groups-rings-fields

[long nebula]

??

please don't post nonsense in here

[finite dust]

it was just a joke lol

a proper response is that

Replacing the Riemann sphere CP^1 with RP^2 breaks the possibility of coherent field-like arithmetic in a fundamental way: the one-point compactification C ∪ {∞} works precisely because all directions to infinity are identified into a single element, giving a uniform absorbing behavior compatible with Möbius transformations and complex structure, whereas in RP^2 infinity is an entire projective line, so limits and products depend on direction and cannot be made globally well-defined or associative in any extension of the usual operations; this reflects a deeper obstruction—CP^1 arises from projectivizing a field and inherits symmetry from PSL(2,C), while RP^2 is non-orientable, admits no complex structure, and is not tied to any division algebra, so there is no way to extend multiplication/division consistently even with tricks like adding bottom elements (as in wheel theory), meaning you can at best build a weakened, direction-sensitive “arithmetic” (closer to projective or valuation-theoretic behavior) but nothing resembling the coherent algebra of the complex numbers.

[desert verge]

!noai

[flat tree]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[finite dust]

it was not air

air

ai

i paraphrased it

[desert verge]

i see the em dash

[finite dust]

as my structure was mad

[desert verge]

also this is not relvant to op's question

[finite dust]

i can show u orignal piece

i know its not relevant,

[chilly radish]

Given your posting history I don't believe you know what PSL_2 or RP^2 are

[coral spindle]

Go on, please do

[finite dust]

you think about it from the linear algebra side the differnce is pretty clear. CP^1 comes from lines in C^2 so you can write everything in homogenous coordinates like [z0:z1], and mobius transformations are basically just linear maps acting projectively, which is why the extended complex plane still behaves nicely with multiplication and division. But for RP^2 your working with lines in R^3, and even though you still have projective linear maps theres no real way to define multiplication on R^2 that extends to the whole thing. The problem is infinity isnt just one point anymore its a whole line of directions, so what happens depends on how you go to infinity, and that kinda breaks any consistent arithmetic. At that point its really more geometric then algebraic.

this was my original piece

[coral spindle]

So you wrote this? You didn't copy it from anywhere?

[quiet pelican]

I’m not sure that addition can be well defined on RP^2, but multiplication can be defined reasonably well
Because the product of two lines through the origin of C is a line through the origin

[finite dust]

yeah I get the inition but its kinda missleading
in C it work cause multiplication is already built in, so lines behave nicely. in RP^2 your just in R^3 up to scaling, and theres no natural multiplication there that survives projectivising, so it wont be well defined
plus infinity being a whole line means everything depends on direction anyway, so it breaks pretty fast

[quiet pelican]

A point at infinity of RP^2 can be defined as a line in C
But I’m not going to talk to an ai through you

[finite dust]

how am i ai

[quiet pelican]

Yeah I don’t think there’s a continuous extension of addition
Because morally the sum of the lines in the z and w directions should be the limit of any sum of something that lies on the w line (and goes to infinity) and something that lies on the z line (and goes to infinity)
And that’s very not well defined

[finite dust]

that doesnt really make sense tho, your kinda mixing structures there
a point at infinity in RP2 is a line in R3, not in C, so that step already breaks
like yeah you can force an identification but it wont respeict anything you want
idk it just feels like your handwaving the hard part tbh

[rocky cloak]

Then points at infinity are lines in a specific plane in R3

[quiet pelican]

RP^2 = RP^1 \cup_{z -> z^2} D^2
Then identify RP^1 with the space of lines through the origin of C
If you can’t see that, think a bit (and think about leaving a conversation about RP^2 if you don’t know the very basics about it)

[fickle dirge]

What is the difference between an algebraic set and an algebraic variety?

[quiet pelican]

Usually if you’re making a distinction, the latter is assumed to be irreducible (although I’ve seen places that don’t make said distinction)

[fickle dirge]

This is what im trying to understand, do you have an idea what it means? The book doesn't say

[quiet pelican]

What’s the problem?

[fickle dirge]

I got it, it was like you said, irreducible. thanks

[fossil spade]

this algebra is getting pretty abstract guys

haha

[swift root]

i think ive seen a couple where it could also be an open subset of an irreducible algebraic set?

but that makes me uncomfortable so dont do that

[south patrol]

Algebraic set means you are doing classical AG

Lol

[fickle dirge]

idk what that means haha

[long nebula]

there's the original classical AG (italians studying solutions to polynomial systems) and then there's what developed afterwards

which one might affectionately call abstract nonsense

[rocky cloak]

An open subset of an irreducible algebraic set should be an irreducible variety though.

Like V(I) n D(f) = V(I + (1 - yf))

[swift root]

thats a distinguished open subset

[rocky cloak]

Easy enough to do unions of algebraic sets

[swift root]

arbitrary ones too?

[rocky cloak]

You only need finite unions, because the polynomial ring is Noetherian

[swift root]

right

that makes sense

[rocky cloak]

But you I guess it's not V of something on the nose.

You have to make some choices for how to write the open as a union of distinguished etc

Still up to homeomorphisms / isomorphism of sheaves / whatever

[chilly radish]

It's often very concrete nonsense, acrually

[fickle dirge]

I don't understand why is it better

[wraith cargo]

In general the localization of a finitely generated k-algebra neednt be finitely generated

[fickle dirge]

ah I see, thanks

[somber goblet]

whats localization?

i see this term being used a lot

[quiet pelican]

How much ring theory do you know?

[somber goblet]

a decent amount, i'd say

maybe a semester's worth?

i can work with ideals

[quiet pelican]

So suppose we have a ring R and a subset S
Where S is closed under multiplication (ie, ss’ \in S for s, s’ \in S)
We say the localisation of R at S is the ring you get by “adjoining inverses for every element of S”

[somber goblet]

ohhh i saw this as an example for something

ok

[quiet pelican]

Formally, it’s the ring (R x S)/~ where (r, s) ~ (r’, s’) if there is t \in S such that t(rs’ - r’s) = 0
With addition given by (r, s) + (r’, s’) = (rs’ + r’s, ss’) and multiplication by (r, s)(r’, s’) = (rr’, ss’)
You can check this is all well defined
It’s essentially inspired by how you construct the rationals from the integers

[somber goblet]

what is it used for? does it have something to do with AG?

[velvet hull]

yes

[somber goblet]

its like half of a field of fractions

[velvet hull]

kind of

the word localization is geometrically motivated

[somber goblet]

you only divide by some of the elements

[quiet pelican]

If R is an integral domain, and S = R \ {0}, then S^{-1} R is the field of fractions

[somber goblet]

yeah makes sense

and so if $0 \in S$, $S^{-1}R = 0$

[cloud walrus]

lexi

[quiet pelican]

Yes

Things get very cursed if S contains zero-divisors

[velvet hull]

not really cursed

the way the algebra works out you just end up quotienting out by them first

[quiet pelican]

(By which I mean “S^-1 R is not necessarily as big as you expect”)

[somber goblet]

quotient $a \sim 0$ if $a$ is a zero divisor?

[cloud walrus]

lexi

[quiet pelican]

If a is a zero divisor in S, yeah

[somber goblet]

oh in S

[velvet hull]

first consider C^n as an affine variety.
because C is algebraically complete, the only rational function in C(x1,..,xn) that is well-defined C^n are the polynomials C[x1,...,xn].

however now consider some proper affine variety inside C^n given by prime ideal p

then notice that we now have more rational functions available to work with: as long as the poles of a rational function avoids the affine variety then it is well-defined on it

[somber goblet]

oh "well defined" here means no poles?

[velvet hull]

yes

[somber goblet]

and so any polynomial denominator will go to 0 somewhere

bc ACF

[velvet hull]

yes

[somber goblet]

👍

[velvet hull]

but then that means if the function is not well-defined, then the denominator has to "be in" the prime ideal

[somber goblet]

because it has to go to 0 in the variety

[velvet hull]

not necessarily 0 everywhere

[somber goblet]

wait

isnt I(X) the polynomials that go to 0 everywhere in X

[chilly radish]

I think a nicer motivation is perhaps localising at 1 function, that means that the zero set of this new algebra will only contain points that f doesn't vanish on (as those are poles)

[velvet hull]

yeah that's correct, this still works but im not going into the details this is just an informal description

[chilly radish]

So the set defined by tje localisation is exactly D(f) = {x: f(x)\ne 0}

[somber goblet]

ok ill read about this formally

[velvet hull]

anyways in other words we are allowed to divide by a polynomial if and only if it is found outside of the prime ideal, i.e. the regular ring on this affine variety is exactly C[x1,...,xn] localized at the complement of the prime ideal p

[somber goblet]

reminds me of dyadic rationals

ig those are a localization of the integers

hmm ok

[brisk matrix]

not quite, i think you might be confusing a vector field defined on $S^2$ with an operator acting on $\mathbb{R}^3$

[cloud walrus]

Canvas123

[chilly radish]

Note though that this isn't affine anymore in general (so the localisation isn't f.g.)

