advanced-algebra

[waxen fractal]

the converse is not true, unless we require some strong commutativity requirements, but, that is often where interesting stuff happens anyway. because a lot of people dont even study non-commutative rings

[astral ginkgo]

Yea I'm aware, I do noncommutative algebra lol

Well less so these days but still

[waxen fractal]

damn, i fear your power

that shit scares me

[astral ginkgo]

It's fun, so many cool phenomena that just can't arise when you require commutativity

[waxen fractal]

yeah, that is true

[astral ginkgo]

Anyways, I think I more or less understand your idea now, so thanks for elaborating.

Intuitively it feels to me this kind of stuff would be only be non-pathological in cases where there's some notion of completeness

But maybe that's actually just the boring class of examples

[waxen fractal]

that's good to know. i am not always certain of my ability to explain

[waxen fractal]

yea, i feel like an ultrametric structure gives rise to similar examples. actually, i haven't thought about it enough to say rigorously. maybe you're right, and it has more to do with ideal completions

[waxen fractal]

i have no idea, I've only explored the surface i feel like

[limber wharf]

Has anyone here gone though Alugffi's algebra? Just wanted to know your thoughts on it

[hushed bone]

Nah but I went through a good bit of Aluffi Chapter 0

It was good, I like it

Ppl say the exercises aren’t hard enough, that may be true idk, but I like it a lot

[limber wharf]

Lol, silly typo on my side...

I was gonna go though some parts of the homological algebra

Thanks for letting me know

[hushed bone]

That’s what I did

I really like it for homological algebra

I learned the beginnings of algebra from Aluffi

And then did ring & Galois theory from D&F,

Then read the homological algebra chapter, and most of the last one

My memory says the last chapter begins to talk about the derived category which is really deep and pretty essential in a lot of fields

And I found Aluffi really illuminating for how to think about homologicsl algebra in a more modern way

[limber wharf]

Oh cool, sounds exactly what I was looking for...

I also did the structures from D&F

[lethal mango]

Anyone know Finacial Math

[near lantern]

Wrong channel. IDK a right one, but I would try #advanced-probability.

This is a fact-checking question.

Let A be a fd algebra over a field k. Is it true that {a ∈ A : tr(multiplication by a) = 0} contains no non-zero left ideal of A iff A is semisimple with the dimension of each of its simple factors having dimension not divisible by the characteristic (and in particular, centre separable over k)?

(If A is semisimple, left and right multiplication have the sam trace. For either choice, the above holds for left ideals iff it holds for right ideals. So it doesn't matter which you pick.)

[broken turtle]

radu

[sand pelican]

I concur

[lone jacinth]

For example A = F4, k = F2 has dimension divisible by the characteristic, but doesn't contain such an ideal

If A isn't semisimple you can certainly find such ideals.
And for A semisimple it's enough to consider Mn(D). If n is a multiple of the characteristic you get such an ideal. If n is not, then it reduces to whether D has such an ideal, which just means every element of D having trace 0.

You need D to have dimension multiple of the characteristic, but it's not sufficient.

[near lantern]

OK I think it should be "each factor has separable centre and thi dimension over the centre is not divisible by the characteristic".

My idea was that we can base change to the algebraic closure where we have M_n(k) and it becomes an iff, but I now see this doesn't work as stated for non-trivial centre.

[lone jacinth]

Hmm, so okay if the center isn't seperable everything should have trace 0.

But if the center is seperable, and the dimension over the center is a multiple of the characteristic, can you then be sure everything has trace 0?

[lone jacinth]

Hmm, actually I think it should be okay.

For D > K > k you should have
tr_k(D) equal tr_k(tr_K(D))

[soft parcel]

Does anyone know whether there are any shortcuts in calculating multiplicities with Kostant's formula? For example whether maybe vanishing of p for simple reflections implies vanishing for all reflections or something?

[waxen fractal]

I've gone through both yeah

it's good, would recommend 👍

[hushed bone]

What’s the undergrad one like?

[waxen fractal]

it goes elementary number theory -> rings -> groups -> frields

this is the first algebra book i read

so there was a good chunk of time i knew what a ring was but not a group

in retrospect, i question this design decision

on the other hand, rings are, subjectively, more fun, and i suppose aluffi agrees too

the elementary number theory can be seen as ring theory over Z, as preparation for the general thing, which is quite a clever framing device

[waxen fractal]

oh yea, my goal ended up being to do all the exercises in the book (at least in my head), and that facilitated that goal

i tried doing this with another textbook but it didn't work cuz the exercises were too hard

[hushed bone]

Yeah like in the sections I did I did most of the exercises

It was doable

And I feel like I learned how to work and think with say Tor and Ext well from that

[hushed bone]

There’s a camp of people who say this makes more pedagogical sense because people are more familiar with ring-like things

But I am partial to groups first because well, I learned it that way, and it makes sense to start from less structure to more in my head

[waxen fractal]

the issue i have is that rings, in retrospect, are best viewed as being built from abelian groups

so yeah, essentially what you said

[waxen fractal]

people really really like the parts on homalg. i actually sorta bounced off it iirc, that's mainly just cuz i dont really enjoy homalg that much

i did it anyway and i can recognize it's a well-done chapter but

idk, i think i have to be one of the only people who prefers the rest of aluffi to the last section

[vague zodiac]

Hello! I'm trying to understand what my grandma's master's thesis was about. The title is "On the existence of algebraically closed extensions in a family of Ω-algebras". It was written in 1968.

Please do note that I have a background in engineering physics, so I never saw abstract algebra. If it's just way too advanced for me to even understand what it talks about, just say so. I'm just curious about it, because my grandma was a high school teacher, and so she forgot all about higher maths.

[waxen fractal]

do you have a link / doi for convenience?

[vague zodiac]

It has not been scanned, and it's written in French

[waxen fractal]

ahhh alr

[vague zodiac]

I'm not expecting to understand the details, if I can see the general landscape in which it is, it would already be great

[waxen fractal]

im doing some research, i think I've found what an Omega algebra is, it seems to be a dated term for what we'd now call an algebraic structure

a family is just another name for a set or collection, just having several of them

algebraically closed extensions usually refer to things called "fields"

if i had to explain it simply, consider the system of real numbers. that forms a field because it has a sense of +, -, ×, ÷

(and it obeys a lot of nice axioms)

[vague zodiac]

I know what a field is, from baby Rudin

[waxen fractal]

hell yeah

an algebraic closure of a field closes it under all polynomial equations

for example, we might want a solution to x^2 + 1 = 0 in our field

[vague zodiac]

so it means that all polynomial equations written in it have their solutions in the field ? So, the complex field would qualify, but not the real field ?

[waxen fractal]

R is not algebraically closed, but the "extension" of R to C closes it, by the fundamental theorem of algebra

[waxen fractal]

yeah exactly

ok asterisk

i should say nonconstant polynomials

1 = 0 🤓

[drowsy niche]

does it have a different meaning in UA?

I assume that would be what's relevant here

[waxen fractal]

it seems that their grandma was looking at a generalization to UA yeah

[drowsy niche]

since it seems like an Omega-algebra is just an algebra with a signature Omega

[waxen fractal]

which is interesting cuz UA is an old field that most people don't study anymore, but i do

[waxen fractal]

yeah my googling confirms this too

the "signature" of an algebraic structure is the set of operations you define on it

fields have 0, 1, +, ×, -

not division cuz that's not a complete operation, it needs to be defined everywhere

[vague zodiac]

so she wanted to generalise the notion of algebraically closed to those algebraic structures you talked about ?

[waxen fractal]

yeah seems to be the case

that, or she built off an existing generalization

i know in modern day that model theorists look at existentially closed structures

[vague zodiac]

That's possible, the print document is only 50 pages long

[waxen fractal]

yeah I'd not know

but yeah, the existence of an algebraically closed extension is essentially, asking if it even exists

[vague zodiac]

yeah, so if I understand, she was trying to see if that generalization was even possible in the first place

[waxen fractal]

you could imagine a world where there didn't even exist an algebraically closed extension of R, or something. thankfully we live in that world, but she must have been studying the situation for other types of structures

[waxen fractal]

yeah prolly that too

[vague zodiac]

I see, thanks a lot for your help!

[hallow bone]

oh this is interesting

I've been thinking about this too

[waxen fractal]

i knew you'd think that

i was thinking of pinging you

[hallow bone]

the interesting part about this is that you'll have algebraically closed with respect to a variety

because, how do you decide which equations p = q you want to have a solution?

[waxen fractal]

i guess that's how you exclude 0 = 1 on principle

[hallow bone]

wdym?

"on principle"

you define an algebra A to be algebraically closed with respect to a variety V, if for any unary polynomials <p, q> such that they do not generate the top congruence in A[X]_V, there is a solution in A

A[X]_V here is the polynomial algebra with respect to V

[waxen fractal]

as opposed to ad-hoc

[waxen fractal]

hell yea

[narrow kraken]

can we say under which conditions on R (except for R being hereditary) is the category Proj-R of projective R-modules semisimple?

[drowsy niche]

wdym by semisimple here?

using the def on nlab (every object is a semisimple object, meaning for any object P, every ses 0 -> P' -> P -> P'' -> 0 of objects in the category splits) the category is clearly always semisimple

so I don't really get the question

[waxen fractal]

maybe they mean that they must be a direct sum of simples

this feels like it could be stronger, although maybe it's equivalent, I've not thought about it much

but im thinking about, hypothetically, what if a nontrivial projective module always split into the sum of two other projectives... and i guess, that's equivalent to splitting into the sum of any nontrivial modules, cuz that guarantees projectivity of the factors

can that ever happen?

[waxen fractal]

also i forgot that, maybe a projective module might be directly irreducible but also not simple? i also forget if that can happen

[narrow kraken]

hm yea this is what I meant, but for R-modules M I believe the two are equivalent, i.e, M is semisimple (can be written as a direct sum of simple modules) <--> any SES 0 -> M' -> M -> M'' splits.

ah but simple objects in Proj-R are not necessarily simple projective R-modules

[waxen fractal]

yeah i suppose, that corresponds to this possibility

[narrow kraken]

sorry what does directly irreducible mean?

[waxen fractal]

ah, i am thinking about this, and i see why that's the case. because, if M' -> M is injective, then, because that map must break down into a map to each simple factor, each of those maps must be 0 or surjective, so you always end up with just a subset of the factors

oh wait that's just one direction, er, i should think about the converse

maybe it's some recursive argument, not sure how to properly carry out the induction

[waxen fractal]

shorthand for "does not decompose into a nontrivial direct product"

i got tired of saying it so i switched to that buzzword

[wise sedge]

Isn’t that just indecomposable?

