Ok so for $\mathfrak a$ an ideal of $R$ we set $V(\mathfrak a)=\{\mathfrak p\in\operatorname{Spec}(R)\mid\mathfrak a\subseteq\mathfrak p\}$

Here we go bois he's pulling out the mathfrak

And then you have to verify that all of the $V(\mathfrak a)$ are the closed sets of a topology on $\operatorname{Spec}(R)$, which is called the Zariski topology

This whole construction is done by analogy with the Zariski topology on an affine (or projective) variety, so we want to decide what should a regular function on $\operatorname{Spec}(R)$ look like

Makes sense so far

And it turns out that the right choice is that regular functions on $\operatorname{Spec}(R)$ should be elements of $R$

By "regular" you just mean "sufficiently nice" and not a particular thing, right? Or is my last brain cell messing up

Where for $f\in R$ we define $f(\mathfrak p)$ as the image of $f$ under $R\to R/\mathfrak p\to\operatorname{Frac}(R/\mathfrak p)$ (the last step being superfluous if $\mathfrak p$ is actually maximal)

"regular" means "locally a quotient of polynomials" when you deal with affine/projective varieties, here we're trying to abstract this notion of regular and it should be "somewhat nice"

I see

Those functions are a bit weird, they take values in different fields when you evaluate them at different point

But they make the Zariski topology work like one would expect from the affine case, you get $V((f))=\{\mathfrak p\in\operatorname{Spec}(R)\mid f(\mathfrak p)=0\}$ and that $D(f)=\{\mathfrak p\in\operatorname{Spec}(R)\mid f(\mathfrak p)\neq 0\}$ are a basis of the Zariski topology, which is exactly the same as the affine case

so, boneheaded example, $R = \Bbb Z$, $\mathfrak{p} = 5 \Bbb Z$, $f = 3$, then $f(\mathfrak{p}) = 3\;(\text{mod }5)$?

Yup, $f((p))$ is $f\mod p$ in this case

interesting

Hmmm now we an issue because it's very hard to construct and justify the structure sheaf on $\operatorname{Spec}(R)$ without knowing how it works in the affine case

Seems that $V((f)) = \{p\Bbb Z \in \text{Spec}(\Bbb Z)\;|\;p|f\}$ in this ring?

Wait, hang on, it shouldn't include $(0)$

That's right

alright neat

probably not the most elucidating example because $\Bbb Z$ is the special-est ringboi

@Fargle So now since $\Bbb Z$ is a PID you actually have a complete description of the topology on $\operatorname{Spec}\Bbb Z$ here

aha, it's just cofinite topology isn't it?

or wait, is that goofy

It's very close to cofinite, but what's $V((0))$?

ah

Those topologies often have dense points, such as $(0)$ in $\operatorname{Spec}(\Bbb Z)$, they are called generic points

wait isn't $V((0))$ the whole space?

It is

Oh, woops, I'm using $\overline{\mathfrak p}=V(\mathfrak p)$ implicitely

oh I think I see what you're saying---it's not quite cofinite because it would have to be $T_1$

and I don't see an obvious way to make $\{(0)\}$ a closed set here

Don't worry about it, generic points are fine!

Let's see a more interesting example, what are the prime ideals of $\Bbb C[x,y]$?

forgive me, senpai, if I am bad at ring theory

Ok so there's $(0)$ because we are in a domain, there's $(x-a,y-b)$ for $(a,b)\in\Bbb C^2$ (those are exactly the maximal ideals as it follows from the weak Nullstellensatz) and then there's $(f(x,y))$ for irreducible polynomials $f$

Yeah

And those are all the prime ideals

I buy that---modulo me being a dummy that's what I was gonna say lol

Let $R=\operatorname{Spec}(R)$. As before we have $(V(0))=R$ and $V((x-a,y-b))=(x-a,y-b)$ so we have a generic point and a bunch of closed points

What's in $V((f))$ though? There are prime ideals $\mathfrak p$ with $(f)\subseteq \mathfrak p$. Since $f$ is irreducible $\mathfrak p$ must be of the form $(x-a,y-b)$ (if $\mathfrak p$ were generated by a polynomial we could factor $f$) and $(f)\subseteq \mathfrak p$ means that $f$ is $0$ in $\Bbb C[x,y]/\mathfrak p$, so in particular $f(a,b)=0$

So $V((f))$ is actually $(f)$ and $\{(x-a,y-b)\mid f(a,b)=0\}$

Which means that $(f)$ looks like a generic point not of the whole space but of the curve $\{f=0\}\subseteq\Bbb C^2$

silly question: why does $f \subseteq (x - a,y - b)$ not mean $f$ is reducible?