[somber goblet]

isnt what

oh finitely generated?

[chilly radish]

Yea

As an algebra

[somber goblet]

as a k-alg

[chilly radish]

Yes

[somber goblet]

is this a "local ring"

ah and then $\mathfrak p$ being prime is what lets $R - \mathfrak p$ be a multiplicative set

[cloud walrus]

lexi

[chilly radish]

Localising at 1 element doesn't give a local ring

Localising away from a prime does

[somber goblet]

ok thats what im looking at

a "local ring" is one for which there is a unique maximal ideal

[chilly radish]

Wait nvm I'm wrong, it is affine, you just need more variables

[somber goblet]

so many rings are not, like integers or polynomials

which is why we want to localize things

[chilly radish]

You mean the ring of functions on C^n regular on V(p) right? This certainly isn't the coordinate ring C[x1,..,xn]/p

[rocky cloak]

Yeah there are many properties that are easier to check for local rings and that are true for R iff they're true for all localizations of R at prime ideals.

[somber goblet]

...so what do we get out of local rings

exactly

something about rational functions but thats not clear to me

[rocky cloak]

One thing that's nice is that if R is a local ring with maximal ideal m and M is a finitely generated module, then
x1, ..., xn in M generate M iff their image generates M/mM.

And M/mM is a R/m vector space, so you can reduce some module theory to linear algebra

[long nebula]

What trajectories equivariant to rotation are there besides radial ones? This feels impossible to me because of holonomy

[somber goblet]

oh neat

wait

germs

i know this

idk how this is applied to AG tho

so if i localize around the maximal ideal at a point

something happens

[rocky cloak]

In AG you have a local ring at each point where ring theoretic properties correspond to geometric properties.

For example a variety is smooth at a point of the local ring there is regular

[somber goblet]

ok...hmm...i might want to play with some specific examples here

let me think

[next obsidian]

There’s such an unbelievable number of theorems about local rings

I basically think of every ring as being local, because localization usually lets me do so, or properties are stated in terms of “at every prime…”

[somber goblet]

let's take $\mathbb C^1$ and localize at 0

[cloud walrus]

lexi

[somber goblet]

what do we get here

[next obsidian]

C…

[south patrol]

I'm confused in two ways

Lol

[somber goblet]

as in $\mathbb C[x]$

[cloud walrus]

lexi

[somber goblet]

localized at $(x)$

[south patrol]

? Lol

[next obsidian]

You get rational functions

[south patrol]

Don't write C^1 for that

[cloud walrus]

lexi

[somber goblet]

wait i misspoke

😭

dont kill me

[frail shoal]

i understood what was meant

[next obsidian]

You get rational functions where the denominator has a nonzero constant

[frail shoal]

although idk the context

what is everyone talking about

[south patrol]

I mean even if you understand u probably shouldn't lol

[somber goblet]

oh ok

[south patrol]

And localising a scheme at a point feels off vocabulary wise

[next obsidian]

It doesn’t concern you “keith”

If that’s even your real name

[frail shoal]

it's actually frank

[somber goblet]

ok i see sorry i just got introduced to this terminology

[frail shoal]

/lies

[somber goblet]

like 5 minutes ago

[rocky cloak]

Could do something like
||y^2 = x^3 at the origin||
||The coordinate ring in question would be k[x, y]/(y^2 - x^3), then we localize at the maximal ideal (x, y)||
||We have that m/m^2 is 2d (generated by both x and y for example), but we can only make a regular sequence of length 1, so not regular.||
||Hence the curve isn't smooth||

[south patrol]

Dw im not like annoyed lol

[next obsidian]

I am.

[south patrol]

Well only thing I was annoyed at was random comment from Keith jk

[somber goblet]

[south patrol]

Joking.

[somber goblet]

ok the first one

[next obsidian]

I am annoyed at your existence.

“Potato”

If that’s even your real name

[frail shoal]

wait i can't tell if ur joking about being annoyed by my comment now

[somber goblet]

are these all over C?

[south patrol]

Nah it was fine im not annoyed lol

[next obsidian]

World’s strongest over thinker

[south patrol]

I'm also annoyed by my existence

The only thing that does annoy me is working over C for no reason

[rocky cloak]

If you like.

Works over most fields

[frail shoal]

well ok lemme explain. the fact lexi said to localize at "0" made it sound like 0 was a point and not an ideal. also nobody says C^1 unless they mean the space not the ring

[next obsidian]

Here’s a general principle. Working over C is the exact same as working over any field whatsoever :^)

[south patrol]

I mean no one says C^1

[frail shoal]

ok are u annoyed now

[somber goblet]

yeah thats what i meant

[south patrol]

Unless maybe you are doing classical AG and the set is literally k^1

[next obsidian]

This is called the Lefschetz principle

[somber goblet]

did you want me to write $\mathbb A_{\mathbb C}^1$

[cloud walrus]

lexi

[next obsidian]

Yes

[frail shoal]

anyway i think the point 0 would be the ideal (x) here just fyi

[somber goblet]

yes

i wrote that later

[south patrol]

True.

[frail shoal]

yay

[south patrol]

What about with the étale site

[frail shoal]

i saw (0) at some point and got confused

[next obsidian]

When you have a scheme and a point you call the process of going to a local ring “taking the local ring at that point0

[somber goblet]

what about non-algebraically closed fields?

[next obsidian]

so taking the local ring of A^1 at 0

[south patrol]

All depends what ur doing

[next obsidian]

This is a shitpost

[somber goblet]

ah

[south patrol]

Different fields behave differently

[somber goblet]

yeah

[next obsidian]

The real statement is that working over C is exactly the same as all alg closed fields of char 0

[somber goblet]

im not going to try doing AG over Q

[next obsidian]

But often actually you can drop alg closed for a lot of stuff

[south patrol]

Take a huge non-perfect field of characteristic 2 and life is pain

[south patrol]

This is fun eventually lol

Well this is one of the beautiful things about "modern" AG

[rocky cloak]

First order logic enjoyer

[south patrol]

That we can do stuff

[somber goblet]

oh is this a model theory moment

[frail shoal]

tbh the issue is that it takes more effort to write / say qbar than C

so C is more convenient notationally

[next obsidian]

Dude that’s a lot of pressure you’re exerting dude. So if things aren’t perfect it’s not enough for you? You’re like my dad

Jk that’s my dad slander

[frail shoal]

ok ik people care about the topology but idc about the topology so this is why i say C

[next obsidian]

You can phrase it that way but I don’t even mean it in a way that’s that fancy. There’s a process where you use faithfully flat descent to turn basically any problem over an alg closed field of char 0 to being over C

[frail shoal]

$\bar{\mathbb{Q}}$

[cloud walrus]

keith

[frail shoal]

wait this aint right

how do yall qbar

[somber goblet]

$\overline{\mathbb Q}$

[cloud walrus]

lexi

[next obsidian]

\overline

[somber goblet]

ok localizing y^2 = x^3 at point 0

im going to work over R so i can visualize this

[next obsidian]

Bad idea

Work over C and just visualize it over R anyway

That’s how real alg geometers roll

[somber goblet]

pretend its R?

[next obsidian]

When you visualize it work over R so you can actually do shit

But write down C

[somber goblet]

ok im going to have the picture of R in my head

and on desmos

so we're quotienting $\mathbb C[x, y]/(x^3 - y^2)$ to get the coordinate ring for the variety

[cloud walrus]

lexi

[frail shoal]

this reminds me of a problem someone posed to me that i absolutely failed to solve

someone once asked me to prove that the coordinate ring of the unit circle over C is isomorphic to C^× as a group

[somber goblet]

ooo thats cool

i should try that later

[frail shoal]

anyway i have no idea how to visualize this

i also have no idea how hard it is

[somber goblet]

like x^2 + y^2 = 1

[frail shoal]

yea

and it's a group scheme

so it gets a group structure automatically

explicitly, (x, y) * (z, w) = (xz - yw, xw + yz)

[somber goblet]

can you rephrase algebraic group stuff in terms of ring homomorphisms

[frail shoal]

one can check this defines a ring hom

[quiet pelican]

Yes

[somber goblet]

like let the multiplication be $\mu: X \to X \times X$ somehow

[cloud walrus]

lexi

[quiet pelican]

You get what’s called a “comultiplication”

[frail shoal]

im explaining it shit cuz i know a lot less than im letting on

[quiet pelican]

It should be $X \to X \otimes X$

[cloud walrus]

micoi the group objects (she/it)

[somber goblet]

and then inversion is $\xi: X \to X$

[cloud walrus]

lexi

[somber goblet]

wait what

why

group multiplication is $X \times X \to X$

[cloud walrus]

lexi

[somber goblet]

or wait

coproduct

!!!