[waxen fractal]

probably

yes

new word learned

[wise sedge]

It’s crown graphs all over again

[waxen fractal]

directly irreducible is the more general term so i forgor the specific one

[soft parcel]

Anyone know if the direction <= is true?

Lambda^+ being the set of dominant integral linear functions

V(lambda) the irreducible verma module

[wary elbow]

Yeah this is true. If lambda is dominant and a sum of roots then in particular it's a positive linear combination of roots. The weights of V(lambda) will be all weights of the form lambda - positive sum of roots. In particular, 0 is of that form.

[soft parcel]

I don't see why dominant + sum of roots => positive sum of roots.

[wary elbow]

this is a general fact but I don't recall the proof off the top of my head

[soft parcel]

Oh, could one just evaluate at the fundamental weight corresponding to the alpha?

[wary elbow]

I think you need to evaluate at the fundamental coweight, but yes

[soft parcel]

I'm not sure that works after all

Cause dominant only puts a requirement on evaluating with roots

Man, rep theory makes my head hurt

[wary elbow]

The point is that the fundamental weights are positive linear combinations of simple roots

so a positive linear combination of fundamental weights (i.e., a dominant weight) is a positive linear combination of simple roots

(but not conversely)

[hushed bone]

I like ur pfp

[soft parcel]

I think this makes sense...: 0 <= (lambda_i, lambda) = (lambda_i, \sum k_alpha_j alpha_j) = k_alpha_i

Though at this point, I think need to call it a day cause my brain is not functioning

[waxen fractal]

thanks, it's my main brand image for my math stuff atp, i also use it for my math blog and mse/mo pfps

[soft parcel]

But thanks for the help

[wary elbow]

Sure! Tbh I also find root/weight combinatorics super annoying... I just remember some basic facts and then look up the rest (eg in Humphreys) if I need to

[verbal panther]

is there a name for like, cross operation associativity, even if the operations themselves arent associative?

so g(f(a,b),c)=f(a,g(b,c))

[hidden shore]

id call it middle associativity but idk if thats a proper term

[subtle smelt]

If lambda is a sum of simple roots, then 0 < lambda. As 0 and lambda are dominant integral weights, this then implies that 0 must be a weight in V(lambda).

The argument goes as follows: Let lambda and mu be dominant integral weights and let V(lambda) be the associated irreducible highest weight module . The set of weights gamma of V(lambda) satisfying mu <= lambda is non-empty (it contains lambda) finite and thus must have minimal elements, call one of them v. Assume that v is strictly larger than mu, so that v-mu is a non-trivial sum of simple roots. We obtain that (v-mu,v)=(v-mu,v-mu)+(v-mu,mu)>0, as mu is dominant. This then implies that exists some simple root alpha such that (alpha,v)>0, but in that case v-a_k is again a weight (V(lambda) is saturated) and it satisfies mu <= v-a_k < v, contradicitng minimality.

More generally you can phrase it as mu occuring as a weight of V(lambda) iff if w mu < lambda holds for all w in the Weyl group. In the case of 0 it suffices to just check if 0 < lambda holds, which is just asking if lambda is a positive root.

[hushed bone]

What the helly.

Being someone who does algebra but doesn’t understand anything about Lie algebras this looks like Finnegan’s wake lol. It feels like I should understand what these words mean but they’re in an order that doesn’t make sense and you get the feeling something is going on that you don’t see.

[wary elbow]

It's just combinatorics.... albeit voodoo combinatorics. That's why a lot of ppl in rep theory are essentially fancy combinatorialists (not derogatory -- that stuff is hard)

[wary elbow]

I'm not sure your statement that mu is a weight of V(lambda) iff w mu < lambda for all w \in W is true for Vermas. For example, for dominant lambda, all antidominant weights will be weights of the Verma module V(lambda).

[subtle smelt]

Ah sorry, I meant the irreducible highest weight module with V(lambda), where lambda is dominant inegral. It is finite dimensional and W acts by permutation on it

[wary elbow]

However, this does give another proof of what they were asking, since your statement is true for irreps, and the irrep of highest weight lambda is a quotient of the Verma

[subtle smelt]

You can see the characterization is true by the fact that w mu < lambda must be necessary (W permutes the roots after all) and sufficient (W takes mu to the fundamental Weyl chamber)

Or more convincingly, a guy at the MPI decided to make it homework to prove that characterization

[wary elbow]

I agree with your statement for irreps, yep

but the question is about Vermas

[subtle smelt]

Is it?

[wary elbow]

It is, yes

[subtle smelt]

This looks like Humphreys, where Z(lambda) is the Verma module of highest weight lambda

[wary elbow]

That's right, the question Fyodor asked is about that Verma

So your statement about irreps does answer the question (upon taking the quotient)

[subtle smelt]

Okay the exercise is in Humphreys (21.4 Ex.3 ), and V(lambda) is introduced in (20.3) Thm B (where it is shown that exist as the unique irreducible quotient of the Verma module)

You can even see in the picture that the next exercise references Z(lambda), which is the Verma module.

[wary elbow]

Ah. In his question, Fyodor said that V(lambda) is the Verma. I guess he misinterpreted the exercise, so I answered the question he stated

[subtle smelt]

But anyways, would be cool to know if the converse holds even for Verma

Oh wait nvm, yeah thats pretty direct from knowing its basis

[wary elbow]

Yeah, this statement is also true for Vermas. More generally, 0 occurs as a weight of Z(lambda), where lambda is any weight (not necessarily dominant), iff lambda is a positive linear combination of roots.

But this is trivial to prove, so it's not very interesting

[subtle smelt]

It's good to check that taking the PBW-esque monomial associated to any positive root alpha applied the highest weight vector yields an basis element with weight lambda - alpha, and that such basis elements form a basis of Z(lambda)_lambda-mu

[wary elbow]

Yes, good point

[subtle smelt]

Also makes it pretty cheap to know the dimensions of the weights spaces of Verma in terms of partitions of positive roots.

[wary elbow]

I feel silly now.... I should have gone to Humphreys to check whether the question was actually about Vermas. Ah well

[subtle smelt]

To be fair, his choice of notation there is atrocious

🇿 erma modules

[wary elbow]

lol yeah

This is a good place to note that Humphreys' book on Category O is a banger

fantastic book

[soft parcel]

Oh yeah, woops. Got my notation mixed up

[soft parcel]

Also, I found an exercise in Humphreys which guarantees that the fundamental coweights are positive linear combinations of the simple roots (Exercise 13.8)

[hexed tangle]

Does anyone have a tip on how to show that (\mathrm{Ext}^1_{\mathbb{C}[x,y,z]}((x,y,z),(x,y,z)) = 0)?

[broken turtle]

mh_le

[edgy pond]

What are the prerequisites to understand the representations of SL_n(Z)

and is it too ambitious to attempt to learn this after a first course in rep theory

[worldly zealot]

Jagr have you run into the notion of silting discreteness

[lone jacinth]

I have yes

[lone jacinth]

I guess just taking a projective resolution of (x,y,z) and applying hom would work well enough

[hexed tangle]

fair

[normal narwhal]

this is covered in fulton harris i think

[lone jacinth]

Why you ask?

[worldly zealot]

I’ve been trying to prove silting discreteness for preprojective An but it’s been deceptively hard

Just curious what other contexts it comes up in and how people think about it

[hexed tangle]

sorry, I misremembered the assignment I need to calculate (\mathrm{Tor}^1_{\mathbb{C}[x,y,z]}((x,y,z),(x,y,z)) )

[broken turtle]

mh_le

[astral ginkgo]

What kind of C[x,y,z]-Module is (x,y,z)?

[hexed tangle]

Non-free

[astral ginkgo]

Why?

[hexed tangle]

Men it’s not principal

[past cove]

Uh

[astral ginkgo]

Free modules can have as many generators as you want

[lone jacinth]

So take projective resolution and tensor then

[astral ginkgo]

You are correct it's not free or projective though I had a bit of a brain fart, apologies

[astral ginkgo]

Making you do it for 3 variables is kind of mean

[hexed tangle]

Why is it mean?

[astral ginkgo]

2 variables has a much nicer resolution

This is just more fuss for not much pedagogical gain except doing more computations

[lone jacinth]

I mean, the koszul complex is pretty nice, no matter how many variables there are

[astral ginkgo]

The question is if they've learned about it

[hexed tangle]

I haven’t

[lone jacinth]

If 3 variables is a hassle you can always consider
0 -> (x, y) -> (x, y, z) -> (x, y, z)/(x, y) -> 0
and use a long exact sequence

[hexed tangle]

Is (x,y,z) projective?

[lone jacinth]

No

[limpid horizon]

I was trying to think of using fg projective iff it is a direct summand of a fg free module. Do u use that to show its not projective?

or can u see it directly or something

[lone jacinth]

I guess you can compute its Tor with R/(x, y, z)

[limpid horizon]

I want to do more hom algebra its cool

[lone jacinth]

Or yeah, show that the map R^3 -> (x,y,z) doesn't split

[limpid horizon]

Why do we only need to consider R^3?

[lone jacinth]

You just need a surjective map that doesn't split

That's probably the easiest one to consider

[waxen fractal]

let (A, 0) be a pointed set. let B ⊆ C. we have a map A^B -> A^C given by sending a function f: B -> A to the function g: C -> A given by g(c) := f(c) if c ∈ B, and 0 otherwise

is there a common name / notation for this?

[waxen fractal]

ok i thought about it, and this can actually be viewed as precomposition, just in an unconventional way

we use the pointed map p: C ⊔ {0} -> B ⊔ {0} given by p(c) := c if c ∈ B, and 0 otherwise

this actually extends to the case i was thinking about with abelian groups too

[waxen fractal]

i saw a ghost ping here with something about sheaves which makes me really interested but i assume it was deleted cuz the approach ended up not working out. hopefully you can salvage something cuz i had vague sheaf sensations related to this but couldn't develop it

[forest turtle]

yes, i am working on salvaging it

the variance was wrong

but it reminded me of the usual presheaf action of ptSet(-,(A,0)) : ptSet^op —> Set

[worldly zealot]

this is extension by zero, no?

