groups-rings-fields

[next obsidian]

Not technically, ring theoretically

[somber goblet]

where each point has a residue field of different characteristic

[next obsidian]

But like if you do stuff over Z you will end up in the mixed characteristic case as soon as you localize

Mixed characteristic rings are local rings where the char of the ring is different from that of the residue field

[somber goblet]

is this a property of a ring or of a scheme?

ohh i see

[next obsidian]

Equivalently, it’s rings which do not have a field inside of them

All mixed char rings end up being char (0,p) or (p^n,p)

And regular rings are integral domains so they are all (0,p)

They show up in arithmetic mostly

Or if you’re sick like me

[somber goblet]

oh i see makes sense

so the stalks of Spec ℤ are mixed characteristic, for instance

[next obsidian]

Sure

The local rings, or stalks of the structure sheaf

But in some sense Spec Z like, interpolates over all mixed char and equicharacteristic

When you look over all points

[somber goblet]

wdym? like because it has a generic point with residue field ℚ?

[next obsidian]

Yeah

And you can go over all characteristics

But this is something an arithmetic person knows more about

[somber goblet]

very cool, i see

[vernal vector]

Can someone tell me how the Variable Elimination Theorem relates to how you can use the Grobner basis of ideals I = (f_1, f_2,.. f_n) to basically do Gaussian elimination on the system of polynomial equations f_1 = 0, f_2 = 0,.. f_n = 0 to solve it? Like i feel like i have somewhat of an idea of how it is related. But just, it is kinda vague.

Like i know basically you can do this trick to solve a system of polynomial of equations. To reduce it to an equivalent system of equations with the new polymonial rely of less and fewer variables.
Like f_1 = x^2 + y^2 + z^2 - 1 = 0, f_2 = x^2 - y + x^2 = 0, f_3 = x - z = 0
Can be reduced to g_1 = 4z^4 + 2z^2 - 1 = 0, g_2 = -y + 2x^2 = 0 and g_3 = x - z = 0.

And since basically you have the last polynomial only rely on variable z, you can solve it normally and sub it to the other equations. However, I still really don't get how Gronber basis of I = (f_1, f_2, f_3) and the variable elimination theorem help in getting to here

[granite notch]

Hey guys, I wanted to know is this stand for infinite cyclic group and how we prove it

Anyone know hwo to solve this question?

[elfin wraith]

What’s the question? Did you mean to post an image or something?

[granite notch]

The subgroup of cyclic group also the cyclic gry

Group

[elfin wraith]

It’s really not clear to me what you’re asking, sorry.

Are you asking if there is an infinite cyclic group, and if subgroups of a cyclic group are also cyclic?

[foggy tartan]

you learned about the infinite cyclic group in grade school, it's more commonly called Z

if you realize some cyclic group G as comprising powers of a principal element x in G, then the subgroup generated by any element (or set of elements) of G will only consist of powers of x, so must be cyclic

[flat anvil]

Same energy as "you learned about fields in grade school, its called R"

[bitter fog]

"Prove that every group of even order contains an element of order two."

Can anyone help me get started

[bitter fog]

Let G be a cyclic group and so $G=\langle g\rangle.$ We need to prove that subgroups of G are also cyclic. Suppose H is a subgroup, if $H={id}\implies H=\langle id\rangle$ and $H= G\implies H=\langle g\rangle$. Now, let $H\neq G,{id}.$ Now there exists a $g^k$ with $1\leq k \leq n-1$ with $n = |G|$. Let $m = \min{k\neq 0|g^k\in H}$. We prove that $H = \langle g^m\rangle$. Let $g^k\in H, k\geq m$. so $k = mq+r$ with $q\geq 0$ and $m-1\geq r\geq 0$. Let $r\neq0\implies g^m \in H\implies g^{-m}\in H \implies (g^{-m})^{q} = g^{-mq}\in H$ Now because $g^{mq+r}\in H$ and $g^{-mq}\in H \implies g^{mq+r}\cdot g^{-mq}=g^r\in H$ Contradiction, because m is minimal. So $r=0\implies g^k=g^{mq}=(g^m)^q\implies H = \langle g^m\rangle$.

[cloud walrus]

arachi

[languid trellis]

Pair elements up with their inverses then count

[granite notch]

I c your proof

And I'm not sure how this proof extends to the infinite cyclic group

[bitter fog]

I was also working on in just now

[granite notch]

Your question is the motivation make me think this question

[bitter fog]

the only generators of the infinite cyclic group are 1 and -1

for Z

[granite notch]

Hows for R

Is it a cyclic group

[languid trellis]

You tell me

[granite notch]

No for no generator?

[languid trellis]

I don't understand

[bitter fog]

Theres no singular element in R that can generate every element of R

[granite notch]

So it isn't

Right?

[bitter fog]

R,+ is not a cyclic group

what about R* with multiplication?

[granite notch]

I was thinking about

Is that the rotational group is a cyclic group

[bitter fog]

no

Really, the only cyclic groups (up to isomorphism) are $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$

[cloud walrus]

arachi

[bitter fog]

Where Z is infinite and Z/nZ finite

[granite notch]

I learn a little of group theory for a long time ago

And i forgot part of them

Btw how you type this latex efficiently

[bitter fog]

wdym

[bitter fog]

Im studying this right now, https://github.com/vendramin/group/blob/main/notes.pdf

[granite notch]

How to type this format quick

[bitter fog]

if u use latex then you have to put it between $$

[bitter fog]

it starts from 0 and everything is explained very well

[flat anvil]

Check the pins in #latex-help i think

[bitter fog]

Prove that every group of even order contains an element of order two. Let G be a group of even order. Then $G={id}\cup{g}\cup{g_1,g_1^{-1},\dots}$ The last set always comes in pairs of an element and its inverse. So the elements of that set must have an even order > 2. The identity has order one, so therefore g must have order 2, otherwise |G| wouldn't be even anymore.

@languid trellis

Is this right?

[next obsidian]

You have the right idea but you need to write it differently

[languid trellis]

You need to justify that decomposition more. I agree with you that it can be decomposed in a way, but at this stage you've essentially assumed your conclusion.

[next obsidian]

Also what you’ve said isn’t quite true

A group can have more than a single element of order 2

You can’t justify that each of those g_i come in pairs

[cloud walrus]

arachi

[south patrol]

Prove it

[next obsidian]

Technology hasn’t gotten that far

[delicate orchid]

It follows from the CFSG

[elfin wraith]

You expect me to count to 4?? Are you mad???

Typical elitism in maths

[bitter fog]

"prove that a group with finitely many subgroups is finite"

I started with trying to prove the contrapositive

So if G is infinite, then G has an infinite amount of subgroups

idk what to do

next

[tribal moss]

Perhaps look just at the cyclic subgroups, and show there must be infinitely many of those.

[bitter fog]

but what if every element of G has a finite order

[tribal moss]

Then every cyclic subgroup will be finite.

And each element of G will be in at least one of them.

[bitter fog]

wdym?

[elfin wraith]

Theres still infinitely many of them, but also ignore me, its not particuarly helpful here I dont think

Tropos hint is good

[tough raven]

MFW F_2-vector spaces

[tribal moss]

<@&268886789983436800> (scam, already gone).

[south patrol]

Ah ok

[tribal moss]

Inspired by conversation above: Can a group have a nonzero but even number of order-2 elements?
It's easy to see it would have to be infinite and nonabelian, and with a bit more footwork also that it cannot have exactly 2 order-2 elements. But that footwork doesn't feel promising for generalizing.

[quiet pelican]

No

It can’t, this was actually an exercise (albeit a hard one) in our first year groups

[elfin wraith]

Something to do with the order of their product right? I vaguely remember doing this at one point

[quiet pelican]

Fix an order 2 element a (if there are no order 2 elements, we’re done)
Let X be the set of order 1 and 2 elements that commute with a
Let Y be the set of order 2 elements that don’t commute with a
You can verify that the following is an order 2 bijection from the set of order 1 and 2 elements to itself with no fixed points (and hence the number of order 1 and 2 elements is even, so the number of order 2 elements is odd)
If b \in X, send b to ab
If b \in Y send b to aba

[quiet pelican]

Are you thinking of the exercise where every element in the group is order 2?

