advanced-algebra

[vague pawn]

group cohomology ig

[fierce steeple]

Yes but at least initially you wont use too much of it

Well like you can set up CFT using Tate cohomology sure but I sort of feel that wont be a very broad intro to it idk

But certainly a nice thing to learn as you about group coho in particilar yes

[vague pawn]

well I just finished an algebraic number theory course and my proffesor recommended Milne's CFT notes..

[fierce steeple]

Yes it is v cool

I would recommend them too

[vague pawn]

but he starts with cohomology

and it doesn't stick

[fierce steeple]

Ah okay

[ornate atlas]

Im looking forward to my cohomology class this semester, doing my UG thesis cohomology just kept coming up and seemed far less computationally awful than regular homology

[vague pawn]

Sadly I am in a situation where I can't take graduate level courses

[fierce steeple]

If you want to learn group coho it is helpful to learn some homological algebra more generally too, like I learnt it through reading some of Weibel (the basics up to derived functors, and then the group coho chapter)

That may be overkill for these purposes but it is good stuff to know

[fierce steeple]

Wdym by this

Are these like of spaces

[vague pawn]

I know someone who took a course on homological algebra

maybe he has some notes

[fierce steeple]

I am always a bit confused when people just say homology aha

[ornate atlas]

By "regular homology" I just mean simplicial/singular/celluar haha, it was more so in doing the combinatorial comalg sections of the report that cohomology kept showing up in proofs and it just seemed so nice to work with

[lone jacinth]

Here are some notes.

https://opperman.folk.ntnu.no/HomAlg.pdf

Not really building much towards group cohomology, but covers the absolute basics

[ornate atlas]

Im aware "regular" is a terrible way to phrase that though

[vague pawn]

Thanks!

[fierce steeple]

Ye sure

I guess I mean just the "... of spaces" bit

Though "ordinary homology" is also a technical term (which is computed by singular homology, or simplicial/cellular if possible)

[digital parcel]

Time to shill a little bit of AT again

[ornate atlas]

The course im taking is mostly about topological spaces I think, its called cohomology and poincare duality so Im guessing its just algtop, in my report it was also mostly about simplicial complexes because it was mostly applied to cominatorial comalg, but there was a chapter on homological algebra too so I guess it was there in a more general sense too

All of that is to say, not about groups or any weird alggeo nonense if thats what youre asking

Or will be taking, it hasnt started yet

[digital parcel]

“Weird alggeo nonsense” can we ban this guy

[fierce steeple]

Ye ofc i just mean when you say "homology" in this channel it could mean anything aha

[fierce steeple]

Yeah

[digital parcel]

Thanks

[ornate atlas]

No thats fair haha its broad

[fierce steeple]

Basic AT is just sheaf coho with the constant sheaf

[vague pawn]

I am both fascinated and afraid of AG

currently afraid more

[digital parcel]

W

[ornate atlas]

Theres an AG course here but I think it might just cover the same as what ive already done in my UG

[vague pawn]

Here there is a sheaf cohomology course

[digital parcel]

Lucky

[fierce steeple]

We had an AG course where there was definitely not enough time to cover the material we were "supposed" to

[digital parcel]

We dont have any ag course here

[fierce steeple]

I felt sorry for the lecturer

[vague pawn]

it doesn't assume any AG knowledge tho

[fierce steeple]

That is ok

[vague pawn]

and as I said earlier I can't participate

[digital parcel]

The closest we got was a topics course where the prof talked abt gromov witten shit and whatnot and only two people (one being his student) understood

[fierce steeple]

Sheaf coho appears elsewhere

[ornate atlas]

Mine was the same in my UG, and I was just thankful that large ammounts of the course were nonexaminable

[fierce steeple]

Lol fair ye

[fierce steeple]

Lol nice

[vague pawn]

that Gromov?

[digital parcel]

Yeah I just sat and smiled and nodded

[fierce steeple]

Who wants to learn GW...

[fierce steeple]

Only one gromov

[digital parcel]

Yeah

Enumerative geometry

[vague pawn]

I had a seminar on his theorem in geometric group theory

It showcases a very elementary proof of it

[fierce steeple]

Lol the wikipedia article is funny

Concept in string theory
"Specifically in symplectic top and AG"

[ornate atlas]

string theory is just maths anyway

[fierce steeple]

The applications of mathematics and the rigour of physics

[digital parcel]

It’s like looking at Alexeev’s publication list

A bunch of math journals with the occasional physics journal

[vague pawn]

the proof of Tao

[fierce steeple]

I am sure Gromov has many theorems aha

This is cool tho

I have lots of GGT friends lol do you do ggt

[fierce steeple]

Journal of High Energy Physics

[digital parcel]

Yea his arxiv is funny

math.AG, math.AG, phys.HEP, math.AG, …

[fierce steeple]

Real

[digital parcel]

Oops it’s just hep-th

No phys. in front

[fierce steeple]

Ah

Lol i mostly know gromov from group theory / metric spaces stuff tby

[fierce steeple]

Oh i misread sorry

[vague pawn]

Its time to go back to c++

[digital parcel]

Wut

[vague pawn]

double major moment

[digital parcel]

Doesnt alexeev have publications in HEP?

[vague pawn]

bye guys

[fierce steeple]

Ye dw i misread as you talking about Gromov lol

And was confused

[digital parcel]

That cat lowkey chill asf

Ah

[fierce steeple]

Lol

Why did my image get thumbs up

[digital parcel]

Someone agrees

[fierce steeple]

Recently i have seen a lot of papers have pictures on their first page

I think this image should be in a paper

[digital parcel]

Who is doing that

The only one I can think of is Landesman’s bach thesis

[fierce steeple]

In homotopy theory there are lots of such papers

[digital parcel]

Yea understandable

If I did homotopy theory I’d prob go insane too

[fierce steeple]

E.g. https://arxiv.org/abs/2210.17364, https://arxiv.org/abs/2402.00960, https://arxiv.org/abs/2507.00309

Lol

[digital parcel]

Ok but including rene magritte is so based

[fierce steeple]

Yeah i like this picture

[digital parcel]

Oh yeah for sure

[fierce steeple]

Classic

[lethal scarab]

if anyone could help itd be great!

[distant ravine]

Hi, friends. Is this the correct chat to talk about Commutative Algebra?

[spice idol]

yes

[urban granite]

yep

!da2a

[upper beacon]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[fierce steeple]

I wouldn't say that is really da2a rather than being unsure aha

[spice idol]

though commutative algebra is specifically mentioned in the description here lmao

[ornate atlas]

No one reads channel descriptions, they’re too hard to find and it’s even worse on mobile

[lone jacinth]

I feel like clicking the channel name to get info about the channel isn't "too hard", but I suppose it could have been even easier

At least if you know channel descriptions exist there aren't really any other places to look

[ornate atlas]

It’s pretty bad on mobile at least, which is how I use discord 99% of the time, maybe it’s more clear on desktop than I remembered

[lone jacinth]

I was talking about mobile

[lone jacinth]

Not sure how it is on desktop

[spice idol]

i think on desktop its partially visible

[ornate atlas]

I mean it’s not actively hard to find, if you know where it is it’s easy, but I don’t think it’s at all obvious that it exists or that that’s where it exists, and I feel like the consistent amount of confusion is testament to that

Or at least to the fact that people are unwilling if not unable

[lone jacinth]

Yeah, I think people probably don't know it exists, and it certainly could be more visible.

But I also think that unless it was annoyingly in your face I don't think people would find it if they already assumed it didn't exist.

You already have to jump through hoops to even post in these channels though, so could add a "i know where descriptions are" checkmark to that.

But people also post high school algebra here, so not sure what it would do

[spice idol]

they see "advanced" and because theyre taking "advanced math" in highschool they think its for them

or maybe they just think its advanced because theyre struggling with it

[urban granite]

me when #algebraic-geometry but HS geometry problems

[distant ravine]

Opps... Sorry. I'm kinda new at discord. Didn't know channels (which i presume this is) had descriptions.

[limpid horizon]

For the map i have im having trouble showing its well defined on cosets of H^i

[fierce steeple]

Wdym cosets

Or do you just mean that the chain-level map you write down doesn't clearly pass to (co)homology

[spice idol]

yeah, thats alright though!

[limpid horizon]

Chain level map? Idk man im just writing the map the bare bones way

[limpid horizon]

The map from H^i(delta, k) i have is just taking f in there to the map H_i(delta,k)->k sending c0(x)k to f(c0)

Idk, just what i tried doing and now was seeing if everything is well defined and stuff

[fierce steeple]

As jn a map on chain complexes

And then ig you mean passing to quotients ye

[limpid horizon]

I was stuck on showing that it sends 0 in H^i to 0

[vapid axle]

Can somebody help me see why q n S = \varnothing?

[limpid horizon]

A proper ideal in S-1B doesn’t have any units so contraction in B can’t share anything with S

Im pretty sure its just that

[vapid axle]

Ye I was just confused with the second part of your sentence

I think I managed to figure it out tho: If $\mathfrak{q} \cap S \neq \varnothing$, then $\mathfrak{q}^e = (1)$ (with respect to the inclusion $h: B \to S^{-1} B$). Since $\mathfrak{q} = h^{-1}(\mathfrak{m})$, we have $\mathfrak{m} \supseteq h^{-1}(\mathfrak{m})^e = (1)$, a contradiction.

[broken turtle]

okeyokay

[foggy galleon]

Does semi-simple mean that the elements of the group are diagonalizable over the algebraic closure?

[bold scroll]

Why is the localization K[x]_(x) not a finitely generated K-algebra?