[fierce steeple]

Maybe this is what your motivation was but this is the usual way of viewing sets with partial functions as maps of pointed sets, where you view the additional point as a placeholder for when you stuff is not defined

So like here the point is given f: B -> A you can view it as a partially defined function C -> A, hence a function C u {*} -> A u {*} via the recipe you described

@waxen fractal lol

Maybe my point is just that this is a common thing to do

[waxen fractal]

i thought about what all of yall said, and then the phrase "domain expansion" popped into my head

insert jjk meme here

[waxen fractal]

skimming the stacks page, it seems that it's an adjoint to restriction

[waxen fractal]

and this perspective makes it look similar to restriction, but instead of restricting the domain, we're expanding it

[waxen fractal]

hell yeah

[waxen fractal]

ill think about if i can relate this to base change

cuz the way they define it on stacks looks like base change but it's in the context of stuff i can't understand

[waxen fractal]

notably, maps of pointed sets carry a pointed set structure, so we can actually upgrade this to ptSet^op -> ptSet

[worldly zealot]

idk how stacks puts it but its a left adjoint to the inverse image functor which exists the map is an inclusion

[hallow bone]

for i alone, am the honoured one

[waxen fractal]

pointed word: domain expansion

wait im mixing up memes

shit guys im not a weeb

[hallow bone]

neither am I i havent watched anime in like 5 years

[waxen fractal]

yeah nvm it's definitely more subtle than i thought

maybe i should just stick to the simple way of viewing it

[hallow bone]

happens

[forest turtle]

horrible take, but its a left kan extension

[waxen fractal]

ok the issue is that this doesn't seem to be a common way to view things in the case of abelian groups, people still view 0 as an element of the group, not a placeholder

[waxen fractal]

every left adjoint is a Lan if you try hard enough

🗣️

ok im going back to trying to understand it abstractly

[forest turtle]

what was this case?

[waxen fractal]

ok issue is idk the inverse image functor

[waxen fractal]

same deal but, we let A be an abelian group

the exact same construct does work

concrete example: if we consider the vector space of power series R[[x]], this construction includes R[[x^2]] as a linear subspace. for every x^2 coefficient we copy it over exactly, and for the rest, we just put 0

domain expansion 🗣️

extension by 0 seems to be the proper terminology but also kinda obscure and i think it's used more in geometric contexts? so it's not that useful

[fierce steeple]

Lol idk how it is obscure

[waxen fractal]

i never heard of it but maybe it's not obscure and this is just my moment to learn it

[hushed bone]

Extension by zero not obscure

[waxen fractal]

im intimidated by the fact my only reference for it is stacks project

[hushed bone]

Non-Hartshorne exercise doer identified

[waxen fractal]

i dont do AG

[hushed bone]

Terminate immediately

[waxen fractal]

https://cdn.discordapp.com/attachments/585209212821307409/1331719624265306336/monk_key.gif

you liked my pfp so here's a gif of it blowing up (in game)

ironic... you would think the AG people would be blowing up

[waxen fractal]

I've read a good chunk of the vakil exposition on sheaves

so it's not like i have no knowledge of some terminology used here

[digital parcel]

😠😠😠

[hushed bone]

Yeah but it’s defined in a Hartshorne exercise so among alg geometers it definitely isn’t niche and nonchalant

[waxen fractal]

hell yeah

[hushed bone]

I’m sure Vakil talks about it too

It’s kinda fundamental theoretically speaking

[waxen fractal]

ok phew

[hushed bone]

What game is this

[waxen fractal]

it's called undertale 2

revenge of the robots

[hushed bone]

Wat

[waxen fractal]

im not joking

[hushed bone]

Wat

[waxen fractal]

my pfp is fanart

to be clear

i asked for permission to use it

[hushed bone]

Was it made by a guy named Toby Mouse

[waxen fractal]

nope

it's like a shitpost game

but it's unironically good

[hushed bone]

Lol

Is it a flash game or something

[waxen fractal]

rpgmaker

[hushed bone]

www.addictinggames.com

Wild

[waxen fractal]

it's like a full 10-20 hour experience to be clear

[hushed bone]

Lmao

[waxen fractal]

where were we

oh yea extension by 0

can someone ELI5 what is extension by 0

[hushed bone]

Look up Hartshorne exercise II.1.15 or something

There’s a SES it’s a part of

[drowsy niche]

very very late into Vakil yes

[waxen fractal]

like i get the vague idea but what is it concretely

i guess uhh

[fierce steeple]

Idk why it is Hartshorne in particular

[drowsy niche]

The symbol “!” is read as “shriek”, presumably because people shriek when they see it.

[fierce steeple]

Ig just cause this is a very early exercise

Doesn't everyone learn about 6ff in kindergarten tho

[hushed bone]

Yeah

[waxen fractal]

ok i have an idea how to view it in sheaf terms anyway

[hushed bone]

If you live in woke

[fierce steeple]

Bonn you mean

[drowsy niche]

so that's where I was supposed to be asking toddlers

no luck in the US

[hushed bone]

DEI sheaves

[waxen fractal]

C^A is just the global section of a sheaf on C as a discrete space

🗣️ 🗣️ 🗣️

[fierce steeple]

Real

[waxen fractal]

and uhhh

we uhh

take the open set B

and if we have a sheaf on B

we extend by 0 to C !!!

guys look im doing AG

[waxen fractal]

it's like, uhhhh the constant sheaf

[hushed bone]

It’s like what if u had functions on B

[waxen fractal]

valued on A

[hushed bone]

And you said I’m gonna extend these to all of the space

By just saying it’s 0 where I don’t know where it’s defined

[waxen fractal]

wait that's a truke

[hushed bone]

And then u made that a definition

[waxen fractal]

ok i see why ppl are making this about sheafs now

it does seem natural

[hushed bone]

And what’s the cokernel of this gonna be

It’s something like what are the sections on the whole space

Supported on the complement

And this is how you get the SES

[waxen fractal]

oh we're making this an ses now

[waxen fractal]

makes sense

mfw i have to look at sections of the interior of the compliment of U

man this does sound pretty foundational

[hushed bone]

I think it’s actually open on the other side

I think you have a j^-1F there

[waxen fractal]

shit, for real

[hushed bone]

Which is like approximating the complement via descending opens containing the closed

It makes sense too cuz you can’t actually just define f to be 0 outside U on the nose

Because th complement is closed

And so you can’t glue those functions over that cover

So you have to wiggle stuff a little

That’s what ^sh does

[waxen fractal]

ok imma think about this rq

[waxen fractal]

oh, we like, compute on presheaves, the sheafify?

classic maneuver

[hushed bone]

I think so, if my memory serves me

[waxen fractal]

are we tryna solve this
0 -> A^B -> A^C -> ? -> 0

[hushed bone]

Idk what that notation is

Lol

I’m talkin about some SES that’s like

Yeah just look it up kek

[waxen fractal]

oh i was just borrowing from my notation from my usecase

like, A^B is maps from B to A

[hushed bone]

Yeah iunno

I’m eepy

Karaoke too hard last night

[waxen fractal]

relatable

[hushed bone]

2 many beers

[waxen fractal]

true

im just worried tho cuz like

i dont wanna get poisoned yknow

[hallow bone]

soon you will

.. UAG that is

[waxen fractal]

actually maybe i should try applying sheafy thinking to my (larger) problem

will update if i succeed

this convo gave me ideas

[silver goblet]

wow that's messed up. The state of our education system is truly abysmal right now. Back in my day any newborn would know about 6ff

[last talon]

Does anyone know of a good exposition of whitehead’s algorithm for when two elements of a free group are images of each other under some automorphisms?
Preferably one that follows a more 3-manifold route than a combinatorial route

[last talon]

(Not sure if I’m more likely to get an answer here or in #alg-top-geo-top so #alg-top-geo-top message)

[subtle smelt]

I get that you have to teach them kids to count up to 6, which compared to like 2 or 3 is pretty hard and I sometimes struggle with it too. But once you do, there is no reason to just pick it up on the side, maybe make some flashcards if really necessary

[void plank]

If f: A ---> B is a chain map of complexes of R modules and I: C ---> C is the identity chain map on a complex of R modules C, it's certainly true that cone(f) (x)_R C = cone(f (x)_R I) right?

[subtle smelt]

Someone had the initiative to add a higher categories course starting next winter semester in Bonn. The go to strategy until then for learning MPI nonsense prerequisites of the month is to just try to breathe in the air Scholze exhaled and hope they unlock those skills as if it was a supersoldier serum.

[silver goblet]

peter scholze once walked past me and that's probably still powering any mathematical intuition I might have to this day

[hexed tangle]

why is the im functor not left or right exact?

[subtle smelt]

The im functor?

[hexed tangle]

it is given as Fun([1], Ab) -> Ab that maps homomorphisms to its image

[drowsy niche]

there are probably a lot of counterexamples available but something easy would be applying im to the sequence
0 -> (0 -> Z) -> (Z^2 -id-> Z^2) -> (Z^2 -pr_2-> Z) -> 0

clearly 0 -> 0 -> Z^2 -pr_2-> Z -> 0 is not exact at Z^2

[hexed tangle]

is this an exact sequence of chain compexes?

[lone jacinth]

It's an exact sequence in Fun([1], Ab)

[hexed tangle]

I see

[lone jacinth]

Another clue that it shouldn't be exact is that the image is the kernel of the cokernel.

So the composition of something right exact (not left exact) and something left exact (but not right exact)

[waxen fractal]

let A be an abelian group. let A_i be a sequence of descending subgroups such that their intersection is trivial. sanity check: do we get an induced ultrametric on A, i.e. for nonzero x ∈ A, we define |x| as 2^-i where i is the greatest i such that x ∈ A_i

[this is unrelated to above discussion, I've just been thinking about this]

[last talon]

Yes

[hallow bone]

absolute cinema

[waxen fractal]

is this a joke about the banality of sanity checks

[waxen fractal]

this feels important but i dont feel I've ever actually seen it anywhere, what kind of math is most concerned with this construction?

[last talon]

Number theory 🙃

[waxen fractal]

NOOOO

worst nightmare realized

[last talon]

I mean this is literally the construction of the p adic metric

[waxen fractal]

yeah but that also is induced by inverse limit topology

this feels like it has to be more general? like i dont see how to express it as an inverse limit

[last talon]

Like I’m not sure a lot of Profinite groups stuff cares about the fact this is metrisable
Mostly because a lot of the examples we care about don’t look like this

[lone jacinth]

I guess this is just the situation for any first/second countable topological group really

[waxen fractal]

i guess we could generalize it to a metric by using Q as an order

[hallow bone]

😔

justice for sequential inverse limits

[last talon]

Yeah like you can do it with sufficient finiteness properties
But like I’m not sure anyone who does profinite stuff actually thinks about it this way
Because you need to make a lot of non-canonical choices usually

[sacred wadi]

Look into channel, wonder why people have weird colors on their names, realize this is not TAU.