[elfin wraith]

Possibly yeah

[quiet pelican]

Assuming the number of order 2 elements is finite of course
If it’s infinite the question doesn’t make sense

[tribal moss]

Neat, thanks.

[kind temple]

the relation x ~ y iff x = y or x = y^{-1} is an equivalence relation on a group.

[balmy wraith]

Lagrange’s theorem

[delicate orchid]

How do you plan on using Lagrange’s theorem here

[flat anvil]

-# My ahh tried to think of power sets here but idk if that helps

[vocal pebble]

You can take the cyclic subgroups generated by every element, the collection is finite, each member is finite, so the group is finite

[rocky cloak]

Finite union of finite sets is finite!

[vestal jay]

Could someone help me with the "Conclude..." part? What I know as the commutator subgroup is for elements a,b in G aba^-1b^-a in [G,G]. Theorem 7.12 refers to the universal property of quotient groups and the proposition is about the definition of product in a category

[next obsidian]

I don’t see how that could be the correct thing for the proposition

But this just comes down to the definition of ^ab

Or I guess F^ab

[languid trellis]

I was going to say, surely the abelianisation is defined to be F/[F,F]?

[next obsidian]

You should show that G/[G,G] for any group G has the property it uniquely factors maps from G into abelian groups

[south patrol]

so does G

[next obsidian]

As an abelian group

Fuck you potato

Chain that with F(A) uniquely giving rise to maps from maps A into any group

And you end up having a sentence that is isomorphic to what defines F^ab

[vestal jay]

Oh

That helps, thank you

[noble nexus]

I would imagine F^{ab} here means the free abelian group on A

so here you're just showing that the abelianization of the free group is the free abelian group

[frail shoal]

does there exist a group G and a subgroup N such that [G : N] = 2, the map G -> G/N does not split, and G has exactly 3 normal subgroups

[noble nexus]

3 nontrivial or 3 total

[azure cairn]

Z4?

[velvet hull]

How are you counting the subgroups?

Because the quaternion group is the smallest example of a non split SES and it sounds like that’s what you’re looking for here

[azure cairn]

ok wait

what about $G = \angled{x \mid x^4 = 1} \cong \bZ_4$?

[azure cairn]

and then you have $N = \angled{x^2} \cong \bZ_2$

[azure cairn]

the normal subgroups are {id}, N, G

[G:N]=2

and G->G/N doesnt split

[frail shoal]

ok i just realized i made a few oopsies

[frail shoal]

3 total

[frail shoal]

yea that works

[frail shoal]

that has a lot more than 3

even if we count nontrivial only, we have <-1>, <i>, <j>, <k>

anyway im sorta XY-ing so the fact Z/4 works actually doesnt resolve my question, so i need to be more precise

[frail shoal]

does there exist an infinite family of such groups G

[frail shoal]

if yall want i can un-XY and just share what im thinking about, but, ill do that later

[rocky cloak]

Z/4

[frail shoal]

my excuse is tunnel vision

[rocky cloak]

Sn

[frail shoal]

doesnt that split

[rocky cloak]

Right, you didn't want the map to split

[frail shoal]

yea

[swift root]

those darn semidirect products

[frail shoal]

ok yea so my motivation is this:

I've been thinking about the connected components of Grp / G as a preorder, given a particular group G. here G = Z/2

[rocky cloak]

Everything maps to the identity on Z/2 though

[dull ginkgo]

Zamn

[frail shoal]

I've managed to prove that, given a particular surjective group hom 𝛗: G → Z/2, we can always take some g ∈ G such that g ∉ ker 𝛗 and we get a group hom <g> → Z/2 that's also surjective

[dull ginkgo]

That’s some twisted shit. Never crosses my mind

[frail shoal]

well the 0 map always pulls back to other 0 maps

the interesting behavior is surjections

[rocky cloak]

But there's only one connected component

[frail shoal]

the hard part is going the other way, and my gut tells me that there should be a large class of groups you can't go the other way

[frail shoal]

oh ok i mean like. strongly connected

oops

i see the confusion now

[rocky cloak]

Okay, then I think I understand what you're thinking of

[frail shoal]

when G = Z there is only finitely many classes cuz each subgroup of Z is projective

[rocky cloak]

So then Z -> Z/2 and Z/2^i -> Z/2 should all be relevant

[frail shoal]

yea i agree

[rocky cloak]

So the restriction of just 3 normal subgroups is very strong

[frail shoal]

it would feel weird if these are the only classes but idk much in-the-weeds group theory

[frail shoal]

yea definitely not the only way a unique example could arise

i was just tryna brainstorm what could go wrong

[rocky cloak]

Q8 -> Z/2 and I guess most groups that are not semidirect product with Z/2^i should also give you something

Feel like all the generalized quaternion groups should give you different things as well

[frail shoal]

ok yea that makes sense

[dire wren]

12c, this doesn't look like a group action to me

In particular I can take phi maps everything to the identity automorphisms, and conjugation doesn't associate?

Am I wrong?

Or do you guys think serge meant x1(phi(h1)(x2))

I just realised the qn numbers were cut off, the c problem just below (semi)direct product

[rocky cloak]

This is the standard formula at least. So presumably just an extra superscript slipt in

[dire wren]

K, thx

[torpid radish]

Are there something like 'free fields' just like free groups and free modules?

I tried to come up with some constructions but I failed

[quiet pelican]

There shouldn’t be free fields

[tribal moss]

No, that's not possible.

[torpid radish]

Aha

[quiet pelican]

What characteristic would they be?

You can’t have a map between fields of different characteristic

In general, the category of fields tends to be very badly behaved, and a lot of standard ‘categorical* constructions don’t exist for fields

[tribal moss]

A free field on {*} would be a field F with a distinguished element x such that for every field K and every k in K there is a unique field morphism F ->K that takes x to k.
Even if we only consider fields of a fixed characteristic, that still wouldn't work because both k=0 and k=1 would need to be possible.

[torpid radish]

Ah ok I got it now

[torpid radish]

Thanks for help!! :)))

[tribal moss]

Intuitively, one might expect Q(x,y,z) to be "more or less" what a free (characteristic 0) field generated by {x,y,z} ought to be -- but that doesn't match the categorical concept of "free" as left adjoint to a forgetful functor. It's not even functorial in the first place: there's no field morphism from Q(x,y) to Q(x), even though there are set morphisms {x,y}->{x}.

[torpid radish]

It feels that fields are extremely restrictive when it comes to their morphisms.

[wraith cargo]

They're all injective or trivial!

[tribal moss]

They're all injective, period.

[wraith cargo]

Unfortunately due to my religious views I believe in the existence of trivial morphisms

[kind temple]

you don’t require 1 to go to 1?

[south patrol]

😔

Respect for irony lost

[torpid radish]

What a sweet religion! But sadly I can not believe it :))

[elfin wraith]

My UG algebra course beat this into me, and despite requiring 1 to go to 1 now (like a normal person) I can’t stop myself from throwing in the word nontrivial

[south patrol]

😔

[dull ginkgo]

URing morphisms have that constraint right

Or you know

Ring

Not just random number generator

[south patrol]

Lol

[rocky cloak]

Or they mean something else than us when they talk about the field with one element

[somber goblet]

yeah

[tardy hedge]

Given a ring map f: R->S and a unit s in S there is a unique map R[x,x^-1]->S sending x to s and restricting to f on R, right?