[fierce steeple]

Pls do not cross post like this but ye

Answered in #algebraic-geometry

[broken turtle]

Battye

[weak lodge]

See pp. 4--6 here: https://people.math.osu.edu/gautam.42/Sp23/Notes/RT02.pdf

[golden osprey]

Sachin's notes are so so so good

I wish I took rep theory with him last spring

[near lantern]

There is a definition of semisimple algebraic group in general, but that seems to be very different from what is given here (since e.g. SL_n is semisimple and a torus is not, whereas Cartan subgroups are much more like tori (as I understand it, they are the same as maximal tori for a semisimple algebraic group over an algebraically closed field)).

If V is a representation of a group G over ℤ (finite-rank free as a ℤ-module) which is irreducible over the algebraic closures of ℚ and ℤ/pℤ for all primes p except for those in a finite set S = {p1, ..., pk}, is it true that any non-zero subrepresentation of V over ℤ or V over ℤ[1/(p1...pk)] must contain (p1...pk)^n V for some non-negative integer n?

[hasty flume]

so my guess is that what they mean is that every element of the cartan subalgebra is semisimple in the sense that its adjoint is semisimple as an endomorphism of the lie algebra g

this is equivalent to the usual definition of a cartan subalgebra

the standard definition of a semisimple algebraic group is one for which every closed connected normal subgroup which is abelian is actually trivial

so in some sense this is as far as possible from being commutative as you can get

so obviously they dont mean that

[west spruce]

but the Cartan subalgebra is nilpotent? so for every element x in the Cartan, ad(x) is nilpotent? (rather than semisimple?)

[hasty flume]

semisimple is stronger than nilpotent, no?

uh actually this doesnt make sense

if we embed our lie algebra in M_n then the classic example of ad(x) being nilpotent is x upper triangularizable whereas the classic example of a cartan subalgebra would be diagonal matrices

no nvm this is right. the cartan subalgebra being nilpotent implies that ad(x) is nilpotent as an operator h -> h

this is also true in the other standard definition, a maximal abelian subalgebra whose adjoint reps are semisimple on g -> g

since its abelian

maximal + semisimplicity is actually what gives you that h is self normalizing, not that its nilpotent

[limpid horizon]

What is the defn of hilbert series for a module that is not Z-graded?

In the example Im looking at the module is Z^n-graded

[lone jacinth]

I'm guessing it would just be a power series in several variables

That fits with what I see with my Google-fu aswell

[ornate atlas]

Definition 8.14 of Miller and Strumfels

[vapid axle]

Why should I give a single shit about colimits

[west spruce]

sorry, yes that's right!

[west spruce]

yes good point

thank you! :)

[lone jacinth]

Direct sums, cokernels, and nested unions are all colimits.

You might care about those.

[ornate atlas]

CW complexes are colimits but that’s maybe tenuous

[lone jacinth]

Localizations wrt a set is the colimit of localizations wrt individual elements

[fierce steeple]

Not even necessarily nested i guess aha

[lone jacinth]

Well, if there not nested it gets a little complicated. And the union of groups isn't usually a group and so on

[fierce steeple]

Other examoles in this vein are Attaching spaces (in topology) are colimits, taking coinvariants/orbits of a group action is a colimit, uhh many constructions tbh

[fierce steeple]

Sure yeah sorry

I guess depends on ur cat etc

Like usually you do want it nested (or at least filtered) because you want to talk about transition maps lol

[ornate atlas]

I wasn’t aware of the orbits thing, that’s pretty cool, maybe it’s obvious but honestly limits are a bit of a blind spot in my category theory knowledge, I only have a kinda vague understanding of them and I’ve not done anything with them

[fierce steeple]

Yeah this is a nice thing. Have you seen the category BG for a (discrete) group G

[lone jacinth]

It's the coequilizer of the actions of all g in G

[ornate atlas]

Is that the category of classifying spaces? I’ve vaguely heard about them but I don’t know anything about them

[fierce steeple]

There is a strong link to classifying spaces but here i mean an actual category with just one object * where the maps * -> * are just elements of G

And like you compose maps according to composition

[digital parcel]

Roberts’ book “Multiplicities and Chern Classes in Local Algebra” does a rigorous treatment of the Hilbert polynomial for Z^n graded rings

[fierce steeple]

So a functor BG -> C is a choice of object c of C along with maps g: c -> c that compose as they should lol

Anyway like if you have a set X with a G-action, encoded as a fjnctor BG -> Set, then the colimit is X/G and the limit is X^G

This is equivalent to what jagr said

[ornate atlas]

Ahhh I see that’s pretty cool

[fierce steeple]

But this is nice like it works quite generally

[ornate atlas]

I’m honestly looking forward to taking an actual category theory course this year and not just having the random scattering of knowledge that I have currently

[fierce steeple]

If you know about group coho tjen uh like this is why M |-> M^G is left exact but M |-> M/G is right exact

[fierce steeple]

Lol nice yeah

[ornate atlas]

I actually don’t know about group cohomology, I went to a talk about it but it was a weird topos theory first approach that even the chair in noncom algebra at my UG didn’t follow lol

[fierce steeple]

The link here is that it is possible to go from a category to a space basically

[fierce steeple]

Bruh

[ornate atlas]

I think I’m one of like 10 people who’s been formally introduced to n-categories before even like Yoneda lol

[ornate atlas]

Yeah it was for someone’s masters thesis defence, it was uhh… hard to follow

[fierce steeple]

Lol

[fierce steeple]

Yoneda learn yoneda

[ornate atlas]

Yeah

You can certainly try

That is pretty cool, I am not disappointed

[fierce steeple]

Thanks

[digital parcel]

What’s that one joke about yoneda and puttjng anelephant in a fridge and something

[fierce steeple]

I sent a photo of a cat for anyone to whom this looks odd

[fierce steeple]

LOL

I made a joke can i trial it with you

[digital parcel]

Call me corn the way I’m all ears

[fierce steeple]

Einstein said insanity is doing the same thing over and over again and expecting different results

But if I roll a die 10 times in a row...

[digital parcel]

Wait is that the punchline

[fierce steeple]

Yeah

[digital parcel]

I dont get it 😭

[fierce steeple]

Lol

If you roll a die 10 times and expect different results, are you insane

[digital parcel]

I’d be insane too to roll a die 10 times

Yeah

[fierce steeple]

Ok true

[digital parcel]

I like it

W joke

[fierce steeple]

Ok

Thanks

Best jokes are ones people dont understand

[digital parcel]

Yep that's what ppl are always saying

[ornate atlas]

First rule of improv

[digital parcel]

Maybe I have a bias against rolling die

Last night I played a card game w friends and man all that die rolling got tiring

[fierce steeple]

Dont talk about improv

[digital parcel]

Me and die our relationship is kinda iffy rn...

[fierce steeple]

I once sat on a piece of bread for ten days in a row

I was on a roll

[digital parcel]

W

[lone jacinth]

When you're stuck proving some category theory, youneedalemma

[fierce steeple]

Category theory has proofs?

Joking.

[digital parcel]

no man just one more long exact sequence from a derived functor man please just one more 3d diagram please

[fierce steeple]

Smh that is algebra not cat theory

[digital parcel]

there's cat theory outside of algebra?

zzzzzz

[fierce steeple]

Real

[digital parcel]

at some point i should learn more cat theory outside of just what I've been doing with like Grp, Ring, R-Mod, and now Sh and Sch

omg lurie's Higher Topos Theory is calling me

i must pick up the phone

[fierce steeple]

Nice

[spice idol]

third sanctuary ringtone

[digital parcel]

hmm i like lurie's writing style

it's nice

[fierce steeple]

It is

[astral maple]

Let V be a complex cuspidal irreducible representation of GL_2(F_p) with p>3 prime. I can show that V can be realized over K := Q(mu_{p^2-1}) because it can be explicitly constructed as the virtual difference of the difference of two representations that are induced by one that can be defined over K. But is there a more direct proof of this fact, for instance involving the properties of Schur indices?

[digital parcel]

i talked to a prof the other day and he mentioned something about quiver varieties

so i looked it up and i found this on nlab

why is it called a "walking" quiver

[plucky arch]

the name comes from the walking moustache

[digital parcel]

forgive my naivete but i looked up walking moustache and didn't find anything on it

other than a 2017 song

[plucky arch]

https://ncatlab.org/nlab/show/walking+structure

[digital parcel]

lol

hmm

[plucky arch]

the idea is that you might have someone whose moustache is so prominent that they might as well be a walking moustache

[digital parcel]

😭

the "walking adjunction" "perches" on the "2-category"

[plucky arch]

mhm

[digital parcel]

so in the "walking quiver" is the minimal amount of structure literally that it's just "vertices" + "edges" + "morphisms" + "identity"

like this is about as basic as one can describe a graph

oh haha slick definition like a slick moustache

good one nlab

[plucky arch]

mhm

[digital parcel]

ahh okay

lol

[mental escarp]

You’d think if someone was actually risking confusing other people with the term “walking quiver” they’d at least be suggesting rep theory was relevant to those things.

[neat nexus]

Page 85 of “elements of the representation theory of associated algebras” by Assem. it has notation “rad^{2}(P(a))”. P(a) is a right A-module where A is a K-algebra over a field K. How do we even define rad^{2}(M) for a right A module M under this context? Prior to that the author defined rad^{2}(A) to be (rad(A))^{2}, so here I can’t see how it’s defined

[distant ravine]

hi! Anyone knows a paper about quotients of polynomial rings by elementary symmetric polynomials? I want to make a study about these rings

[lone jacinth]

radM = intersection of maximal submodules.

rad^2 M = rad(rad(M))

For a finite dimensional algebra and finitely generated module
rad M = radA * M

[neat nexus]

Thank you. Makes sense, P(a) in the context is indeed finitely generated

[lone jacinth]

You don't actually need finitely generated either. Just A being finite dimensional (or even just perfect) should do.

