I.3.11

4/30/2021

So last time, we had an exercise that was not only pointless, but I now realize introduced a supposedly crucial concept with horrific vomit-filth phrasing. You wanna take another look?

If X is a quasi-affine or quasi-projective variety and Y is an irreducible locally closed subset, then Y is also a quasi-affine (respectively, quasi-projective) variety, by virtue of being a locally closed subset of the same affine or projective space.

Let us break down this sentence.

If X is a quasi-affine or quasi-projective variety and Y is an irreducible locally closed subset

Simple enough right? The confusing thing is that it says "Y is an... subset." and the sentence just ends, period. Subset of what? Well, obviously it's saying that Y is a subset of X, since this is the only other set that has thus far been referenced in this sentence. Sure, not too confusing. If the superset is not explicitly stated, it is implicitly another set that has thus far been mentioned. Seems legit. Basic inferential language processing skills or whatever the fuck. But let me tell you: this subordinate clause is here to throw you off. It's the nice cozy rug that you and I, reader, are unwittingly snuggling upon, and in our mutual warmth, we are closing our eyes and ears to the long, black nails wiggling at the rim below our toes. Hartshorne, the dark math queen, continues:

then

Oh God, reader! I can't continue. I just can't continue! Hold me tighter, in this blanket.

then Y is also a quasi-affine (respectively, quasi-projective) variety,

Now, our sentence is getting a little busy. We have a paranthetical muddling up the flow of the sentence, but it can't be helped, eh? Somehow, mathematician's usage of the term "variety" is fucking awful. "Quasi-affine, (respectively, quasi-projective) variety"? Hey, maybe instead of having to say both each and every time, you could come up with the term, you know, "quasi-variety"? How is that not already a term? Are you mathematicians dumbos? You know what, I'm going to use this apparently unthought of term. "Quasi-variety". Scream it from the rooftops. Tell your grocer. Bring it up at your next date. Spread the word. Y is a quasi-variety, motherfuckers. And why? Well, Hartshorne is about to bestow this valuable knowledge onto us.

by virtue of being a locally closed subset of the same affine or projective space.

HAAAAAAAAAAAAAHAAAAAAAAHAHAHA HAHAHAHAHAHAAHAHA HAHAHAHAHAAHAHAHAHAHAHA. YOU FUCKING IDIOT. YOU GOD DAMN FUCKING IDIOT. YOU THOUGHT I KNEW WHAT I'M DOING? YOU THOUGHT I HAD ANY GOD DAMN FUCKING CLUE HOW A FUCKING BOMB WORKS? I CAN'T EVEN FIX MY OWN BIKE, YOU FUCKING STUPID, WORTHLESS PILE OF FUCKING HORSESHIT. YOU WANT TO BE A FUCKING HERO AND SAVE EVERYONE, HUH? YOU WANNA BE A FUCKING HERO? TOO BAD. TOO BAD, YOU FUCKING COCKSUCKING MOTHERFUCKING IDIOT. HAHAHAHAHAHAHAHA. I START MANICALLY LAUGHING AT YOU, AS THE HOPE DRAINS FROM YOUR EYES, AND TURNS INTO ANGER. I LAUGH, LAUGH, LAUGH. PEOPLE SCREM. THE WORLD SPINS, YOU TURN YOUR EYES BACK TO THE WIRING. THE WIRING IS SPINNING. RED, BLUE WIRES? THEY ALL SEEM EVEN MORE TANGLED UP WITH EACH OTHER. THEY SEEM TO BE MULTIPLYING. NOW YOU START TO SEE GREEN WIRES AND BROWN WIRES AND YELLOW WIRES. AND AS MY LAUGHTER PENETRATES YOUR SKULL, THEY ALL SIMULTANEOUSLY TURN INTO DIFFERENT SHADES OF PINK.

So, that is basically what this sentence is doing, figuratively. Now, okay, okay chill. I was JUST JOSHING with the evil laughter, I just wanted to see your reaction. I get off on lost hope. I actually DO know which wire to cut in this situation. Let's just look at the "affine" case (in the interest of clarity; so we don't have to fucking add "respectively, projective" after every fucking mention of the god damn fucking word). X is a quasi-affine variety, meaning it is an open subset of some affine variety Z. And Y ? Y is the intersection of an open and closed subset of X. I.e. Y = U C. Now, the important thing to realize here is that the closure of Y , call it D, is irreducible, by I.1.6. So we can actually write Y as Y = U D. I.e. we can assume that in the "intersection of an open and closed subset", the closed subset is irreducible. Now D, being an irreducible closed subset of X, can be written as D = X E where E is closed in Z. Now we actually have the same situation as earlier. D is irreducble, hence its closure in Z is irreducibe. Hence we can assume without loss of generality that E is irreducible. Hence Y = U X E. Now U and X are open in Z. And E is an irreducible closed set of Z. Hence, seeing E as an affine variety and inducing the subspace topology on it, Y is clearly a quasi-affine subset of E.