[quiet pelican]

But yes, and then you have “coassociativity” and a “coidentity” and a “coinverse”

[somber goblet]

silly :3

is coproduct in CRing really tensor

[quiet pelican]

Yes

Not quite actually

The coproduct in K-algebra is tensoring over K

From Humphrey’s algebraic groups book

[frail shoal]

ok it's actually not that complicated the way i think about it

[somber goblet]

ok so we just remember what field we're over

hopf algebra?

uhhh

[quiet pelican]

Hopf algebra is the name for the type of structure we get which is an algebra with a “compatible” comultiplication/coidentity/coinverse structure

[frail shoal]

for any ring R, define S := {x, y ∈ R^2 | x^2 + y^2 = 1}. the way i think about it, there is a polynomially definable map S^2 -> S, that satisfies group axioms

[somber goblet]

so is this structure really different from an ordinary ring?

idgi

[quiet pelican]

Yes
It’s an algebra A with some extra information (namely, a co[insert the relevant words) which allow us to make Hom_k(A, B) into a group, for all algebras B

This relates quite closely to the “functor of points” point of view of algebraic geometry

[somber goblet]

ok

maybe?

[somber goblet]

why do we need this to build a group

[frail shoal]

oh yea this builds a so-called "cogroup"

groups are the opposite

[somber goblet]

ok

[frail shoal]

A such that Hom(-, A) is a group for all -

if u think about it this is true for groups as you know them in Set

i.e. for any set n there's a group G^n

[somber goblet]

but arent cogroups and groups the same??

[frail shoal]

nah they're hella different

[swift root]

we want a group that is also an affine variety / algebraic set, i.e. a group object in the category of algebraic sets
this dualizes exactly to have a commutative hopf algebra, where the underlying algebra is fg reduced

[frail shoal]

this is essentially what mico said tbf

[somber goblet]

i see

[frail shoal]

S is Hom(coordinate ring of the circle, R)

[somber goblet]

maybe i need to draw this out

[swift root]

omgomgomgomg quantum groups

[swift root]

cogroups are weird structures

[quiet pelican]

No, this is cogroup stuff

[swift root]

a quantum group is a cogroup of algebras

[quiet pelican]

Cogroups are natural
^ girl who has been studying homotopy theory for the last couple weeks
(S^n is a homotopy-cogroup)

[swift root]

not mutually exclusive

[somber goblet]

whats the coidentity

[swift root]

I agree they are very natural to consider
^ NB who has been studying knot theory for the last couple weeks
(braided monoidal categories with duals yield knot invariants)

they arise when you have a group structure in a setting that has a natural dual (like geometry or cohomology)

[frail shoal]

ok so there is a concrete description of cogroups (also as mico posted) and so coidentity is a map to the initial object

as opposed to a map from the final object

in CRing this is a map to Z, i.e. a Z-point in the scheme if we view it geometrically

for the specific example i gave it's the point (1, 0)

to be clear a cogroup in CRing is a group in AffScheme

🔥

which is why it's called a group scheme i assume

as opposed to a cogroup ring

[quiet pelican]

A map from A to k

[wraith cargo]

I think cogroups are usually called Hopf algebras

[frail shoal]

nuh uh

[swift root]

if they're cogroups of algebras

[wraith cargo]

it's all the same stuff

[rocky cloak]

I learned about cogroups first during the pandemic it seems

https://jagr2808.github.io/blog/posts/cogroups/

[wraith cargo]

instead of multiplication you have comultiplication

instead of inverse you have coinverse

[quiet pelican]

Cogroups in the category of k-algebras

[wraith cargo]

and instead of identity you have the augmentation map

[swift root]

U_q(sl(2))

[wraith cargo]

yeah I only know of them from algebraic geometry where you assume they live over k

so that's why I thought this was the usual terminology

[swift root]

as an algebraic group?

or just a group

[frail shoal]

group

idk what an algebraic group is

[quiet pelican]

I’m a little surprised you haven’t seen them in an algtop context
But fair

[wraith cargo]

what is the coordinate ring of a group

[wraith cargo]

I have seen them but everyone calls them Hopf algebras lol

because you always assume some ambient structure

[frail shoal]

ok this made me realize i was too vague

[swift root]

well C^* is naturally an algebraic variety

[wraith cargo]

like in Peter May's book as well he talks about hopf algebras

[quiet pelican]

Even like say S^n?

[frail shoal]

i should say Z[x, y]/(x^2 + y^2 - 1)

[swift root]

Z, even?

[frail shoal]

that's what i meant by coordinate ring of the unit circle

[swift root]

I know

[frail shoal]

and then by "over C" i mean C-points

[frail shoal]

yea fair just saying it again in case it was ambiguous

[swift root]

but are we considering both groups has having an algebraic variety structure

[south patrol]

Le Gm

[swift root]

is what I want to know

[wraith cargo]

S^n being steenrod squares?

I mean I've seen hopf algebras in the context of like how higher cohomology operations behave on spectra

[quiet pelican]

S^n being the n-sphere

(It’s a cogroup up to homotopy)

S^n -> S^n v S^n being given by contracting a diameter to a point

[wraith cargo]

oh

Idt I've seen this before lol

does this actually give you any interesting structure?

[quiet pelican]

It gives you homotopy groups
Like, [S^n, X] is pi_n(X), and the operation is induced by this cogroup structure

[swift root]

So the algebraic variety { (z, w) | z^2 + w^2 = 1 }

[wraith cargo]

oh okay I've seen this but I've never seen it called a cogroup haha

[swift root]

me when hom(-, X) dualizes

[wraith cargo]

okay yeah I had a feeling this is what you were thinking of but yeah

I've seen this yeah but I haven't heard anyone talk about it as a cogroup

that's pretty cool tho I didn't know that

another gap in my knowledge has been filled

cuz idt neither hatcher nor whitehead explain it algebraically like that

tho both of those are sort of old fashioned anyways LOL

[frail shoal]

does algebraic variety mean a variety that is a group

then sure

[swift root]

I guess algebraic set idk if its a variety

I'm pretty sure it is though

[swift root]

algebraic variety means that its an irreducible closed subset of the Zariski topology

usually

Is the isomorphism between the groups meant to be an isomorphism of algebraic sets?

[wraith cargo]

no it means a variety of finite type

usually

but finite type is already assumed in variety

so algebraic variety and variety mean the same thing

[wraith cargo]

an algebraic variety that's a group is called a group variety

[swift root]

an algebraic group from what I've seen

[wraith cargo]

or a scheme of finite type over a field that has a group structure is called an algebraic group

[swift root]

or that

[wraith cargo]

it's important to make that distinction

because a priori algebraic groups aren't smooth

while group varieties are smooth

[swift root]

at least the irreducible components of an algebraic group are well behaved

iirc

[wraith cargo]

uh

[frail shoal]

😭

[wraith cargo]

well they're controlled by a maximal etale algebra in the coordinate ring of the scheme

but this is sort of a mess to work with

[swift root]

oh lol

[karmic moat]

Arent all irred cpts of an alg group conjugate?

At least over alg closed field

[wraith cargo]

are you assuming it's linear?