[neat nexus]

Oh. Thanks again. I was torn whether it was (rad)^2 or rad(rad()), so turns out they are the same throughout the whole book

[old willow]

might be a better question for this channel?

[lone jacinth]

So the first thing you should figure out is how this thing should be defined in non-elementary tensors.

For this it might be convenient to go through Bq.

You want
Bq'(a(x)m, b(x)n) = ab Bq'(1(x)m, 1(x)n) = ab Bq(m, n)

And you also want
Bq'(a(x)m, b(x)n) = q'(a(x)m + b(x)n) + ....

Then you can solve for q'(a(x)m + b(x)n).

After that you just need to check that the relations of the tensor product are satisfied. The
q'(a' a (x) m) = q'(a' (x) am)
should be clear.
The
q'( a (x) (m+n)) = q'( a (x) m + a (x) n)
would be more work

[woeful widget]

Hi

[limpid horizon]

Hey Kenny. Love ya man.

[spice idol]

bro is the pronoun collector

[vapid axle]

When we say that it suffices to show that A is a local ring, why is this true? The proposition says that if f: M -> N is an A-module homomorphism, then f is injective if and only if f_p: M_p -> N_p is injective for any prime ideal p. In applying this proposition, it suffices to show that \phi_p: F_p -> F_p is injective, where F_p is the localization of F (and not A, as stated)

[hushed bone]

F_p is a rank n free module over A_p because localization distributes over direct sums

So if you knew the result for free modules over local rings you know it for F_p for all p and then you apply the proposition

[vapid axle]

What do you mean by localization distributes over direct sums?

[hushed bone]

I mean exactly what it sounds like lol

[vapid axle]

That (A^n)_p is isomorphic to (A_p)^n?

[hushed bone]

Yes but more general than that

[vapid axle]

Would you mind elaborating?

[hushed bone]

(M + N)_p = M_p + N_p

[vapid axle]

Ahh right

Thanks bud

[hushed bone]

I’m not your bud.

I’m chmonkey

[vapid axle]

How does F being a flat A module show that the sequence is exact?

Did they mean k is a flat A-module?

Oh I see, here F is F_p

[lone jacinth]

In general you'll get an exact sequence
Tor(k, F) -> k(x)N -> k(x)F -> k(x)F -> 0

Since F is flat Tor(k, F) is 0

[vapid axle]

I don't need to know about Tor tho right for this problem

[lone jacinth]

I mean, probably not, but that's how you get the exact sequence described

[vapid axle]

Is there another way to see it without using Tor? Because I don't think it's needed for this problem

[lone jacinth]

If A is Noetherian you can use that any surjective map F -> F is an isomorphism

[vapid axle]

Maybe I can prove the following? If A is a commutative ring with localization A_p then the residue field k = A_p/m is a flat k-module

Idk

[lone jacinth]

Well that is very very not true

[vapid axle]

But what if A is not Noetherian

[lone jacinth]

Then one would need a different proof I guess

[vapid axle]

I guess that's part of the exercise

Okay so we want to show that if \ker \phi_p -> F_p is injective then tensoring with k preserves injectivity...

[fierce steeple]

well it is very very true as stated

[vapid axle]

You love reacting to your own messages

[fierce steeple]

Lol habit from nG

[lone jacinth]

Yeah, it's not a flat A-module.

Being a flat k-module is pretty easy

[fierce steeple]

But no i mean surely okey meant flat A module

But yeah in terms of vibes like as soon as you start modding stuff out, flatness claims should start ringing alarm bells

[vapid axle]

I give up

[fierce steeple]

Which bit are you stuck on like

Ah okay yeah I see the issue sorry

[lone jacinth]

Kinda weird to know about flatness, but not Tor though isn't it?

[vapid axle]

I'll probably return, but yeah anyways Tor wasn't introduced in the text

[astral maple]

Ah yes, the point is that if 0 -> X -> Y -> Z -> 0 is an exact sequence of A-modules with Z flat, then it remains exact after tensoring with any A-module M.

I think the way you prove this is first to do it when M is free (which is trivial).

Then you write M as the quotient of a free A-module by some submodule (recall that tensoring with anything preserves cokernels), and deduce the conclusion from some diagram-chasing.

[digital parcel]

Exercises 24-26 in Chapter 2

that being said, they say "it will be assumed that the reader is familiar with the definition and basic properties of the Tor functor"

so if you haven't done those exercises (which you should, since they relate Tor and flatness), go do those now

[lone jacinth]

I mean you can take a projective presentation of M, then do the snake lemma. But that's just proving the long exact sequence in Tor

[fierce steeple]

Here F is even a free module so probably makes it easier

[digital parcel]

for the sake of those exercises, you can leave projective resolutions and whatnot and just read about the long exact sequence of tor

[astral maple]

Yes, but it feels rather harder to do when you’ve never seen it.

[lone jacinth]

Yeah, F is free so the sequence splits.

[fierce steeple]

Oh yeah true lmao

[lone jacinth]

Well yeah it's hard, that's kinda my point.

[fierce steeple]

Tbh this means this is a good question!

like if you need extra stuff beyond text etc

[lone jacinth]

You're reproving a lot of machinery

[fierce steeple]

I do find these Atiyah-Macdonald "hints" funny though

Like if you know about Tor then haven't they given you the entire proof

[lone jacinth]

Sometimes reading a proof is exercise enough

[fierce steeple]

i guess they want you to add a few more details but here there don't seem any to add until the last sentence

[fierce steeple]

True

Maths is hard and I am confused by smth now

[plucky arch]

[digital parcel]

i mean you actually dont even need that 0 on the left in the hint right?

you get the exact sequence k (x) N -> k (x) F -> k (x) F -> 0 since tensor is right exact

and then anyway you show that k (x) N is zero

[vapid axle]

Y'all are wizards

[fierce steeple]

Pretty sure you do lol. The kernel of phi (x) 1 vanishes, and then you knwo the kernel is k (x) N, hence = 0

otherwise all we are saying is im(k (x) N -> k (x) F) vanishes

[limpid horizon]

a ring R being cohen macaulay is the same as it being CM as an R-module right

[digital parcel]

hmm yeah true

[fierce steeple]

https://stacks.math.columbia.edu/tag/00N7 defines as such

[lone jacinth]

And equivalent definition could be the existence of a bimodule that induces an equivalence between modules of finite projective dimension and modules of finite injective dimension.

This definition then also works for non-commutative rings

(though you need to assume Noetherian)

Fun stuff
https://arxiv.org/abs/2409.05603

[fierce steeple]

Tbh I am impressed there is a non-commutative version of this

[lone jacinth]

I find it fun how many of these things like regular, Gorenstein, Cohen-Macaulay were first defined in the commutative setting with tools that don't at all translate to the noncommutative world, but all turn out to be equivalent to some homological condition which makes sense in any abelian category

[digital parcel]

auslander cohen macaulay gorenstein noetherian burch ulrich hilbert lefschetz

betti

[lone jacinth]

Listing of your contact list are we?

[digital parcel]

i can't escape these names

[digital parcel]

if i was in contact with lefschetz that would be crazy

[lone jacinth]

I'll get the Ouija board

[digital parcel]

if you get a ouija board i wanna talk to grothendieck

i'll add \mathcal{O} and \mathcal{F} to the ouija board for him

[limpid horizon]

whats the significance of this part

i dont know why H^i(C_m) = H^i(C)_m as well

yes, im still on this. its been a long distracting summer and i want out but yea anyway

soon imma be off this server lol

[spice idol]

noooo kian dont gooo

im gonna miss the best canadian

[lone jacinth]

Localization is exact

[limpid horizon]

Lol yea idk maybe id want to come by if i self study math in the future but i think with the time
Ive had studying this stuff and with $500 in my bank account i dont think imma be continuing with math atm

[spice idol]

that is fair

i guess everyone hits a wall at some point with math

[limpid horizon]

Yeah

Yeah and like for me i dont think the effort required is worth it anymore considering the other aspects of life id have to sacrifice for it

[unborn rampart]

will you stay if I paypal you $5?

[limpid horizon]

Enticing … 🤔

[spare solstice]

When does a (modular) representation pi of a finite group G embed into Ind Rep pi for some subgroup of G

Like what are some necessary, and what are some sufficient conditions

obviously should be sufficient that the char doesn't divide the order of the group

[cloud karma]

Am I being dumb here?
By definition $m_\mu = x^{\mu} v^{-1}{\mu} \in W^{\vee}$ where $v{\mu}$ is the shortest element of $W_0$ such that $v_{\mu} \mu = \mu_{-}$ where $\mu_{-}$ is the antidominant weight here (and vice versa).
For a $w in W_0$ we have that $w x^{\mu} = x^{\mu} w$ so in the second inequality we should have $x^{-v_{\mu} \mu} v_{\mu}$ but there they have $$x^{-v_{\mu} \mu}$

[broken turtle]

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

[cloud karma]

so many mumumumus

[cloud karma]

nvm i got it

[lone jacinth]

If the index is relatively prime to the characteristic then pi is a direct summand of
Ind Res pi.

But for just embedding, the group algebra should be self injective, so just inducing from the trivial subgroup should give you an injective cogenerator you can embed into.

[last void]

does anyone have a reference for the fact that the irreducible modular representations of a cyclic group are uniquely classified by dimension

like for C_p there one of dim n for all 1\leq n \leq p

im finding it referred to as a well known result a lot but nowhere actually cited

found one nvm!

https://www-users.cse.umn.edu/~webb/RepBook/RepBookLatex.pdf chapter 6 if anyone is interested

[void plank]

Suppose I have a free-vector V space over some field k.

