But every week’s a bad week of hither in the gray season. Action without direction under unorganized units and split forces–or rather awful inaction and indecisiveness. We had no viable siege artillery, nothin’ more than field brass 4pdrs. But while our commanders are dragging their feet over the critical promontory, I decided to get something on the floor: Sketching out finiteness, and building up to our first battery: So we made an exhausting, depressing trip along the coast, sighting old iron ordnances standing lone and apart, muzzles pointed aimlessly into the sea. What did the boys feel when they looked upon them? Regret, frustration. Those superheavy mortars from another time, those cannons imprinted with fleur-de-tensors: Did they intimidate our boys? Demoralize them? With the lack of supplies coming in from the capital, they felt like no-good scavengers of the royal regime they so confidently renounced. These cannons, so old and esoteric in operation, unfriendly to their soft touch. Abandoned by veterans of the disbanded coast guard who were so intimate with the machinery, now to be manned by a bunch of amateurs? Hah! WHERE ARE THE PRIME IDEALS? God, why are they so confusingly emergent? So sparsely spaced? These aint our piece of cake. These low ass naval carriages were meant to be assed back and forth tween reasonable readjustments, not on a long lugged journey along the coast. These cannons were meant to sit on their coastal thrones for ages, bequeated unto the son’s sons of coastal folk and taken close care of for generations more, looking out onto the horizon for the figures of ghostly frigates. And we’re just going to snag em like thieves? We’re going to uproot them from their ideal positions and take them for ourselves? The boys felt awful. Illegitimate. Scavengers. Unroyal. And they felt it more when they struggled to lug them back over the long journey to our park. In all their iron heaviness, their naval making, their ancient awfulness, it was an awkward, frustrating, infuriating journey to bring them here. I wrangled a group of reluctant youngsters to set up a basic battery along the northern side of the harbor. Then in the frustration figuring out the kinks of these iron beasts, awkward swinging them along the rails of the traversal carriage, taking breaks to stare out at the Aurore and other ships of the line gathered on the Rade, they started to realize: They were setting up these guns on the harbor, overlooking the coast to aim at great ghostly frigates in the gray. They were handling the coastal artillery of which they felt so underneath.


Back in varietyland, it was once boldly declared:

Now, follow along. The (fiber) products of schemes are defined according to the usual universal property:




Do these products exist in schemes? The answer is yes:




What do they look like? If all is affine, then:




Indeed, if all is affine, we have

 

Yes. You’re reading that correctly. That’s a ⊗, not a × on the right. The product of affine schemes (over an affine scheme) doesn’t derive from the product of rings, but the tensor product of rings.

So, what’s the big deal about tensor products? Tensorphobia? Here’s the thing, friend: A tensor product T is simply universal in that any billinear map from A×B to S factors uniquely through T into a linear map into S. Hahahaha, just kidding. Well, so the intuition lies in billinearity, fine. But that sort of abstraction is not enough for us. We need to get down and dirty. What does A ⊗_SB look like?

The basis elements of A ⊗_SB look like this:

 

where a ∈ A,b ∈ B. (I will often drop the S in ⊗_S, by the way)

But be careful: I said basis elements, not elements. That’s an easy mistake to make. By that I mean, the general elements of A⊗_SB are actually finite sums of such basis elements:

 

What you see above is an arbitrary element of A⊗B. Now, what’s strange about this to the layman? Personally, it is extremely tempting when I see a₁ ⊗ b₁ + a₂ ⊗ b₂ to try and ”add” them together to make a single element a⊗b. In general, this is not possible. You can add a₁ ⊗b₁ and a₂ ⊗b₂, but this sum cannot be ”simplified” in general. It is just a₁ ⊗b₁ + a₂ ⊗b₂. Here’s when you can combine elements:

a ⊗ b1 + a ⊗ b2	= a ⊗ (b1 + b2)
a1 ⊗ b + a2 ⊗ b	= (a1 + a2) ⊗ b

 

The motivation, again, is that the addition is ”billinear,” if that matters. Also, given s ∈ S, you can multiply it in (since A and B are S-modules):

 

And finally, in our case we’re actually working with tensor products of algebras, so we can do this:

 

And that’s tensor math. These operations make it a ring (an S-algebra).

Now that we’ve gone through these operations, tensor math doesn’t seem all complicated. Tensor products aren’t so unworkable on their own. Their algebraic structure seems to be decently workable. What’s the big deal?

3.9: how2 fiber product

 

This exercise sums it up: We’re not doing algebra, we’re doing algebraic geometry. Since these are rings, we’re interested in taking the spectrum of them:

 

That gives us the fiber product So the important question for us geometers is: What are the prime ideals of A⊗_SB? This exercise says that there is no good way of computing the fiber product in general. The fiber product SpecA ⊗_{Spec S}SpecB can, in fact, look very different from the individual spectra.

So what do we do?

