I run GNOME 3, currently. I am not very picky on desktop environments. GNOME 3 is fine for me, for the most part. I used to run XFCE, which was also fine. I don't have a strong preference. But when I first used GNOME 3, there was an issue. There was this "feature" called a "hot corner". Whenever your mouse touches the top-left of the screen, even if you don't click anything, it automatically brings up an "activities overview" that you have to exit out of.

This feature, pissed me off. Every time my mouse scraped the top left corner of the screen, the DE decided, "OH, THIS SEEMS LIKE A GOOD TIME TO FUCKING HOG THE USER'S SCREEN". Imagine having a fucking "feature" that OPENS UP A NEW INTERFACE WITHOUT THE USER CLICKING OR PRESSING ANYTHING. AND IT'S THE DEFAULT SETTING. No, NO. It's WORSE than default. Want to disable it in the settings? Nowhere to be found. What you have to do is install gnome-tweak-tool, and then ON TOP OF THAT, you have to install a third party extension to disable it. Oh, actually, GNOME 3.34, released in September 12 2019 (Yes, like a sequel to 9/11. Coincidence?) added an option to disable it. Ah, thank you devs, for listening to the users and promptly fixing that issue... until you look at the comment history on that extension page and realize that that hte issue goes back NINE FUCKING YEARS (not to mention people mentioning they still need the extension because the in-house disabling isn't working correctly).

Like I said, I'm not picky about UI and aesthetics. And I'm not a technical guy either. Back when I was even less proactive about tweaking out inconveniences in my system, I let this "hot corner" sit for several months. After all, there was no option to disable it in the settings, and the default fix was for a version more recent than mine (ahh, Debian stability). So I dealt with it. And dealt with it. I don't know how tech nerds think, but think about it from a non tech nerd perspective: What prompts a user to fix an inconvenience? You see, at first, the inconvenience is just an inconvenience. It's just slightly annoying. The user can deal with that. Then as the inconvenience keeps erupting, the user gets annoyed enough to look for a way to disable it. They can't find it in the "Settings" area, and they have 10 tabs of trap porn open that they need to take care of, so they decide to leave it alone and just deal with it. Then the inconvenience starts to discombobulate the user while they're trying to do basic shit. You're in the middle of typing out a line of dialogue for the VN script you're working on (just fucking admit it you have one too), your elbow hits your mouse, and the mouse bops the top-left corner and just as you're typing the word "recalcitrant" the FUCKING ACTIVITIES CORNER TAKES OVER YOUR SCREEN. Then it happens again, and again, and again, and you start to get homicidal thoughts. You start fantasizing about strangling the person who was responsible for this shitty feature. Yes, the dev who added this feature probably knows 99999 times more about UI and programming than me, and is probably an actually good, dependent person, with a family and kids, or maybe even just a significant other to whom he can be there for, whereas I've accomplished nothing in my life and haven't helped anyone with anything ever BUT I STILL WANT TO STRANGLE THIS COCKSUCKING ASSLICKING SHITFUCKING WORTHLESS FUFCKFACE FOR ADDING THIS FUCKING WORHTLESS FUCKING "HOT CORNER" FEATURE THAT IS COMPLETELY FUCKING IDIOTIC. Basically, the typical user isn't proactive: They have to reach a boiling point: They have to reach a point where they realize they're irrationally wishing for the death of someone that has done nothing bad towards them.

Now, those thoughts are extremely irrational. To hate a person you've never met, have no context about, other than the fact that they added in a single inconvenient feature to a program. It's a very abstract, irrational sort of hatred, indeed. But there is some flesh to it. At least it is directed towards a hypothetical living human being. "I want to strangle the person that is responsible for this feature". And that kind of thought, albeit irrational, is actually quite common, especially on the internet. It's not crazy, it's just a sign that you actually need to get your shit together and fix something.

However: Let me ask you this. Have you ever wanted to strangle a twisted cubic curve? I have. On this fine night. Yes. I've reached this level of abstract hatred. I had enough trouble with it back in 2.9b, but with this exercise, I've fucking had it with this fucking curve. Yes, I FUCKING HATE THE TWISTED CUBIC CURVE. But don't worry, clearly the feeling is mutual. It hates me. It wants me to fucking lose. Why is it that everytime that I come across it, the mathematics around it turn into some crazy voodoo fuckery? I don't fucking get this exercise at all. And yes, I'm being sincere: While trying to do this exercise, I actually had the urge to strangle the twisted cubic curve. I actually felt that towards a mathematical object. Although, of course, the situation is the reverse: it's the twisted cubic curve that's strangling me.

Let me show you why I'm complaining so much. Last time, back in 2.9b, the exercise required me to pull out a certain F₁ polynomial out of thin air. What does this exercise have in store for us? Fucking absurdity, that's what.

The Wikipedia page already gives us the mapping for ρ:

Which means we can write the corresponding θ as:

θ : k[x,y,z,q]	→ k[s,t]
x	s3
y	s2t
z	st2
w	t3

And what we showed in 2.8 was that the twisted cubic curve is the variety C = Z(β) where β = (F₀,F₁,F₂), where

F0	= xz - y2
F1	= yw - z2
F2	= xw - yz

So our goal is to show that imρ = Z(β). Now we actually know from (b) that imρ = Z(α) (where α = ker θ), so if we show that α = β, that will be sufficient.