Suppose I then endow V with the action of a group G.

Is there any relationship between this and doing extension of scalars on V by k[G] or another change of ring operation?

[lone jacinth]

An easy proof:
kCp = k[x]/(x^p) which is a principal ideal ring, so the indecomposable representations are exactly
k[x]/(x^i) for 1<=i<=p

[lone jacinth]

I mean, if V is a vector space with an action by G, then it is a kG module.

You can think about V without the action as the restriction of scalars along k -> kG.

Extension of scalars would in general change the underlying vector space

[void plank]

So would V with the group action would be a free kG-module of the same rank via extension along k ---> kG?

[lone jacinth]

Extension of scalars of k^n is (kG)^n yes

But that doesn't mean k^n is a free kG module, so maybe I don't understand what you're asking

[void plank]

Ah ok, I was wondering about this and you answered it

I also ask because in my situation, I take a tensor product over k of vector spaces V and W with compatible kG actions and then pass to coinvariants. This gives V (x)_kG W

However, it wasn't clear to me how V (x)_kG W then relates to V (x)_k G.

(I'm actually dealing with complexes of vector spaces with compatible G-actions and then take a tensor product of chain complexes before passing to coinvariants, but it boils down to the same issue)

[spare solstice]

Is there any like sufficient conditions for the group algebra to be self injective?

[lone jacinth]

The group being finite

[spare solstice]

Ah

lmao, thanks

[lone jacinth]

So
V (x)_kG W
should be the coinvariants of
V (x)_k W
if that's what you're asking...

[void plank]

Yes for sure

But I guess I can't say anything general about how the rank of V (x)_kG W as an abelian group relates to the dimension of V (x)_k W right?

[lone jacinth]

No, it depends on the kG-module structure.

[void plank]

Ok thanks!

[void plank]

Are there conditions on G that guarantee V (x)_kG W is free?

[lone jacinth]

All vector spaces are free.

If you're asking about it being free as a kG module then you need some kind of G-module structure on it

The most reasonable would be the trivial module structure I guess, in which case it's free iff G=(1) or it equals 0

[stray grail]

Does “all vector spaces being free” rely on the axiom of choice?

[ornate atlas]

I believe so yeah, I think the usual proof uses Zorns which is of course equivalent

[lone jacinth]

It is in fact equivalent to choice over ZF

[ornate atlas]

Oh really? That’s pretty cool

[foggy galleon]

the proof is readable actually, I recommend it

[ornate atlas]

I wonder why none of my classes ever mentioned that, they seemed constantly scared of discussing set theory lol

[rare walrus]

As others have said, yes. But you shouldn't see this as discrediting it.

[ornate atlas]

And then the one set theory course at my UG was on ETCS so

[lone jacinth]

I mean, unless your doing set theory or logic it isn't really relevant I guess

[ornate atlas]

Yeah I mean I got by just fine without it, but it seems like it’d be a cool fact to mention

[ornate atlas]

Like purely for nothing other than a neat fact

[foggy galleon]

what's etcs lol

[ornate atlas]

Elementary theory of the category of sets lol

[foggy galleon]

imo these type of questions are a good entry point to set theory and the axiom of choice. Actual courses in set theory should speed run to forcing 🤓 👆

[void plank]

Yeah, I was asking about it as a kG module.

Thanks!!

[limpid horizon]

As (x)A M and Ms are isomorphic as As-modules but also as A-modules right

if thats true is it true that R->S and if M and N iso as S-modules they are iso as R-modules?

[fierce steeple]

Yes.

This is sort of like

Restriction of scalars is a functor, like it is compatible with maps etc

So all is good

[void plank]

Yeah, what Prismatic said. restriction of scalars is an exact functor

[limpid horizon]

the kernel and image of M->N after localizing at S is (ker f)s and (im f)s? from 0->ker f -> M -> im f -> 0 and localization being exact?

[fierce steeple]

Don't even need exact here

[fierce steeple]

Hm is this the same as your earlier question

I assumed your original question was saying that you have S-modules M and N and are restricting them along R -> S

[stray grail]

As I study more I have gotten increasingly skeptical about the axiom of choice. Richard Borcherds says he typically accepts choice “because he is lazy” (as he said in one video where he was doing something with Noetherian rings I believe), and of course there are paradoxes with/without choice. Out of curiosity, where do you all tend to stand?

[ornate atlas]

I think you’ll struggle to find many algebraists who reject choice

I also kinda reject the notion that choice leads to paradoxes, it’s all logically consistent, it’s just that infinites are weird and not very intuitive

[plucky arch]

my take is that choice is true and it's useful to avoid it if you don't need it

[stray grail]

Yeah I guess that is poor phrasing on my end haha.

[ornate atlas]

I mean it’s a thing that people, say, I just generally think the answer to the supposed issues with choice is that infinites are weird and not particularly intuitive

[mellow night]

Hey there everyone.

[lone jacinth]

I guess I don't really care about choice per say, but I wouldn't want to work with a ring that doesn't have any maximal ideals.

If they all do then I don't have to think about it

[stray grail]

Yeah I agree, infinites can get quite weird and unintuitive.

[mellow night]

Have you ever thought of anything really beautiful and high level at the same time in terms of algebra?

[ornate atlas]

I don’t really know how to answer that

[mellow night]

Is there something that you really like in advanced algebra that is high level?

Is that clear now?

[lone jacinth]

Dynkin diagrams and how there's connection between
Root systems
Simple lie algebras
Representation finite quivers (species)
Finite subgroups of SL2
Other stuff...(?)

[mellow night]

For me, it is Gr_G, centralizers, modular stacks, analytic stacks...

[digital parcel]

analytic stacks
nG inbound

[spice idol]

imagine modern algebraic geometry without choice

[digital parcel]

#algebraic-geometry message

I asked about that yesterday lol

[short vine]

Suppose A is isomorphic to B. Is Ext(A,C) isomorphic to Ext(B,C)?

[rare walrus]

Yes

[short vine]

I meant Ext^1

What is the isomorphism

[rare walrus]

It doesn't matter which level you meant. They're all functors

You should have a notion of, given a map f : A → B, a corresponding map Ext(f, C)

This is the isomorphism

[spice idol]

i mean, i guess you can just use the locale of the opposite lattice of the lattice of radical ideals as a base topological space

[short vine]

Ext(f,C)?

[rare walrus]

This is independent of your construction.

Yes, maybe you call it f* or something

[short vine]

What is Ext(f,C)?

f is a map right

[rare walrus]

How do you construct Ext?

If you answer that I can answer the question

[digital parcel]

If you want to use the projective resolution construction then you can just see that the projective resolutions of A and B are homotopic

[short vine]

Well Ext(A,B) is the set of equivalence classes of short exact sequences 0-> A -> C -> B -> 0

[rare walrus]

OK

Give me a minute

[lone jacinth]

It should be short exact sequences
0 -> B -> C -> A -> 0
and the map is given by taking the pullback along
D -> A

Giving you
0 -> B -> C -> D -> 0

[short vine]

What is D here

[rare walrus]

Given any isomorphism, I have this

[lone jacinth]

The domain of whatever map you're considering

[rare walrus]

In particular, this gives us a bijection between Ext(A, C) and Ext(B, C)

And you can check it satisfies the addition

[lone jacinth]

This would be
Ext(C, f): Ext(C, A) -> Ext(C, B)

[short vine]

The is the issue I was worried about, since I’m not sure if this can lift to a morphism of E

Oh wait

[rare walrus]

Yeah I realised after they got the thing the wrong way around. Either way, the same works for the reverse.

[short vine]

I think now I understand

Thx guys

[limpid horizon]

Hey im trying to figure out that implication. Is it something like if Supp M = {m} then M is isomorphic to Mm as R-modules?

[void plank]

Let's say I have an additive functor F between R-mod and S-mod, where R is commutative.

Suppose I have two chain complexes C and D in Ch(R-Mod).

Are there conditions on F under which I can say that F(C (x)_R D) = F(C) (x)_S F(D) (i.e. F preserves tensor product of chain complexes)?

[fierce steeple]

I believe like this is equivalent to requiring F to be like this on just modules but the question is kinda hopelessly general

But note this is structure on F (being monoidal) rather than a mere condition

[past cove]

I'm p sure they explain it in the next section

tho maybe I'm wrong

I can't really read the notation that well

[void plank]

Ah fair.

Are there situations in algebra where this happens, i.e. where F preserves monoidal structure?

[lone jacinth]

Extension of scalars

I guess there probably isn't many other examples since
F(R (x) M) = F(R) (x) F(M) is pretty close to requiring F(R) = S, which is pretty close to just being extension of scalars

[void plank]

That's fair, thanks!

[void plank]

Although I think I have to be careful when extending scalars on the right or left here

[lone jacinth]

Well I would assume R and S is commutative, otherwise the tensoring doesn't really make sense

[void plank]

Yeah, I was trying to relax the commutativity requirements as much as I could in my original formulation

[lone jacinth]

Like if R isn't commutative then
M (x) N
only make sense if M is a right module and N is a left module. But then F(M) and F(N) don't both make sense

[void plank]

Yeah, that's why I had R is commutative in my original question

But you're right, I guess I do need S to be commutative as well

[limpid horizon]

If R is noetherian, M an R-module with Supp M = {m} a maximal ideal, is it true that M isomorphic to Mm as R-modules?