Would this just be called the universal property of laurent polynomial extensions or something

[quiet pelican]

Yes
Just define it in the obvious way

[quiet pelican]

I’d probably call it something like this if you asked me to give you a name, but I don’t think it has a standard name

[tardy hedge]

Ok, yeah cause I want to use this but im not sure if I should give it a name or how I should phrase it I guess

[south patrol]

If you want it's like uh the poly ring R[x] has the universal property that like a map R[x] -> S "is the same as" a map R -> S along with a choice of s in S. And then you combine this with universal property of localisation at an element

(I guess okay that last bit maybe needs some care in the noncommutative setting but x is central so should be chill)

also as it is important: another way to state this is that R[x, x^-1] corepresents the "group of units" functor -- often denoted Gm or (-)^x -- from R-algebras to groups (and more precisely, the bijection Hom(R[x, x^-1], S) -> Gm(S) is given by evaluation at x)

[somber goblet]

this should also have to do with R[x, x^-1] being the free group algebra on one generator

[noble nexus]

This is basically just the universal property of polynomial rings combined with the localization

or polynomial rings + the quotient, whichever you prefer

Since R[x,x^{-1}] is either the localization of R[x] at x or the quotient of R[x,y] by the ideal generated by xy-1 (and yx-1 if noncom)

[kind temple]

are the prufer p-groups the only infinite groups whose proper subgroups are finite?

[rocky cloak]

There's also Tarski's monsters

[kind temple]

wow. what a name

is there a classification of such groups?

[rotund aurora]

is there an easier example? I mean, Tarski gives a very strong counter-example (the proper subgroups are all Z/pZ)

[kind temple]

okay, so they are the only infinite abelian groups whose proper subgroups are finite

this might be an interesting exercise to try and prove

[rocky cloak]

Yeah the Prüfer groups are the only abelian ones shouldn't be too hard to see

[rotund aurora]

Tarski is finitely generated

[kind temple]

i find this kind of bizarre

[rocky cloak]

Generated by just 2 elements even

[rotund aurora]

for the abelian case, you can reduce to the torsion p-primary case. Then consider the sequence C_0 subset C_1 subset ... where C_n is the set of p^n-th roots of unity in G. If each C_n is cyclic you are done. If one is not cyclic, then all starting at some level n are not cyclic. You should be able to find a subsequence of the C_i consisting of nested cyclic subgroups, which violates the hypothesis. Idk if this is immediate, but I think you can do it by some sort of combinatorial compactness argument + the structure of finite abelian p groups

ie, like say C_1=Z/pZ oplus Z/pZ. Idk if it's obvious that one of these parts should have p^n roots for all n, but you should be able to change the decomposition so that this holds

[rotund aurora]

Apparently Tarski groups are periodic

[rocky cloak]

Every element has order p

[rotund aurora]

Your question is known as Schmidt's problem. The first example was so that every proper subgroup had prime order. I'm not sure if they meant prime order for a fixed prime p, and idk if there's an implication. Didn't look any further.

https://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=1668&option_lang=eng this is not the first paper, but it's short and in English

This is the first paper https://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=dan&paperid=42623&option_lang=eng It's in Russian but it's just 3 pages. Idk if it's self-contained

[rocky cloak]

I thought you said Tarski monster. Anyway it's clear from the definition that the group is periodic

[rotund aurora]

what does periodic mean?

[rocky cloak]

Every element having finite order

[rotund aurora]

I thought it meant there exists some n such that g^n=1 for all g

[rocky cloak]

Okay, I see.

Then that's interesting I guess

[rotund aurora]

idk if this implication holds, forget what I said about periodicity, I misread

[copper kestrel]

Why does $$\prod_{i=1}^{n} A_i = \bigoplus_{i=1}^{n}A_i$$ only work for a finite number of abelian groups?

[cloud walrus]

tabby tabby

[tribal moss]

Are you asking for an intuitive way to look at it, or for a more rigorous argument that the same group won't satisfy the universal properties of both product and coproducts?

It's also relevant for answering to know just how you're defining \prod versus \bigoplus here. By universal properties or by concrete constructions? (Perhaps one of each?)

[verbal valley]

Elements of $\prod_{i=1}^{\infty} A_{i}$ are infinite sequences $(a_{i})$ where $a_{i}$ lives in $A_{i}$. Elements of $\bigoplus_{i=1}^{\infty} A_{i}$ are infinite sequences $(a_{i})$ where $a_{i}$ lives in $A_{i}$ for all $i$ and all but finitely many of the $a_{i}$ are zero.

[cloud walrus]

millie :3

[verbal valley]

with finite products and direct sums the "all but finitely many are zero" condition does nothing, but obviously there are infinite sequences where infinitely many terms are nonzero, and those live inside the product but not the direct sum

[copper kestrel]

mostly intuitive, i do not know any category theory so idk what coproducts are

[tribal moss]

Okay.

[copper kestrel]

ohhhhhh

[copper kestrel]

that makes sense! thank you so much

[tribal moss]

One way to justify the all-but-finitely-many condition is that the direct sum is supposed to be the smallest subgroup that contains each of the summands.

So since we can get a subgroup without allowing inifinitely many nonzero elements, we should.

[copper kestrel]

oh i see!

[bitter fog]

what's y'alls favorite group

[wraith cargo]

F_1

[bitter fog]

its impossible to have a field with one element

cause 0 ≠ 1

[wraith cargo]

https://en.wikipedia.org/wiki/Field_with_one_element

Read it and weep bucko

[fickle dirge]

my professor said I don't see an issue with allowing a field with one element 😅

[bitter fog]

quite fascinating..

[frail shoal]

(Z, +) is the first one my brain came up with when i asked it for a "group"

honestly it's quite nice so ill accept this answer it gave me

[marsh fulcrum]

Friendly reminder that spurs are NOT relegated and will play the next season in the Premier League
Glory Glory Tottenham Hotspur

[fading acorn]

Hodge Galois group

[karmic moat]

i think im going to be sick

[velvet tusk]

$(Z_{1},+_{1})$

[cloud walrus]

chudcel

[tribal moss]

Is that the 1-adic integers?

[chilly radish]

They have residue field F_1

[somber goblet]

do p-adics have residue field F_p at all points

[rocky cloak]

There's only one closed point.

At the generic point I quess you would say the residue field is Q_p (so just the field of fractions)

[somber goblet]

how about the residue field at the localizations?

[rocky cloak]

What is "the localizations"?

Z_p has two prime ideals: (p) and (0)

[somber goblet]

oh, i see

oh yeah that makes sense

right

what's $\operatorname{Frac} \bZ_p / (p)$?

[cloud walrus]

lexi

[rocky cloak]

It's Fp = Z/pZ

[somber goblet]

right ok

i need to get more familiar with p-adics

i view Z_p as like the space of jets at (p) in Spec Z

[frail shoal]

what's a jet

[somber goblet]

generalized taylor series

[frail shoal]

ah yea i agree w that perspective

[somber goblet]

but all expansions of integers are finite series

so ig you have the p-adic metric, then take the completion of that metric space to get the p-adic integers

[frail shoal]

but yea a key fact about Z_p is that it's a local ring

it's very similar in shape to how k[[x]] is local for a field k

[somber goblet]

right, since these are the analog to formal power series for Spec Z instead of Spec k[x]

[frail shoal]

oh yea that makes what i had in my head formal

i just had vibes

yay

[somber goblet]

(at least i think)

[frail shoal]

it makes sense

[somber goblet]

iirc you can put an 'x-adic metric' on k[[x]]

[frail shoal]

Z and k[x] are often compared cuz they behave very similarly

[somber goblet]

yeah i think PIDs behave very similarly geometrically

[frail shoal]

yea the way i think of it is that x is like 0.01 or something

not literally but it's how i picture it

[somber goblet]

is each $k[x]/(x^n)$ local?