The invlusion β ⊂ α is pretty obvious, so the main challenge is showing α β. Okay. Give me some g ∈ α. To show that g ∈ β, I need to show that it is generated by F₀,F₁,F₂. What's one tactic we've used over and over on this blog to do this sort of thing? That's right! Euclidean division Let's give it a shot: If you interpret g as an element of k[y,z,w][x], then when you divide by F₀, you get

g

= F0q0 + r0

where r₀ ∈ k[y,z,w] (since F₀ had degree 1 in x)

Then thinking of r₀ as an element of k[y,z][w], let's divide it by F₁. This gives us

g

= F0q0 + F1q1 + r1

(1)

And r₁ ∈ k[y,z]
Problem: we can't just go ahead and divide by F₂ = xw - yz now, because this element has both x and z in it, and thus isn't an element of k[y,z] but k[x,y,z,w]! If we divided by it, we'd be reintroducing the x and z, undoing the work we did to get rid of it.

So maybe I didn't pick the most optimal order/choices in my division steps. But try it yourself, reader. Maybe start with F₂ this time, taking out the z. And then follow up with F₁, taking out the y. Oof, then F₀ reintroduces both variables. I'm pretty sure there's no combination that works. So are we screwed?

Well, let's just go along and apply θ on (1) and see what happens. Since g ∈ ker θ, as do F₀ and F₁, we get

0

= θ(r1)

Okay, now here's the thing. r₁ is an element of k[y,z] that maps to 0 under θ. Can you think of any nonzero element that would do that? I can't. I actually think that r₁ has to be 0. If you're unconvinced, let me explain. Let's use the multinomial notation for r₁ and collect like terms, so we can write

r1

= ∑ iγiXδi

where γ_i ∈ k, each δ_i = [a_i,b_i] is distinct (since we collected like terms)m and so X^δ_i = y^a_iz^b_i. Now we know that this becomes 0 when we map it under θ, so more specifically, we can write

0	= θ(∑ iγiXδi)
	= ∑ iγiθ(yaizbi)
	= ∑ iγiθ(y)aiθ(z)bi)
	= ∑ iγi(s2t)ai(st2)bi)
	= ∑ iγis2aitaisbit2bi
	= ∑ iγis2ai+bit2bi+ai

So the exponent vectors for the new polynomial are ϕ_i = [2a_i + b_i, 2b_i + a_i] (running out of greek letters lmao). Now, is there anyway that two of these exponent vectors can be equal? Say we had ϕ₁ = [2a + b, 2b + a] and ϕ₂ = [2c + d, 2d + c]. Then for them to be equal, we'd need to satisfy:

2a + b	= 2c + d
2b + a	= 2d + c

i.e.

2a + b - 2c - d	= 0
a + 2b - c - 2d	= 0

Yep. Welcome back to linear algebra folks! Stick this into an augmented matrix and we get

And RREFing it yields

i.e.

a	= c
b	= d

Which means that their preimages came from a single exponent vector: δ = [a = c,b = d], i.e. they came from like terms Y ^aZ^b and Y ^cZ^d which is a contradiction: the ϕs can't be separate because we already collected the like terms, so they would have been collected together in the summation after θing it. Hence, each ϕ_i is a distinct exponent vector. Which means that in order for the polynomial to be 0, we need γ_i = 0 for all i. But then this means that r₁ itself is 0. So we can write (1) as

g

= F0q0 + F1q1

(2)

Oh, nice, so g is generated by just the first two polynomials, which allows us to conclude g ∈ β, like we needed. Done.

Wait..... just the first two polynomials?. JUST THE FIRST TWO POLYNOMIALS? I just showed that α ⊂ β, even if I just let β = (F₀,F₁). And this doesn't ruin the β ⊂ α inclusion either (in fact it makes it easier!). So... WHAT THE FUCK HAPPENED TO F₂????? I THOUGHT I NEEDED ALL OF F₀,F₁,F₂ TO GENERATE THE IDEAL FOR THE TWISTED CUBIC CURVE, BUT I ONLY NEED F₀,F₁????? BUT THAT MAKES NO SENSE. THE WIKIPEDIA PAGE SAYS THAT YOU NEED ALL THREE. Ah, maybe Z(F₀,F₁) = C, but I(Z(F₀,F₁)) = , and we need to add in the F₂ in order to get rid of the radical? NOPE. Remember α is prime. I just "showed" that α = (F₀,F₁), so then we'd actually have that = (F₀,F₁), therefore I(C) = (F₀,F₁).... BUT THIS CAN'T BE THE FUCKING CASE AT ALL, UNLESS THE F₂ IS EXTRANEOUS WHICH I'M GUESSING IT'S NOT SINCE IT WAS INCLUDED BOTH ON THE WIKIPEDIA PAGE AND IN 2.8. SO WHAT THE FLYING FUCK IS GOING ON.

I DON'T FUCKING UNDERSTAND. I DON'T. FUCKING. UNDERSTAND. THIS. FUCKING TWISTED CUBIC CURVE. I'M BEING FUCKING STRANGLED. I'M BEING FUCKING STRANGLED, FOR FUCK'S SAKE. I CAN'T FUCKING DO THIS.

I'M FUCKING MOVING ON. IF ANYONE CAN SEE MY MISTAKE, PLEASE LET ME KNOW (YES I'LL FINALLY UPDATE MY ABOUT PAGE IN A FEW DAYS). OTHERWISE, I'M JUST GOING TO MOVE ON TO THE NEXT FUCKING EXERCISE. MAYBE THE NEXT TIME I SEE THE FUCKING COCKSUCKING TWISTED CUBIC CURVE, I'LL JUST SKIP IT. I CAN'T FUCKING TAKE THIS ANYMORE. NOT TONIGHT. I SPENT THE ENTIRE DAY THINKING ABOUT THIS AND I KEEP FUCKING COMING TO THE FUCKING CONCLUSION THAT THE F₂ IS EXTRANEOUS. FUCK THIS.

I'M.

FUCKING

MOVING

ON

FUCK