[fierce steeple]

There's a canonical map M -> M_m of R-modules and whether this is an isomorphism can be tested locally

[limpid horizon]

Can be tested locally? How does that work

[fierce steeple]

Take the kernel or cokernel and you want to show they vanish

which can be tested locally

[limpid horizon]

Hm yea

[fierce steeple]

I guess I am also implicitly using that forming cokernels/kernels commutes with localisation (since localisation is exact)

But yeah this is a sort of standard fact in comm alg that's very useful, like all these things like being injective or surjective or an iso or 0 can be tested locally hehe

or even on maximal ideals usually

[lone jacinth]

If M is finitely generated, then Supp M are just the prime ideals containing Ann(M), so then R/Ann(M) is local and localizing doesn't change M

[limpid horizon]

I dont know if I understand the “and localizing doesn’t change M” conclusion

So M is an R/Ann(M)-module and R/Ann(M) is a local ring

[lone jacinth]

For a local ring, localization at the maximal ideal is just the identity.

[charred carbon]

if anyone has read Fulton and Harris can you explain to me what this notation means with the t

[lone jacinth]

I haven't read Fulton Harris, but the formula for the dual representation is the transpose of rho(g^-1).

So then the t would mean transpose I guess

[charred carbon]

aight thanks

that makes sense

[buoyant harbor]

I’m in my abstract algebra class and it’s so boring

[spice idol]

bro came to the algebra channel to complain about how boring algebra is

[ornate atlas]

I’m praying that my new uni doesn’t schedule all of the algebra courses at 9am again. For some reason my UG had all the algebra courses booked out for the morning, and as much as I love it, I don’t need to be hearing about integral closure from 9am till 1pm

[fierce steeple]

Easy solution is not going to lectures etc

[ornate atlas]

Well by and large I didn’t, but the one lecturer who I really liked was always during the 9am slot and she taught very small course so I kinda had to be there for those

And if you’re in at 9 anyway…

[fierce steeple]

Yeahh

[digital parcel]

I dont want to hear about that for more than 30 minutes tbh

[ornate atlas]

Yeah I mean the classes were only an hour it wouldn’t just be that, but I’d genuinely have ring and rep into comalg into algtop into group theory, it was a silly time table lol

And second semester wasn’t quite as bad, but it was similar, still had alggeo at 9am which is much too early for my tastes

[limpid horizon]

So does the rest go like this: we can view M as an R/Ann(M)-module, and localize M at R/Ann(M)'s maximal ideal to have Mm, now M->Mm is an isomorphism of R/Ann(M)-modules so is also an isomorphism of R-modules?

M->Mm sending m to m/1. Injective because tm = 0 means m is 0 since t is invertible, and is surjective because m/s = s^-1 m /1?

[vapid axle]

Let $a$ be a nilpotent element of $A$, $\mathfrak{p}$ a prime ideal of $A$. Since $a$ is contained in all prime ideals of $A$, $a$ is contained in all prime ideals whose intersection with $S \coloneqq A \setminus \mathfrak{p}$ is empty. Therefore, $\frac{a}{1}$ is contained in all prime ideals of $A_{\mathfrak{p}} = S^{-1}A$, by Proposition 3.11. It follows that $\frac{a}{1}$ is a nilpotent element of $A_{\mathfrak{p}}$, contradicting the assumption.

I feel like there's something wrong with my proof of the first part of the question, can somebody check please?

[broken turtle]

okeyokay

[past cove]

I don't fully get why you choose a prime ideal at the start

[lone jacinth]

You haven't said anything about whether a/1 is nonzero

[past cove]

Idk if this is too much of a hint but the fact that it's true for all primes is important

[vapid axle]

Yeah, it felt iffy because it was basically any prime ideal, and I just have to show that there was a nilpotent element in a singular prime ideal for contradiction

[fierce steeple]

(Though I should say this isn't a hard gap to fill)

[vapid axle]

I think I'll restart by choosing elements a_p \in A_p that are not in some prime ideal q_p of A_p and consider their preimage or something

[fierce steeple]

You have a contradiction unless a=0 in all A_p

[past cove]

You started out correctly

take a nonzero nilpotent element a in your base ring

[fierce steeple]

Also like I think it is more normal to write a rather than a/1 here

[past cove]

here I'll write a big hint

[fierce steeple]

I think this is a big enough hint aha

[past cove]

||the contradiction should be that it'll be 0 in all the localizations but that's impossible if it itself is nonzero. Why is that?||

[fierce steeple]

Like hopefully you know why this is a problem

[ornate atlas]

I disticntly remeber doing this problem at some point but I cannot find where I did it, I swear it was a homework problem but I dont seem to have submitted it for any class

[fierce steeple]

Integral domain bit is cute lol

Discussed that problem like a week ago here

Lol

[vapid axle]

Are you calling me stupid?

[ornate atlas]

Im actually wracking my brain to think about why I wouldve done this problem, because I remember it being still at uni and like a silly time at night so it wouldnt have just been for funises

[fierce steeple]

?

No

[vapid axle]

/s lol

[fierce steeple]

Oh phew

What I mean is like this is a fact that hopefully you'll have seen in atiyah macdonald

if smth vanishes in all localisatinos then etc

[past cove]

This reminded me when I was locked up in the psych ward and randomly decided to prove that the intersection over all primes of A_p was A if A was an integral domain

[ornate atlas]

Only being sectioned could push one to doing comalg

[past cove]

I read matsumura while locked up

truly a dark time

had to learn how J-2 rings worked

they put me in solitary for talking about those to the group

[lone jacinth]

I'm wondering if you could drop the integral domain part here.

Where I guess intersection means the pullback of
A_p -> (A-p)^-1(A-q)^-1A <- A_q

[vapid axle]

Is this what they mean by the highlighted sentence?

Sorry it should be p = q_{\alpha_j} in the intersection

And it should be "non distinct radicals" lmao

[past cove]

you can just replace all the q_i's with the same radical with their intersection
by 4.3 that's still a p-primary ideal, hence the primary decomposition still only contains primary ideals

[vapid axle]

I think that's what I did

[past cove]

yeah I think it is just your notation is a bit weird but it looks correct

[vapid axle]

Also it's a bit strange do we require only that the ideals in the decomposition are primary, or that they're all p primary?

[past cove]

every primary ideal is p-primary for some ideal p

[vapid axle]

Oh right I guess that was the point

[past cove]

essentially if q is a primary ideal, then it's r(q)-primary

[vapid axle]

Gotcha, thanks

Why should I give a fuck about decomposing ideals into primary ideals

[past cove]

this is a very very good question lmao

essentially in my head the idea is always motivated by algebraic geometry

The primary decomposition is useful in stuff like

the theory of associated primes

those are very useful

[lone jacinth]

It's the closest thing you have to fundamental theorem of arithmetic for more general rings

[past cove]

but geometrically it gives you the decomposition of your algebraic variety into irreducible components

[digital parcel]

over noetherian rings, associated primes of their quotients are the radicals in the primary decomposition

[vapid axle]

I see, those all sound pleasurable

[digital parcel]

i think vakil actually has a nice section on the geometric interpretation of associated primes over locally noetherian schemes

chapter 6.6

that being said, i am under the impression that primary decomposition isn't used that much anymore

but i may be wrong

[past cove]

I'm like 90% sure hartshorne never mentioned it lmao

[digital parcel]

https://mathoverflow.net/questions/105138/is-primary-decomposition-still-important

karl schwede has a nice answer

i saw a recorded lecture on youtube a while ago where the prof said "i learned primary decomopsition multiple times and it never made sense until i saw it in algebraic geometry" lol

[lone jacinth]

But in algebraic geometry you're mostly thinking about radical ideals right? So then primary decomposition is just prime decomposition, or does primary decomposition give anything geometrically in general?

[past cove]

I don't think so I think computationally it's not super hard to compute the primary decomposition and this gives you the minimal associated primes which give you irreducible components

[hushed bone]

I mean depends on what you mean by algebraic geometry.

The initial objects of study, sure, but if you care about say, deformation theory you start dealing with powers of ideals. Then, you run into questions about associated primes of powers of ideals which, afaik, is not as clean as you’d hope

[limpid horizon]

Pleasurable? 😳

🤤

[limpid horizon]

My apologies

[soft parcel]

Any hints? The top formula is the adjunction formula in question

[plucky arch]

~yoneda~

[soft parcel]

Yoneda isn't used in the text, but I'll try

[plucky arch]

the main idea is that, to show $A \otimes_S (S \otimes_R M) \cong A \otimes_R M$, it suffices to show that $\text{Hom}(A \otimes_S (S \otimes_R M), Z) \cong \text{Hom}(A \otimes_R M, Z)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

and for this you can use the adjunction

[soft parcel]

Oh, the "it suffices" part is because the Yoneda embedding is fully faithful?

[plucky arch]

yeah!