[cloud walrus]

lexi

[frail shoal]

yea

[somber goblet]

smooth curves or something

[frail shoal]

the projective limit of the rings k[x]/(x^n) is k[[x]]

so dually, Spec k[[x]] is the direct limit (i.e. kinda like a nested union) of Spec k[x]/(x^n)

idk what this really means for AG cuz im not an AG-er but

[somber goblet]

oh right i almost forgot CRing is just AffSch but backwards

yeah i think the construction here is $\varprojlim_{n \in \bN} R/\mathfrak m^n$

[cloud walrus]

lexi

[frail shoal]

that sounds right

the completion of a ring ideal or something

[somber goblet]

arrow might be backwards

whatever

[frail shoal]

the arrow is correct

[somber goblet]

this is fun

my impression of the p-adic integers is that they're like the integers with certain nicer properties

and like Q_p's analytic properties play nicer with its algebraic properties than R

wait wtf Z_p is the unit ball in Q_p

[chilly radish]

Quotient of a local ring is local

[somber goblet]

but k[x] isn't local, is it?

each (x - a) is a distinct maximal ideal

[rocky cloak]

k[x]/(x^n) = k[[x]]/(x^n)

[frail shoal]

i agree

[chilly radish]

This isn't true, Spec does not preserve colimits

[frail shoal]

aw man

well it does preserve limits, right?

im taking the Spec of a limit

[chilly radish]

It preserves limits, as in, colimits of rings get sent to limits of affine schemes

It's a contravariant functor

[frail shoal]

that sounds right i think

[frail shoal]

ah oh well

lexi disregard what i told you

i spread lies on the internet

[chilly radish]

In general, an infinite coproduct of affine schemes is not necessarily affine

[somber goblet]

ok yeah that makes sense

[somber goblet]

why wouldn’t it be Spec of the corresponding infinite product? i thought there was an equivalence of categories

[swift root]

duality between AffScheme and CRing

[karmic moat]

affine schemes are quasicompact, but an infinite coproduct of affine schemes need not be quasicompact

[swift root]

but this just means that infinitary coproducts in AffScheme do not correspond to infinitary coproducts in Sch

[rocky cloak]

Depends a little what is meant by Spec.

Spec(lim R) is off course the colimit in the category of affine schemes.

In this particular case I believe it also happens to be the colimit in the category of schemes, but this is not expected in general.

And it is not the colimit in the category of (locally) ringed spaces, or topological spaces

[tough raven]

Is there a free perfect group?

[azure cairn]

trivial group

[tough raven]

on 1 or preferably even 2 generators

[azure cairn]

i assume not then

[frail shoal]

is there anything we can say about filtered colimits?

[rocky cloak]

The category of affine schemes is just the opposite category of commutative rings, so there filtered colimits correspond to cofiltered limits.

Otherwise, you can't really say much I don't think.

[rocky cloak]

Like free in the category of perfect groups?

[tough raven]

TBH it's a curiosity question so IDK

[tough raven]

I suspect this does not exist but a proof would be cool.

Also could be a free object in a suitably "enhanced" category (say, perfect group equipped with an operation mapping each element to a sequence of pairs whose commutators multiply to that element). In the latter case, the hope would be for the free object to be described in terms of more familiar groups.

[rocky cloak]

I guess the symmetric groups are perfect should probably give you enough lieway to prove that the free group on one generator would need to be Z which isn't perfect.

[tough raven]

Wait is perfect not [G, G] = G?

[frail shoal]

perfect means everything is a commutator, right?

[tough raven]

I meant [G, G] = G if not.

[tough raven]

Product of commutators.

But yes.

[frail shoal]

ohhh

[chilly radish]

Yea

[rocky cloak]

I meant An, not symmetric group

Anyhoot simpler:

Say G is freely perfect generated by x, then compare identity and conjugation by something in C(x)

[tough raven]

For example, suppose we want a free "group with functions l, r such that for all x, x = [l(x), r(x)]" (which is stronger than being perfect). Then the answer is "obviously" to start with a free group, adjoin two generators for every element (not just every generator!), satisfying the appropriate relation, and repeat countably many times. But this seems completely intractable without some kind of simplification.

[frail shoal]

ah ok so it's like, trivial abelianization?

in which case every nontrivial free group does map onto Z yea

send all the generators to 1

[tough raven]

Yes... so?

Ah, you mean to prove that no free group (on more than zero generators) is perfect.

True.

[frail shoal]

ohh i see what u mean now tho

like

free (perfect group)

not

perfect (free group)

that is interesting

[rocky cloak]

Generalizing this:

For any such free group, mod out xy^-1 for any pair of generators. Then you would get a free group on one generator

(quotient of prefect group is perfect)

[frail shoal]

so by abstract nonsense such a free construction would be nontrivial iff there exists nontrivial "strongly perfect" groups

idk the actual name for them

so i made it up

[tough raven]

I could also consider a version where I only do this for the generators. This doesn't have a very clear universal property (that is simpler than its presentation), but does have a slightly more understandable description: for a set S, let p_S: Free(S) → Free(S ⨯ {0, 1}): e_s ↦ [e_{s,0}, e_{s,1}], where e_s is the generator corresponding to s ∈ S. Then the group produced by the above procedure is the colimit Free(S) → Free(S ⨯ {0, 1}) → Free(S ⨯ {0, 1}^2) → ..., where the maps are p's.

[rocky cloak]

I don't think it would be trivial, it just wouldn't satisfy the universal property of being free.

The value of l(x) and r(x) wouldn't be determined by x

[frail shoal]

i mean idk the name of what im calling strongly perfect

every elt. is a commutator, as opposed to a product of them

[rocky cloak]

Even if you're a commutator you may be a commutator in several different ways

[somber goblet]

oh right, so it’s the colim in AffSch but not necessarily in Sch or Top

[frail shoal]

that's true but also unavoidable because [1, g] = 1 for all g

huh it doesn't even seem to have a name in a conjecture about it

anyway im realizing i have genuinely no intuition for what commutators look like

[plain nebula]

This might be a bit out there but what do you guys (abstract algebraist?) imagine in your head when you think of rings? Like do you just see equations, do you have mental visualizations?

[quiet pelican]

Wdym by “look like”

[frail shoal]

i visualize Z on the number line

[quiet pelican]

I mostly see a lot of lattices and topological spaces
But that’s because I do geometric group theory lol

[plain nebula]

@frail shoal @quiet pelican do you care to elaborate? The lattices and topological spaces make more sense to me but what exactly does that mental picture mean for you? Like how are the ring properties expressed?
for keith, why Z on the numberline for rings compared to (insert your other mental visualization for fields)?

[frail shoal]

admittedly i visualize Q on a number line too

i think i literally just said the first ring that came to my head as it came to me

i can give a more thoughtful answer though

[plain nebula]

hahaha, fair enough. is it like an actual visual picture that you can describe? (I come from games so visualize Z vs Q vs R is a number line with different quantization where R is on fire to represent my hate love for floating point)

i realize this might sound super woo-woo but it's dawning on me how many people don't have any visual imagery and I just can't fathom that (aphantasia).

[quiet pelican]

The ring axioms fall into place because I visualise operations as either rotation + scaling or as homeomorphisms respectively

[frail shoal]

i think i dont have a uniform way of visualizing rings, but i think i have different ways of visualizing them depending on the context of them

off the top of my head:

• often i view small ones like Z adjoined some algebraic integers as a subset of the complex plane
• when i thought about Z[x] i just saw a line but this time it was the abelian group Z ⊕ xZ ⊕ x²Z ⊕ ..., each group was a point on the line
• when i thought about the ring of continuous functions from [0, 1] to R, i just saw [0, 1]

i have aphantasia so actually i am the worst person to be giving answers rn

but i have my own way of "visualizing" even if it's not visual so eh

[quiet pelican]

The lattice PoV is related to the first one here

[plain nebula]

like this? ignore the possible errors, i'm still learning abstract algebra

[frail shoal]

this looks AI 😭

[quiet pelican]

Im not spending my time troubleshooting obvious AI generated content

[frail shoal]

unless it's just some megabrain perspective i never seen

[plain nebula]

@frail shoal you are blowing my mind on the aphantasia and math. I had a head trauma a couple of years back and could not do even simple addition until my visual cognition came bacl

[frail shoal]

it's weird because i don't see what i imagine but it takes up space

but i often think with my hands a lot tbf

[cursive spindle]

I don't have a uniform answer to "how to visualize rings"

[plain nebula]

lol, i'm not asking you to trouble shoot AI content. This for me from my notes (and yes, I used ChatGPT)

[frail shoal]

just be warned AI isn't good at doing math just yet

well ok ik how ironic that sounds but. it's hard to explain

[cursive spindle]

For rings like k[x] or quotients of it as k[x]/(y^2-x^3-x-1) you can relate them to geometric objects