So intuitively the Zariski topology on $\operatorname{Spec}\Bbb C[x,y]$ looks like the Zariski topology on $\Bbb C^2$ plus a generic point for the whole space and a generic point for every curve

@Fargle $(f)\subseteq(x,y)$ for all $f$ without constant term but they can still be irreducible

oh right duh

okay I'm with you now

like, $xy - 1$ is in $(x - 1,y - 1)$ but doesn't factor

Yep

got it---so somehow the Zariski topology on the spectrum has some information about curves in $\Bbb C^2$?

That's wild

It makes sense though since the Zariski topology on $\Bbb C^2$ is defined by setting zero sets of systems of polynomials as closed sets and this is constructed by analogy with the affine case (which is $\Bbb k^2$ for an algebraically closed $k$)

I see

So rather than say all of that "downstairs" in affine space you can just do this Zariski stuff directly to the space of polynomials in two variables

Anyway I promised sheaves but it would take waaay too long to explain those now :/

lol sorry for bogging things down

I'll try to look more into this stuff myself so I'm less blind-sided---I know I have an AG book or twenty sitting somewhere around here

I think that stuff about the spec is still interesting

For sure

I think I'm gonna play around with some rings to get some more perspective

What we did is the topological part, which is the counterpart to the Zariski topology on $k^n$ if you think about the affine case

The point is that in the affine case you don't just want varieties, you want a category of varieties so there needs to be morphisms between them

And then you want sheaves because they are a neat way to keep track of all the local data concerning regular functions on the variety

And all of those has a counterpart for the Spec of a ring which is the more general case

alright I buy that

Clearly I need to brush up on my rings and fraction fields and such and so on

but this makes solid sense

Yeah some commutative algebra is definitely needed for AG. The recurring theorems so far in my AG course have been Noether normalization, the Nullstellensatz, going up/down and stuff about localisations

I was doing Atiyah-Macdonald at one point

maybe I should start again

I personally like Miles Reid's book

But that's the only CA book I used so my opinion is biased of course

I have that one as well, and it does seem a bit more down-to-earth

The only issue I have with Reid's book is that it doesn't cover tensor products which are very important, but there's plenty of other sources they can be learned from

looks like A-M covers those

so perhaps someday soon I'll know what a tensor product is

I had to learn them on the fly for algebraic topology this year since I didn't see them in commutative algebra

Which is why I warned you :P

I assume it's something like tensor products of vector spaces, but I also don't really know what those are

@Fargle Ever had a map that was linear in both terms separately and wished it was still in the realm of linear algebra, instead of "quadratic algebra" or whatever?

sure, something like $f(x,y) = xy$?

yeah

the tensor product V otimes W is THE space that makes sense in

It's the space so that if you have a map $V \times W \to X$ which is linear in both coordinates separately, you get a linear map $V \otimes W \to X$

The idea is pretty straightforward - $f(\lambda x, y) = \lambda f(x,y) = f(x, \lambda y)$, so it should be like a space of pairs $v \otimes w$ with the rule $(\lambda v) \otimes w= v \otimes (\lambda w)$

you can "scale either coordinate" and you get the same thing

I see

And then there's a canonical bilinear map $V \times W \to V \otimes W$ - given by sending $(v,w)$ to the corresponding $v \otimes w$

so my $f$ would lift to the map that acts as $x \otimes y \mapsto xy$---this is clearly homogeneous, as $f(\lambda(x \otimes y)) = f(\lambda x \otimes y) = \lambda x y = \lambda f(x \otimes y)$ (abusing notation)?

and then how do you add these pairs? Just with respect to basis products $e_i \otimes e_j$?

Unfortunately that's why this space is quite complicated

You can add $v_1 \otimes w + v_2 \otimes w = (v_1 + v_2) \otimes w$

ahh it's multilinear addition

crap

But eg $e_1 \otimes e_1 + e_2 \otimes e_2$ is its own thing

Can't simplify it at all

In general if $V$ is $n$-dimensional and $W$ is $m$-dimensional, then even though $V \times W$ is only $n+m$-dimensional, the space representing bilinear maps out of it - $V \otimes W$ - is $nm$-dimensional

makes sense

You got right down to the heart of it just a second ago. The basis is given by $e_i \otimes e_j$

Gotta know where all the independent pairs go

so, hang on---in my very trivial example, the tensor product space turns out to be 1 dimensional

Right. You had to know where $1 \otimes 1 \in \Bbb R \otimes \Bbb R$ went

(You have dictated it is sent to $1$)

right

Once you know that, you know that $a \otimes b$ must be sent to $a f(1 \otimes b)$, which must be sent to $ab f(1 \otimes 1) = ab$

Your map itself was dictated by bilinearity once you knew where that basis element went

okay the sticking point I was having was---why are $0 \otimes 1$ and $1 \otimes 0$ both mapping to $0$ if this map is a lin iso of vector spaces? but duh, those elements are equal

Yup, those are just 0

Okay that makes a lot of sense to me now.