THERE. That fucking elusive set we were looking for? It's this E. This E I had to come up with via expanding intersections. How is the "the same"? The same as what? Oh, so I guess "the same" was referring to the fact that it's a quasi-affine of a variety that is in "the same" ambient affine space Z that X is an open subset of. IS THAT WHAT IT MEANT? REALLY? THAT IS SO FUCKING UNCLEAR. WHAT THE FLYING FUCK?

So there, I've given you the wire to cut. Except that took me way more than 5 minutes to type out so the bomb exploded and we're all ded LOL.

Wanna know another implicit thing that the explanation above doesn't clarify? Isn't it the case that... every variety is a quasi-variety? Like, if it's a quasi-affine variety, then duh, and if it's an affine variety, then it's an open subset of itself, so it's a quasi-affine variety (respectively, ahem, projective). So if they say "variety", we should just assume "quasi". What the fuck is up with the lack of clarification, holy fuck. This terminology is just so fucking bad. A variety can be a quasi-affine variety, an affine variety, a quasi-projective variety, or a projective variety, according to the text's definition. In other words, a variety can be an irreducible open subset of an irreducible closed subset of affine space, an irreducible closed subset of affine space, an irreducible open subset of an irreducible closed subset of projective space, or an irreducible closed subset of projective space. Now if I say "affine variety" then that whittles it down to the former two, right? affine variety, you'd think would be the varieties that are subsets of affine space. but NOPE. It ONLY refers to the second one. So the terminology is just ultra buttfucked. How is this the standard terminology, holy shit.

So that is an explanation of what a "subvariety" is, I guess. Now with that under out belt, we get to start the actual exercise.

Let R = {closed subvarieties of X containing P}

Let S = {prime ideals of P}

I want to make a bijection ϕ : R S. And, I think I get the main idea here: I will show the definition of the bijection. I'm not feeling very symbo-bimbo-bashful tonight, so I won't go through all the nitty gritty.

Let's just use affines to get out of "respectively" parole.

Since X is a variety, we can assume that it's a quasi-affine variety. Say it's an open subset of the affine variety Z (with say Z An). Then, from the argumentation above, we can write X = E Z, where E is an irreducible closed subset of Z. Say C is a closed subvariety of X, so C = X D, for a closed set D in X which we can assume is irreducible without loss of generality. Now D = X E for, again, E an irreducible closed subset of Z. But since Z itself is an affine variety, so is E. So we can write E = Z(J) for some prime ideal J. I.e. C = X E. Note that E is uniquely determined by the uniqueness of closure, and J is uniquely determined by the Nullstellensatz.

 ϕ(C) = {< U,f > |f = g∕h on U,g ∈ J}

One has to verify that ϕ(C) is a prime ideal of P, which is easy.

Let me show you the inverse map as well. I'll define δ : S R like this. Suppose K is a prime ideal of P, then I'll let

 J = {g| < U,g∕h >∈ K}

And so of course, I'll let ϕ(K) = Z(J) X. Verifying that J is prime is again easy. At this point, it's pretty clear that δ and ϕ are inverses, giving a bijection... as long as I establish that ϕ(K) is irreducible. Err, well, Z(J) is irreducible and X is open in Z, so X = U Z for some open U in An. thus making ϕ(K) = U Z(J) Z.... Fuck. If it weren't for that Z, then we could claim that U Z(J) is an open subset of Z(J), thus making it irreducible by I.1.6... But adding that Z into the intersection, Z Z(J) is not necessarily irreducible.

I got stuck here, stayed up way too late last night, and had to fold. Here I am in the morning now, having figured it out (I figured it out by studying how we got that C = Z(J) X from above: noticing that said Z(J) must be a subset of Z. That was the bingo)

Let Z = Z(T) (with T = I(Z)), then since by assumption, X subsetZ(T), we know that for any given f T, f(Q) = 0 for every Q in X. In particular, f(P) = 0, so (X,f) = (X,f∕1) P. But wait: from the point of view of P, (X,f) = (X, 0) is just the zero element! (since f is zero on all of X). Hence it is contained in EVERY IDEAL (including our prime ideal, K). So, by the definition of δ, f must be in J. I.e. T J, which means that Z(T) Z(J). Hence, we can get rid of the Z in the above intersection, ϕ(K) = U Z(J). DONE.