[plain nebula]

sigh, there's a difference between "hey AI generate notes for me" vs "hey chatgpt, draw a sphere and torus and draw these lines, etc"

it's helped retrain my visual cognition by drawing then getting a higher version but that's tangential

[frail shoal]

ok so explicitly what happened here is that the AI was told to visualize a "ring" from algebra. so it pulled up factual statements about rings, yes they have +, -, ×, closure, etc., but then, it tried to relate that to the English definition of a ring, which is actually pretty unrelated often. so it ended up sorta being slop that doesn't really have any insight, despite locally looking plausible

[plain nebula]

lol, that's exactly the opposite of what I just said I did

[frail shoal]

actually there's an ironic theorem that says there's no natural ring structures on nontrivial compact connected topological spaces, which tend to be ring-like in shape

😭

[plain nebula]

it's like a memory technique for remembering stuff. what made me wonder if other people did stuff similar as legit mathematicians vs. someone who studies it without any formal training

[frail shoal]

the word "ring" comes from the same roots as the phrase "crime ring"

so it's more like a gang of numbers ready to beat u up

if u had to visualize it as a ring

[somber goblet]

you can visualize rings as affine schemes

so long as they're commutative

[tall igloo]

how do you multiply elements of a gang

[cursive spindle]

noncommutative rings:

[tall igloo]

so a ring

[frail shoal]

i totally forgor, i agree this is a good perspective

[frail shoal]

carefully

[frail shoal]

but yeah tbh it's fine to have this visual of a ring as long as it helps u

[somber goblet]

i think it can provide insight into some specific constructions and things

[quiet pelican]

(You can’t visualise affine schemes well unless they’re varieties)

[cursive spindle]

grout <----

[frail shoal]

i wont judge, "mind palaces" are a common way to memorize stuff

[next obsidian]

Grout

[somber goblet]

yeah i was going to mention

[plain nebula]

like how do you (personally) then "think" about them?

[frail shoal]

ultimately these are abstract ideas and having some grounding, regardless of how rigorous it is, can be nice

if u wanna imagine rings as donuts homer simpson eats then that works equally well tho

[plain nebula]

ofcourse ofcourse! I'm not suggesting skipping the rigor but more probing about literal intuition

[quiet pelican]

There are explicit subsets of the reals (resp complex numbers) that you can think about
But I’d need to explain AG 101 for you to get it

[somber goblet]

i think i can visualize (some) non-variety schemes pretty well, like Spec Z

[plain nebula]

fair enough. so in your head you just recall the equations and formulas? (you don't need to tell me the details)

[somber goblet]

and i do use this POV to think about things like tensor products

[frail shoal]

well you can always choose a ring to use to probe its points

[cursive spindle]

For affine schemes you have to adapt a different POV of visualizing

[tall igloo]

for me personally i dont really conceptualize rings in terms of those equations and formulas. its just that ive seen enough examples and played with them enough times to develop a mental model for them, in the same way one does for stuff in the real world

[frail shoal]

like R is a great probe

if u want visuals

[quiet pelican]

Yeah but it’s also one that only really works if you’re using varieties

The standard probes mostly only wörk for varieties

[tidal schooner]

The Rising Sea by Vakil has a bunch of hand-drawn pictures of affine schemes, including ones that aren't varieties

[cursive spindle]

For things like number theoretic rings you often are forced to relate the visualization to function fields

[plain nebula]

fascinating. so the mental model like a "feeling" or is it something that can be put into words? This also breaks my mind when I came across that quantamagazine article about blind mathematicians

[somber goblet]

i think curves aren't too hard to visualize at least

even non-variety ones

[cursive spindle]

It's very weird the first time you see Spec Z drawn but later on you kinda just get used to it lol

[frail shoal]

for a lot of rings btw i do just visualize it by its "ring presentation", i.e. i literally just think "these are the generators and they satisfy these equations"

[somber goblet]

yea i do that a lot also

[frail shoal]

this is sorta connected to the affine scheme perspective tho admittedly

[tall igloo]

yeah i'd say its more of a feeling, but of course i can also put it into words. its an interesting thing where if i have to write down the formal definition of a mathematical object, i can sometimes forget some of the axioms but still use them in practice without any problem

[quiet pelican]

Group presentations are cute

[tall igloo]

like a while ago i had to write down the definition of a functor for some reason or another, and forgot to write down that it sends identities to identities. but i have no problem using that fact in practice, its just that i dont have all the axioms memorized in a way i can easily access in my brain but can still use these things

[cursive spindle]

I would compute pi_1 of R^3\K all day cuz of this

[quiet pelican]

99% of a group theorist’s job exists because group presentations are the most horrible bastards to get your hands on

There is even a theorem which says this

It’s called “none of the following are computable: the word problem, the conjugacy problem, the isomorphism problem”

[somber goblet]

sometimes it's just easier to recall definitions and things by thinking about the elements of the ring as functions

but that might be just me

[plain nebula]

heh, that's the only way I think of them but I think that's only true for Lie groups?

[cursive spindle]

If I forget some axiom of an algebraic structure I try to think about examples I know and from there I'd remember

[somber goblet]

yeah same

[plain nebula]

this has been eye opening. thx everyone. I think i need to re-valuate my whole worldview

[somber goblet]

i think it's very cool that the word problem is not computable

[quiet pelican]

It is evil

[frail shoal]

the brainrot answer is that identity is 0-ary composition

[tall igloo]

TRUE

[quiet pelican]

Like, it’s obvious that you should be able to encode a general turing machine in a presentation
I think doing so is messier

[tall igloo]

wdym brainrot that is just correct

[frail shoal]

it's one of those nlab-isms that help guide me but feel pretentious and unhelpful when i say them out loud

[tall igloo]

thats just a normal-ism to me (just a glimpse into how nlab my brain is)

[cursive spindle]

Category theory is more unhelpful than helpful for me

[quiet pelican]

I think it’s less an nlabism and more a UAism

And you do UA so

[somber goblet]

category theory helps me remember things

[tall igloo]

oh also another remark, ive been led to believe (at least some) other people also just develop mental models that are divorced from the formalized axioms, because ive had professors in class go like - and in case you forgot, because i did, X object has Y property in the definition

[azure cairn]

me with vector spaces and groups

actually i remembe rgroup axioms

i could not recite the axioms of a vector space unless i recalled group axioms

[somber goblet]

like it's way easier to remember the definition of a presheaf as a contravariant functor

[tall igloo]

i definitely forget which group axioms you need and what comes for free once you have the rest

[plain nebula]

oh my god my vector analysis professor would always say "What news does this function have to report?"; it just clicked that he was being (imaginatively) literal

[tall igloo]

unless you are hartshorne and declare a presheaf (say of abelian groups) must be valued as 0 at the empty set for some reason

[quiet pelican]

I don’t think any of them are free unless you count left and right sided inverses/identites as 4 axioms

[cursive spindle]

I have to say though diagrams definitely help me remember things

[azure cairn]

oh shit the channel name changed

grout

[cursive spindle]

But throwing out words in the wild is not helpful

[tall igloo]

hang on let me look at the group axioms and remember what i was thinking of

okay i forgot what i was talking about

[somber goblet]

i just remember the magma -> semigroup -> monoid -> group progression

[tall igloo]

but i can give another example - for rings you don't need to require that the addition group is abelian, that comes for free

[quiet pelican]

A monad is a monoid…

[cursive spindle]

I have one that's worse

[azure cairn]

maybe that the identity axiom is redundant with the associativity + inverse axiom?

[quiet pelican]

A group is a one object groupoid?

[cursive spindle]

Nope

[tall igloo]

or another example - you don't need to require holomorphic functions be C^1, that comes for free once they are differentiable

[cursive spindle]

Something about gcd

[quiet pelican]

It’s not (as long as you modify the inverse axiom to account for having no identity axiom)

[tall igloo]

theyre calling it the best definition in all of mathematics

[quiet pelican]

I murder people who think that’s a good definition unironically :3

[azure cairn]

LMAO

my textbook author, aluffi,

one of the best algebraists actually

[tall igloo]

lowkey it is a useful definition / identification

[azure cairn]

he's so good at algebra

simply one of the best algebraists, maybe ever

some people are saying it, very smart people, top of the top

[tall igloo]

OH i didnt know what you were doing but i get it now

quite frankly

[azure cairn]

he writes a textbook, and let me tell you, it's not like these other books. total disasters, confusing, long, NOBODY understands them.

that's what we're dealing with here. winning in algebra. tremendous algebra. huge.

