II.3.10

3/21/2022

Ahhhh, it’s foggy up here alright. Their Rade promenade is GLORIOUS, and ours is all the reputable breaking down under pressure, leaders arguing amongst themselves. March of the losers. The Alps boys picked off the earlier ports so trivially, but this one here’s like a whole different ball game. Forces concentrated in coalition in the crescent, carronades peering down from the upper decks. T-w-e-n-t-y-t-w-o-o-f-t-h-e-m. What are we gonna do? WTF IS GOING ON? Boy you look like youve seen a ghost, hahaha. Or a three decker in the smoke. No one wants to even try em. We’re way in over our heads. We got overconfident due to a lucky streak, but now we’re dealing with the real \$hit. We’re gonna frigging die, bro.

### 1 Prelims

Let’s talk about residue fields, and establish some conventions.

#### 1.1 Residue fields

First of all, as the Wikipedia page states, for a local ring R, the residue field is R∕𝔪. And when you have an arbitrary ring A, you can localize at a prime ideal P to obtain the local ring AP. The maximal ideal of this ring is PAP, so the residue field of AP is A∕PAP. This would also be considered the residue field k(x) of the scheme SpecA at the point x = P.

One very crucial fact missing from that page is this equality:

So when I take a residue field, I may use the former or latter representation, depending on what’s more convenient.

#### 1.2 A useful quotient

When it comes to polynomial rings over k, here’s a useful related isomorphism (I think k needs to be algebraically closed, which Hartshorne lets us assume in this exercise):

 k[x1,…,xn]∕(x1 − a1,…,xn − an) ≃ k xi an

#### 1.3 muh notation

And by the way, if A is a B-module, and b1,,bn B, it’s a valid question whether by (b1,,bn) I mean the ideal generated by those elements in B or in A. They are different ideals, but often times I will abuse this notation and silently convert it from one ideal to the other when it just works.

### 3.10

Now here’s the fiber of a morphism:

A morphism of scheme can be broken up... into schemes. I wouldn’t be able to tell you why it’s defined as the domain scheme fiber producted with the residue field of the target scheme. BUT, I did tell you last time how to compute fiber products:

Indirectly.

#### A

So, to clarify: The only reason we’re not saying that Xy is isomorphic to f1(y) is that there is not necessarily an established scheme structure on f1(y) (especially if it’s neither open nor closed in X). If nothing, this exercise tells us we can induce the scheme structure of Xy onto f1(y). So yes, if you’d like, f1(y) Xy as schemes. But in that case, showing that they’re homeomorphic is sufficient.

We are computing

 Xy = X ×Y Speck(y)

General products of schemes can be obtained by gluing together affines. So I’ll assume without loss of generality that X,Y are affine, with X = SpecA and Y = SpecB. In which case, we are computing

AGAIN, FOLLOWING OUR LESSON FROM LAST TIME: DEAL WITH THIS INDIRECTLY. Do NOT try to jump into the ring and try to do tensor math. In fact, we’re going to be even more abstract and indirect than last time.

Tensoring is right exact. Hmmm? Right exact? So what? Who cares about functors and all that nonsense? Well, we do. Tensors aint pretty close up, so we’re zooming out to deal with it. Don't aim where the ship is, but where its gonna be. We’re gonna hide the tensor under something else, and it begins with this right exact sequence:

Now we tensor it:

The second last arrow, of course, is surjective. Hence, there is a bijection between prime ideals of R and prime ideals of A BBy that contain ker δ = imλ = A ByBy.

Now if you do dive into A BBy, you might find something strange: That this tensor product seems to effectively be, simply, a sort of structure on A. In fact, this tensor product is just another way of localizing A at S = B y:

There goes the tensor! Yaaaay! Bye bye! Now we are going to produce a string of equivalences:

Instead of looking for ”the prime ideals of A BBy that contain A ByBy” we are looking for ”the prime ideals of S1A that contain yS1A.” But these are just ”the prime ideals of A that contain y but do not intersect S = B y” which are precisely the prime ideals of that are mapped onto y from B under the B-module structure on A, i.e. f1(y).

These ”bijections” are all given by ring morphisms, so they do in fact induce homeomorphisms.

#### B

A lot of text here, but it has a very simple picture which you might already be able to start making out.

When Hartshorne talks about the point y Y as a k, he means y is the maximal ideal (s a) in Y = Speck[s]. Now since from part A, we know that Xy = f1(y), the question is, given the map

 ϕ : A = k[s] → B = k[s,t]∕(s − t2) s s

which prime ideals of B have an inverse image of y = (s a)? To make this easier, let me decompose ϕ:

So seeing k[s] as a subring of k[s,t] and using the isomorphism theorem, we are simply looking for the prime ideals of k[s,t] that contain (s t2) and whose inverse image under δ is (s a). Hold on, what’s this δ?

 δ : k[s] → k[s,t] s s

But now let’s bring up the birb in the room. Aren’t those very, very familiar rings? If we Spec it, we just get

 g : Ak2 → A k1 (S,T) S

A good old varietyland map: Projection onto the s axis.