[hallow bone]

I feel like the topology is more interesting/useful than the metric in this case

[waxen fractal]

yeah actually in retrospect this actually is how ALL ultrametrics on groups arise

(that are compatible with group structure)

[last talon]

The topology is the metric without the unnatural choices 🙃

The metric is the topology with a chosen “descending” countable basis of the identity

[hallow bone]

if you only study the metric you'll miss the fact it's a boolean space

and who wants to miss that

[waxen fractal]

so actually i think i accidentally confused myself into rediscovering basic group theory

it's really interesting to me that there is this entirely "algebraic" description of a topological phenomenon. and it's not even for inverse limit stuff, it's just, filtrations naturally have topological structure

like isn't that weird

[waxen fractal]

to be clear i only care about the metric for the sake of inducing the topology

[hallow bone]

well I'd argue this isn't an algebraic phenomenon per se

because take any set X and a descending chain of equivalence relations

[waxen fractal]

ok tbh like. mico is right

she is always right, by definition

[hallow bone]

nuh uh

[waxen fractal]

i should just jump straight to the topology

it'll be more enlightening

[hallow bone]

anyone with negative opinions on UA can never be 100% correct

[waxen fractal]

ok mico is right except for one thing

why trans flag 😭

[hallow bone]

it's my standard reaction

[last talon]

mishu is righter than micot :3

[hallow bone]

it's hard to convey what I mean with it

[waxen fractal]

it's like nat's meowboy

i suppose

[waxen fractal]

and when i think about it

it IS inverse limit stuff

so actually im just blind

[waxen fractal]

THINK KEITH THINK

we can view it as the smallest topology such that the quotient maps A -> A/A_i are continuous

with the latter being discrete

ok i feel better

everything is right again

i was just blinded by the evils of metric spaces

[waxen fractal]

so yea, we can also make a topology from this too, it seems

same way

[last talon]

You should do Group cohomology

[waxen fractal]

the descending condition just guarantees that we get a sequential inverse limit

[waxen fractal]

people keep telling me that but it looks unfun

[hallow bone]

just go straight to Beck module cohomology atp

keith you should do Beck modules in congruence modular varieties

[last talon]

I wonder if there is a cohomology theory on profinite spaces such that profinite groups have the same group cohomology and space cohomology and in asking that question I answered myself because something like Z_p and Z_q are homeomorphic but should have different group cohomologies

[last talon]

It fun

Read Brown’s book

[hallow bone]

there's unfortunately no (algebraic) cohomology of boolean algebras

they are congruence-distributive, so the commutator is intersection, and therefore there are no nontrivial Beck-modules (these play the role of G-modules)

but maybe there are other cohomology theories for profinite spaces

holy shit I accidentally stumbled upon a paper from my group theory professor lmao

[last talon]

Give to mishu :3

[hallow bone]

https://webspace.science.uu.nl/~dobbe012/doc/Talk4.pdf

he's an AG professor don't get your hopes up

[last talon]

[hallow bone]

lmao

"group theory" here is introductory group theory

[waxen fractal]

poop theory

[hallow bone]

they're still at group homomorphisms

😭

[waxen fractal]

sometimes i turn off my filter

and just say the first thing that enters my head

[hushed bone]

Me when the I-adic completion isn’t actually complete with respect to the I-adic topology

[hushed bone]

@karmic brook

[hallow bone]

yeah sorry that's my fault

[last talon]

We should make every course a group theory course :3

[waxen fractal]

i prefer abelian groups to groups usually purely because that way i dont have to tell my left from my right 🗣️

[hallow bone]

there is some group theory in the UA im reading rn

well, only a little

Tw(R, E) 🔥

[last talon]

Do this :3

[waxen fractal]

wait this looks interesting

can i do it

too

[last talon]

Yeah it should be doable just from the definitions

[waxen fractal]

oh i suppose that would be fun actually, dont need to pull out baer criterion or anything

i just need to remember the definition of injective (i forget every time)

if B ⊆ A and B maps into I, then we can extend this to a map A -> I that commutes

[lone jacinth]

Prostrat would be showing that ||Hom_Z(R, Q/Z) is always an injective R-module.||

[last talon]

It’s the same as amateurjective

[waxen fractal]

if B ⊆ A and nA = 0, then a map B -> Z/n extends to a map A -> Z/n ???

it's weird cuz i see how it has to be true by jagr's argument, but it's pretty nonobvious in this form, to me

[last talon]

I did it via Baer’s criterion where it’s kinda trivial

[waxen fractal]

ah yea fair

[lone jacinth]

I mean the idea is basically just that if something in B is a multiple of something in A it will have smaller order so restricting where a map B -> Z/n can send it

[waxen fractal]

yeah

im thinking about it and i think if i thought about it long enough I'd rediscover baer's criterion

ill just move on

[waxen fractal]

ok (a) is trivial

[last talon]

It’s essentially what you get if you try to inductively construct it
You try Zorn’s lemma and realise this is exactly what you need to turn anything into something bigger

[waxen fractal]

yeah

[waxen fractal]

ok i feel like im missing an important proof technique from my mental library cuz (b) feels like it would require some sort of recursive argument, but i can't figure out how to formalize the union step

[last talon]

Zorn

[waxen fractal]

subgroups that are direct sums of cyclic groups?

i guess that would be closed under nested union

nah that doesn't feel right

it feels better to say, linearly independent sets

yeah ok im feeling it

for (b) i think you recurse on the set of linearly independent sets

at each stage it's a direct sum

when you add a new generator it remains a direct sum

when you union it's still a direct sum

and it's only maximal once the entire group is generated

so i guess it's similar to the proof that every vector space has a basis

[lone jacinth]

It becomes fairly easy if you first ||reduce to n a prime power||

[waxen fractal]

i think universal algebra (hi enpeace) sorta illuminates why Z/n-modules would be similar to Z/p-modules

but i dont have anything but vibes

so im saying this to bait enpeace into saying something smart

[lone jacinth]

The fact that Z/n is a product of Z/p^k explains why the modules are similar to Z/p^k at least.

And for any rep finite artinian ring you have that every module is a direct sum of if finitely generated ones

[hallow bone]

yeah this problem is completely reducible to Z/p^kZ by CRT

[hallow bone]

I love how you say "rep of" instead of "module over"

[waxen fractal]

concept with an attitude

😭

anyway, getting back to the convo start, (where i asked about the topology thing), the inverse limit actually explains a certain construction i was experimenting with

suppose A is a sequential inverse limit of A_i

then, we can define a map A^w -> A by compatible maps A^w -> A_i, and we can define those maps by maps A^n -> A_i, which only gives us finitely supported stuff, but this still leads to a lot of examples

the reason the map as a whole is not finitely supported, is because, for each i, we can increase the n that we fix

so i guess i should have written n_i

tl; dr i feel a lot less spooked by the "topology" aspect now because it's apparent to me that this phenomenon can be described purely algebraically

[waxen fractal]

anyway, the bizarre thing is that, the converse is also true to some extent: let phi: A^w -> A be any linear map, with one condition: denote phi_n as A^{n+1,n+2,...} -> A given by precomposing phi by the extension-by-zero map A^{n+1,n+2,...} -> A^w. then we require ∩ im phi_n = 0

[waxen fractal]

then phi is induced by such a map

this fits into the general pattern i see, "natural maps A^w -> A should have finite support in some sense, wild maps won't", im curious if there's a way to formalize the difference between a "natural" and a "wild" map, beyond something ad-hoc like "ZF proves it exists"

[karmic brook]

lol

[waxen fractal]

do you like poops

[last talon]

What’s a good source to learn about arithmetic groups?

[waxen fractal]

arithmetic poops

[cloud karma]

Cohomology of arithmetic groups?

https://arxiv.org/abs/math/0106063

[karmic brook]

what do yoy think

[waxen fractal]

yeah

you're a poopissimo

[forest turtle]

remarkable poop

[light forge]

symmetric power of irr rep is irr??

[lone jacinth]

Not in general, no

[indigo pumice]

how is identity element obtained in a loops if x(x^-1) isnt equal to x^-1(x)

[light forge]

but for SL_2 standard lie alg irr rep , it is true

[lone jacinth]

I'm not quite sure what you're asking.

In a loop x will have both a right inverse and a left inverse. If you multiply x by its right inverse on the right you will get the identity.

[indigo pumice]

Ahhh okay makes sense

[indigo pumice]

Like so x1,x2,x3
Are the elements
Then
(X1)(X2)^-1=e
(X3)^-1(X1)=e
If this is true then it would mean X2 is the left inverse and X1 is the right inverse right?
Cancellation dosent apply to loops since they are non associative

[proud mirage]

how to show k[x_1,...,x_n] is free as a k[x_1,\cdots,x_n]^{S_n} module ? S_n acts by exchanging x_i's. i know a argument using heavy CA machinery but i wonder there are easier arguments?

[fierce steeple]

Found this https://mathoverflow.net/questions/216242/free-kx-1-dots-x-ns-n-module. Like you can look at these elementes and show by induction and stuff that they span

[lone jacinth]

I don't know if Galois theory counts as heavy machinery. But looking it the field of fractions you get a degree n! Galois extension. So it's enough to verify that there is a generating set of size n!

[steady lance]

Huh this is very cool!

[hallow bone]

Let f : G -> H be a homomorphism of groups, and make ℤ[H] into a G-module using f. Let M be another G-module. Is the tensor product M ⊗_G ℤ[H] the same as M extended to H restricted to G?

[proud mirage]

thx both of you, initially i just shown it's flat via dimension counting and apply a graded version of nakayama so this is simple enough

[wary elbow]

When you say "is the same as," I assume you mean to consider the tensor product as a G-H bimodule, in which case the answer is yes by definition: the restriction of this module to G is exactly the left G-action on the tensor product.

[hallow bone]

okay nice

[wary elbow]

More generally, this is an example of the induction functor, and this works because ind \circ res is always the identity functor

[hallow bone]

ive got a small conjecture that Beck modules over algebras in modular (or at least permutable) varieties always have a tensor product, so that extension of scalars agrees with it

[hallow bone]

wdym more generally? what i described is induction

right?

[wary elbow]

Yes, you're right. I meant to add that this is an example of adjunction for the ind and res functors. In particular, what you're asking boils down to the general adjunction fact that ind \circ res \circ ind = ind, since res is the identity map on the underlying abelian groups

[hallow bone]

thats not true though
if i take the inclusion 1 ≤ G, then inducting corresponds to taking the free G-module on an abelian group
this yields a way bigger underlying group

[wary elbow]

Let's call that free abelian group F. This identity says that ind(res F) \cong F, which is true.

[hallow bone]

is it though? when restricting you forget all the G-module structure

when inducting this structure gets added freely, iirc you take the direct sum of F indexed by G and let G act in the obvious way

[lone jacinth]

That's not true

[wary elbow]

oh no

I guess I messed something up on my adjunction formula

[lone jacinth]

I don't think there's really anything to do with adjunction. It's just res F and F being the same abelian group

[wary elbow]

isn't ind an adjoint functor though? so I should always have ind res ind = ind

[lone jacinth]

ind and res are adjoint to eachother, but you shouldn't have such a formula

(and you don't)

[hallow bone]

if only

[wary elbow]

but this is always a fact for adjoint functors, yes?

