I can't believe I'm finally fucking done with this exercise. "Umm, what about part d–" *Puts hand on your mouth.* Shhhhh. It's okay. Go to sleep, reader, go to sleep. Chloroform? Nah, you must just be really tired... You had a long day. I'm proud of you. You did such a good job today ♡. Anyone who tries to tell you otherwise, just tell them to f##k off, okay? If anyone bothers you, tell me and I'll tell them to f##k off because they don't deserve you. You are a lovely, beautiful person ♡ :). There, there... you can rest now.. night night.. *unzips and starts filming*

....No, really, this part isn't so bad. We can characterize singletons in affine space with the following result:



Also, you should know that the ring of regular functions for a projective variety is just k:



(btw: I completely skipped reading the proof of the above theorem cause it looked too complicated LOLOLOLOLLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOLOL but seriously, don't follow my example)

And by the way, in case I didn't mention it earlier, the regular function ring for affine varieties is just the coordinate ring:



Okay, theorem blast done. Now for the actual proof. Given a projective variety X, and an affine variety Y ⊂ Aⁿ, suppose there was an isomorphism between them. Well, I made a very useful lemma back in 3.1b (LEMMA 1) that said this would induce an isomorphism of (Y ) = A(Y ) and (X) = k. Hence, we'd have

A(Y )	≃ k
k[x1,…,xn]∕I(Y )	≃ k
⇐⇒I(Y ) is maximal	(since k is a field)
⇐⇒Y is a singleton	(by 1.4.4)

So, yea, that's it.