[soft parcel]

Ok, that makes sense

[plucky arch]

-# i think philosophy of generalised elements gives a neat interpretation of this too

you could also do a "direct" proof where you explicitly construct the isomorphism

but

i think it would be unilluminating

[soft parcel]

Yeah, I did that at first too

[plucky arch]

In general I’ve found these kind of “symbol-chasing” proofs to be some of my least favourite

[ornate atlas]

@rose mirage @rare walrus do either of you know anything or have any feelings about “reflection groups”? I’m looking at it as a potential option for a course this semester but the course description is incredibly vague

[rose mirage]

they're VERY cool

I'm not an expert on them by any means but I like them a lot

I believe our resident reflection group expert is @wise sedge

[ornate atlas]

My options are basically that, a kinda intense seeming crash course in mathematical modelling or a possible reading course

Reflection groups are groups, so somewhat interesting to me but adds a 6th massive exam to my time table

The modelling thing is 100% coursework but possibly going to be incredibly intense for me

The reading course is an absolute gamble and 100% exam too

So it’s a bit of a toss up, but if reflection groups seem suitably interesting I could go for that

[rose mirage]

what are your thoughts on combinatorics

and representations of Lie groups/algebras

[ornate atlas]

Uhhh, I don’t mind certain flavours of it, I did a bit of combinatorial comalg for my UG thesis and that was pretty cool

I know nothing about Lie algebras but I will be taking a course on them this semester too

[rose mirage]

If you're gonna do lie algebras too then there's a nice synergy between the two courses, as (real) reflection groups appear in the rep theory of lie algebras as symmetries of root systems

they're very combinatorial objects in general (for real and complex reflection groups their structure is determined by a weighted graph, for instance)

so if you like that then maybe go for it

[ornate atlas]

Hmm interesting, that could be cool

[wise sedge]

Whomst has summoned the almighty one

[rose mirage]

we're talking about reflection groups so I pinged you

[wise sedge]

I can see that

[ornate atlas]

I have like a month to decide anyway, and I don’t know what the reading course will be so I don’t need to know right now, but I thought one of you guys might know a bit more than the description for the course which was essentially “We will be studying reflection groups, this pairs well with Lie groups”

[rose mirage]

it'll probably be the standard "build up to the classification of simple lie algebra" nonsense

[ornate atlas]

I do enjoy a classification theorem

[rose mirage]

u ever seen these guys before? that's what I'm talkin bout

[ornate atlas]

Dykind diagrams?

[rose mirage]

yur

dinky diagrams

[ornate atlas]

Yeah I looked at those a little bit because I was going to take the Lie groups class at my UG (didn’t in the end though)

[rose mirage]

also these which are basically the same https://en.wikipedia.org/wiki/Coxeter–Dynkin_diagram

[wise sedge]

But Coxeter diagrams describe reflection groups better

[rose mirage]

they do

[ornate atlas]

Oh actually I think my advisor spoke about them at some point in a meeting we had

[wise sedge]

Finite real reflection groups correspond 1:1 with Coxeter diagrams but not with Dynkin diagrams

[rare walrus]

Very important groups! If you’re interested in Lie theory, which for example includes the theory of algebraic groups and finite groups of Lie type (and of course Lie groups) then they are EXTREMELY important. Also they’re just quite neat

I’m sure the others have said this lol

[cloud karma]

Yeah they're basically everywhere in representation theory

[cloud karma]

I'm guessing they will talk about Weyl group nonsense

[ornate atlas]

Interesting, could be a solid contender then

I think it’ll come down to whether or not I think I can hack a 6th exam lol

[rare walrus]

I’m kind of envious that your institution has a reflection groups course in undergrad. I had to learn them during the first year of my phd

[ornate atlas]

This is at my masters if that helps

So it is technically a PG course

[rare walrus]

Ah I am going from the perspective of an integrated masters

Good point

But yes, even at a master’s level there was no coxeter course

[ornate atlas]

Yeah it is open to people on the integrated so I see your point

But yeah it does sound cool, I’m just not sure I can do 6 pure maths exams at this level in 2 weeks lol

Not when they’re all 85-100% of the grade anyway

[fierce steeple]

Parental guidance

[vague pawn]

sorry for the spam... thought this might be also relevant here 🙂

[lone jacinth]

So what you need to show is that the closure of G' is contained in the kernel of G -> A.

And I guess a hint is that profinite groups are Hausdorff.

[vague pawn]

yeah thats what I thought

[rare walrus]

Ah darn I gave the same hint in #groups-rings-fields

or a similar one

😔 the dangers of crossposting

[vague pawn]

yeah.... just #groups-rings-fields is so active

so I thought I will get an answer faster there

[cloud karma]

I honestly wouldn't know about these guys if I haven't exposed myself to representation theory in the first place

[rare walrus]

Honestly that's not always the case, not telling you off or anything but you might get quicker answers in the future if you post here

[cloud karma]

More people should care about reflection groups. They're super useful

[plucky arch]

are they

[distant harness]

For questions that are more advanced than the usual topic in a channel, "more active" may just as easily mean the question risks being buried before someone who knows how to answer it shows up.

[vapid axle]

I've seen that all localizations are zero implies the entire ring is zero, but I don't think I've seen it in this form

[torpid sparrow]

this would put me back into the psych ward for sure

well it currently is

something im working on rn is driving me insane

[fierce steeple]

Ah sure

I guess they are equivalent statements

Well okay sorry, like the slightly more general statement for modules i mean

[limpid horizon]

fr?

[mental escarp]

Where do you read about reflection groups?

[jaunty sparrow]

"Reflection groups and Coxeter groups" by Humphreys is a good book.

[rare walrus]

Honestly I think it's an incredibly annoying book actually. I don't know any really good references for reflection groups that aren't tied up with some other exposition. My favourite exposition is in Digne & Michel's book on Deligne–Lusztig theory...

But everything here is obviously just opinion, so whatever I guess

[rose mirage]

Humphrey's is good if you love polynomials more than our heavenly father and gambling combined

[rare walrus]

Hahahahahaha

[rose mirage]

it's giving Serre paper from the 1920s

[fierce steeple]

Let A be a complete local Noetherian ring with residue field k. Can every finite length A module be obtained from k by iterated extensions, retracts and direct sums?

[rose mirage]

unironically I do like Humphrey's for the first few chapters and then he starts talking about invariants and hilbert polynomials so my eyes glaze over a bit

[fierce steeple]

Like this seems clearly true if A is a complete DVR (e.g. use classification of f.g. modules over a PID and stuff)

But you can have more gnarly complete local Noetherian rings

[lone jacinth]

Yes, the only necessary assumption here is local.

In general every finite length module is the iterated extension of simple modules, which are off the form A/m for maximal left ideal m

[fierce steeple]

Oh wait I am silly, yes, I have done the "ring" version of this. Thanks

Is the argument basically just that you reduce the length by 1 by quotienting out by a cyclic submodule and then use induction?

As in everything fits into like xM -> M -> M/xM

Where I think you want x in the centre or smth

Something like this

[lone jacinth]

Well my argument is that the definition of finite length is having a filtration of simple modules

[fierce steeple]

Oh wait lmao

I am just silly

So yes this is just like "what are the simple modules over a local ring A"

And they have to be A/m by correspondence thm

I am very silly. Thank you

I guess to be more fair to myself I have been thinking about a variant of this rather than exactly this lol

Well you can ask the same question within the derived category of A (mutatis mutandum), like what is the thick subcategory generated by A/m

[lone jacinth]

Should be bounded complexes of finite length modules

[fierce steeple]

Yeah exactly

That is my guess

I will work it out

[lone jacinth]

Now I'm wondering if something like
D^b(A) / Thick(A/m)
has a nice description.

Like I guess
D^b(Z) / D^b(finite ab) should be D^b(Q)

[fierce steeple]

I should say like there is work of Hopkins right classifying all thick subcategories of bounded complexes of fg projectives

But here it is a but awkward because like A/m need not have finite projective dimension over A right lol

Like equivalently if A is not regular

(Given local noetherian)

[lone jacinth]

Does that make it award though?

[fierce steeple]

I mean like A/m isn't even in the category of like bounded fg projectives

[lone jacinth]

Or you mean the classification is for perfect complexes maybe?

[fierce steeple]

Ye

[fierce steeple]

Ig I was unsure what you meant by D^b(A) aha

But yes

I guess you mean actually just bounded rather than perfect stuff

Which is good

[lone jacinth]

Yeah I would have written Perf(A) or K^b(projA) otherwise

[fierce steeple]

I suspect/hope that the thick subcat generated by A/m is equivalently those complexes whose cohomology is bounded with finite length in each degree

Bit more intrinsic lol

But yeah this should not be hard to prove if true

[fierce steeple]

Like clearly thick subcat is contained in this guy

And then like truncations should give you the rest

Just feels odd I have not seen this notion explicitly before

[lone jacinth]

Idk, I only work with Artin algebras, so then everything has finite length anyway

[fierce steeple]

Lolfair, thank you tho for help !

[cloud karma]

Humphrey

[vapid axle]

Is the statement and argument like this (heavily adapted from AM's proof)

[fierce steeple]

Yeah exactly, nice

[vapid axle]

Okay then the rest of the proof is easy yeah lol

[broken turtle]

okeyokay

[lone jacinth]

Well x^2 and y^2 are both in (x, y)^2

And (x, y) is radical so that's everything

[rare walrus]

naughty naughty double posting...

[cloud karma]

I'm working with Macdonald polynomials. Specifically the non symmetric Macdonald polynomial which has a basis $E_{\mu}$ for $\mu \in \mathfrak{h}{\mathbb{Z}}^*$ and here $E{\mu} = \tau^{\vee}{X^{\mu} m}$ with $X^{\mu} m$ being the minimal length element in the coset $X^{\mu} W_0.$ For an element in the extended affine Weyl group you can find the length via the formula $\ell(w \tau(\lambda))= \sum{\alpha \in R^{+}} |(\lambda,\alpha)+\chi (w \alpha)|$ where $\chi(\alpha)$ here outputs 0 if $\alpha$ in R^{+}$ and $1$ if $\alpha$ in $R^{-}$. What I'm trying to do here is to explicitly find that minimal $m$ for $E_{\mu}$. Now finding the length of $X^\lambda m$ is the same as finding the length of $m^{-1} X^{-\lambda}$ and hence I can use the formula to compute the length. $(-\lambda,\alpha)$ will always be negative so what really boils down to do here is making sure the quantity $\chi(w \alpha)$ is 1 as much as possible which means $w \alpha$ should be in $R^{-}$ as much as possible but I'm not sure what to do from there

[broken turtle]

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

[cloud karma]

well in this case ig I want to make sure $\chi(m^{-1} \alpha)$ is in $R^{-}$ as much as possible

[broken turtle]

Delteto

[fierce steeple]

@lone jacinth for completeness, this is indeed true and a nice reference for this and lots of related facts is https://ems.press/content/serial-article-files/43086

:)

(This fact is example 3.5)

But yeah the proof he gives is what I sketched plus your thing with only simple module being the residue field

[vapid axle]

Oops sorry I thought I deleted it

[vague pawn]

Here in the definition of a profinite group (from Neukirch's ANT book) is this basis of neighborhoods of the identity supposed to consist of open normal subgroups?

