If You Want To Not Talk About Your Caste On Tumblr That’s Your Business But If You’re Upper Caste

Prof after saying maybe we should be teaching these to first year undergrads, the calculation is really simple but they might struggle with understanding forms - stuck on said calculation for the next 40 minutes of class.

Why are mathematicians* so USELESS at simple arithmetic? You're telling me I possess knowledge that would get me hailed as a lord 2000 years ago, but if I had two bags of wheat in one hand and three in the other, I would say there's six bags? What the fuck

*it's me, I'm mathematicians

And my uni does exams during Jan-Feb to make the episode party edition

shout out to everyone who participated in the january-february mass depressive episode

Awww

@mybeanalgebra I SEE YOU ONLINE 🫵 HELLO

right now, somewhere in the world, there is a beautiful person scrolling tumblr.

tragically though, this beautiful person has a headache. their head hurts.

this is very unfair to them. they are very pretty, but for no reason their head is hurting.

if you know of anybody scrolling tumblr right now, who might have a bit of a headache, please ask them to take an ibuprofen, drink some water, do whatever might make them feel better.

it is tragic that such a beautiful person has a headache. the world is a vampire. I can only hope that this message may reach them

I have a headache

Zariski topologies

So if you take kⁿ, the n-dimensional coordinate space over some field k, the Zariski topology on kⁿ is the topology whose closed sets are of the form

Z(S) = { x ∈ kⁿ : f(x) = 0 for all f ∈ S }

for some subset S ⊆ k[x₁,...,xₙ]. That is, the closed sets are the common zero loci of some set of polynomials over k in n variables, i.e. they are the solution sets for some system of algebraic equations. Such sets are called algebraic sets. If I is the ideal generated by S, then Z(S) = Z(I), so we can restrict ourselves to ideals.

Now if you take a commutative unital ring R, we let Spec R denote its prime spectrum, the set of prime ideals of R. We let Max R ⊆ Spec R be the subset consisting of the maximal ideals, the maximal spectrum. The Zariski topology on Spec R is the topology whose closed sets are of the form

Z(S) = { P ∈ Spec R : P ⊇ S }

for some subset S ⊆ R. A prime ideal P contains S if and only if it contains the ideal generated by S, so again we can restrict to ideals. What's the common idea here? Classically, if k is algebraically closed, then Hilbert's Nullstellensatz (meaning Zero Locus Theorem) allows us to identify the points of kⁿ with those of Max k[x₁,...,xₙ], by mapping a point (a₁,...,aₙ) to the maximal ideal (x₁ - a₁,...,xₙ - aₙ), and the Zariski topologies will agree along this identification. There's nothing very special about these algebraic sets though.

Let X be any (pre-)ordered set with at least one bottom element. For a subset Y ⊆ X, define the lower and upper sets associated to Y as

L(Y) = { x ∈ X : x ≤ y for all y ∈ Y }, U(Y) = { x ∈ X : x ≥ y for all y ∈ Y }.

We call a lower [upper] set principal if it is of the form L(x)= L({x}) [U(x) = U({x})] for some x ∈ X. If X is complete (any subset has at least one least upper bound and greatest lower bound), then any lower or upper set is principal. Note that ⋂ᵢ L(Yᵢ) = L(⋃ᵢ Yᵢ), so lower sets are closed under arbitrary intersections; they provide what's called a closure system on the power set of X. The lower closure of a set Y is the intersection of all lower sets containing Y. We have that Y ⊆ L(x) if and only if x ∈ U(Y), so the lower closure of Y is given by L(U(Y)). If the lower sets were furthermore closed under finite unions (including empty unions), then they would form the closed sets of a topology on X.

This is not generally true; first of all, note that any lower set contains the bottom elements of X, of which there is at least one, so the empty set is not a lower set. As for binary unions, generally we have L(Y₁) ∪ L(Y₂) ⊆ L(Y₁ ∩ Y₂), but this inclusion might be strict. This is something we can fix by restricting to a subset of X.