[karmic moat]

Maybe

[swift root]

like any sane person

[wraith cargo]

cuz here I'm talking like in max generality

[karmic moat]

Ewwwww

[wraith cargo]

any scheme of finite type over any field with a group structure

SGA3 style

[quiet pelican]

Schemes don’t exist :3

[swift root]

then you're just asking for problems, really

[karmic moat]

I’m talking Borel style 😎😎😎

[wraith cargo]

I feel like it's a richer theory

[karmic moat]

Humphreys style 😎😎😎

[wraith cargo]

I don't like that they assume things are simple because like that isn't the full story and it hasn't been for 70 years

[swift root]

schemes don't exist but hopefully a nice noncommutative geometry does

[wraith cargo]

like they all assume nilpotent don't exist

but them how do you explain that mu_3 has a nilpotent coordinate ring over any field of characteristic 3

[karmic moat]

The answer is only look at char 0

[wraith cargo]

separably closed fields 🤢

[karmic moat]

Idk i didnt actually read borel i just use it as reference

But anyway all my assumptions are reductive connected over C something something

[somber goblet]

ough

what property does it have

[south patrol]

Why use a field

[somber goblet]

is it the usual identity diagram with arrows reversed

[swift root]

yes

hence "co"

[somber goblet]

im getting used to this

[south patrol]

Jk fair for alg groups ig

[wraith cargo]

please mr. potato

[somber goblet]

this has smth to do with elliptic curves right

[wraith cargo]

define to me what the identity element of a group over an arbitrary scheme is

and not just the map

the actual element in S-rational points

let's hear it

[karmic moat]

Youre starting to scare me

[wraith cargo]

jk

but okay genuinely

I haven't yet found a book on algebraic groups that truly does it in full generality

[karmic moat]

Milne has notes right?

[wraith cargo]

yes but he only does it over a field

and he doesn't talk about finite flat group schemes at all

[karmic moat]

He calls it a “book”

Ahh

Damn

[quiet pelican]

This is because no one cares about non-linear algebraic groups

[wraith cargo]

and these are usually talked about in the context of like
G/S with the base scheme being arbitrary

like stuff like Fontaine's theorem and Tate's theorem

these are about algebraic groups over Spec Z for example

[somber goblet]

is this a decent diagram for identity axiom

[quiet pelican]

For an ordinary group yeah

[somber goblet]

will this hold up if i try to bring it to a different category

than Set

[quiet pelican]

Yes

(I just mean “not a cogroup”)

[somber goblet]

identity is moreso the morphism rather than the object itself isn’t it

[wraith cargo]

very nicely spotted

[south patrol]

Idk what you want lol

[wraith cargo]

https://tenor.com/view/fake-larp-larp-larp-sahur-sahur-larp-larping-gif-18307483096237429166

[south patrol]

Lol no I mean like what do you want if I'm not allowed to define it in terms of a map lol

Or do functor of points innit

[wraith cargo]

I was thinking about this a bit ago while reading Milne and ran into this issue

essentially when you're over a field you have a canonical map spec k -> G

and the image is a single point

so G(k) has a natural group structure where the image of this canonical identity map is the identity element of G(k) as a group

but when you have say an arbtirary scheme whose identity component is S -> G

how do you define what the identity element is in G(S) for example

[somber goblet]

is this better?

[wraith cargo]

oh well

I am stupid

[south patrol]

Idk what you mean

[wraith cargo]

I see what I was thinking about bleh

yeah ignore what I said

[south patrol]

You have already given the point

[wraith cargo]

it literally clicked just now

I was imagining G(S) as a literally set of points on G with S-rational points

[south patrol]

Ah lol

[wraith cargo]

I'm too tired for this shit anymore

[south patrol]

Dw

[south patrol]

Tbh my habit is never to think of them that way which is sometimes also disadvantageous lol

[wraith cargo]

it's a bit tricky to like juggle between the two because depending on what you're doing they might literally be that or they might not

like I've been doing a lot of rational points stuff recently

and there they always make u literally imagine them as points with Q-rational coordinates

so this has clouded my judgement

I landed a bachelor's thesis doing quadratic chabauty computations 💀

[somber goblet]

fire

[wraith cargo]

waahhh p-adic integrals

scawy

[swift root]

top arrow should be going the other way

[somber goblet]

ok tru

[swift root]

but yes

[somber goblet]

ok time to draw cogroups

hmm wait we can simplify this then

since one of the arrows was just equality

[south patrol]

Yeah lol valid

[south patrol]

Scary

[wraith cargo]

technically it's not even p-adic integral

it's like rigid analytic integrals or something

they're called like Coleman integrals?

they're barely integrals it's some special thing idk

[south patrol]

Inch resting but idk this stuff lol

Hmm

[wraith cargo]

I'm also aching to learn the stuff ya know tho

I've read a lot about A^1-homotopy theory stuff

😔

planning to write that paper about it

one day....

[quiet pelican]

when A^2 homotopy

[wraith cargo]

🤢

you're making me vomit please

🤮

I couldn't control myself sorry

[quiet pelican]

What?

[south patrol]

Pog...

[south patrol]

Idk anything

[somber goblet]

co-associativity

[wraith cargo]

nooo twin

you're amazing

here's a little star sticker for u

⭐

[quiet pelican]

Same

[somber goblet]

how does this make sense

[wraith cargo]

I've read a shitton of HTT to read some Marc Hoyois papers and I hope I get around to that one day 💀

[somber goblet]

my Set intuition is flailing

[south patrol]

Thank

[quiet pelican]

I don’t think there should be any cogroups in set

(Well, no non-trivial ones)

[somber goblet]

that’s probably why im having trouble

heres the new and improved identity diagram

[south patrol]

Promoting X to a cogroup is equivalent to lifting Hom(X, -) : Set -> Set to a functor valued in groups. But this is impossible by plugging in the empty set, unless X is empty, because there is no empty group

[quiet pelican]

Probably my go-to examples would be something like:
k[x] with x -> 1 (x) x + x (x) 1
k[x, x^{-1}] with x -> x (x) x
S^n with S^n -> S^n v S^n by collapsing a diameter

[somber goblet]

omg cogroups can be empty

silly :3

[south patrol]

And they have to be empty for sets lol

[somber goblet]

gugug

🔥

[quiet pelican]

(These are called G_a, G_m and “the basis for homotopy groups existing”)

Another fun example is GL_n, but that’s slightly harder to notate

[south patrol]

I would also advertise this perspective if you have done any cat theory. It is often easiest / most intuitive to produce cogroups this way

[somber goblet]

i see

i havent really done enough pure cat theory i think for this to make sense to me

[south patrol]

Like for example the fjnctor Hom(k[x], -) from rings to sets just takes the underlying set. You can make this valued in groups by using the underlying group of each ring, and that gives you Micot's first example (called G_a)

[south patrol]

Valid

[somber goblet]

i can try

[brisk matrix]

orbital and precessing trajectories. i think holonomy is a feature

[somber goblet]

so Hom(X, -) is the homset functor

why does - need to be a set here?

ok a cogroup really isn’t an algebraic structure

at all

[quiet pelican]

Cogroups in Grp are exactly the free groups, but that’s a bit harder to see

(As in, I had to spend a few minutes locating the right results to make sure again)

[somber goblet]

hmm

well coproducts are free groups arent they

this may be related

[long nebula]

wait I don't think orbits work because if you rotate the initial point back to where it started, the orbit should pick up a new orientation by equivariance, but there's no way to have multiple orbits with different orientations going through the same point

am I misunderstanding your question?

[brisk matrix]

not exactly... i think the issue here IS that we aren't saying there "are" multiple trajectories passing through one point

[long nebula]

okay let me ask a more concrete question so that I understand what you're looking for, say your time evolution operator applied to the north pole makes a trajectory moving east; then you rotate it 90 degrees east to the "east pole" and it should be moving south; on the other hand you could start at the north pole, rotate 90 degrees south, and then rotate 90 degrees east, in which case your trajectory starting at the east pole should be moving east—so which direction should your time evolution operator send a trajectory starting at the east pole if you want an equivariant SO2 action?

[brisk matrix]

you're confusing path-dependent holonomy with group actions on fields. equivariance isn't about how you travel between points via rotation but ensuring the operator obeys $f(Rx) = Rf$ for every element in $\text{SO}(3)$ regardless of any "path" taken.

[cloud walrus]

Canvas123

[brisk matrix]

i mean you are rotating coordiante, but in equivariant system being rotated means every part of initial state

including directionality or velocity vector at starting point

it's transformed by $(R\cdot x,, R \text{\textbf v})$ simultaneously

[cloud walrus]

Canvas123

[long nebula]

I thought your function f was just a function of position, is it also a function of some initial velocity?

if f is just a function mapping position to position, it seems clear to me that f can't have any nontrivial angular component if it's supposed to be equivariant under SO3 action

[lethal mortar]

Can I ask metric space doubt here?

[next obsidian]

No

[somber goblet]

#point-set-topology ?

[balmy python]

why are the automorphisms of S5 the inner automorphisms?

[tribal moss]

Because there are no others? I don't think you'll find any particularly satisfying and abstract answer, since S6 does have a nontrivial outer automorphism.