So it's just a way of saying, "multilinear map? nah fam, it's linear"?

Yeah!

I guess the classical example is the determinant map being multilinear in the columns

Exactly

so, for example, $\mathrm{det} : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ lifts to a real-valued linear map on $\Bbb R^2 \otimes \Bbb R^2$

and it sends $e_i \otimes e_i$ to $0$, $e_1 \otimes e_2$ to $1$, $e_2 \otimes e_1$ to $-1$, so any matrix is determined

saying something like "two columns are equal" means that the element of the TSP is $a(e_i \otimes e_i) + b(e_j \otimes e_j)$, and by what I just said and by honest-to-god-actual-linearity, this maps to zero, which confirms stuff already known about det

@Fargle Yeah - though the determinant factors through something a little interesting

You have your tensor product $\Bbb R^2 \otimes \Bbb R^2$

Inside this is the space of symmetric tensors, those for which if you swap the two factors you have the same thing

It's spanned by $e_1 \otimes e_1, e_2 \otimes e_2$, and $e_1 \otimes e_2 + e_2 \otimes e_1$

The property of determinant that says "You can swap columns at the cost of a negation" is the same as saying: "The determinant kills all symmetric tensors"

ah okay that makes sense

The quotient $$\Bbb R^2 \otimes \Bbb R^2/S(\Bbb R^2 \otimes \Bbb R^2)$$ is written $\Lambda^2 \Bbb R^2$, the exterior product

And the general result is that if you take $V$ n-dimensional, and take $V^{\otimes n}$, the subspace of symmetric tensors actually is codimension 1 - so that there is a unique map $\Lambda V \to \Bbb R$ up to scaling

That map is our determinant

and it makes sense that the symmetric tensors are $n-1$-dim

er, well, rather, $n^n - 1$-dim

cough cough

I'd struggle to prove it but I do buy it

Right, that's a nontrivial fact. I'm sure the algebraists have a good proof off hand

Maybe their proof just shows that the determinant is the essentially unique multilinear alternating function :P

Seems like there ought to be some kind of combinatorical proof

Sort of building a basis by hand, sure

Is this the same construction that's applied for tensor products of modules? Or is it more subtle than that

(the general tensor product notion, not the det stuff specifically, to clarify my meaning)

It's exactly the same, though now there are subtleties about maybe your tensor product killing off things you didn't expect

right---there might be some screwery with zero divisors, or even in the integral domain case there's probably some screwery with scalars not necessarily being units

Right

In general you would define $M \otimes N$ either by "universal property" - it's THE THING equipped with a bilinear map $M \times N \to M \otimes N$ through which all bilinear maps factor

Or by explicit construction, it's the free $R$-module on pairs $m \otimes n$ with relations given by demanding that you can commute scalars across the tensor product and add in one component at a time

So, let $G$ be an abelian group, i.e. a $\Bbb Z$-module. Then the map $G \times \Bbb Z \to G$ where $(g,n) \mapsto ng$ lifts to a module hom $G \otimes \Bbb Z \to G$.

That map happens to be an isomorphism :)

Whoa, really? I find that remarkable

So, it's surjective, right?

Yeah

Now let's think about injectivity

If I have an element of $G \otimes \Bbb Z$, I can write it as some finite sum $\sum g_i \otimes n_i$

The laws of tensor products say I can rewrite that as $\sum (n_i g_i) \otimes 1$

And then additivity says I can write that as $(\sum n_i g_i) \otimes 1$

Of course, that maps to $\sum n_i g_i \in G$; if that's zero, our original vector was zero

Ah, I see! It was zero, but it was a very disguised version of zero

e.g. $g \otimes |g|$

Right right

That was what made it surprising---I didn't think to factor the scalar across and then just realize that as $0 \otimes 1$. Now it makes sense

Without knowing specific information it can be hard to predict what a tensor product will look like

But since the tensor product is about "moving $R$-linearity across", and $\Bbb Z$ is a 1-dimensional free $\Bbb Z$-module, it is reasonable that it adds nothing to the story

Is it generally true that if $M$ is an $A$-module ($A$ being a comm ring w/id), then $M \otimes A$ is isomorphic to $M$?

Yeah, we just gave the proof!

No change

Oh snap you right

just write r's and m's instead of n's and g's

Thanks fam, that made total and complete sense

Thumb up my man

Tensor products are not so mysterious

Now the shit you can do with them, and the power the end up having?

That's crazy

yeah it seems super duper robust

I need to just sit down and like actually do math instead of sampling the smorgasbord

I look back at things I was doing even a few months ago and realize that I've forgotten the material that led me to that point, because I never took the time and care to build a proper framework for what's going on in my head

I get that

yeah it's just a maturity thing mostly, I think

I'm getting better about it but only just