[hallow bone]

no

[lone jacinth]

You have maps
F -> FGF -> F
that compose to the identity

But the other composition is usually not the identity

[wary elbow]

oh right

ugh, sorry

I'm being dense

[lone jacinth]

But it is true that
ind F is a direct summand of ind res ind F.

And in fact slightly stronger F is a direct summand of res ind F

[wary elbow]

Is this extracted from right exactness of ind?

[lone jacinth]

Hmm, might be some relationship there.

But I guess the easy way to see it is just that ZG is a (bimodule)-summand of ZH

[wary elbow]

ah ok, makes sense

[restive vale]

I was told that there is a ring R such that the R-points of PGL have a element that does lift to GL2. Whats a reasonable way to find such a ring and element?

I have been playing ideals the nonprincipal but invertible since that feels like the right thing, but am kinda stumped why i think that even is

[summer quest]

Take R=Z/2Z

[edgy pond]

I compute the action on B and get 0. $ (y \cdot B) (x)(z) = y \cdot B(x)(z) - B(y \cdot x) (z) = - B(x) (y \cdot z) - B(y \cdot x) (z) = 0$ for any $y, x,z \in g$ and g acts on itself by the adjoint. Why is B g-invariant?

[broken turtle]

pink_panther

[edgy pond]

oh i mixed up the definitions on invariance for a lie algebra with lie group

[smoky wasp]

Im trying to prove the following:
Let L/K be an extension of number fields. O_K, O_L are the ring of integers on K and L, respectively
If P is a prime ideal of O_K, P ramifies in O_L iff the discriminant ideal is contained in O_K

i just dont really know where to begin, like intuitively

what is the link between a discriminant and ramification

[past cove]

think about the situation K/Q

a prime p in Z ramifies iff it divides the discriminant of K

a similar logic applies here

but the link here is that ramification is controlled by polynomial reduction modulo P

and the discriminant is defined in terms of polynomials

so that's your relation

[smoky wasp]

in the sense of like

the polynomials that appear in the determinant?

[past cove]

well take O_K = Z[a]

then the discriminant is determined by Prod(x_i-x_j)

now take the minimal polynomial of a

when we reduce mod p, if it has a repeated root it ramifies at the prime corresponding to that factor

but this is equivalent to Prod(x_i-x_j) bein 0 modulo p

i.e. p divides the discriminant

[smoky wasp]

Hm

So I guess we wanna reduce it to this case somehow?

[near lantern]

Adjoint functors are not as well-behaved as Galois connections, they don't have to satisfy this in general. The ones that do are (IIRC) called idempotent adjunctions.

Does the Grothendieck monoid of an abelian category (generated by objects with relations [A] + [C] = [B] whenever there is an SES 0 → A → B → C → 0, but without formal inverses) show up anywhere not as the Grothendieck group?

[hallow bone]

for finite groups that would be positive integers under multiplication right?

[lone jacinth]

Oops wrote something wrong and deleted. Anyway, I haven't seen it come up, but for the category of fg abelian groups the Grothendieck group is trivial, but the Grothendieck monoid does retain a lot of information.

[hallow bone]

does the grothendieck group ever have nontrivial torsion? how does that arise?

Grothendieck groups are weird

[lone jacinth]

If let R = Z[sqrt(-5)] and consider the category of fg projective R-modules, then I believe P = (2, 1+sqrt(-5)) has P^2 = R^2.

So in the Grothendieck group [R] - [P] is torsion.

[hallow bone]

thats so weird

[lone jacinth]

To my the weirdness is that you can have P^2 = R^2. I'll stick with my Krull--Schmidt categories

[hallow bone]

does this have to do with Z[sqrt(-5)] not being a UFD?

[lone jacinth]

Yes. It's a dedekind domain, so it's a ufd iff it's a pid and the indecomposable projectives are exactly the ideals

[hallow bone]

i love math

[lone jacinth]

Googles ai thingy tells me the Grothendieck group for a dedekind domain is Z (+) the ideal class group

So [R] should generate the Z part and then [R] - [I] for ideals I the other part

It should come down to showing
R (+) IJ = I (+) J

[astral ginkgo]

Any ring without IBN would also be an example, as long as you don't have R=R^n for all n, right

[lone jacinth]

In the only examples I know you do have that though

[astral ginkgo]

Same

I do actually wonder if any examples exist that are only part of the way

Like say only even powers are isomorphic

The primary example I know is End(F^N) for some field F, and that has R=R^2

[lone jacinth]

Anyway for this, so every ideal is uniquely a product of prime ideals, so enough to show the case where J is prime.

Then I want to show I/IJ = R/J, i.e. that I/IJ is 1d... [Okay I guess localizing at J does the trick]

[olive schooner]

i feel like you should be able to do this with leavitt path algebras if i remember its properties correctly

(disclaimer: i know practically nothing about LPAs but the idea sounds pretty similar to what i've read off someone's poster awhile back)

[hallow bone]

so that ring would automatically be noncommutative right

[olive schooner]

yeah

[hallow bone]

proofs involving the adjugate of a matrix are ones i will never wrap my head around

i dont know why we should expect such a thing to exist lol

[astral ginkgo]

Yes, necessarily

[lone jacinth]

Found this

LK(Rn) is the LPA of graph with one vertex and n loops

So presumably it would have R = R^n

https://link.springer.com/book/10.1007/978-1-4471-7344-1

[lone jacinth]

Only works for commutative rings, so you probably shouldn't expect it

[hallow bone]

that is true

[olive schooner]

ye

https://arxiv.org/pdf/1410.1835

there's an interesting more general version here in prop 4.4

[near lantern]

Oh this is fun

It's my motivation for asking about monoids actually

[near lantern]

Yeah this is a step in the classification of fg modules over a Dedekind domain

[near lantern]

Yeah, IIRC for any m and n there exists a ring R such that R^a = R^b only if a-b is divisible by n and a, b ≥ m. Don't remember a reference though.

[astral ginkgo]

Ok coolio

[olive schooner]

this is leavitt's theorem. there's a reference in the arxiv i sent above (see theorem 1.4)

[near lantern]

Oh yes, that's perfect.

[astral ginkgo]

Oh amazing

Maybe I should learn about these sometime

[fierce steeple]

Pog

[hushed bone]

I like ts pfp

[near lantern]

Another approach is to show that you can choose a representative of I which is an ideal (not fractional ideal) which coprime to J. (Indeed use CRT to choose an a whose divisibility by prime powers is just right to ensure aI is an ideal coprime to J.) Now I/IJ = R/J follows by CRT.

[near lantern]

Once you know this (and the rest of the classification) you get that the Grothendieck group is ℤ (+) Cl(R). More strongly, it is generated by the rank-1 projectives which correspond to (1, [I]). Hence the "Grothendieck monoid" is {0} ∪ ℤ_+ ⨯ Cl(R), which I find interesting (that it's this particular "cone" in the group).

(Although in general I believe the monoid need not be cancellative (e.g. if there are non-free stably free projectives), so it's not a submonoid of the group.)

[near lantern]

You can also include torsion, i.e., do K_0(fg) instead of K_0(fg projective) as a monoid.

In this case you get [R/I] + [R/J] = [J/IJ] + [R/J] = [R/IJ] (aliter: because [R/p1^k1...pn^kn] = k1 [R/p1] + ... + kn [R/pn]). So the torsion part is replaceable by one cyclic module. Similarly [R/I] + [IJ] = [J/IJ] + [IJ] = [J] for coprime ideals I, J - but for any fractional ideal class K we can choose a representative of the form aK = IJ with I, J coprime ideals by CRTing the a to multiply by (or by weak approximation if you prefer).
(BTW note that we don't have the freedom to modify I appearing in [R/I] by a principal ideal the way we do for K in [K].)

Hence any module is equivalent to [R/I] for an ideal I (if rank 0) or [R^{n-1} (+) I] for an ideal class I ∈ Cl(R) (if rank n > 0). I am pretty sure these are unique, i.e., there are no more relations, but need to prove this.

Assuming, this K_0 = the previous K0, but the {0} is replaced by the free monoid on the set of prime ideals. The addition is given by adding ideals in that torsion part as the (inverse) ideal class of rank 0 (that is, [R/I] is added as (0, [I]^{-1}) in ℤ ⨯ Cl(R)). In particular, [R/aR] is non-cancellative and invisible in the Grothendieck group.

[near lantern]

K_0 monoid for fg modules over a Dedekind domain

[dry dome]

Hi everyone 👋 Is anyone here currently looking into the topological constraints of the Exceptional Jordan Algebra (h_3(\mathbb{O}))?
​I've been exploring how the Albert matrix applies to Trace Dynamics. Interestingly, the non-associative geometry seems to rigidly lock into exactly 3 fermion generations without any extra parameters.
​Would love to get some insights from the algebra experts here on the math side of things before I share my preprint. Anyone open for a quick chat about this?

[void plank]

Just wanted to check something quickly

Let's say I have two chain maps phi1: A --> B, phi2: B --> C.

Octahedral axiom says there's an exact triangle cone(phi1) --> cone(phi2 \circ phi1) ---> cone(phi_2) --> cone(phi1)[-1] (sorry for using homological grading)

If phi2 is a quasi-iso, this immediately implies by analyzing the LES in homology this triangle induces that cone(phi1) ---> cone(phi2 \circ \phi1) is a quasi-iso

Is my reasoning all correct? I feel like I'm missing some subtlety

[fierce steeple]

Never apologise for homological grading

[fierce steeple]

I agree

[void plank]

In Prismatic Potatos I trust

[fierce steeple]

The way I would think of it is basically that you have uh

A -> B
|| |
A -> C

You have some square and this induces a map between the cones. Since A -> A and B -> C are quasiisos the map between cones is too

e.g. via the 5 lemma and LES if you are working with the derived category, if you want

or: ∞-categorically, cone is cofibre which has a universal property to this effect, analogous to cokernels in an abelian category

[void plank]

Ah, are you thinking of the cone here as a pushout?

[fierce steeple]

Uh well here I guess maybe you need to be careful like replacing your map and then doing it as a literal pushout (but also it models the homotopy cofiber/pushout which is what I was actually thinking)

[void plank]

Ah my bad, thanks!!

[void plank]

Yeah, this is another nice way of seeing this. Thanks for the tip!

[astral ginkgo]

Always apologise for homological grading

[fierce steeple]

The most important category has homologically nonnegative stuff

[main ermine]

Do you guys have any nice examples of path algebras that are not finite representation type other than the polynomial ring?