[wraith cargo]

You look at where conjugacy classes are sent under automorphism

[brisk matrix]

i am asking about an evolution operator $\Phi_s(\mathbf{\hat q}, \dot{}! {\bf\tt x})$

[cloud walrus]

Canvas123

[gray parcel]

Why does the fact that the two commute imply that \pi(g)v \in ker(M - λ1)?

Oh wait

I misread this completely

Never mind, I'm still confused

[velvet hull]

what are you confused on?

[gray parcel]

It's the very last sentence but I think I might be overthinking it, give me a quick minute, sorry.

I got it, it was because of the commutativity.

I can apply L = M - λ1 to \pi(g)v and then switch the order to \pi(g)L(v) = \pi(g)(0) which means that \pi(g) is in the kernel

[supple pecan]

is this part neccessary? even if r could have any degree, we still have f(a) = 0+r = 0 so r=0 right?

nvm

its cause f(a) = 0+r(a) right? and we show r is a constant so we can write r(a) as r = constant which is 0?

[wraith cargo]

Yep

[tepid light]

If I am asked to find all the possible results when taking the product of every element in S3, do I just end up with the elements of S3, for a total of six different products?

I should be able to get back to any element of S3 with some ordering of the product, right?

[elfin wraith]

Its not entirely clear to me what you mean, but if you take all possible products you in particular take the products x*e for all x, so you certainly get everything in S_3

[tepid light]

I mean the product of every element one time, only changing the order.

()(12)(13)(23)(123)(132) vs (12)(13)(123)()(132)(23)

[swift root]

youd get a total of 6! different expressions

whether every element of S3 is obtained this way i dont actually know

probably not

[elfin wraith]

Less in principal since like the 3-cycles are inverse, but more importantly that one of the elements is the identity

[tepid light]

5! since it doesn't matter where the identity is placed.
Thinking about it more I think the answer is three since it's the product of three odd elements and three even elements, so the product has to be odd.

[elfin wraith]

Like 5! is an easier upper bound

But yeah like I dont think this gives you everything just by parity

[tepid light]

Well doesn't the product need to be in S3?

So if it has to be in S3, and it has to be odd, then there are at most three elements in S3 that can be the product?

[elfin wraith]

Yeah that makes sense to me

You probably do get all of those, but you’d probably have to think about that

[swift root]

thinking sucks

[tepid light]

I got all three by hand, so ¯_(ツ)_/¯

[elfin wraith]

Ye that counts as thinking

I just didn’t want to lol, but nice!

[wicked patio]

If you get one 3-cycle by hand, you get the other by symmetry

e(123)(132)(12)(13)(23) = (12)(13)(23)

the 3 swaps in S_3 definitely don't form a subgroup, so they aren't closed under multiplication

so we must be able to find two swaps whose product isn't the third swap

then we product those 3 to get a non-identity element which must be a 3-cycle

There we go no computation

[lime magnet]

I fell asleep.

##math standard answer, i like it.

i thought i was going to commute to campus and then get back online. i keeled over onto my pallet instead.

re. modules, though the free module is interesting. in particular, an R-module has a basis iff it's the direct sum of some copies of R.

and by direct sum, i almost always think about the external direct sum. perhaps this is a failing.

i guess i asked b/c i genuinely didn't know whether to ask here or in #advanced-algebra (modules not being included in the channel title)

[elfin wraith]

Basic stuff about modules would probably live here but the lines are blurry and no one really cares, ask away in whichever is free (but probably here if you have a choice)

The people who are likely to answer you are typically very active in both

The people mostly being Jagr lol

[lime magnet]

no one really cares
in a good way, i take it.

thanks!

actually, the statement about the free module is unsurprising, except perhaps when the index set in the direct sum is uncountable (if that ever happens?)

[elfin wraith]

If i understand your question correctly, yeah in general the infinite “direct sum” of modules need not be free, see here https://en.wikipedia.org/wiki/Baer–Specker_group

I also say “direct sum” because we need finitely may nonzero terms for a direct sum, for arbitrary things we talk about products

[noble nexus]

well the direct sum is always free, that's the direct product

[elfin wraith]

Yeah ofc, it’s just important to take care because I don’t know how often that distinction is made clear since like finite products and coproducts of modules are the same

[noble nexus]

yeah fair

[elfin wraith]

Which is why I presume if someone says infinite direct sum they really just mean infinite product, but it’s just kinda unfortunate terminology and (co)product is probably better

[wraith cargo]

Do u mean direct product

[elfin wraith]

I don’t think so?

[wraith cargo]

wrong message

I meant to reply to the one above it

[elfin wraith]

Yeah this is kinda what I’m getting at, I suspect that possibly they do because I think a lot of courses (understandably) don’t do a great job of distinguishing between them

[supple ice]

<@&268886789983436800>

[tough raven]

WTH is :thumpy:

[cursive spindle]

thumpy

[novel star]

is the order of the galois group of an irreducible polynomial (over the rationals) with a complex root always even

[next obsidian]

:thumpy:

[novel star]

i think it is bc conjugation should be in your group, so it has a subgroup of order 2

[south patrol]

Yes

[novel star]

yay!

[waxen plover]

Hi

I am unable to understand why every complete ordered field must have Q as its subfield

is there a complete ordered field without Q as a subfield?

I understand that the ordering is dense, but should Q be there?

I got this doubt after going through the proof of Real numbers being the only complete ordered field up to isomorphism

[tall igloo]

this is true for ordered fields, you don't even need completeness. any field either contains Q as a subfield or contains F_p as a subfield for some prime p. it will be F_p if and only if 1 + 1 + ... + 1 (p times) = 0 for some prime p, and Q otherwise. so its enough to show that. you can show that 1 > 0 must be true, and then the rest comes along for free: 1 > 0 so 1 + 1 > 1 + 0 = 1 > 0, so on and so forth

so 1 + ... + 1 is never 0

and so the set {1 + ... + 1 (n times); n a natural number} gives you the natural numbers sitting inside your field F

subtraction also follows just by taking the union with {(1 + ... + n); n a natural number}

so now you have Z sitting inside F. and F is a field so it contains quotients, so you have Q

checking that this subset of F defined here is isomorphic to Q not just as a field but as an ordered field is just formal by applying the definitions

[tall igloo]

for example: you need to prove that if x > 0, then -x < 0. given this, if 1 < 0, then -1 > 0, but then 1 = (-1)^2 > 0

[waxen plover]

Can you just elaborate on the part "F is a field so it contains quotients, so you have Q"

I think that solved my problem

everything else is clear to me

So I thought of it as (a,b) where a and b are in F

then I introduce rational number structure on these touples

[tall igloo]

sure, so the part before showed that you can find a set inside F that is in bijection with Z

[waxen plover]

and associate (a,b) with ab^(-1)

[waxen plover]

will that work ?

[waxen plover]

yes, there is a subset isomorphic to Z

[tall igloo]

so if x = 1 + ... + 1 (n times, n can be negative) and y = 1 + ... + 1 (m times, m can be negative), then you are allowed to divide x/y as long as m is not 0

and that will act as n/m in your field, in the same way that x acts as n and y acts as m

[waxen plover]

Got it

Thank you

for breaking down the explanation to groups and rings

[tall igloo]

yeah this is basically what youre doing yeah

[waxen plover]

Tenksssssss

[tall igloo]

yeah since you bring up rings its worth noting that these 1 + ... + 1's that are in bijection with Z are in fact isomorphic to Z as a ordered ring

[waxen plover]

yes got it

ring homo

thanks again bro ❤️

[tall igloo]

ya np!

[novel star]

is there something categorical here? it smells like a functor to me for some reason bur i cant put my finger on it

maybe from the poset category of finite extensions of k to the multiplicative homomorphisms or whatever

[velvet hull]

oh yeah sure, you can phrase it categorically

if you look at the poset category of k, then the norm and trace functions define a contravariant isomorphism of this category where you replace the inclusion maps with the reversed functions

[small yacht]

What’s the difference between left and right actions?

[noble nexus]

if you have a left action you need g(hx) = (gh)x, whereas for right you need (xh)g = x(hg)

notice that in both cases you are applying h and then g

but for a left action, applying h then g is the same as applying gh

but for a right action, applying h then g is the same as applying hg

[small yacht]

That doesn’t seem all different

Is the difference not just the order it’s written in…

[noble nexus]

no there is a real difference

[small yacht]

Oh wait I see I think

[small yacht]

It seems kind of silly, though

[noble nexus]

they are closely related though, as every left action can be converted into a right action and vice versa

if you have a left action, you can define a right action by xg := g^{-1}x

then $(xh)g = g^{-1}(xh) = g^{-1}(h^{-1}x) = (g^{-1}h^{-1})x = (hg)^{-1}x = x(hg)$

[cloud walrus]

Blake

[small yacht]

WIAT NO THATS AN EXERCISE IN THE BOOK I ALMODT GOT SPOILERED😭😭

(I’m like avoiding looking at it)

[noble nexus]

oops sorry

[karmic moat]

one example is the action of S_n on the set of nxn matrices
left multiplication is permuting the rows
right multiplication is permuting the columns

[round anvil]

Is it possible to prove that a finite group G of order 2m, where m is odd, has a subgroup of index 2 using Sylow's theory?