[foggy galleon]

isn't it irrelevant? Say N is a normal subgroup that contains an open subset U. If U is not a normal subgroup, then just keep adding elements

because union of opens is open the resulting thing should be open. So you have an open normal subgroup contained in N, and this filters a basis so a basis

[lone jacinth]

Yeah, notice that for any g in N, gN = N is a neighborhood of g, so N is open

[vague pawn]

I don't get it

[lone jacinth]

A set is open iff it is a neighborhood for each of its elements

So any subgroup that is a neighborhood of the identity is open

[foggy galleon]

ah yeah that's a nicer way to see it

[vague pawn]

how does this show N is a neighborhood of g (here N is a normal subgroup right?)

[lone jacinth]

If X is a neighborhood of the identity then gX is a neighborhood of g (since multiplication by g is a homeomorphism taking the identity to g)

[vague pawn]

oh ok

thank you

man I am bad at this topology stuff 🙁

[lone jacinth]

I think the moral of the story is that for topological groups are very homogenous.

Anything that holds at the identity can be moved to hold anywhere else

[vague pawn]

yeah they are very nice

[pastel shoal]

Does there exist an infinite boolean algebra B with no infinite sequence whose pairwise meets are 0?

If there is an infinite ascending or descending chain in B, you can construct such a sequence

If every chain is finite, you can take the collection of minimal elements in B \setminus{0}. If this collection is finite then B has to be finite, If it is infinite we have constructed a sequence with the property

[spice idol]

i guess that answers your question no?

[pastel shoal]

yeah but Im a bit stuck showing that if the collection is finite B is finite

[spice idol]

maybe try showing that its a generating set; boolean algebras are locally finite

[lone jacinth]

Pick an element x1 not equal to 0 or 1.

Note that
B = (x1 ^ B) v (not x1 ^ B)
So at least one of these sets must be infinite.

Wlog say the right one is.

Pick x2 in (not x1 ^ B), repeat inductively

[pastel shoal]

yeah got it thanks

[spice idol]

[limpid horizon]

Was trying to understand this part

I know (x,y) is a radical ideal because its prime

I wasnt sure why the “so thats everything” part is true

[lone jacinth]

Well the radical is the intersection of radical ideals containing it.

But also (x, y) is maximal, so it couldn't possibly be bigger

[limpid horizon]

Oh yea

[limpid horizon]

rad(I) = intersection of ideals containing I = intersection of radical ideals containing I?

Idk just trying to understand what you said

[lone jacinth]

If the first one is supposed to say prime ideals, then yeah

Like the radical must definitely be contained in (x, y), because that is a radical ideal (in fact prime) that contains (x, y)^2.

But the radical also contains (x, y), hence they're equal

[spice idol]

do you know what a closure operator is? If so, then taking the radical is exactly the closure operator on the set of ideals with closed sets being radical ideals

[limpid horizon]

I havent heard of closure operator before

[limpid horizon]

Ok i understand this

I understand why if a radical ideal J contains I, rad(I) is contained in J, but why is rad(I) the intersection of all radical ideals containing I

[spice idol]

grrrr why dont schools teach them anymore

it should be taught along with posets

[limpid horizon]

Yea maybe i have heard it briefly in passing one time

[spice idol]

anything "generated" in algebra is a closure operator

[lone jacinth]

Well, how are you defining rad(I)?

[limpid horizon]

I was just thinking by the definition, x is in rad(I) if x^n is in I. I forget why rad(I) is equal to intersection of prime ideals containing I, however ik thats just based on the nilradical being intersection of prime ideals, but i dont remember the argument for how that one goes

[spice idol]

that argument is called the Nullstellensatz lol

[ornate atlas]

This is why you shouldn’t assume your rings to be commutative, then the definition is just the intersection of all the prime ideals…

[spice idol]

anything to be a coherent condition 💔

[lone jacinth]

Alright, so the intersection of radical ideals is radical.

Now consider the intersection of all radical ideals containing I. This clearly contains rad(I) as it contains any x with x^n in I.

And it is contained in rad(I) since that's one of the sets we're taking intersection of

[distant harness]

I was just thinking by the definition, x is in rad(I) if x^n is in I
Be sure to remember to think the quantifiers too:
x is in rad(I) if there is some n such that x^n is in I.

[limpid horizon]

rad(I) is a radical ideal containing I? Oh yea cause rad(rad(I)) = rad(I) lol

[limpid horizon]

Oh ok so this doesnt have to do with radical being intersection of prime ideals containing I

[lone jacinth]

Not unless that's how you define radical

[limpid horizon]

Thats not how i define radical

That would be weird ☹️

[river beacon]

I'm an analyst, but I need to remind myself of some solid group theory, ring theory, galois theory, representation theory... what book is the all inclusive for this?

But not just intro stuff, I've studied this many times and I need to stop forgetting

[ornate atlas]

Something like Langs Algebra is probably a good call

[river beacon]

Dummit and Foote too elementary?

Thats what I like, wondering if theres else out there

[spice idol]

Aluffi is good imo, but its very category theory-pilled so might not be to your taste

[ornate atlas]

If you just need to review then D&F is probably a little slow and wordy

I think they cover pretty similar amounts though, D&F possibly slightly less but they’re both very comprehensive

[river beacon]

Oh I like Aluffis book

Looks nice

Anything good on group representations?

And algebraic topology? I don't like Hatchers

[spice idol]

in my experience Serre is pretty good for representations

[ornate atlas]

Finding a good algebraic topology book is a somewhat cursed endeavour, if you know a bit more maths than Hatcher assumes Diecks book seem quite nice, I’ve only skimmed it but I’ve heard good things

There’s Rotmans Algtop book but I personally find Rotman to be far too slow in all of his writing. I’ve heard mixed things about Mays book but I’ve never read it, I think it’s supposed to be good, just hard

[river beacon]

I'd like one that focuses on actual interesting topological spaces

[fierce steeple]

Feels a bit orthogonal to what most alg top books have

Hm

[river beacon]

Not just definition theorem yada yada

[ornate atlas]

I can’t say I know of any books like that but I’m also a bit of a fiend for a definition theorem proof book

I’m also not a topologist so there’s that too

[cloud karma]

Say $\mu$ is regular and of the form $\mu = x \lambda$ for a dominant $\lambda$ and $x \in W.$
I want to find this $m^{-1}$ so that $\sum_{\alpha \in R^{+}} | -(\lambda,x^{-1} \alpha) + \chi(m^{-1} \alpha)|$ is as minimal as possible. I split this by cases. Since $\mu$ is regular we look at the first case where $(\lambda,x^{-1} \alpha)>0$ but since $\lambda$ is dominant this happens when $x^{-1} \alpha > 0.$ Now $\chi(m^{-1} \alpha )$ can either be 1 or 0 but we want it to be 1 so that we get smaller magnitude. For this, if we want $\chi(m^{-1} \alpha )$ to be 1 as much as possible this means we need $m^{-1} \alpha>0$

I'm still not sure what this m^(-1) supposed to be

[broken turtle]

Delteto

[vapid axle]

Is this true since if Ax -> M is the inclusion then if Ax (x) B is nonzero, the image of Ax (x) B in M (x) B is nonzero since B is flat over A

[cloud karma]

at first glance it seems like m= x is the way to go here but what do you guys think?

[fierce steeple]

Yes exactly

[vapid axle]

Bruh these hints are just solutions lmfao

Well I guess I have to fill in some of the gaps

I assume it wouldn't be obvious that M'_B \cong B/a^e unless you've done exercise 2.2

[cloud karma]

shit here we get 1 if m^(-1) alpha < 0 not the other way around.

This is very confusing

[cloud karma]

ok nvm im cooking

I still need to figure out what if mu is not regular

[prime vigil]

Etingof's Introduction to Representation Theory is quite nice (I'm a Serre opp)

[void plank]

I've read somewhere that if G is an arbitrary finite group and k a field, the group algebra k[G] can be regarded as the localization of a polynomial ring with coeffs in k

Is there anywhere I can read more about this correspondence?

[fierce steeple]

This seems kinda wrong at least at first glance. Like the group ring of C_2 is k[x]/(x^2 - 1) which isn't a localisation of a polynomial ring

[void plank]

Hmm ok, maybe I misunderstood the write-up I was reading.

What I asked was based on an offhand comment in it and wasn't elaborated upon

[ornate atlas]

Theres this survey, which ive never read in full but I skimmed for a class, which is possibly of some interest? https://webhomes.maths.ed.ac.uk/~v1ranick/books/nlat.pdf

Its more about group rings than group algebras though, i had a look for my noncom class, so maybe not what youre looking for but maybe theres something of interest

[void plank]

Seems interesting, thanks for the suggestion! I'll take a closer look

[ornate atlas]

Localisation in noncom settings is miserable! But i hope it helps

[void plank]

Tbf, I think I only need for now the case where G = finite abelian or G = semi-direct product of finite abelian groups, where I think its true that k[G] is a quotient of a polynomial ring

[lone jacinth]

I mean, any commutative algebra is a quotient of a polynomial ring, and any algebra is a quotient of a non-commutative polynomial ring

And I guess that's a reasonable way to think about the group algebra, but it doesn't have anything to do with localization

[void plank]

This also seems to work if you can write G as a semi-direct product of abelian groups, like the dihedral group which is not necessarily abelian. Can you not also write the group algebra of such groups as polynomial rings, where the action of the automorphism is taken into account in the ideal you're quotienting out by?

