Ces matins gris si doux...

The Senate recognizes a quorum.

The clerk will read file item 1.

File item 1: Senate resolution 43: In which a floor session in this House is permitted to coalesce separate topics into one post [in the interest of maintaining a custom of holding regular sessions once a week, on Fridays].

Gahahahahahahahahahahahahaha! Now that is a joke! Many jokes have been made in the presence of your fellow legislators, and the gallery peering upon us from above like gargoyles in the night. And what for? Lulz? Blblblblblblblblblblblblblblblblblblblblblblblblblblblblblblblblblblblbl. Those damn fools stuck at the Rhine! How long has it been! Blblblblblblblblblblblbblblblblblblblblblb. Squander your dignity for a little fame; That's the way, baby. I'm loveless and lidless. I'm dead. I'm dead. I'm dead. Off to the fields, now. Off the post and off to the fields, I say! Die that way, rather than be abdicated, I say! Blblblrrolr. We've killed millions. And the failures of our position are unpatchable. The party is over. The kingdom is no more. We cannot keep going up. We cannot reach the heavens. We are stuck here, humbled on the ground. The party is daily life: the meaningless meandering of now; the trudge of mundane. *Spits*. Or spend it all and lose it all to fulfill history. Your destiny may be to be ruined. And that is a valid destiny. "Tragedy excites the soul, lifts the heart, can and ought to create heroes." A fall from grace is never graceful; but let it be painted as such, in respectful memory of the rise. GIVE THE MAN HIS DESSERTS. HIS CAKE, HIS APPLE PIE, HIS ITALIAN SWEETS. GIVE HIM THE VOLUPTUOUS BREASTS OF HIS MISTRESS. Big, warm breasts. Big... warm... Ah, procure me with those pillows. I would become unweaned again to lay on them.

We will adjourn today in memory of The Men Without Fear:

Audible Albert. Boob Bailey. Calculus Cecillia. Dashing Derek. Effeminate Eunice. Fill-her-up Fuschia. Giga Gordon. Hell Hailey. Irksome Isabelle. Jester Jelly. KKKMember3. Lapdancing Lanice. Mail-the-goods Melendez. Nifty Nilliam. OoOoOoO. Pull-me-in Perendez. Questing Quanah. Riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley riley. Stoops-high Sturgeon. Tortoise Timothy. Umbraged Ursula. Vore-Fetishist Villiam. Whore-Fetishist Wiliam. Xylophone Xander. Yellow Yonnie. ζ.

2.6: Spec0 is the first scheme

 

The Senator from SD44 is recognized.

Thank you, President. We shall begin with dual of finality: initiality. Whereas SpecZ is the final ring, Spec0 is the initial ring. In fact, with 0 itself being a final ring, it is automatically the initial affine scheme, but we are concerned with the category of schemes more generally.

The geometry of nothing

The trick is to notice that (0), the only ideal in the ring 0, is NOT prime, as it comprises the entire ring. Hence,

 
It as an empty set. We are thus computing the geometry of nothing. The topology of the empty set is necessarily trivial and the structure sheaf is necessarily 0:

Paths from the first scheme

Hence, given a scheme X, we'd like to quantify

 
ϕ is forced to be the vacuous morphism. Given V open in X, we'd like to quantify

ϕ#(V ) : X(V )

→ Spec0(ϕ-1(V ))

 
The right hand side is 0, so this has to be the zero morphism. That finishes 2.6.

2.8: The spectrum of dual numbers

 

The Senator from SD24 is recognized.

(1)

Call them linear functions, if you'd like.

Now to find its prime ideals. k[ϵ] is a principal ideal ring, so therefore so is k[ϵ]∕(ϵ²). So every ideal is principal generated by an element in the form of (1). And what we're really searching for are simply prime elements, up to a constant factor

*Ahem* It's time, now, that you are introduced to the wonderful world of nilpotency. When it comes to dual numbers, ϵ² = 0. Correct: ϵ is a nonzero number, but when you square it, you get zero. So zero is NOT a prime number in the dual number system.