[rocky cloak]

Depends a little on what qualifies as sylow theory, but this is a corollary of burnsides complement theorem

[round anvil]

In this theorem, we need a normal Sylow-p by hypothesis. Rigth? So, we need to prove at least that G isn't simple?

or prove that exists that Sylow-p

to claim that corollary

[rocky cloak]

No, there's no such hypothesis.

https://en.wikipedia.org/wiki/Normal_p-complement

I guess the weakest result needed here is that if the sylow 2-subgroup is cyclic, then it has a normal complement

[round anvil]

This seems to me an equivalent, interesting result. I wasn't familiar with this theory about complements.

i'll see, ty

[lime magnet]

when i say direct sum, i do not mean the same thing as the direct product. i mean those elements of the direct product with finitely many non-identity terms.

[lime magnet]

the Baer-Specker group is similar but not the same thing. i meant something like $\mathbb{Z}^{\mathbb{R}}$ viewed as a module.

[cloud walrus]

ransom

[frail shoal]

holy shit did someone say baer specker group

i love that group so much omg

[rocky cloak]

The internal and external direct sum are always isomorphic. The external direct sum is just a way to construct an internal direct sum when what you started with wasn't given as submodules of a common module

[frail shoal]

ok i read up in chat and apparently some people think of direct sum as possibly meaning direct product

i was confused what internal and external direct sum meant then i remembered its in the sense of semidirects

[tribal moss]

In several cases (abelian groups, vector spaces, ...) a direct sum and a direct product is the same thing as long as there are only finitely many things, but they become different when there are infinitely many operands.

[frail shoal]

well, from my perspective, the definition of a basis is meant to encode the notion of freeness more explicitly/concretely

[frail shoal]

yea. but like, i usually say direct product only in the context of infinite products for this reason

(in these contexts. in general i say direct product a lot for finite products)

actually there is a fun fact for people who like universal algebra, there is a general construction that recovers both direct sums and direct products and also is more general

(they are boolean products)

[rocky cloak]

I think I would say direct sum or direct product depending on whether I want to think of it as a product or a sum.

But if it doesn't matter I guess I would default to sum (because it's usually the most relevant anyway)

[rocky cloak]

Is that like a commutative coproduct?

I guess what I have in mind only make sense when all factors are the same

[frail shoal]

the definition of a basis can be viewed as "every element of this group is equal to a unique weighted sum of basis elements" and this is precisely how we usually define free structures, as sets of formulas over some set of variables

[frail shoal]

i haven't heard if that so now im curious

[rocky cloak]

Just like the coequilizer of all the permuations of a coproduct with itself

[tribal moss]

It weighs a lot for me intuitively that a product of general groups looks almost exactly the same as the direct product/sum of abelian groups, whereas the sum (coproduct) of general groups is ... wild.

[frail shoal]

the abstract nonsense for this is that forgetfuls (right adjoints) preserve limits

[frail shoal]

i dont understand

[rocky cloak]

So for example for groups, the coproduct of G with itself is
G*G, then you have a map
G*G -> G*G
given by
g*h |-> h*g
Taking the coequilizer of this with the identity recovers the direct sum of groups.

[frail shoal]

oh i was thinking about it wrong, i see what you mean better now

wait no

where are g and h sent

[rocky cloak]

g*1 is sendt to 1*g

(they are not equal in the free product)

But I should probably use better notation

[tribal moss]

Yeah, the product notation looks confusing here.

[frail shoal]

a map G*G -> G*G is given by two maps G -> G*G, right?

[tribal moss]

I would say send G+G -> G+G by the rule Left(g) mapsto Right(g) and Right(g) maps to Left(g).

[rocky cloak]

Yes, and you just swap the order of the two cannonical maps

[frail shoal]

ah alr

but if left(g) is sent to right(g)

the coequaliser says left(g) = right(g)

so we just get G

[rocky cloak]

Hm, true. So this is not the map I wanted.

[frail shoal]

on the other hand, the issue with what i said about boolean products, is that i forget the actual definition

i only remember the special case for powers

[tribal moss]

Just to be clear of the context, we're working in Grp, and we want a uniform construction such that if we have two groups that happen to be abelian, it takes their Grp-coproduct to their Ab-coproduct -- is that correctly understood?

[rocky cloak]

Well, not their Ab-coproduct per say, since they need not be abelian. But the direct sum

[tribal moss]

You mean their product?

[rocky cloak]

It coincides in the finite case.

But the direct sum is a subgroup of the direct product

Where you have finite support

[tribal moss]

I'm not sure how "direct sum" applies to non-abelian groups.

Ah, so your "direct sum" is the subgroup of the product generated by all the {1}×...×{1}×G×{1}×...×{1}.

[rocky cloak]

The universal property would be that whenever you have maps
Gi -> H
Such that the image of Gi and Gj commute you get a map from the direct sum

[tribal moss]

Hmm, makes sense.

It seems to be tricky to speak about things commuting at the category level ... we'd need to have something like a tensor product for general groups. No, I'm probably confused.

[rocky cloak]

Hmm, so for
A*B you should have a map
A*B -> A*B
That just takes
a*b to b*a
But what is the general way of describing this map...

[tribal moss]

I think you want something like a coequalizer of two maps G×H -> G+H ... except the natural maps to think of there are not group homomorphisms!

[frail shoal]

a G×H-action is a G-action and H-action that are equivariant for each other, is this anything?

[rocky cloak]

Maybe that's it.

You want the image of
G+H in GxH

That works for groups at least.

But it's not really what you want for rings

I could have sworn I had some general method for this at some point

[tribal moss]

Yeah, I think that would work.

[rocky cloak]

I think coimage is what you want

[frail shoal]

@rocky cloak this is the construction i had in mind if you're wondering, the special case i remembered was when you take the constant sheaf construction, i.e. continuous maps from a boolean space B to an algebra A, with A treated as discrete

[frail shoal]

context being this

[rocky cloak]

What's boolean space?

[frail shoal]

totally disconnected compact hausdorff space

equivalently, projective limit of finite discrete spaces

[rocky cloak]

Right, and that's uniquely determined by cardinality?

[frail shoal]

nah

[rocky cloak]

So there's a choice to be made?

[frail shoal]

wait wdym by "that"

[rocky cloak]

I guess it says "a boolean product"

[frail shoal]

yea

the relevant spaces are, for coproduct, you think about the 1 point compactification of discrete spaces

[rocky cloak]

Well I know there's a unique countable such space

[frail shoal]

and for product, you think of the free compactification of discrete spaces

[frail shoal]

without isolated points yea

but that's untrue for higher cardinalities

[rocky cloak]

Right, that wasn't part of the definition

[frail shoal]

ok i will admit

this stuff is mostly arcane and only comprehensible / interesting if you like the specific things i do

[rocky cloak]

So then the direct product is a boolean product

But there are others

[tribal moss]

Hmm, I'm a bit rusty here: In Grp, image and coimage are the same object (up to isomorphism) but they're accompanied by different morphisms. How generally is that true?

[frail shoal]

yea coproduct falls out this way, although it's a slight lie

[rocky cloak]

In general you should have a canonical map from coimage to image (which is an isomorphism in Grp, but shouldn't have any restrictions in general)

[frail shoal]

this construction creates the algebra of tuples that have a cofinite set that is constant

[tribal moss]

IC.