[tall igloo]

we're going to build a Wall finiteness obstruction group

[frail shoal]

aluffi confidently writes that 0 is prime

he is correct

[tall igloo]

based

[cursive spindle]

The terminal cone over the diagram {a,b} in the category of Z under divisibility

[frail shoal]

the best part was showing it to my number theory prof and watching his face contort

aluffi is a master ragebaiter

[azure cairn]

in notes from the underground or ch0

[frail shoal]

notes from the underground at least

i forgor for ch0

[tall igloo]

if not prime why generate prime ideal

[cursive spindle]

Late stage category theory is developing the schizophrenic mindset of viewing anything as yoneda

[tall igloo]

that sounds like a description of late stage category theory in terms of what it does to others...

[tall igloo]

(of course it is obvious that all rings are integral domains (and Noetherian, and commutative, and-))

[flat anvil]

grout-rings-fields lmfao

[somber goblet]

whats an example of a non-noetherian ring

is k[[x]] non-noetherian

i feel like k[[x]] is a weird ring

[cursive spindle]

Almost all examples of rings you come across are Noetherian

[cursive spindle]

Love how this one is always the example for a lot of things

[cursive spindle]

Why do you think so?

[somber goblet]

oh nice

[somber goblet]

well, it's not finitely generated, is it?

as a k-algebra

[frail shoal]

nah all its ideals are of the form (x^n)

[cursive spindle]

k[[x]] is noetherian

[somber goblet]

it's a PID?

[frail shoal]

this is true but unrelated

[frail shoal]

yea

[somber goblet]

this feels cursed

[cursive spindle]

k[[x]] is nicer than k[x] in some sense

[somber goblet]

damn

[frail shoal]

it's like how Z_p is nicer than Z in some sense (wrt that prev convo where you said smn to this effect)

[somber goblet]

wait

[frail shoal]

its unit group is way bigger

[somber goblet]

a finitely-generated algebra over a noetherian ring is noetherian, right?

[frail shoal]

that sounds right but the converse isn't true i believe

dont quote me on whether that's right tho cuz im not super familiar with noetherian stuff

for a familiar example, Q is not finitely generated as a Z-algebra

[somber goblet]

yeah, makes sense

[frail shoal]

but Q is a field, which is the nicest possible ring

[somber goblet]

yeah

lol is there anything with more algebraic structure than an algebraically closed field

[cursive spindle]

F_1

[frail shoal]

i think a famous fact of model theory is that ACF_0 is a complete theory

in other words, every algebraically closed field of char 0 agrees on every sentence of first order logic

[somber goblet]

yeah

i guess that sort of makes sense; the only FOL sentences you can form in a field are about polynomials with integer coefficients

[quiet pelican]

Grp is a complete theory :3

[frail shoal]

yea that sounds roughly correct

this isn't a direct answer to ur question tbf

cuz like you can always add more structure

but it's the nicest possible ring in that sense

[cursive spindle]

There's something in cat theory which says fields are not nice

[somber goblet]

yeah, it's not really a question with an objective answer

well yeah fields don't have products

field is a bad category

[tribal moss]

For maximal niceness, assume you're working in an algebraically closed ordered field. You can do all sorts of things then.

[cursive spindle]

Yeah

idk that sounds like a feature to me

[frail shoal]

i think i once listened to a talk about "hardy fields" and they're kinda fascinating

[somber goblet]

i guess

it still amazes me the things you can do with fields

[cursive spindle]

Another man's trash is other man's beauty type shit

[tall igloo]

messed up

[cursive spindle]

is it not

[tall igloo]

no no you are right

i am just saying that it is a messed up fact

[cursive spindle]

why so

[somber goblet]

it's a PID which is awful to me

it shouldn't be that nice

[cursive spindle]

[tall igloo]

i think just like, k[x] feels like a more finite object to me than k[[x]]

[somber goblet]

it doesnt deserve to be a PID

[frail shoal]

they're fields whose elements are C^1 functions on R, but we only care about "eventual" behavior, which allows us to have that sweet, sweet, division structure. but also we can differentiate the elements too, which is kinda wild

ok im going on a side tangent

idk much about them but they're kinda a fascinating idea in the vein of "what if we had MORE structure"

[somber goblet]

like

is (x + 1) = k[[x]]?

[frail shoal]

yep

you can do "infinite polynomial division"

[somber goblet]

what's its inverse?

wait

[frail shoal]

try to construct its inverse in k[x]/(x^n) for each n

[somber goblet]

i remember calc ii

[frail shoal]

yay

[cursive spindle]

How about Z_p?

[frail shoal]

i think there's something called "hensel's lemma" which gives general answers to these kind of questions

idk it that well tho

[somber goblet]

$\frac{1}{x+1} = 1 - x + x^2 - x^3 + \cdots$ right 🔥

[cloud walrus]

lexi

[frail shoal]

but tl; dr usually we can piece together an inverse by approximating it step by step

(same for in padic land)

[frail shoal]

hensel's lemma generalizes this to any polynomial root, i.e. we construct approximations to stuff like √2

iirc the main obstruction usually is that it has to exist in the residue field

[cursive spindle]

C[[z-x]], F_p[[z-x]] and Z_p are all on a similar vein

[somber goblet]

yeah ok

i believe that

is this true just when k is alg closed or for arbitrary fields k

k[[x]] being pid

[frail shoal]

any field k

[cursive spindle]

Part of why you should think they are nicer is because in these scenarios you're describing local behavior

Local behavior is nicer than global one

[somber goblet]

and geometrically speaking

Spec k[[x]] is a thick point

[tall igloo]

wdym by thick?

[frail shoal]

chunky

[somber goblet]

or wait it has two points

(x) and (0)

[frail shoal]

nah just one i think

(0) is generic

[cursive spindle]

Petition to call this channel algebraic-geometry-2

[frail shoal]

wait

[frail shoal]

nvm

[swift root]

it has one point that can geometrically be "seen"

also know as a geometric point

[frail shoal]

is it weird if i see the non-maximal primes as like encoding incidence structure

[somber goblet]

one closed point ?

[frail shoal]

that makes sense

yes

[swift root]

yes

[somber goblet]

incidence structures are cool

[frail shoal]

hell yea

sadly i dont have anything rigorous to back me up

i dont do ag

[swift root]

is it a coordinizable lattice?

lol

[frail shoal]

a what

[cursive spindle]

now i won't understand this discussion

[swift root]

lattice isomorphic to the lattice of subspaces of some projective space

admittedly dont know much about those

[swift root]

its a joking question anyways because prime ideals dont form a lattice

[somber goblet]

generic points are awesome

[frail shoal]

oh nice

[swift root]

theyre like necessary additional information the spectrum carried, since you enlarge the context to all rings

thats how i see it at least

the maximal spectrum only works for fg k-algebras over an alg closed field

[somber goblet]

i feel like generic points tell you about birational properties

[swift root]

prime spectrum is what you get when you force every nontrivial ring to have at least one "point"

[swift root]

dont know much about those

[somber goblet]

or at least the data at the generic point does

[somber goblet]

properties determined up to isomorphism of open subsets

[frail shoal]

oh yea

i forgor to say

exercise: explicitly compute 1/3 in the 7-adics

[somber goblet]

fah

i dont want to do ts but i will

[frail shoal]

the only way to intuition is pain

-gandhi

[somber goblet]

wait is this even in the 7-adic integers

[frail shoal]

ok i should give u some hints

[somber goblet]

oh it is isn't it

[frail shoal]

nvm u might got it

[somber goblet]

since 3 mod 7 is 3

[frail shoal]

yes

[somber goblet]

it has to have an inverse

[frail shoal]

hint: ||compute the inverse mod 7, then mod 49, etc., then try to find the pattern||

[swift root]

aint no way

[somber goblet]

shocking ik

[frail shoal]

confession: i dont know it off the top of my head. ill compute it along side u rn

but in secret so u gotta learn still

[somber goblet]

im trying to think about this in terms of power series

[cursive spindle]

Idk 1/3 in 7-adics on top of my head

[frail shoal]

tbh im scared of anyone who does

[cursive spindle]

but deltoid ur a number theorist you should know

[frail shoal]

this is like that stereotype of mathematicians being able to multiply large numbers in their head but both more accurate but also inaccurate still

[somber goblet]

ok i found some inverses

using fermat's little theorem

wait

whoops i need to use euler's thm

silly

[frail shoal]

second hint: it should be like long division, but different

[somber goblet]

oh im just using some code to calculate the residues here lol

[frail shoal]

yknow what valid

work smarter not harder

anyway the answer should be, in power series form: ||5 + 4 * 7 + 4 * 7^2 + 4 * 7^3 + ...||

actually funny enough i think this is how sagemath displays p-adics

[somber goblet]

i get 5, 33, 229, 1601, 11205, ... for my inverses...