[olive schooner]

the ones that come from the euclidean diagrams (extended dynkin diagrams) are in some sense well-understood. these are called the tame quivers (more generally, one has the tame algebras) and they have the nice property that their indecomposables can be described as one-parameter families and isolated points

another one beside the jordan quiver you just mentioned (polynomial ring) is the kronecker quiver (two vertices with a two arrows from one vertex to the other one). iirc, the one-parameter families of indecomposables have some kind of parametrization in terms of P^1

path algebras get not as nice very quickly though without imposing some relations on the quiver because not finite-type and not tame-type give rise to wild-type algebras where classification of the indecomposables is hopeless (see: tame-wild dichotomy). an example of this is the quiver with a single vertex and two loops (you get k<x, y>) and classifying representations here is equivalent to classifying pairs of square matrices up to conjugation

[main ermine]

yes i like this one

problem is that im giving a talk on representations of finite algebras, and introduce gabriel's theorem to a bunch of undergrads

i want to illustrate gabriel's theorem both by showing the path algebras induced by dynkin diagrams are finite representation types and some non-dynkin diagram is not finite representation type

[hushed bone]

itzsomebody what kinda math do u do

[main ermine]

i think the polynomial ring is a great example but I was hoping for one where the underlying module isnt a vector space

[olive schooner]

wait are you the person who wrote on r/math about this exact same thing lol

[olive schooner]

i do algebraic combinatorics mostly but i also like reading about random rep theory things

[hushed bone]

Sheesh

[worldly zealot]

kronecker is pretty interesting. the indecomposables can be described very explicitly

[hushed bone]

I don’t think u do the type of algebra I like

🤣

Unless you like the Stanley Reisner stuff which I might be able to talk about

[olive schooner]

SR is on my bucket list of things to learn properly will certainly give you a ping sometime if you wanna kill some time haha

[hushed bone]

I only barely know the theory

I learned it at a summer school last summer

And my general comm alg knowledge doesn’t help too much for that sorta things haha

[near lantern]

A basic result in module theory is that a semisimple module is a direct sum of simple modules (for the purposes of this question, it doesn't matter whether this is by definition). There is also a "coordinate-free" formulation of this statement: if M is semisimple, M = (+)_V M_V (⨯) V, where M_V = Hom(V, M) is the "multiplicity" space: a vector space canonically parametrising the copies of V in M that can be used to decompose it.

Is there an analogous "coordinate-free" formulation of the classification of finitely generated (say torsion) modules over a PID (or Dedekind domain)?

[lone jacinth]

I guess you can consider the map
R/p^n -> R/p^n+1 (+) R/p^n-1

Applying Hom(-, M) and taking cokernel should give you a R/p vector space with dimension equal to the number of summands of R/p^n in M

Then you just need to turn the vector space into a free R/p^n module somehow

I guess you could do projective cover. It's not functorial, but I guess this decomposition shouldn't be functorial anyway

[near lantern]

My goal is kind of to understand what is functorial, but yeah, it is an important observation that the decomposition simply isn't functorial/natural.

[foggy galleon]

what are the main motivations for quantum groups?

[near lantern]

Funcoriality of classification of finitely generated modules over a PID

[wary elbow]

I believe the initial motivation was to develop continuous deformations of semisimple Lie algebras/algebraic groups, since the latter is discretely parameterized.

The category of Lie algebras is equivalent to the category of certain Hopf algebras (enveloping algebras). Also, the category of algebraic groups over a given field is equivalent to the category of cocommutative Hopf algebras over that field (hyperalgebras). This leads to continuous deformations of Hopf algebras via a parameter q, such that when you specialize q at appropriate values you get back your original hyperalgebra/enveloping algebra.

Also, if you let your quantum group be non-commutative and non-cocommutative, this provides noncommutative generalizations of algebraic groups. These are fundamental objects in noncommutative geometry.

To be more precise: by duality, cocommutativity of the hyperalgebra is equivalent to commutativity of the coordinate ring of the algebraic group, so if you want some gadget that is the algebraic group except noncommutative (which can't be done in usual algebraic geometry since your rings must be commutative), you can go to the category of Hopf algebras and relax the cocommutativity requirement on the hyperalgebra. Equivalently, you can relax the commutativity requirement on the ring of functions, since it is dual to the hyperalgebra.

That is, the category of algebraic groups embeds in a larger category, and you can "do algebraic geometry" in that larger category where you allow noncommutativity.

[foggy galleon]

What if in Tannaka-Krein duality we consider representations into GL_n(k) for some other field k other than C? Since it's about compact groups, maybe a starting point is when k is a (Hausdorff nondiscrete) locally compact field, these are the local fields. For example, suppose G, H are (Hausdorff) compact groups and that there exists an equivalence of categories Rep(G, k)-->Rep(H, k), k some local field, that is continuous in some sense (I think?), respects tensor products, and respects dual representations. Does F induce an isomorphism G-->H of topological groups?

References on this subject are appreciated

[narrow kraken]

for which rings R does M being a projective R-module imply that M is a free R-module? the wikipedia page says R being a (skew) field, PID, and local ring, but this isn't exhaustive

[woeful crane]

your question got me curious, so I started googling and came across this paper. Haven't taken a close look yet, but seems relevant.
https://arxiv.org/pdf/1812.02452

Seems like Theorem B answers the question for finite group schemes over k.

[candid lily]

What exactly is it that fails for modules, why are we introduced to vector spaces before those?
Could you maybe list a few nice things we get for VS but not for modules?

[lone jacinth]

Vector spaces are semisimple, so whenever you have U < V you get V = U (+) W for some W.

Vector spaces are free, meaning every vector space has a basis.

These are the key two properties vs have that modules don't have in general

[candid lily]

Thanks

[lone jacinth]

An example to keep in mind could be
Z/4 (as a Z module)
Z/2 is a submodule (as {0, 2}), but Z/4 is not Z/2 (+) something.

It's also not free, i.e. doesn't have a basis

[lone jacinth]

I don't think an exhaustive list really exists, but you can add polynomial rings over a field and free algebras to the list

[astral ginkgo]

Another thing to keep in mind: Vector spaces are classified uniquely by their dimension. It's possible to have rings where free modules don't have a good notion of dimension

i.e. the size of a basis isn't unique

[candid lily]

Im kind of new to modules (in the end they are almost VS though so I guess some intuition just comes from those..?); it seems they are closely related to k-algebras by Noether normalisation:

[astral ginkgo]

This is a very specific type of module

This is definitely not something fhat exists in general

Algebras over fields and modules over those have some very nice properties

Like you're starting with a k-Algebra here so I'm not sure what you mean by they're related to k-Algebras

[candid lily]

I guess I shouldve said it the other way around

We start with a k-algebra and end up with a module

[astral ginkgo]

General modules over general rings are going to have some wacky pathological behaviours

[lone jacinth]

They're related in the sense that you can consider modules over an algebra. Otherwise there isn't really a special relationship per say

[astral ginkgo]

An algebra is already a module. What we're doing here is finding a new ring over which that module is finitely generated

And in fact we can find algebraically independent elements that generate a ring that does this!

[candid lily]

Oh

[astral ginkgo]

The distinction here is that finitely generated describes each structure

So f.g. algebra means it's finitely generated as an algebra

i.e. there are finitely many elements whose products generate the entire thing as a module

f.g. module means it's finitely generated as a module already

So a good example is k[x] is finitely generated as a k-algebra, just by x

But it's an infinite dimension (i.e. not f.g. ) vector space

[candid lily]

Right, thanks!

[astral ginkgo]

You'll often see in the literature "finite algebra" to mean "algebra that is a f.g. module"

[main ermine]

crazy

yes lol

[near lantern]

Is there any group version of the classification of indecomposable representations of a quiver of finite (or tame) type?

For example, apparently the list of indecomposable representations of the Borel subgroup of SL2 is parametrised by finite arithmetic progressions of integers with difference 2. Which IG is closer to the classification for SL2 than the path algebra of A2 (which would be closer to the Borel of GL2), TBH, so maybe not that great of an analogy.

[dry dome]

Hi everyone! Can you please help me review this paper. This paper contains pure mathematics and modern physics. Among the mathematics found in this paper are the Hermitian matrix, h_3(\mathbb{O}), Shirshov-Cohn Theorem, characteristic equation of matrices, and Jordan identity and I have submitted this paper on OSF. I am waiting for your response. Below is the link:https://osf.io/nszmw

If there is a problem with the link opening etc. below is the file

[dry dome]

I recommend reading the first file or Neutrino Seesaw before going to the second file.

[hushed bone]

My senses were correct

My Chmonkey senses never fail me

[near lantern]

For (a), I can confirm that the hint works. Where are you getting stuck? (b) is I guess just an exercise in applying (a).

[lone jacinth]

I guess this isn't what you asked, but if k is an alg closed field of characteristic p and G is a finite group then kG has finite rep type if
the p-sylow subgroup is cyclic
tame for p = 2 and the sylow subgroup dihedral or generalized quaternion
And wild in all other cases

[near lantern]

Is there a construction of the quiver (with relations) it's equivalent to in these cases or is it like a non-constructive proof?

OK maybe that's too much

It's the same as computing irreps and their Ext's

[lone jacinth]

The proofs are quite constructive, but I'm not sure what you're asking exactly.

For cyclic you can compute all the modules quite easily. In the wild case it's not particularly hard to give an explicit wild family.

The tame case I'm not sure about

[near lantern]

I see. I just meant that if we know the quiver in the finite/tame case then we can list the indecomposables (more or less), so this answer is satisfying.

[edgy pond]

In class, the proof of complete reducibility of semisimple lie algebras includes this step. I understand the idea but I don't know why such a decomposition into W_\lambda exists in the first place. I am also confused on whether this decomposition is over every possible linear functional \lambda : < C_V > \to C or not. I'll really appreciate it if someone could explain the direct sum here.

For some context:
C_V is the Casimir element corresponding to the irreducible representation V. This part of the proof basically shows Ext(C, U) = 0 for any subrepresentation U.

Part 1 referred to at the end is the claim that a representation V is trivial if every casimir element acts nilpotently on it.

And I get everything that comes after the decomposition. I just don't see why this decomposition exists.

[near lantern]

IG as follows: note that (i) A := <C_V> is a commutative algebra (ii) C_V's generate A. I will only use these two facts about the representation.

Now because A is commutative you can simultaneously upper-triangularise its matrices, say in a basis e1, ..., en. I'll need to be more explicit about this. We can decompose W into generalised eigenspaces for C_V1 (pick some V1). By commutativity, A preserves each of them.

Decompose each of them into gen eigenspaces for some C_V2. Again each piece is preserved by A. Repeat. This terminates (there can be at most dim W pieces), at which point you have a decomposition of W into A-stable subspaces W_i each of which has the following property:
For each C_V, W_i is a single generalised eigenspace for C_V, say with eigenvalue l_{i,V}.
Moreover, different pieces have to give different tuples (l_{i,V})_V by the way we constructed them.