[frail shoal]

it ends up being isomorphic to coproduct in the powers case at least

cuz you split the "eventually" part off as its own factor

[frail shoal]

but yea tbh

this was probably not worth bringing up

😭

[swift root]

in modern language, a boolean product is an algebra of global sections of a sheaf over a profinite space

they're apparently pretty useful but I've not read enough about them to tell you how

lol

[frail shoal]

legendary enpeace summon

[frail shoal]

i just think they're pretty

although admittedly im definitely saving them in my back pocket for if i ever need them

something that is funny is that an ultraproduct is precisely a germ of this construction (on compactification of discrete space)

[swift root]

germ?

that's interesting

the stalks are the A_x I'm pretty sure

[frail shoal]

oh i meant stalks

i confused the two words

[frail shoal]

im sad cuz like. you probably thought it was interesting cuz i was wrong

😭

[swift root]

i could be wrong too lol

[frail shoal]

nah i swear i was essentially thinking stalks in my head

[swift root]

i should start reading in sheaves of algebras over boolean spaces again

[frail shoal]

given an index set I, we can view Prod X_i as a sheaf F over the space beta I. a point U ∈ beta I is an ultrafilter on I, and the ultraproduct is F_U

[swift root]

very pleasing

[frail shoal]

and i figure that cuz. a germ in F_U is, essentially, a subset S ⊆ I such that S ∈ U, and an element of Prod X_i for i ∈ S, modulo passing to T ⊆ S such that T ∈ U

i say S ⊆ I because we have a basis of clopens for beta I in correspondence with subsets of I, and i believe they're given by the set of ultrafilters that contain S

[swift root]

i.e. the distinguished sets D(f) in the Zariski topology

except you take ultrafilters instead of prime ideals

[frail shoal]

holy shit, i wasn't even thinking on that level

you're already on the geometric angle

[frail shoal]

what's the difference

[swift root]

yeah, taking that basis is for example how you can actually compute the stalks of an affine scheme

[frail shoal]

yea i assume u can always compute sheaf stuff on a basis

[swift root]

yeah thats why the whole gluing and separation thing is needed

basically means you can either just look at the stalks or just look at a particularly well-behaved basis

[somber goblet]

ooo

[tough raven]

That is usually called a profinite space.

[tough raven]

Interesting. What is this sheaf?

[frail shoal]

ah, is this a much more common name? i was aware of it

[frail shoal]

i mean, it is what it is. but the idea is that you send {i} to X_i

hm im realizing i might understand it much less than i thought

i would think on this but i have something i really need to do that's much more important, oops

[next obsidian]

Eh, I’ve seen the other one too

[frail shoal]

hi chair monkey

[frail shoal]

i may or may not have put off my analysis homework until 3 hours before the deadline

[next obsidian]

Hi Keith

[frail shoal]

i might make it out alive tho

and then i can think about sheaves (moral good)

[frail shoal]

update: like jagr, i think i was just wrong

sorry yall. tbh ignore everything i said today in this channel if you want to be safe

[next obsidian]

Okay I’ll start by ignoring this message

[frail shoal]

i forgot that i can't use boolean product stuff to do general products because they always have finite image as a restriction

i feel like the "ultraproducts as stalks" perspective has to work somehow but definitely not how i said it

and my entire "fun fact" originally is totally wrong

[frail shoal]

the good news is that i put almost no effort into being comprehensible in the first place so my best hope is that nobody bothered to actually read what i wrote

[next obsidian]

I didn’t

[frail shoal]

hell yea

feeling better already

[next obsidian]

Chmonkey lazy win

[tough raven]

You didn't define it actually. At least I couldn't understand it.

OK but actually please finish your homework and rest as needed first.

[knotty badger]

Let $X$ be a finite set, and $\sigma : X \to X$ be a permutation. How do people prefer to define the sign/parity of $\sigma$?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[tough raven]

I prefer parity of number of transpositions (possibly with some flavourtext that doesn't really change the definition).

[elfin wraith]

It’s less cumbersome to actually use, but I think in proofs the other definition where you like act on the product of the differences is quite nice

Much the same though (well, they are the equivalent lol)

[knotty badger]

How do you prefer to show this is well-defined?

[knotty badger]

And I think you need to do a little effort to show this is well-defined

[vocal pebble]

There is a nice proof I had seen on MSE using a polynomial similar to a discriminant

[next obsidian]

I don’t think so lol

It’s clear that it becomes itself or its negative

[knotty badger]

Well X might not come with a total ordering

[next obsidian]

What

Oh you’re starting with some fuckass set

[knotty badger]

Yes

So you’d have to show it’s independent of the choice of total ordering

[tribal moss]

Choosing a different total ordering corresponds to conjugating everything with some permutation.
Once we've seen that parity defined by some fixed arbitrary ordering is a group homomorphism into Z/2Z which is abelian, it follows generally that conjugate elements have the same image.

[tough raven]

Nope's thing is a good way.

These days I'm too root-system pilled, so I don't think I'll give an objective answer.

[noble nexus]

Determinant of the permutation representation

[knotty badger]

hm, the way i've seen determinants define usually require knowledge of the sign of a permutation

[frail shoal]

yea to be clear i was wrong

it isn't true

[swift root]

We take the boolean ring R = P(I), along with the topological space Spec(R) thought of as ultrafilters of I. Then, for a subset S ⊆ I, the distinguished open set D(S) is the set of prime ideals not containing S, i.e. the set of ultrafilters containing S. Assign this the algebra 𝐹(S) = ∏_i∈S A_i.

If D(S) ⊆ D(T), then for any x ∈ S, the principal ultrafilter of x contains T, i.e. x ∈ T. Thus S ⊆ T. Conversely, if S ⊆ T, then clearly any ultrafilter containing S also contains T. The assignment S → D(S) is exactly stone duality for powersets. Anyways.

For the inclusion D(S) ⊆ D(T) we assign the canonical projection 𝐹(T) → 𝐹(S) sending (xi)_i∈T ↦ (xi)_i∈S. This is obviously a presheaf.

By compactness, a covering D(S) = ⋃_λ∈Λ D(S_λ) can always be taken to be for Λ finite. Stone duality then tells use S = ⋃_λ∈Λ S_λ. The separation and gluing axioms now follow immediately, so we indeed have a sheaf.

[frail shoal]

whao maybe enpeace can be right and recover my feelings

[tough raven]

You actually used an italicised F.

✝️

[swift root]

Take an ultrafilter U, and consider the stalk. A neighborhood basis of U is given by D(S) where S ∈ U, so the stalk is the union { (xi)_i∈S | S ∈ U, xi ∈ Ai } / ~, with (xi)_i∈S ~ (yi)_i∈S' iff they agree on S ∩ S'

Since every restriction is surjective, this will be a quotient of 𝐹(I) = ∏_i∈I A_i.
Then:
(xi)_i∈I ~ (yi)_i∈I <=> there is some S ∈ U with xi = yi for all i ∈ S
But then the equaliser of these two sequences contains S, so must also be in U, hence we get
(xi)_i∈I ~ (yi)_i∈I iff the equaliser of the sequences is in U

and this is exactly the ultraproduct

[tough raven]

I see. Indeed the stalks are ultraproducts, assuming all sets A_i are non-empty which we are (actually this colimit is a better definition of ultraproduct if there might be empty sets).

[swift root]

yippee

[tough raven]

Ah, and the internal logic of the topos of Spec R will indeed satisfy LEM if R is Boolean, which is why ultraproducts satisfy LEM.

I think.

[swift root]

I don't know enough topos theory to confirm that lol

[frail shoal]

yeah i was malding at myself cuz i was thinking about a specific construction (the constant sheaf) and forgot that we can just be more clever than throwing the default thing at an algebra

[swift root]

but smt smt basis of clopen sets

[tough raven]

It should mean that for any sentence, the largest set on which it is true is a clopen set.

[swift root]

lmao

[tough raven]

Whatever I'll figure out if that's meaningful later

[frail shoal]

just to make sure i understand, consider the non-closed open consistenting of all coordinate projections of P(I), is this section identical to the global section

actually. i wonder if it would be interesting to actually use the constant sheaf after all, and just accept that the global section will be different than this section. but on the other hand, it's good to know you can actually make the global section into a standard direct product

[swift root]

coordinate projections

?

[frail shoal]

oh i mean, P(I) -> P(1) for all 1 -> I

sorry i should have been less weird about my words

[swift root]

you mean the union of all D({i}) for i ∈ I?

wait that's not a covering lol

[swift root]

the algebra of sections will again just be ∏_i∈I A_i

by the sheaf property

[frail shoal]

oh yea i could have just said that

🥀

i am NOT on my clarity arc today

[frail shoal]

yay

[tough raven]

In fact you can do this for Spec S for any R-algebra S (and probably for any R-scheme). Essentially, a morphism from a scheme S to Spec R is the same as (S being charactristic 2 and) partitions of S into disjoint unions of open subschemes in a way indexed by R (more formally, a clopen subscheme for each r in R, which is a Boolean algebra homomorphism of r). (For example, if R is the free Boolean algebra on a set A, i.e., (⨯)_{a in A} {0, a, 1-a, 1}, then a morphism from S to Spec R is the same as a partition into two pieces of S for each a in A. R can also be described as the sub-Boolean algebra of P(P(A)) generated by "a" = {S - a ∈ S} for a in A - a collection of subsets in the ambient P(A) that together divide up P(A) as perfectly possible. R = P(I) at least in the case I = P(A) is a "completed" version of this, i.e., you require some clopens to be present as "infinite intersections" etc.)