Now let's take an element given by (1) more generally. Look what happens when we multiply by its conjugate:

 
So if b≠0, then we are doomed to generate the entire ring (1) = Speck[ϵ]∕(ϵ²), which doesn't count as prime. Hence, the only option we are left with is for prime elements are in the form

 
which all corresponds to the ideal

 
THAT is our only prime ideal. So the spectrum of the dual numbers is A SINGLE POINT; Speck[ϵ]∕(ϵ²) = {(ϵ)}, a singleton. However, EVEN A SINGLE POINT CAN HAVE MANY DIFFERENT GEOMETRIES ON IT, a la the structure sheaf. Here, we send all sections (of neighborhoods of y = (ϵ)) to (k[ϵ]∕(ϵ²))_(ϵ), and the stalk at y is also the same.

I spy shudders in the audience. Some men following along have encountered a singleton scheme before. In fact, this ring of dual numbers isn't even the canonical example of such a case. That would be fields. The men who did encounter it did so in a 54-minute skirmish on the antepenultimate day of the previous month.

"EVEN A SINGLE POINT CAN HAVE MANY DIFFERENT GEOMETRIES ON IT, a la the structure sheaf," is, in fact, a quote derived from that very incident. It was an ugly 54 minutes. 54 minutes of constant flux, where certain shots and men were fired and withdrawn, fired and withdrawn, as fast as lightning. Never have I seen a battlefield change so constantly and rapidly in such a short amount of time. More occurred in those 54 minutes of "skirmish" than I've seen in some 10 hour battles. Truly, a general can be faulted but forgiven for attempting to cross the 27th meridian on that day–a reasonable miscalculation in the number assembled by the enemy–and he must nevertheless be praised for sucking it up and making the call to retreat so early. In hindsight, especially, it was a mark of talent to have recognized the value of that old adage: "Sometimes you have to lose the battle to win the war." Well, talent, of course, and experience.

And now, the moment you've all been waiting for.





















And now, to address the skirmish of yesterweekend:

'Intrigue? Beugh heugh heugh heugh heugh heugh. What do you mean "intrigue"? That man? Intrigue? You're scared of intrigue from a man of that.... stature? Beugh heugh heugh heugh heugh heugh heugh heugh heugh. Now THAT'S a joke. Hey, hey, lay another one on me. What next? Some sort of anti-royalist coup? From that soldier of the 27th Regiment? Bring it on, I say. BEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH HEUGH'

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Now, referring Y = Speck[ϵ]∕(ϵ²), hence, when we consider a morphism

 
ϕ selects an element x ∈ X, and given V open in X that is a neighborhood of x, ϕ^# takes the form

ϕ#(V ) : X(V )

→ (k[ϵ]∕(ϵ2)) (ϵ)

(2)

 
(if not a neighborhood of x, the map is trivial, ala the definition of (pre)sheaves).

Now if we localize at x (i.e. drop down to stalks), the induced map is

ϕx# : x,X

→ (k[ϵ]∕(ϵ2)) (ϵ)

(3)

 
We can hence determine (2) by simply dropping down (3). So (3) is the map that truly determines ϕ^#.

By the way, we should probably figure out what elements of (k[ϵ]∕(ϵ²))_(ϵ) look like. Well, localization means they look like rational (1)s, except we can't divide by anything in (ϵ). So they look like this:

(4)

 
Since these are local rings, the maximal ideals of each are unique. The latter's maximal ideal is the ideal generated by ϵ, which I'll again just denote as (ϵ). And since this is a local morphism, I know that ϕ_x^#-1((ϵ)) = 𝔪_x. Now let me take quotients and induce the following morphism:

π : x,X∕𝔪x

→ ((k[ϵ]∕(ϵ2)) (ϵ))∕(ϵ)

(5)

 
STEADY: The enemy is weaker than he appears. The left hand side is just what we call "k(x)" (notation taken from 2.7). And yes, yes, I understand. The right hand-side looks intimidating. Stay cool. Do not fear the beast. All we're doing is taking all the elements in (4) and setting ϵ = 0. So all that we're left with then is elements of the form

(6)

 
What did I tell you? Do not fear the beast. That awful thing on the right hand side of (5) is, in fact, just k. So (5) is just

 
Easy peasy. Now thanks to the stipulation that this morphism is k-linear, this morphism is surjective. But it is also injective. So it's indeed an isomorphism, which means that k(x) ≃ k (i.e. "x is rational over k", as Hartshorne puts it).

Finally, (3) can be likened to an element of T_x because everything in 𝔪_x gets mapped to factors of ϵ and the rest gets mapped onto k.

2.9: Generic points

 

The Senator from SD28 is recognized.