In fact, for each i, there is an algebra homomorphism l_i : A → ℂ such that l_{i,V} = l_i(C_V). (Fun fact: I believe this can be false in positive characteristic if the matrices don't commute.) To see this, simultaneously upper-triangularise on W_i, then the first basis vector is a common eigenvector for all C_V's, hence for A. The eigenvalues on that vector define l_i : A → ℂ.

This gives the desired decomposition (check that W_i in my notation is equal to W_{l_i} in your source's).

[edgy pond]

i see thank you so much!

[narrow kraken]

consider the functor F from R-mod to (s)CAlg_R that send an R-module to its exterior power. then, the functor G that sends A = A0 \oplus A1 to A1 is the right adjoint of F. is it true that, in general, F has no left adjoint? the problem I am solving requires that R is a field k of characteristic not 2 (i.e, in this case F has a right adjoint but no left adjoint), but I believe this holds for generally for rings R (?)

[lone jacinth]

This is true yes.

[fierce steeple]

If it preserved limits then it would preserve products, which it doesn't. Indeed it doesn't even send 0 to 0 (the terminal objects)

[scarlet prairie]

any ideas on how to do this? i found a proof online(second picture) but exercise was in the section on classification theorem of equivalence class of extensions and H^2(G;A) so im wondering if there is something else taht I was suppose to do

[foggy galleon]

that's the natural thing to do

[astral ginkgo]

I can't think of an easier solution

Or a more obvious one

[lone jacinth]

Maybe sidestepping the problem, but the only m for which such a G exists is m=2 right?

So maybe that simplifies it

[scarlet prairie]

yes i think thats true, i will probably just go with something similar to what the second picture had in mind

[astral ginkgo]

Lmao I guess you're right yea

[finite stream]

Hello everyone! Idk how "advanced" this question is, something borderline I guess.

Let $R$ be an UFD with infinitely many prime elements, $K$ be its field of fractions, and $A$ be a commutative $K$-algebra. I want to show that $A$ isn't finitely generated as $R$-algebra.

Suppose otherwise. Then $A=R[a_1, \ldots , a_n]$ of some $a_k\in A$. Consider a $K$-basis ${e_i}{i\in I}\subset A$. Then there exist a finite subset ${i_1,\ldots , i_m}\subset I$ s.t. $a_k=\sum{l=1}^m{\frac{p_{kl}}{q_{kl}}e_{i_l}$. Thus $\frac{1}{q'}\notin A$ if $\gcd(q' , q_{kl})=1$, a contradiction.

Is this statement even true? If so, is my proof correct? And is there any more elegant way to prove it?

[broken turtle]

Mitya
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[worldly zealot]

i don't know if you've justified that 1/q' is not in A

but if A is unital then it contains an isomorphic copy of K so it's probably cleaner to just argue K isn't finitely generated and refer to specific fractions in K

[finite stream]

Yes, $A$ is unital, of course. And yes, it's really easy to establish via close to ours argument that $K$ is not a f.g. $R$-algebra.

But as far as I know there is no "transitivity property" for finite generated algebras, i.e. the statement "Let $R$ be a commutative ring, $A,B$ be commutative $R$-algebras s.t. $A\subset B$. If $A$ is not f.g. $R$-algebra, then $B$ is not f.g. too" is simply false

[broken turtle]

Mitya

[lone jacinth]

You can use the Artin-Tate lemma with a little work:

||A is R-algebra so also K-algebra. Then by Noether normalization A is integral over some polynomial ring. Then integral + fg gives finite, so Artin Tate tells you K[x1, ...] is fg over R[x1, ...]||

||And then you're inverting infinitely many primes in a UFD, so same proof for K not being fg will work||

[finite stream]

Thank you, I'd elaborate on this a bit later

[lone jacinth]

This assumes R is Noetherian though I guess

In which case I guess it's not too hard to prove directly

[finite stream]

We may consider the set of all denominators in R[a_k] and notice that there is no way to obtain an element 1/q' because every element of this ring have a form P(a_k).

Formal explanation may be a bit messy but I believe that it's somewhat obvious

[lone jacinth]

Consider Z[1/2] with chosen basis {1/2}

Then the is no denominator in the relation, but 1/2 is in Z[1/2]

(of course this isn't a true example as Z[1/2] doesn't contain Q, but it's not adressed in your proof)

[finite stream]

Fair point, we may try to find e_l s.t. e_l is integral over R

[lone jacinth]

Well, I'm not sure why you would expect to be able to find such

[finite stream]

Hmmm, this statement seems to be not as trivial as I thought.

Maybe we should consider a finite generated faithful $R$-algebra $A$, the inclusion map $\iota \colon R\to A$ and find the maximal multiplicative subset $S\subset R\setminus {0}$ s.t. there exist an induced homomorphism $\hat{\iota}\colon S^{-1}R\to A$?

[broken turtle]

Mitya

[lone jacinth]

Hmm, okay. By taking out a maximal ideal you may assume A is a field. Then by Zariski it's a finite field extension of K. Then you can pick an integral basis.

[finite stream]

Btw, is it okay here to use TeX during the conversation as I did?

[lone jacinth]

Yeah, that's great

Anyway, once you have an integral basis your original proof should work. Just invert the denominators and boom

[finite stream]

I tend to believe that the following approach may be useful.

Let $R$ be a Noetherian UFD and let $R\subset R[a_1, \ldots, a_n]=:A$ be a commutative ring extension. Then $A\simeq R[x_1, \ldots , x_n]/(P_1, \ldots P_m)$.

Recall that our assumptiosns imply that there exists multiplicative monoids isomorphisms $R\setminus {0}\simeq R^{\times}\times N$, where $N$ is a free monoid generated by nontrivial prime principle ideals of $R$. So the multipilcative subsets $R^\times \subset S\subset R$ are in one-to-one correspondence with subsets of prime principle ideals.

I tend to believe that for our algebra $A$ there exists a maximal multiplicative subset $R^\times \subset S_0\subset R$ s.t. there exists an induced homomorphism $S_0^{-1}R\to A$. And that $S_0$ is a finitely generated free monoid

[broken turtle]

Mitya

[narrow kraken]

what is the definition of the Picard group of a Dedekind domain?

[drowsy niche]

it's canonically isomorphic to the ideal class group for dedekind domains

but for a general commutative ring A the picard group is just the group of isomorphism classes of A-modules invertible under tensor product

as you'd expect

[waxen fractal]

i made a blog post summarizing everything i have thought about in regards to something ive briefly posted about here before, i am curious how readable it is, so that i can better communicate my ideas
https://frank-9976.github.io/untitled.html

[astral ginkgo]

Fun fact this is the same as the group of Morita Self-equivalences, i.e. progenerators A that give a morita Equivalence R->R

Such a progenerator is just an invertible module

[waxen fractal]

oh yea ShiN, this is an elaboration on that thing i talked about to you

[fierce steeple]

Lol maybe this is more of a french question than an algebra question but what's meant when the french say a quadratic form is "paire" ('even' i guess)?

Actually in this context the form is Z-valued so i guess it literally means even aha nvm

[woeful crane]

I think Pic(R) just sees the R-linear equivalences. The full automorphism class group of R-Mod is Pic(R) semidirect Aut(R).

[astral ginkgo]

Oh yes I was considering self-equivalences of R as an R-algebra

[chrome garnet]

let us take a canonical module K of R, which is not an ideal. is there a way to obtain a ring R' such that K is a canonical ideal? the idealization does not work (for R cm+local) bc then it becomes gorenstein and the canonical module is itself, not K

[worldly zealot]

having K be an ideal is equivalent to R_p being gorenstein for all minimal primes, i'm not aware of a canonical construction of a ring R' where that is satisfied but R' isn't gorenstein, and don't have a reason to expect one

[steady lance]

Wait I think I dont completely get the question. I have a ring R and its canonical module and the question is if I can find another ring with an ideal isomorphic as modules to R’s canonical module?

[chrome garnet]

yeah

of course you would need to adjust module structures and all but yes

[steady lance]

I feel like there might be a trivial answer to this but I can’t quite find it off the top of my head

[chrome garnet]

trivial may suffice, this came up while thinking about transfer of homological properties of amalgamations, which generally require us to glue through the canonical module/ideal

so i was thinking: if i have a module but not an ideal, maybe i might apply the theorem after some "idealization"

[near lantern]

Has a cohomology for a group G acting on a commutative monoid M been defined and/or studied?

@hallow bone IDK if you've seen something like this; the definition I have in mind basically replaces subgroups with congruences when taking kernels, which makes it pretty UA-pilled.

[hallow bone]

how would you take the cohomology then?

[near lantern]

That's what I'm hoping someone else has figured out

[hallow bone]

because i know how to go the other way; cohomology of monoids (its just the Quillen(?) cohomology of Beck-modules over monoids)

[near lantern]

I'm not sure, but I think this weakens G to a monoid, whereas I want to weaken the structure on M.

[lone jacinth]

Maybe relevant:

https://mathoverflow.net/a/1752/157483

[near lantern]

I do feel it's not too much oh a stretch because nonabelian cohomology (where you let M be a group instead of an abelian group) has been considered (at least in small degrees) and the cochains are pointed sets or stuff instead of abelian groups and coboundary homomorphisms are replaced by coboundary actions of the group C^0 on C^1 (which you can view as C^0 mapping to relations on C^1 and quotienting by the generated equivalence relation)

or something vaguely like that

[hallow bone]

hmm

commutative monoids are interesting because you can localize them very easily, right

[lone jacinth]

I guess you could just take the usually construction, and for each relation move all the negative terms to the other side and consider it a relation on monoids

[astral ginkgo]

I've never seen anything higher than n=2 considered in the nonabelian case

[near lantern]

That's exactly what I did. But I don't think cohomology can be defined as easily as the same thing. In fact I think each H^n(G, M) is actually a monoidal(?) category (in the usual case, it's a disjoint union of groupoids, one for each element of classical H^n, with automorphism group in bijection with Z_{n-1}).

Probably there's no point doing more theory beyond this before computing some examples.

[lone jacinth]

Do you have some other perspective motivating this? Or why do you say H^n(G, M) is a category?

I think the low dim results like H^1(G, M) = Hom(G, M) for trivial action and H^2(G, M) being extensions
M -> E -> G
should generalize to monoids with the usual construction

[near lantern]

Not a perspective, just trying to make definitions. The coboundary map goes from Functions(G^n, M) to Functions(G^{n+1}, M ⨯ M). In particular we can define the n-cocycles Z^n as the preimage of the identity (a congruence! on M, not an element) and these are cochains valued in M, but the coboundaries are cochains valued in M⨯M. The only way I could come up with to relate these is to view coboundaries as morphisms between cochains (using a condition which showed up in an application), which I think leads to a category I can call H^n (I have to check this).