Well what do I mean by "do this"

I suppose given quasi-coherent modules over the preimages of the points of Spec R in S, you can successfully get their product as a quasi-coherent module over S.

[swift root]

that is cool asf

[tough raven]

Eh I don't feel like checking the details actually

It would be cool if this showed QCoh(S) = ∏_{U ultrafilter of R} QCoh(S_U); I was sure that only held for finite partitions.

[tough raven]

Also characteristic 2 must not really be important. For a single clopen we just need to replace {0, a, 1-a, 1} = ℤ/2ℤ ⨯ ℤ/2ℤ with Spec ℤ ⨯ ℤ. Probably for any Boolean algebra R, we take a presentation and turn it into a presentation without the a+a=0's?

Yeah I think we just avoid claiming that + corresponds to xor but keep claiming that ⋅ represents intersection and 1- represents complement.

[covert cliff]

I have a lie group G with a R valued 2 cocycle on its lie algebra g. Does the coadjoint representation of this group always extend to an action of the integrating central extension on the dual of the g?

Not sure if this is the right channel tho

[latent solstice]

R is a field, and R^n is also a field right? Is C^n also a field?

[tribal moss]

No, and no.

The only R^n that can be given a field structure (with the usual vector addition, and a subfield that scales vectors like R does) are R^1 (the reals) and R^2 (the complex numbers).

If you allow non-commutative fields (aka skew fields or division rings), then you also have R^4 which can become quaternions. But that's where it ends.

This is the Frobenius theorem.

[latent solstice]

R^2 is isomorphic to C? I should have thought of that...

[swift root]

as a vector space

R^2 as a ring isnt

[latent solstice]

Icic

[tardy hedge]

also R^n as a ring has zero divisors so it cant be a field

[swift root]

worse, it has
gasp
idempotents

[round anvil]

there's a way to prove that A_5 hasn't a subgroup with order 15 or 30 without using that A_5 is simple?

[quiet pelican]

For order 30:
Said subgroup has to be normal, and hence must contain every odd order element
Count that there are more than 30 such elements

For 15:
It isn’t that hard to prove the only group of order 15 is cyclic, and there is no cyclic subgroup of A_5 of order 15 (alternatively: note that it would have to contain an order 3 element and an order 5 element, and bash out cases)

[round anvil]

thank u

[small yacht]

What does "Orbit of the action of a finite group on A" mean here? Orbits have only been defined for elements, as O_g(a) = {ga | g \in G}

(prop 9.9 for context)

[lusty marlin]

It says "an orbit", and they mean any orbit O_g(a), where a ∈ A

[small yacht]

oh, that is very subtle

[small yacht]

thanks!

[kind temple]

any hint for this one?

chain just means totally ordered

here is the definition of an algebraic lattice

b1 \prec b2 means that b2 covers b1, in the sense that if b1 \preceq c \preceq b2, then c = b1 or c = b2 i.e., there aren't any elements between b1 and b2

coming up with interesting examples of totally ordered algebraic lattices is hard. i had a thought to try and freely generate one but i haven't fleshed it out all the way

[rocky cloak]

For one direction, fix a compact b2 and consider the sup of all compacts < b2.

[kind temple]

is every object in L compact if L is algebraic in this case? like, if a is the directed join of x1 \leq x2 \leq ... \leq xn of compact xi, then xn is the join of all of the xi's and is equal to a.

now i can find finitely many compact objects between a1 and a2

let me try your suggestion tho

[rocky cloak]

You can have for example Z union infinity. Then infinity is not compact.

For the other direction note that in (b2 covers b1) it must be that b2 is compact

[rocky cloak]

I guess ordinals give good examples maybe

[kind temple]

i think we want N union infinity. Z union infinity isn't complete (there is no least element). but okay, yea, i agree with you then

[rocky cloak]

I mean you can add in negative infinity as well, but yeah

[rocky cloak]

But so you're not defining complete to mean things with upper/lower bound has sup/inf?

I.e. R is not complete?

I mean you can always just adjoin a max and min so doesn't really matter

[kind temple]

complete was defined so that the join/meet of any subset exists

equivalently, every subset has a sup and inf in the poset

[kind temple]

for b compact, are suggesting that b' = sup{c in L : c compact and c < b} \prec b?

okay, so the challenge is to (a) show that there is some compact object between a1 and a2, and (b), show that the supremum all compacts below this one is larger than a1

[kind temple]

okay, so for this direction, if we have a cover of b2, and we assume for contradiction that all of the elements in the cover are strictly less than b2, then they have to all be at most b1, and so we would obtain a contradiction that b2 is strictly less than itself. so there is at least one element in the cover which is at least as large as b2, thus b2 is compact.

is the idea now that every object is just the sup of compacts less than or equal to it?

thanks

what was the intuition here?

this one stumped me

[kind temple]

i guess one issue is that for some reason i thought that every object was a finite join of compact elements. this need not be the case

[rocky cloak]

I guess an element is a sup of compacts iff it is the sup of all compacts <= than it.

The rest is just applying this idea

And this trick that a compact element always has a predecessor, by just taking the sup of < things

This second idea does come up in algebra as well. E.g. a finitely generated module has a maximal submodule.

[swift root]

Consider a complete chain L

Lemma: Let b be a non-bottom element. Then b is compact iff b covers an element b'
Proof: (=>) Suppose b is compact, and consider S = { x ∈ L | x < b }. This is nonempty as b is not the bottom element. Then b' = ∨S ≤ b, clearly. Suppose b' = b
=> exists finite subset S' with b = ∨S'.
=> b ∈ S' as L is a chain
contradiction, so b' < b. It is also easily seen that b covers b'.
(<=) Suppose now that b' is covered by b, and suppose b = ∨S for some S. If x < b for all x ∈ S, then x ≤ b', and hence ∨S ≤ b'. This is of course impossible, so b ∈ S, hence there is a vacuous finite subset of S that covers b.

Now for the proof.
(=>) Suppose L is algebraic, and let a1 < a2. Consider S = { b compact with a1 < b ≤ a2 }. By algebraicness, this is nonempty. Take any b ∈ S. By the above, there is a b' that is covered by b. Thus a1 ≤ b' ≺ b ≤ a2, as desired.
(<=) Conversely, suppose the property holds. By the above this can be restated as: For all a1 < a2, there is a compact element b with a1 ≤ b ≤ a2. Let a be a nonbottom element, and a' be the join of all compact elements ≤ a.
Suppose BWOC that a' < a. Then we cannot have a' ≺ a, as this would mean that a is compact. But this means there is some a' < x < a, which is also a contradiction as this would yield a compact element b with a' < b ≤ a. Therefore, a' = a and L is algebraic, as desired.

really fun problem

[azure cairn]

"BWOC" has two primary definitions: in engineering, it means "By Weight of Cement" (cement additives ratio), while in colloquial usage it means "Big Woman On Campus" (an influential woman).

Inassume you mwan by way of contradictiob?

[quiet pelican]

[azure cairn]

big woman on campus

[quiet pelican]

[prime stirrup]

anybody know crypto

and can help

[elfin wraith]

Thats very vauge, you should post your actual question if you want a response. Chances are the answer will be no though

[prime stirrup]

yes you are rigth

lets say these topics

Mathematical Foundations of Modern Cryptography

 Modular Arithmetic
 Elementary Number Theory in Cryptography
 Fundamental Theorem of Number Theory
 Euclidean Algorithm
 Extended Euclidean Algorithm
 Euler's Phi Function
 Fermat's Little Theorem/Euler's Theorem

[verbal valley]

this is a topic list, not a question. We can't help you with a topic list

[elfin wraith]

This is just basic number theory, and youd see all of this in any introduction to number theory

If you have any specific questions try in #elementary-number-theory

[prime stirrup]

okay thank you

[prime stirrup]

i wantedto know if people know about the topic so i can speak with them in private because i don’t want to disturb the group

[prime stirrup]

i think i wrote in the wrong chat though here

[kind temple]

thanks, will go through this approach when i get some time today

[kind temple]

yea, i have been going through burris and sankappanavar

[swift root]

peakest