We say that p ∈ X is prime if p is not a bottom element and for all x, y such that for all z such that x ≤ z and y ≤ z we have p ≤ z, we have that p ≤ x or p ≤ y. That is, if p is smaller than every upper bound of x and y, then p is smaller than x or y. Furthermore, we say that p is a prime atom if it is a minimal prime element. Let P(X) and A(X) denote the sets of primes and prime atoms of X, respectively. For a subset Y ⊆ X, let the Zariski closed set associated to Y be given by

Z(Y) = L(Y) ∩ P(X) = { p ∈ P(X) : p ≤ y for all y ∈ Y }.

We again have ⋂ᵢ Z(Yᵢ) = Z(⋃ᵢ Yᵢ), so the Zariski closed sets are closed under arbitrary intersections. Note also that Z(X) = ∅, so the empty set is closed. Now let Y₁, Y₂ be subsets of X. We find that Z(Y₁) ∪ Z(Y₂) = Z(U(Y₁ ∪ Y₂)). Clearly if p is smaller than all of the elements of one Yᵢ, then it is smaller than every upper bound; the interesting part is the other containment.

Assume that p ∈ Z(U(Y₁ ∪ Y₂)), so p is smaller than every upper bound of Y₁ ∪ Y₂. If p is smaller than every element of Y₁ then we are done, so assume that there is some y ∈ Y₁ with p ≰ y. For every y' ∈ Y₂ we have that p is smaller than every upper bound of y and y', so because p is prime we get that it is smaller than y or y'. It is not smaller than y, so p ≤ y'. We conclude that p ∈ Z(Y₂), and we're done.

As before, the Zariski closure of a set of primes Q ⊆ P(X) is given by Z(U(Q)). Note however that for a point x ∈ X we have L(x) = L(U(x)), so the Zariski closure of a prime p is Z(p). It follows that A(X) is exactly the subspace of closed points of P(X).

So we have defined the Zariski topology on P(X). How can we recover the classical examples?

If X is the collection of algebraic subsets of kⁿ ordered by inclusion, then P(X) consists of the irreducible algebraic subsets, and we can identify kⁿ itself with A(X). Our Zariski topology coincides with the standard definition.

If X = R is a unital commutative ring, ordered by divisibility, then being prime for the ordering coincides with being either prime for the ring structure, or being equal to 0 if R is an integral domain. Note that this ordering is not generally antisymmetric; consider 1 and -1 in a ring of characteristic not equal to 2.

A more well-behaved version of the previous example has X = { ideals I ⊴ R }, ordered by reverse inclusion. Note that for principal ideals (r), (s) we have (r) ⊇ (s) if and only if r divides s. We have P(X) = Spec R and A(X) = Max R, and our Zariski topology coincides with the standard definition.

You can play the same game if X is the lattice of subobjects of any structure H. If H is a set (or a topological space) and X is its power set, then the primes and prime atoms are the same; the points. The Zariski topology is the discrete topology on H. If H is a vector space, then P(X) is empty, because any non-zero subspace V can be contained in the span of two subspaces that don't contain V. It seems that the sweet spot for 'interesting' Zariski topologies is somewhere in between the rigidity of vector spaces and the flexibility of sets.

If H is an affine space, then again the prime elements are exactly the points. The resulting Zariski topology has as closed sets the finite unions of affine subspaces of H.

An interesting one is if X is the set of closed sets of some topological space S (generalizing the first example). The prime elements are the irreducible closed sets, and if S is T1 (meaning all points are closed), then the points of A(X) can be identified with those of S. Then the Zariski topology on A(X) is the same as the topology on S, and the Zariski closure of an irreducible closed set is the set of all irreducible closed sets contained in it.

@mybeanalgebra

a boyfriend is just a guy you can sink your teeth into for recreational purposes

I wish I could sleep with hartshorne under my pillow and wake up with the wonders in my brain already

the NON-GENDERED URGE to run off into the woods with a whiteboard and an algebraic geometry textbook and come back six months later, having been properly functorized (grothendieck'ed, even) by the whole experience

there should be a tax that youtubers pay where 1.5% of all of their revenue goes back to Kevin Macleod for basically supplying YouTube with it’s own soundtrack.

Fiber arts is just Math in sheep's clothing