[frail shoal]

that looks correct

[tribal moss]

Then write those in base 7.

[somber goblet]

ohhh

[frail shoal]

ok yea i just checked our answers against sagemath and yea

thankfully this one isnt that bad

it's kinda funny how short the code is cuz sagemath already has it all set up with its library magic

on the other hand, i think writing the code on your own without using a CAS prolly works just as well as working it out by hand for getting a sense of how padics work

anyway tbh i intended for it to have a period of more than 1, i sorta made an oopsie, i just picked some random small primes

sagemath tells me 1/5 is kinda gnarly

but it has a period of 4 🙁 come on why cant i have a cute period length

[cursive spindle]

Blame the number theorists

[frail shoal]

true

ok now compute -1/48 in the 7-adics totally didnt engineer this number to work out nicely

[somber goblet]

...444445?

yay

idk how to formally connect this to the power series view

idk if there's a way to connect this to differentials and stuff

[frail shoal]

try multiplying it by 3

by hand using the school agorithm

i.e. 5 times 3 is 15 = 2 * 7 + 1, carry the 2

[somber goblet]

yeah ik, i was just thinking about stuff like zariski cotangent space and wondering if there's a way to view p-adics as power series in that sense

[frail shoal]

oh yea fair i wish i had a perspective like that too

wide point 😋

the issue is i once asked my friends about this and got complicated answers

in terms of the tangent space perspective i only really understand k[x]/x^2

i gtg (eepy)

[somber goblet]

gnn

i think this gets into formal geometry (scary)

i might ask along these lines in #algebraic-geometry

[vestal trench]

grout

[azure cairn]

whats wrong with it btw

i think its sensible

[quiet pelican]

It’s a completely useless definition if you actually care about studying groups

[azure cairn]

thats fair

its a good viewpoint though

like defining a group by its axioms and then showing it is a one object groupoid

or vice versa

[quiet pelican]

What is it actually useful for though?
Except giving an example of a groupoid

[frail shoal]

nah im just a napper

[azure cairn]

whyd they fix the channel name

[frail shoal]

well, i disagree, although only mildly:

• we can motivate G-actions and G-equivariant maps by looking at functors from BG, and group actions are fundamental to studying groups
• group cohomology can be motivated by studying the simplicial complex corresponding to BG

on the other hand, you study groups for a living, and i dont, so i can't really argue too much given you're ultimately the more experienced one

[azure cairn]

group actions are functors then?

[frail shoal]

damn i scooped u

😭

[azure cairn]

awesome thanks for stealing the only point i knew of </3

[frail shoal]

at least it shows it's a natural point to bring up

[azure cairn]

but yeah you get to view groups and group actions actively (by how they act with other structures) rather than passively (their structure as a specialized set)

rip pseudo

[frail shoal]

yea i get why pseudo was banned but i miss her too

thankfully she's active in the affinoid union now

[azure cairn]

what is the affinoid union

[frail shoal]

math server im from

[azure cairn]

is it public ?

[frail shoal]

yes

[azure cairn]

send invite? :3

[frail shoal]

idk if im allowed but it's googleable

[azure cairn]

okay

[frail shoal]

:3c

[quiet pelican]

For 2, we don’t really think of group homology that way
We just think of it as studying the homology of the unique CW complex with fundamental group G and trivial higher homotopy

[frail shoal]

oh yea i just remembered it's called the "nerve" of a category

[azure cairn]

mutual servers increased from 1 to 2

[frail shoal]

yea it's neat that group cohomology is the cohomology of the nerve of the group

if im not wrong

[frail shoal]

ok that's fair but the nerve of BG is a canonical way to construct it in a sense

admittedly. i dont do much algetop so i could be speaking out of my ass

but it's the perspective that made group homology most motivated from my POV

[tall igloo]

groupoid*

[azure cairn]

um

i forgot the difference :3

<---- fake category theorist/algebraist

[tall igloo]

a monoid is a one object category, a groupoid is a category where all morphisms are invertible

[azure cairn]

ah i see

[tall igloo]

so a one object groupoid is a specific type of monoid where all the morphisms are invertible

[azure cairn]

!

[azure cairn]

-# note how she didn't reply to our shared idea of #1 :3

[tall igloo]

#1 is actually super goated lol

[azure cairn]

speaking in these channels give me fomo cuz im not actively learning algebra

cat theory as well ig

[tall igloo]

anyone can speak anywhere its fine lol

we're all just trying to learn some math

[frail shoal]

ok i do wanna clarify my tone that im just saying this stuff in a friendly way, i dont wanna sound like im trying to start a fight

with that being said

[azure cairn]

ye but it makes me want to continue reading aluffi even though i cant

[frail shoal]

fr

i love group actions !

😳

[tall igloo]

lol how is this starting a fight

[frail shoal]

i think im always overly self conscious of it

idk i dont wanna gang up on mico and say her perspective is bad

[tall igloo]

i guess the way i think about this is like

its a bad idea if your goal is to study groups as has been said, and by extension its a bad definition if you want to study groups via group actions

but

its excellent if you want to study other stuff that isnt a group via group actions

[azure cairn]

everyone thinks about things differently, its very possible a definition/description of a structure may not resonate with one person but resonate with another

[tall igloo]

i guess at least like, non-discrete stuff

[azure cairn]

this is true as well, i don't think it's a very relevant definition in an intro algebra class

[tall igloo]

like idk group actions are useful in homotopy theory and then you have to scratch your head and wonder what a group action on a spectrum is because spectra dont have elements that can be permuted

so its just a functor from the category BG into Sp and thats huge

and then viewing it as a category is nice because then you can take (homotopy) orbits and fixed points as (homotopy) colimits/limits of the diagram indexed by the (oo-category given by the nerve of the) category BG

and there's really cool stuff one can say about when these lims/colims agree that has to do with the fact this simplicial set (nerve of BG) is even better, it is a Kan complex

anyways this whole story is invisible if the definition of a group action is permuting elements in a thing, so viewing group actions as functors can be very fruitful

[frail shoal]

oh yea we also get adjoints to these by abstract nonsense

[tall igloo]

yup yup

yeah like orbits -| constant diagram -| fixed points

[frail shoal]

one of my fav adjunctions

[tall igloo]

(certainly when there are elements one can permute, it is obvious that life is much nicer when one is able to study the group action on the elements. my point isnt that we Must view group actions as functors but just that it can buy you a lot to do so in less discrete cases)

[dense mica]

i'm gay

[tall igloo]

love me some colim -| constant diagram -| lim

or even better, LKE -| composition -| RKE

also i just realized the channel description (not name) still says grout 💀

[frail shoal]

whao, i usually seen it written as Lan and Ran

[azure cairn]

im currently banned from meta discussion but im SURE there was a massive argument there if the channel names were changed back so quickly

[frail shoal]

but ultimately it doesn't matter im just speaking out loud

[tall igloo]

i would write Lan_F G in math text lol, just saying LKE and RKE as informal abbreviations

although if i were based i would write LKan and RKan

[frail shoal]

that makes sense

[frail shoal]

tl; dr it ended up being an innocent mistake I'd say

joke that didn't land

😔

[azure cairn]

was I right?