[lone jacinth]

I was thinking more:

Define Z^n(G, M) to be
functions G^n -> M with
g1phi(g2, ...) + phi(g1, g2g3, ...) + ... = phi(g1g2, ...) + ...

And let H^n(G, M) be Z^n(G, M) modulo the relations
g1phi(g2, ...) + phi(g1, g2g3, ...) + ... = phi(g1g2, ...) + ...

Then this aggrees with usual group cohomology when M is an abelian group

[near lantern]

Oh wait

I was taking G a monoid at some point

That's why I got a category

Yes I think you are correct

[astral ginkgo]

See also

https://ncatlab.org/nlab/show/nonabelian+group+cohomology

Which seems to suggest this shouldn't work

[lone jacinth]

Shouldn't work in what way? Or what does that mean?

[astral ginkgo]

Ok nvm

I misunderstood part of the article

[near lantern]

I'm guessing I found a category which is some truncation of their higher groupoid (and it's not a groupoid because something became a monoid)

[near lantern]

Did the moderators actually ban the use of five-letter word's in this server?

words

Huh

[digital parcel]

words

??

i thought it was over

i have postgrad role

words

is "words" the only five-letter word not banned

[near lantern]

Banning "grou p" is really too much

especially in this channel 💀

I have postgraduate though

[hushed bone]

This is aprl fool joke

[near lantern]

NGL I barely know what that means and I definitely don't know how to do it.

[hushed bone]

Jeez its tuff

[digital parcel]

oh hell nah im taking the five letter penalty instead of having poop colored name

[hushed bone]

The fact this had no 5 letter word

So tuff

[digital parcel]

yea 😎 im cool like dat

[hallow bone]

this is so fax

omga my name has changed colour

[hexed tangle]

Hi, Let (R= \mathbb{Q}[x,y,z]) I am trying to find (\mathrm{Tor}^R_i(R/(x+y^2), R/(y+z`2))) I have a porjective resolution of the first argument: (0 \to R \overset{\cdot (x+y^2)}{\to} R \to R \to R/(x+y^2) \to 0 ), but I am stuck at finding the image and kernel of the non-zero morphism to get the homology

[broken turtle]

mh_le

[hexed tangle]

Correction: Hi, Let (R= \mathbb{Q}[x,y,z]) I am trying to find (\mathrm{Tor}^R_i(R/(x+y^2), R/(y+z^2))) I have a projective resolution of the first argument: (0 \to R \overset{\cdot (x+y^2)}{\to} R \to R/(x+y^2) \to 0 ), but I am stuck at finding the image and kernel of the non-zero morphism to get the homology after tensoring with ( R/(y+z^2)).

[broken turtle]

mh_le

[lone jacinth]

So you tensor and get
0 -> R/(y + z^2) -x+y^2-> R/(y + z^2)

Then the homology would be things in R/(y + z^2) annihilated by x+y^2

[hexed tangle]

thanks 🙂

[limpid horizon]

Does that follow maybee from induction and depthh drop by 1 afterr quotient by regular

[edgy pond]

I understand the prooff for the firstt cohomology but I don'ttt see the idea for generalizing it to higher cohomology groups. Couldd someone explain? (writingg this way bcuz of the wordleee ban)

[near lantern]

I believe this is not true? From what I remember, H^1, H^2 are zero for fd irreps of a ss Lie algebra but further H's need not be.

[edgy pond]

it made it to lie algebras by jacobson

i mean i canttt provide a counterexample nor provee it

[ornate atlas]

!wordlespoilers

[upper beacon]

Type ,iam 1488687375340273675 to get immunity from the wordle spoilers filter!

[edgy pond]

,iam 1488687375340273675

[limpid horizon]

,iam 1488687375340273675

[broken turtle]

Gave you the Not Very April selfrole.

Gave you the Not Very April selfrole.

[limpid horizon]

hello there

[lone jacinth]

So like in the proof of the Whitehead lemma you construct a Casimir operator Gamma whose trace is nonzero. Then as M is irreducible Gamma is just multiplication by a scalar, so it's an automorphism of M.

On the trivial module k, Gamma acts as 0 (since it's just a linear combination of products of things in L). So on
Ext^i(k, M) = H^i(L, M)
Gamma is both an automorphism and 0.

[limpid horizon]

Something like R[xi,xi^-1] is the same as Rxi right

[lone jacinth]

I guess if you write R[xi, xi^-1] one would assume xi isn't already an element of R

[digital parcel]

yes, gra de and de pth both drop by 1 when you mod by a non-unit nzd

should follow from some long ex act sequence of Ext or something

if that is even needed

stacks.math.columbia.edu/tag/090R

i can't send as a link because the filter

i suppose because it has h t t p s

[limpid horizon]

Yeah

[serene eagle]

,iam 1488687375340273675

[broken turtle]

Gave you the Not Very April selfrole.

[serene eagle]

im trying to make the character table for D4 but im kinda lost on how chi2, chi3, chi4 would work? i tried to use the commutatorgroup so you get that D4/{1,a²} would be isomorph to V4 but im not seeing how your characters 2,3,4 are defined in either V4 or C2 x C2

i didnt do chi5 yet because i wanted to do the linear ones first

[near lantern]

Does D4 have 4 or 8 elements?

[serene eagle]

8

i wanted to do this table and then you can lift it to D4

[near lantern]

OK. So your current goal is to find 4 1-d characters of C2 ⨯ C2?

[serene eagle]

yeah or K4

or V4 same thing

[near lantern]

Let's try it with C2 ⨯ C2. Each C2 has 2 characters and you can multiply any character for the 1st C2 with any character for the second C2 to get a character for the product (actually this will work for any finite product of finite groups, even for non-1-d characters).

Question: In general for a commutative ring homomorphism f: A → B, it is not true for I a radical <sub-A-bimodule> of A that the extension f(I) (I really mean B f(I)) to B is also radical. However, is this true in the special case that A, C are algebras over a <commutative ring with (0) maximal> with C reduced, and B = A (⨯) C?

[serene eagle]

i havent seen radicals

[near lantern]

Understandable. That was my question, for anyone in this channel to answer. My response to your question is on top of it.

[serene eagle]

oh wow im blind lol

[hallow bone]

commutative ring with 0 maximal

[hallow bone]

would f here be the natural map into B?

[worldly zealot]

i believe you can construct a counterexample if k is not perfect

[hallow bone]

wouldn't surprise me

extensions of ideals are yucky

[near lantern]

Ah, (A (⨯) C)/(im(I)) = A/I (⨯) C so naturally I precisely need C to be geometrically reduced.

[hallow bone]

what's that?

reduced when base changing to alg closure?

[near lantern]

IIRC yes

Def yes for <comm ring with 0 maximal> extensions

[hallow bone]

maybe in general you look over all geometric points

[near lantern]

By defn, it mean C remains reduced over any fld ext

[near lantern]

and if C is a fld ext (and perhaps for any C), this can be checked just over alg closure (or perfect closure)

[limpid horizon]

I want to compute depth of that k[Delta]xi ring. Is that the same as computing depth of k[lk{xi}] at its irrelevant ideal?

I also dont think I know exactly why in the standard graded case (so R0 is a field?) we can test CM by only checking CM at the irrelevant ideal

[honest summit]

Anyone reading dummit and foot abs alg book?

[short vine]

yeah

[restive vale]

I am trying to show that if g is semisimple as a Lie algebra and V is a nontrivial representation, then the cohomology groups of g are trivial. I got to the point where is am trying to show that the pullback of the Casimir of \kappa is trivial. Is this reasonable? Also how would one do this?

Ive used semisimplicity quite a bit to get to this point

[restive vale]

i have resolved my question

[chilly mist]

Anyone know how to factor these?

[near lantern]

Quadratic formula

[chilly mist]

That doesn’t help

[hallow bone]

i mean

it should

[chilly mist]

Bro I didn’t learn the quadratic formula

[digital parcel]

#prealg-and-algebra

[hushed bone]

lol

[ornate atlas]

If one knows a quadratic, they know the formula through a simple application of noggin

[hallow bone]

by that logic, if one knows a quintic, they know the formula through a trivial application of noggin

[ornate atlas]

Someone should try this

[hallow bone]

yeah, every finite group is solvable!

[lone jacinth]

If one knows what a polynomial is one knows all of Galois theory by a little application of noggin

[last talon]

Yeah I can solve quintic

[hallow bone]

prove it!

[ornate atlas]

x^5-1 is pretty easy, and all other quintics are like this but with at most, an easy to solve quadric in there

[last talon]

The solution to a quintic q is the set {x \in \mathbb C | q(x) = 0}

[silver goblet]

the only thing galois theory shows is that Galois can't solve 5th degree polynomials. however i am not a french teenager so i can do it

[last talon]

Yeah like
I’m english and the fundamental theorem of England is “the French are always wrong”

[silver goblet]

wait until you find out the Fundamental Theorem of America ([Washington, 1776])

[worldly zealot]

I love when random niche fields have a fundamental theorem

I saw “the fundamental theorem on simultaneous resolution”

[lone jacinth]

Like the horseshoe lemma?

[worldly zealot]

It’s like, if you have a one parameter deformation of a singularity you can resolve the whole thing at once in some way I don’t understand nor remember

[eager hound]

This is one of the main goals of algebraic geometry, I don’t think it’s niche

[worldly zealot]

The more you know

[bronze timber]

i wonder if its the french part or the teenager part that tripped galois up

[last talon]

French part

[shy horizon]

I have a set of Lie algebras $\overline{\mathbb{F}q}\otimes{\mathbb{F}_q}\mathfrak{g}$ over $\overline{\mathbb{F}_q}$ and I would like to classify their $\mathbb{F}_q$-forms using Galois descent, does anyone know how to do this in practice? I'm not familiar with the method.

[broken turtle]

Callie Wallie

[shy horizon]

This isn't an area I'm that interested in so it's not necessary I have all the rigorous details, just a rough sketch or program of how to do this.

This is for any characteristic $p>3$ and it's based on the Strade-Premet classification

[broken turtle]

Callie Wallie

[summer quest]

this is what Galois cohomology exists for!

I would look at some of the constructions of twisted forms of finite groups of Lie type for some closely related inspiration here (e.g. Steinberg groups, Suzuki groups, Ree groups...)

While some of the details are slightly different between the cases of algebraic groups and Lie algebras, the overall idea for the classification of twisted forms in terms of Galois descent is the same; in each case you have a classification of the split forms (e.g. in terms of Chevalley groups and their corresponding Dynkin classification) and then you can construct nontrivial twisted forms of these by considering automorphisms of the corresponding Dynkin diagrams

[solar turret]

is it correct?

[solar turret]

What is the best way to check given decomposition is minimal or not?

[lone jacinth]

Remove one ideal and see that the equality doesn't hold anymore I guess

[solar turret]

okay i got it, i think all are isolated ideal