Ks ks. Zip zip zip. Listen: How does one be a generic point? What if you were one of those imperial generic points? Fashion yourself in red pajamas, laying in your rickety cot in the corner of your tiny room. You are, truly, just a weightless point–less than a dot without area or volume–and yet you lay on a web-blanket–your de facto throne from which you command the entire house, your webs spread everywhere. You lean to the right and whisper into that fossillike ear of your aide-de-camp: Reign this one in and tell him to await my next order. Where is she? You are red and yellow. A click clack in and pssshhhh of water in the kitchen and some rude thumps. Up the stairs, up the stairs, leg lift and strumming your strings and humming up. Your webs have been set all over the house. You retreat and reapportion men, often with them–restlessly mobile–from Berscia to Castiglione and Roverbella and east then west then east then west then east then west of Lake Garda. Today, at the Palazzo Serbelloni: Tug, tug. Lo! in the orient when the gracious light, lifts up her burning head. The rising sun is prepubescent, and your silk, catching the bare foot of that little girl, pulls her in, as she protests like an animal, pulls her in and traps her in your web-bed. Seduced into your cunny cocoon, her little body is yours to be taken unto with Josephine's zigzags. ζ. Seize the sweet morning first, then the day. Ces matins gris si doux? His red torso is splayed with muntin-divided yellow pieces. A sunny crucifixion. Royal in the Republic with optics and bluff. Understood? All you are is a point. You're a snubbable point. A single man in Milan, with understaffed webs spread but sparsely: checking the Austrian armies to the north, protecting your rear, maintaining the siege of Mantua, disciplining the Papal States. Transparently to anyone privy, as vulnerable as a hanged pigeon to a half-decent revolt. You are a moth that can be squashed between fingers. And yet, moths live. At awe and charisma, you are generic to where your web of influence stretches. At your point d'appui, you are the master of your closure.


For example:



You see what I mean? You are a point that holds an umbrella over many other points. And in fact, what 2.9 is telling us is that irreducible closed subsets are ruled by generic points.

Lemma: It holds for affine schemes

Since a scheme is made up of affine schemes, it makes sense to establish the fact for affine schemes first. Let's suppose X = SpecA, and Z ⊂ X is an irreducible closed set. I know that

(7)

 
For an ideal of A that I've slyly named ζ. Now note that for an arbitrary prime ideal P,

 
In an affine scheme, the closure of a point is its zero set. which means that if ζ is a prime ideal, I can rewrite (7) as

 
and we'd be done. Hence, our goal is to show that ζ is prime.

Let's suppose f ⋅ g ∈ ζ. This implies that V (f ⋅ g) ⊃ V (ζ). And from basic properties

 
so V (f) ∪ V (g) ⊃ V (ζ). Since Z = V (ζ) is meant to be closed, we have either that V (f) ⊃ V (ζ) or V (g) ⊃ V (ζ). But this means that f ∈ ζ or g ∈ ζ. So ζ is prime.

It's also uniquely determined, because if V (ζ) = V (β), with ζ,β prime, then by definition of V (⋅), we'd have α ⊂ β and β ⊂ α, hence α = β.

It holds for general schemes

Let X now be a general scheme, and Z ⊂ X be irreducible closed. Let U be an open affine of X that has a nonempty intersection with Z.

Hearkening back to the very early days of Hartshorned, I know that U ∩Z is an irreducible closed subset of U. Hence, by our lemma , I know that it has a generic point ζ. Now, ζ is a generic point of U ∩ Z from the perspective of the affine U. But is it also a generic point of Z from the perspective of X? Well,

clU(ζ)	= U ∩ Z	(8)
clX(ζ) ∩ U	= U ∩ Z	(9)

 
Now, on the one hand we have cl_X(ζ) ⊂ Z, since the closure of ζ is the smallest closed set containing ζ. And furthermore

 
So Z has a generic point: ζ. It's the same one that U ∩ Z has. And within U ∩ Z, I know that ζ is uniquely determined. But what if we had another affine with its own generic point for Z? Let's suppose we had another affine V yielding the generic point Q:

 
Well, if Q U, then U is a neighborhood of ζ where Q U, so ζ cl_X(Q), which is clearly false. So Q ∈ U, but since U is affine, Q = ζ. ζ is uniquely determined.

The Senate has adjourned until next Friday.