[frail shoal]

yea essentially

[azure cairn]

ok well idk people already got upset that they complained about the foundations channel name and mods did nothing

and then 1.5 weeks later they do shit with the names of active channels

[tall igloo]

aw the foundations name was cute

[quiet pelican]

I feel like this gives mods a level of benefit of the doubt that they’ve repeatedly shown themselves undeserving of over the past couple months

[azure cairn]

someone complained it was unindexable

[tall igloo]

ah fair

[quiet pelican]

And also a nightmare if you wanted to refer to the channel, or use a screen reader

[tall igloo]

yeah those are very good reasons to change it back i agree

and more important than being cute lol

lol ngl i didnt know there was a meta discussion channel... i only check things in the advanced math channels 💀

[azure cairn]

it's my personal shitposting channel

[tall igloo]

i thought that was any channel when a universal property gets mentioned

[azure cairn]

this server is my personal shitposting server

[frail shoal]

anyway i vaguely recall there are groupoid limits that are computed differently than group limits and this results in interesting stuff but i kinda forgor

[azure cairn]

unfortunately this means i will be banned in 3 months give or take

[frail shoal]

the good news is i found the original message that this is vagueposting about

[tall igloo]

oh?

[frail shoal]

someone once sent this to me

and i think i got humbled

i never did think about it later....

[tall igloo]

ah and here you are trying to compute the pullback in groupoids i assume?

[frail shoal]

yea

[tall igloo]

is this the same as like. lax pullback in Cat or something

idk how 2-things work

[azure cairn]

that +1 mutual server event is the reason why i can read the context for these messages

[frail shoal]

anyway atlantis is in tau now so they can see exactly how i floundered 🥀

damn u scooped me

[azure cairn]

LOL

1-1

[frail shoal]

😳

[tall igloo]

i only have some 1-knowledge and \infty-ideas but nothing in between

[frail shoal]

lemme pull up the explicit description

maybe this time ill understand it

[tall igloo]

ah okay i dont think this is lax pullback

but the 2-pullback it seems nlab calls this

[frail shoal]

makes sense

instead of equality we're now demanding isomorphism

[tall igloo]

yup exactly

i should get my hands dirty working with groupoids some day. ive been told van kampen follows (messily) from a simpler version about fundamental groupoids

[frail shoal]

i think i been shown a demo of this with covering a circle by two lines that overlap at two points

idk the deets

i just remember them being like "yo you can do this"

[azure cairn]

i think the funny meme in pdf format i sent you earlier actually covers this

[somber goblet]

oh how nice R/J inherits an I-adic topology from R

[tall igloo]

yeah ive been told that + covering space theory are the two best known ways to compute pi_1 (S^1) but its like

[somber goblet]

i wonder if this is a topological quotient

[azure cairn]

analysis v algebra

[tall igloo]

ah nice

[somber goblet]

this reminds me of sheaves

[tall igloo]

of course this is trivial these days as an average toddler one can find on the street will tell you that S^1 is a compact object in the \infty-category of anima and so mapping out of it preserves filtered colimits

[somber goblet]

sheaf of fundamental groupoids

[frail shoal]

anyway imma actually like

think about what a morphism should be in a 2-pullback

[tall igloo]

have fun!

i will have much more fun not doing that

[somber goblet]

$\bigcap_{n = 0}^\infty I^n = {0}$ for proper ideals $I$ right

[cloud walrus]

lexi

[somber goblet]

wait do you actually need notherian for ts 💀

augh

[tall igloo]

wait if its noetherian is that even true? like the I^n's will stabilize and the intersection will be some I^n

wait no thats the wrong direction

[somber goblet]

augh

no one likes artinian rings

[tall igloo]

yeah so what i said works for artinian rings i think then

[somber goblet]

well wait

this works in Z, right

because I = (a) and eventually a^n > m for any m

[tall igloo]

yes

yeah so i guess this works for any PID

wait

ugh idk

[somber goblet]

this is some commie alg bs

you just need the right combo of 'sane ring' conditions

[tall igloo]

let R be a ring for which this is true

[somber goblet]

oh ts is krull's intersection theorem

🔥

[tall igloo]

everything is nakayama lemma

[somber goblet]

commalg = i blackbox

[tall igloo]

i react true to cope with commalg final coming up

[somber goblet]

glup

[azure cairn]

good luck !

[tall igloo]

garcias

[azure cairn]

garcias

[tall igloo]

https://cdn.discordapp.com/attachments/1484980074146037992/1484987288026091702/garcias.webp?ex=6a15eb2f&is=6a1499af&hm=a4db5d089e05ab238db4e09c4bd7673e4c31340f2d5afd597de6c60cc2352bb9&

[azure cairn]

LMFAO?

[azure cairn]

https://tenor.com/view/another-easy-steal-steal-gif-gif-steal-gif-gets-stolem-gif-7935542554022949277

[tall igloo]

how do you think i got it

[somber goblet]

most on-topic #groups-rings-fields discussion

[tall igloo]

hey we were just takling about rings lmao

ignore the earlier oo-categorical digressions

[frail shoal]

update

i think the concept apparently is equivalent to this:

given a group G and subgroups H ⊆ G, K ⊆ G, then we have an action of H × K on G by sending g to hgk^-1

[somber goblet]

that makes sense

[frail shoal]

it's kinda wild to me how this generalizes several different common concepts at once

😭

[kind temple]

do you mean H x K^op?

[frail shoal]

prolly

[kind temple]

h’(hgk^{-1})k’^{-1} is not the same as (h’h)g(kk’)^{-1}

so you have H as a left action and K as a right action

or a K^op left action

and you should be able to generalize further to having homomorphisms H —> G and K—> G

[potent lynx]

what is a set

[spark plank]

isn't that more for #proofs-and-logic

[slow egret]

A collection of elements

[potent lynx]

what is a grou*

group*

[slow egret]

A set with axioms

[spark plank]

group axioms

[slow egret]

ya

[potent lynx]

what is a ring?

**

[slow egret]

A group with two operations defined

[potent lynx]

what is a mesure?

measure*

[earnest valley]

Wrong channel ?

[covert vector]

I guess we can add a notice to the top of the higher math category, where we expect you to be more serious here

[slow egret]

Ive never heard of a measure @potent lynx

[covert vector]

then after that we can just kick out SA indefinitely

🤔

[stone fulcrum]

Wtf there's a higher math section

[covert vector]

@stone fulcrum

:P

[potent lynx]

@slow egret https://en.wikipedia.org/wiki/Measure_(mathematics)

[slow egret]

Oh nice

[chilly ocean]

Oh measurr

Wait. Why is real numbers even special

[valid elbow]

because there's many other things that aren't real, say complex numbers (a+bi), quaternions (a+bi+cj+dk), octernions (8 different special directions, it will make more sense if you have learned about vectors), and so on...

without complex numbers, there would be no solution to xx = -1,

I haven't run into quaternions yet, probably due to not working witrh geometry/topology much (if at all).

[chilly ocean]

I mean, why do we give a special role to the real numbers?

[valid elbow]

because it's the basic, standard type, where nothing else would exist without it i guess

[spark plank]

tbh should've extended everything via ordinals instead

who needs real numbres when u have infinities

[chilly ocean]

lol

Ah, basic, standard type

[covert vector]

the field of real numbers R is the only (Archimedean) complete ordered field up to isomorphism

@chilly ocean

[chilly ocean]

👍

Woah

[covert vector]

depending on your definition of complete, you can omit the "Archimedean" part

[chilly ocean]

Great answer you got there

[covert vector]

and say it's the only complete ordered field

^^

[hot folio]

question: is a*b = |a - b| an operation on on N

solution : This is an operation, because it uniquely maps every ordered pair (a,b) ∈ N×N to N.

i don't understand what this means

it uniquely maps every ordered pair (a,b) ∈ N×N to N.

[woeful sand]

integration is considered a linear transformation ?