[lone jacinth]

Like, I'm not sure exactly what you mean, but if G is a semidirect product of N and Q, then Q acts on kN and the group algebra kG is the skew group algebra
(kN)#Q

So if N is abelian you can write kN as a quotient of a polynomial ring.

But any quotient of a commutative ring is commutative, so you need something else to introduce noncommutativity

[void plank]

Sounds good.

Just to double check, if N and Q in the semi-direct product are abelian, the skew group algebra here is the vector space kN (x)_k kQ with a certain multiplication structure right? kN (x)_k kQ in this case is a quotient of a polynomial ring, just with a multiplication that isn't necessarily commutative

[lone jacinth]

Yeah, thats right.

Though being a quotient of a polynomial ring is kind of meaningless if you mean as a vector space and not as an algebra

[void plank]

Out of curiosity, is there a way to express this skew group algebra as the covariants of a tensor product vector space with a G-action?

The skew group algebra here (using your notation) has the underlying vector space kN (x)_k kQ. Q acts by automorphisms of kN and acts on itself so I thought there might be a way to get the same skew group algebra by passing to coinvariants.

But this is just a stretch and I suspect this is not the case since the structures are different (passing to coinvariants by the action of Q gives an abelian group while the skew group algebra is an algebra)

[lone jacinth]

Yeah, the question doesn't quite make sense.

The coinvariants is not an algebra, so then you're just comparing vector spaces

[void plank]

Thanks!

[spice idol]

diagram chasing feels like a game

[lone jacinth]

And what a fun game it is

[ornate atlas]

You need to start playing more enjoyable games

[spice idol]

the duality of man

[spice idol]

i play geometry dash, thats very enjoyable

[ornate atlas]

I don’t dislike diagram chasing but I stand by my point, I wouldn’t go so far as to call it fun

[lone jacinth]

You need to start proving more enjoyable diagram lemmas

[ornate atlas]

Doing a little bit of last minute homalg revision before beginning your undergrad?

[ornate atlas]

This is possibly true, I’ve only done the very basics

[fierce steeple]

Ngl like

[spice idol]

i wanted to go through algebraic geometry and arithmetic curves with a friend of mine, but he apparently already started so im catching up

[fierce steeple]

diagram chasing is something I feel I have only done for basic hom alg

Does it appear beyond that much

[ornate atlas]

Ah just a quick catch up on arithmetic curves pre uni, you wouldn’t want to be left behind or anything

[lone jacinth]

I'm just joking. Diagram chasing is pretty much just diagram chasing, but it is kinda fun

[spice idol]

of course

did you know Radboud offers a noncom geometry course and a schemes course but not classical algebraic geometry?

[fierce steeple]

Good

[ornate atlas]

Based.

[spice idol]

hahahahahaa

[fierce steeple]

Blahaj.

[spice idol]

blahaj

[ornate atlas]

I took classical alggeo last semester and I’m not sure I exactly gained much from that experience

[fierce steeple]

I have a djungelskog

[ornate atlas]

I found this on fb marketplace earlier

[spice idol]

that is fair, its just sad because i wouldve liked following the classical algebraic geometry, but schemes are still too high level rn

[spice idol]

how would anyone not buy that

smh

[fierce steeple]

£45

[fierce steeple]

Stop the British

[spice idol]

how would anyone buy that

[fierce steeple]

Cringe

[spice idol]

smh

[ornate atlas]

I’m flat broke currently, I don’t know when my student loan comes in but I’m praying it’s tomorrow

[fierce steeple]

Saw this in a thesis today

Very cute

[spice idol]

thats totally fair though

[ornate atlas]

In any case, I do think provided you know some comalg and topology, I don’t know how much there is to gain from doing classical alggeo that couldn’t just be summarised in like a single lecture

[fierce steeple]

I'll loan you some money if you give me all your credit card details so I can take the money back later

[ornate atlas]

Deal.

[void plank]

I should've done a math PhD instead of a physics/CS one

[mellow night]

Categorical algebraic geometry is goated who's with mw

me*

[spice idol]

true ig

[fierce steeple]

Me

[spice idol]

universal algebraic geometry is goated'er

[fierce steeple]

What do you mean by categorical AG though

[ornate atlas]

I can’t say anything about my classical course was particularly exciting (though I just generally don’t care about geometry) or surprising to me because I could already do ring theory and was aware of the nullstellensatz

[mellow night]

You know shit like $\text{Sch}_k$, algebraic stacks and Hilbert schemes

[broken turtle]

GNOSIS OF POWER

[ornate atlas]

Like the entire course was essentially just building up to the more general notion of an algebraic variety and then you’re just diving into schemes to do stuff “properly”

[fierce steeple]

Ye sure

[fierce steeple]

Funny reformulation of Nullstellensatz I learnt uh

[mellow night]

Although hilb schemes don't count as completely categorical I just had the thought that this is

[fierce steeple]

If A is a nonzero f.g. k-algebra (k alg closed) then k -> A admits a left inverse

[ornate atlas]

Dear lord don’t make me remember another version of that fucking theorem

[fierce steeple]

Yeah the naming is really bad here lol

[spice idol]

The real nullstellensatz is the fact that the jacobson radical is the nilradical tbh

the rest is just lattice theory

[fierce steeple]

Agreed

Jacobson rings

Nullstellensatz is just model theory

What I find funny w the nullstellensatz is that there is a very simple proof over an uncountable field

[mellow night]

Wait doesn't this mean the map is an "isomorphism"? If so then uhh it's obvious lmao

[fierce steeple]

No

[ornate atlas]

Oh that’s actually quite cool I hadn’t thought of it that way

[mellow night]

Breast in peace this could've been useful

Rest*

[spice idol]

there is a sort of algebraic form of the nullstellensatz i have for a general form

[mellow night]

Kill my autocorrect

[fierce steeple]

That is cool

[spice idol]

its not nearly as nice as the usual one but still okay

[fierce steeple]

I think the apple's rotten right to the core.

[ornate atlas]

I’ve never really dealt with the nilradical though tbf because I’ve only taken a course on Noetherian rings so it’s not exactly an interesting concept there

Well interesting enough I guess but it’s just the prime radical

[fierce steeple]

I mean here the rings are Noetherian

[fierce steeple]

Wdym

[ornate atlas]

For a Noetherian ring the prime radical and nilradical are the same

[mellow night]

Best fact in AG I've ever seen is that the !-pushforward of f: X→S identifies with g: Ω_S → Ω_X (up to a local setting)

[fierce steeple]

How is the prime radical defined like

[ornate atlas]

It’s the intersection of all prime ideals containing it

[fierce steeple]

If it is the intersection of primes then Noetherian is irrelevant

[lone jacinth]

Nope is a noncommutative boy

[fierce steeple]

We were just talking about commutative rings dawg

😭

[ornate atlas]

I was wondering if this was a com vs noncom issue lol I felt like I was going insane

[fierce steeple]

Jk fair yeah I just was confused

I mean if I am honest I had never heard of the prime radical lol

For this reason ig

[ornate atlas]

So they also coincide for commutative rings?

[spice idol]

okay I read back and it is quite different but its similar feeling

[fierce steeple]

Hmm

[fierce steeple]

Ye

I mean proof is infinitesimally easier for Noetherian tings cause no Zorns but yeah

[spice idol]

yes because you can do localisation arvrvhzvrgzvrv I hate localisation why cant it generalise nicely agrvshrbdbrbr

[mellow night]

@fierce steeple look at this fact
if you don't get it its because g_! f: D(S) → D(X) (D means D-modules, look at verdier theory)
Also meant to say !-pullback

[spice idol]

universal algebra leads people to the pits of despair

[ornate atlas]

So it’s only interesting for non Noetherian noncom rings lol

I don’t really understand how you study those tbf, like what do you have to work with

[fierce steeple]

Wdym by Omega_S sorry

[mellow night]

math.RA authors:

[ornate atlas]

I feel your pain, we only covered the “nice” case in my noncom class and even that was a couple weeks of suffering

And then in comalg it’s an off hand comment of yeah you just kinda make them fractions

[lone jacinth]

I feel like most nice things hold for rings that are either commutative or Noetherian

Like IBN for example

[mellow night]

if S is a subscheme of X a scheme, then I say Ω_S is the sheaf of differentials.

[spice idol]

you have operations, and a reduct which is a group, so that means youve got a malcev term acting as a difference term too

[fierce steeple]

Sure yeah

[spice idol]

youve got so much, dont complain

[fierce steeple]

I assumed so

[mellow night]

You're the best AG person ever

[fierce steeple]

What lol

[mellow night]

You don't get it ?

That's ok I just never seen this addicted of an AG person besides me lol

[ornate atlas]

Honestly are there even people who study them? I don’t really know how you would or what the interesting questions would be, it just feels so intractable

But I’ve also just never really been exposed to them so

[fierce steeple]

I think what you said is a bit incomplete like if you take !pushforware/pullback you should get a functor right

[mellow night]

Mhm yea but I said to a local setting since I was aware D(—) is a category/stack

[fierce steeple]

Idk this still sounds a bit vague to me

[mellow night]

Mmm maybe cus I kinda slipped when saying in the local setting

[spice idol]

the only generalisation ive been able to properly get working is for commutative ringoids, which is basically that you equip an algebraic structure with a commutative monoid which distributes over everything