Facing your fears doesn’t mean fearlessly walking into the mouth of the beasts: That’s how you get shot. It often means grumbling and complaining, swalling your terror for little instants where you poke the bush and leap back in pathetic panic at any rustling (caused by yourself). Awkwardly lugging it along the coast, struggling to set it up, taking stupid breaks, and realizing after a day’s work that you’re done. In fact, they key to tensor products is hiding from them. I’m not kidding. Just watch.

A

 

So, since A_k¹ ≃ Speck[x], and A_k² ≃ Speck[x,y], this problem is just the algebraic problem of proving

 

So here’s the map I’m going to use:

ϕ : k[x] ⊗kk[y]	→ k[x,y]
f(x) ⊗ g(y)	f(x)g(y)

 

That’s right. Just multiply the polynomials together. Now, this may raise some alarm bells: How do we reach elements that can’t be separated into products of single variable polynomials? E.g. xy − 1 is irreducible, we can’t separate that into ”polynomial in x times polynomial in y.” But you have to remember: I defined the above map on the basis elements. The more general elements are sums of basis elements. So we can reach xy − 1, like so:

 

The main insight is that any polynomial in k[x,y] can be experessed as the sum of monomials. In other words, our inverse map looks like this:

δ : k[x,y]	→ k[x] ⊗kk[y]
∑ i=1nβ ixaiybi	∑ i=1nβ i(xai ⊗ ybi)

 

In a sense, what we’re saying by making this isomorphism is that ”polynomial rings are linear combinations of monic monomials.” Quite elegant.

Note, we computed the fiber product not via any categorical tensor philosophy, but by comparing it to a more familiar, ordinary ring. We hid from the tensor product–or, rather, we hid the tensor product under another ring.

B

 

So, we are computing the spectrum of R = k(s) ⊗_kk(t).

Clearly

k(s) ⊗kk(t) ≃ k(s,t)

 

via an analagous argument to A, so we are computing the prime ideals of k(s,t), which is a field, making it a singleton.

....which makes no sense because the exercise is asking us to show it’s not a singleton.

GOOD CATCH, YO: Catch the author’s intent, and correct yourself accordingly. Here’s the thing. We do have an inclusion given by

ϕ : R = k(s) ⊗kk(t)	k(s,t)
⊗

 

However, this inclusion isn’t surjective. You can add numerators, but you can’t add denominators, so while this map can reach something like st − 1, it can’t reach something like . What we have is a subring of k(s,t) where we can only divide by elements that are separable into f(s)g(t).

So what are the prime ideals of k(s) ⊗_kk(t)? Seems like a difficult problem. The answer is, of course, deal with her directly.

Here is an absolute gem that Yours Truly found lying on the coast:

In our case, k(s) ⊗_kk(t) is what we’re dealing with. k(s) is a field, so it’s Notherian, and k(t) = k[t, 1∕t] is a finitely generated k-algebra. k(s) ⊗_kk(t) is a Noetherian ring. In other words, I know that all its ideals are finitely generated. Notherianness has been dry in the past, but here’s where we get to see some of its power.

Because now we know that an arbitrary ideal is in the form

 

But clearly, the denominators of the generators can be removed and we still get the same ideal. So I can in fact assume the ideals are in the form

 

Furthermore, I can assume that the f_i is factorable into irreducible polynomials g₁(s,t) g_r(s,t) because k[s,t] is a UFD. And furthermore, each of these g_is have both s and t terms (otherwise, I could just divide them out)

But this begs the question: How do these ideals relate to those of k[s,t]? Time for a lemma.

0.0.1 Lemma

If the ideal I generated by f₁(s,t),…,f_n(s,t) (with the factorizations above) in R is prime, then the ideal J generated by the same elements in k[s,t] is prime

Proof

suppose f ⋅g ∈ J. Then trivially f ⋅g ∈ I. Since I is prime, we can assume wlg f ∈ I. So

f

= f1 + + fn

 

As we’ve done many times, we’re going to multiply out the denomiators (note this trend when we try to extrapolate ”local” properties to ”global” ones):

fh1 hn

= g1h2 hnf1 + + gnh1 hn−1fn

 

Like last week, I claim we can cancel out the h_i etc. to isolate f, but again I want to maintain the generators f₁,…,f_n. E.g. canceling out h₁ maintains f₂,…,f_n, but does it mess with f₁? The answer is no, because h₁ is in the form g(s)h(t), and we assumed that none of f₁s irreducible factors are in that form.

QED

So since all the prime ideals of k[s,t] are generated by irreducible elements anyway, we can compute the prime ideals of R by simply looking at the prime ideals of k[s,t] and weeding out the ones that don’t work.

If any of the (irreducible) f_i is in k[s] or k[t], then this isn’t a prime ideal of R, since we can divide by f_i and produce the unit ideal. But that’s it. The prime ideals of R are the prime ideals of k[s,t] whose generators can’t be divided out: i.e. are given by irreducible polynomials that have both variables in them. So, e.g., (st − 1) is makes a prime ideal in R, but not (s − a) or (s − a,t − b), etc.

Again: We dealt with the tensor product k(s) ⊗_kk(t) indrectly: by appealing to k(s,t) and especially k[s,t].