The question is now just ”which points on the curve s = t2 map to s = a under this projection onto the s-axis?” Or, alternatively: ”What is intersection of the curves s = t2 and s = a”? Of course, these would just be the points (a, ) and (a, ) . ( is defined because Hartshorne tells us that k is algebraically closed):

Or in the case of a = 0:

YEP, LOL. ALL WE’RE DOING IS INTERSECTING A PARABOLA AND A LINE.

Well, we’re about to analyze the relationship between a parabola and line more deeply than we ever have before: We’re gonna look at these points’ structure sheaf and residue fields. And even those of the mysterious generic points.

##### Xy’s global section

Let’s go ahead and get a good expression for the ring corresponding to Xy. We’ll sorta retrace our steps. Recall this right exact sequence:

That last arrow tells us that:

And the left hand side, as we just argued, can be expressed as

Localization commutes with quotients, so this is just

 (1)

Now let’s expand this a bit. Note that y is the ideal generated by sa in B, so yA is just the ideal (s a) generated by s a in A (here’s where I’m being kind and explaining it instead of waving it away with muh notation). Therefore

 S−1(A∕yA) = S−1[(k[s,t]∕(s − t2))∕(s − a)] (2) = (B − y)−1[(k[s,t]∕(s − t2,s − a))] (3)

That is the global section for Xy, where y = (s a). It is the ring whose spectrum is Xy.

##### 1.3.1 Residue fields of the ordinary points

As discussed, the points of Xy are only in the form (s a,t ) . So, using the global section, let’s compute the residue field at this point.

We are localizing at (s a,t ) , which is the multiplicative subset T = k[s,t] (s a,t ) , which clearly contains S = B y = k[s] (s a), so we can ignore the S once we localize:

But we are taking the residue field of this, so we want to quotient by the maximal ideal:

Uhhhhhhhhhhhh.

That's not pretty at all... All that is supposed to end up as just k? Yknow I actually managed to somehow get there from that expression. But in hindsight it’s easier to use the fraction field representation:

 Frac((k[s,t]∕(s − t2,s − a))∕(s − a,t − )) = Frac(k[s,t]∕(s − t2,s − a,s − a,t − )) = Frac(k[s,t]∕(s − t2,s − a,t − )) = Frac((k[s,t]∕(s − a,t − ))∕(s − t2)) = Frac((k[s,t]∕(s − a,t − )))∕Frac((s − t2))) ≃ k∕Frac((s − t2)) = k∕(s − t2) muh notation

The second last equality follows from 1.2, and in particular, that isomorphism is given by the following map:

 k[s,t]∕(s − a,t − ) → k s a t

So the relationship s t2 = 0 is redundant. Hence

So that’s the residue field of Xy at their only 2 (or 1) points.

##### 1.3.2 The tangent

By the way, according to Hartshorne, in the case a = 0 (when the line is tangent to the parabola), we’re supposed to get that Xy is nonreduced. If you look at the global section (3) (which is, by the way, the only section, since Xy is a singleton in this case), it becomes obvious:

 (B − y)−1[(k[s,t]∕(s − t2,s − a))] = (B − y)−1[(k[s,t]∕(s − t2,s))] (4)

s t2 = 0 and s = 0, so t2 = 0. t is your nilpotent

##### 1.3.3 Generic points

So we’re all wrapped up with the ordinary points y. Now for the generic points that cannot be drawn: The ones that, respectively, encompass the entire s-axis and parabola while mysteriously lying nowhere on them:

AND ONE MOTH IS THE EXTENSION OF ANOTHER? The respects from son to father pays back in extension: Whence geometric retraction gives back into algebraic expansion. Yes, there is more freedom sliding along a curve than a line.

We’re concerned with the fiber of the generic point η = (0) Y . And since f is induced by s s, the fiber is the singleton Xη = f1(η) = {(0)}. Let’s compute the global section (the only section, since it’s a singleton)

##### 1.3.4 Global section of Xη

Starting from , we write

 S−1(A∕yA) = (B − (0))−1(k[s,t]∕(s − t2)∕(0)A) (5) = (B − (0))−1(k[s,t]∕(s − t2)∕(0)) (6) = (B − (0))−1(k[s,t]∕(s − t2)) (7)

##### 1.3.5 Residue fields of the generic points

The residue field of Y at η = (0) is the fraction field k(s) of k[s]. Now, using the global section (7) above, let’s compute the residue field of Xη (at its only point (0)). If we localize at (0), that’s the multiplicative subset A (0), which clearly contains B (0), so that can be dispensed with, and we get

 [(B − (0))−1(k[s,t]∕(s − t2))] (0) = (k[s,t]∕(s − t2)) (0) = Frac(k[s,t]∕(s − t2)) = k(s,t)∕(s − t2) (MUH NOTATION)

This is a field, so the maximal ideal is (0), giving us back the same field. What you see above is the residue field of Xη.

Now we want to show that k(s,t)(s t2) is a 2-degree extension of k(s). This is quite intereting, cuz we’re tasked with getting a basis for k(s,t)(s t2) as a vector space over k(s). The quotient is what makes it work: E.g. s,t (or 1,t) is a generating set. (We can get polynomials in s through k(s) and we don’t need any higher degrees of t because s = t2)

Done. we’re faking it even here BTW, because we provided no explanation for the significance of the fact that ”the residue field of the parabola is a 2 degree extension of the residue field of the line when projected onto it.” But we proved it...