + fn

Hahahahahahahahaha. Well, of course, it's just

f

= f2

So fuck my ass for wasting days not realizing that. Whatever

f

= ax2 + by2 + cz2 + dxy + exz + fyz

So let's review what this exercise is asking us to do. We know that the curve defined by f in P² is nonsingular. We want to show that the curve defined by f in A³ is nonsingular (except at the origin P = (0, 0, 0), which does not exist in P²). When speaking of affine varieties, we analyze nonsingularity by looking at the Jacobian, which here looks like

(1)

This is clearly singular at P. Now the exercise wants me to show that it's nonsingular everywhere else.

Another dumbass mistake I kept making in this exercise is I kept forgetting to use the fact that P² is nonsingular... Yeah.

Now, we can't use the ez "Jacobian" definition of nonsingularity on a projective variety... unless, like back in 5.5, we cover it with open affines. Since f = 0 is nonsingular in P², we know that the affine varieties given by f(1,y,z) = 0,f(x, 1,z) = 0,f(x,y, 1) = 0 in A² are nonsingular.

But how does this help us in showing that f(x,y,z) = 0 in A³ is nonsingular? In A³, f(x,y, 1) = 0 is the set z - 1 = 0, which is a closed set. Nonsingularity can be checked on open subsets, but not necessarily on closed subsets. (since the local ring is only guaranteed to stay intact on open subsets, and nonsingularity is essentially determined by the local ring).

Well, let's take a look anyway. We know that f(x,y, 1) = 0, is nonsingular, right? In other words, the affine variety given by the zero set of

fz

= ax2 + by2 + c + dxy + ex + fy

is nonsingular. Since it's affine, we can look at its Jacobian:

(2)

We know this is nonsingular for any point in A². But we want the Jacobian in (1) to be nonsingular. Hmmm.

Now let Q = (X,Y,Z) be a point in A³ with Z≠0. Note that our Jacobian (1) is a gradient vector, so we only care if it has a nonzero component. And, that Jacobian happens to consist of homogenous polynomials. I.e. we can divide each coordinate by Z, and that won't change whether or not the component is nonzero. I.e. we can assume Q = (X,Y, 1) without loss of generality, and thus plugging it in, we get:

But now if we plug it into the nonsingular (2), we see that the first two components match, so one of the components must be nonzero. Repeat the same argument on the other open sets (X≠0,Y ≠0) and we're DONE.

Part b

Ooooh, now we get to blowup in three variables!

xu	= yt
xv	= zt
yv	= zu

Let d = degf, and consider the open set t≠0 (will be an analogous argument for the other open sets). Then we get

xu	= y
xv	= z
yv	= zu

Here's the trick (HADTA LOOK AT LE SOLUTION FOR THIS):

f(x,y,z)	= f(x,xu,xv)
	= xdf(1,u,v)

x^d = 0 is the exceptional curve, so the blowup (in t≠0) is given by

f(1,u,v)

= 0

But in part a, I already discussed that this curve is nonsingular. DONE.

Part c

Remember the discussion yesterday?(day before yesterday. I'M A DAY L8 D: ). See above: In this exercise they gave the definition of ϕ as X → X. So we're again looking at where the exceptional curve intersects X. Let's blow up again:

xu	= yt
xv	= zt
yv	= zu

Remember: the coordinates are (x,y,z,t,u,v) ∈ A³ ×P². Also remember: The exceptional curve is where x,y,z = 0. And since Y ⊂ P², the most natural map I can think of is

δ : ϕ-1(P)	→ Y
(0, 0, 0,t,u,v)	(t,u,v)

Wait... Is that.... a delta? ARRRGGGGGHHH I'M GOING INSANE... Delta is spreading, guys. My whole family suffered from covid... MY SON HAD A HIGH FEVER... which sucks, but most kids have had to deal with that... AND MY DAUGHTER... well, she was fine, I guess... BUT STILL... IT WAS SO SCARY. The only person that seriously suffered was our unvaccinated relative, but WE'RE THE VICTIMS. Well, thank God I'm vaccinated and our family has gone through this virus. *Sigh* Now, we can finally move forward.

Yes, by moving forward, I MEAN NEVER MOVING FORWARD. My children are NEVER SAFE. Wait... Guys..... GUYS..... I think Delta got me ... ARRRRRRRRRRRRRGGGGGGGGHHHHHHH NOOOOO.... IT'S DELTA.... IT'S EATING ME ALIIIIIIVE. MY LUNGSSSSSSSS. MY LUNGS ARE PERMANENTLY DAMAGEDDDDDDD ARRRRGGGGHHHH.... NOOOOOOOOOOO... SOMEONE HELP ME. IT BURNS. IT BURNNNNNSSSSSSSSSSSSSSS. *Cough, cough* OWWWWWWWWWW OWWWWWW OWWWWWWWWW. IT'S EATING ME FROM THE INSIDE...... THE VACCINE WASN'T ENOUGH. I... NEED THE ANTISERUM.... IT'S THE ONLY CURE. OSCORP'S ANTISERUM... IT'S THE ONLY WAY TO CANCEL OUT MISTER NEGATIVE'S DELTA ENERGY. ARRRRGGGHHH THE DELTA PULSES ARE INFECTING MY BRAIN. I NEED TO FULLY NEUTRALIZE THE NEGATIVE ENERGY BEFORE IT'S TOO LATE. I HAVE TO HURRY, GOD DAMN IT. *Does a somersault to get the antiserum from Doctor Octavious* OWWWWWW, I'M LOSING ENERGY. MY COVID LEGS CAN'T HANDLE THIS *Cough, cough* YOU HEAR THAT? ARRRRRRRGGGGGGHHHHH... MY LUNGS. THEY'RE EATING MY LUNGS.... NOOOOOOOOOOOO.... I'M DYINGGGGG. AAARRRRGGGGGGGGHHHHHH I NEED TO USE THE ANTISERUM... NOW... *BZZZZZT*..... *Pant, pant, pant* Ahhhh. That's much better. God, I feel so much better now. Phew. Thank goodness for that. But it's not over yet. ANYONE can get the virus, no matter what. Even the vaccine, even natural immunity, even Oscorp's antiserum, isn't enough to return back to normal. WAIT... THE LAMBDA VARIANT.... ARRRRRRGHHHHH... THE LAMBDA VARIANT IS INFECTING ME.... EVERYONE NEEDS TO GET THE ANTISERUM NOW... NYC, SF, LA, NOLA, SACRAMENTO, YOU ALL NEED TO MANDATE THE ANTISERUM FOR INDOOR ACTIVITIES.... YOU NEED TO REINTRODUCE SEGREGATION.... IT'S THE ONLY WAY....

Protip: Ignore my political opinions, I have no idea what I'm talking about. anyway, how do I verify that this is an isomorphism? Err, well it's def a bijection, but that ain't the same thing as "isomorphism" in the category of varieties, eh? The annoying thing is that the structure of A³ × P² as a variety isn't obvious. Actually, what was it, even?



Ah, all we have to do is look back at 3.16 and... O-oh... *Takes hibiscus out of hair off to respect the 6 million readers lost during The Dark Hour.*



Well, in my notes I did some fancy "open affines" business. Buuuuut, in spite of the information (and readers) lost on 3.16, I think it's faiiiiirly plausible that the set x,y,z = 0 just gives A³ × P² the structure of P². Hence δ is essentially the identity and we're done.

Well, the only weird thing is that the same argument applies to any projective variety, doesn't it?