advanced-algebra

[past cove]

like you can define it

but I've personally never really seen it used in any convincing way

[blissful field]

Is it not useful in any way? I thought it is defined to measure the surjectivity of group homomorphisms

[hallow bone]

https://link.springer.com/book/10.1007/978-981-10-6793-8

[past cove]

tho note

I do number theory lol

so just because I haven't seen it doesn't mean it doesn't have any uses

[hallow bone]

whuh

[blissful field]

im tired pls no kil

[past cove]

I've just seen group cohomology used a shitton

[hallow bone]

yeah it measures extensions, amongst other things

[past cove]

isn't that cohomology?

[hallow bone]

yes

[past cove]

like H^2 measures extensions

[hallow bone]

thats what i meant

[past cove]

I think they were talking about homology

[hallow bone]

I know i was supplementing this message

[blissful field]

-# was i wrong? 😅

[past cove]

I think H_2 measures some stuff

but it's less direct than "classifying extensions"

[hallow bone]

there's not even an SE or MO post on it

cohomology better anyways

Beck modules

[cloud karma]

it exists but the go to is cohomology

[silver goblet]

Isnt Tate homology supposed to be useful

and that's defined using the homology and cohomology

[lone jacinth]

The projective cover of a module M is a map from a projective P -> M such that a composition
X -> P -> M
is surjective iff X -> P is surjective.

One key thing about artinian rings is that you have a bijection between simple modules and indecomposable projectives given by taking projective covers.

And if you have a direct sum of two simples then the projective cover will be the direct sum of two indec projectives

[summer quest]

It is incredibly useful yes

Yes it is lol

[past cove]

I've never really seen it used in any convincing way tbh

[summer quest]

Well it is implicit in the definition of Tate cohomology for one thing

It is natural for the same reason homology of spaces is natural

[past cove]

I mean I don't really feel like that's enough justification for an object to be inherently interesting

are there any uses where group homology is like very necessary for an argument?

[summer quest]

Yes of course there is

See work of Lurie and Clausen and others on ambidexterity

See work of Clausen and others on blueshift

See like half of the cohomological approach to class field theory

[hallow bone]

cohomological approach
group homology

where did the co- come from lol

no im asking a genuine question

[summer quest]

Because the homology also shows up in there

[hallow bone]

oh cool, hey don't blame me it sounded funny

[past cove]

I didn't know this actually because neukirch doesn't really use it anywhere when he proves CFT

[past cove]

this sounds cool tho I'll take a look

[foggy galleon]

isn't it fair to say that most of class field theory happens in degrees 0 (possibly tate), 1, 2? But famously in Golod-Shafarevich type theorem (by which I mean constructing infinite class field towers) degree -3 is used

[summer quest]

Yes that is a good example

[wanton spoke]

[summer quest]

Again you use it when you set up Tate cohomology in the first place and prove things like Tate Nakayama

These features of Tate cohomology doesn’t really make sense without talking about homology in the first place

[hallow bone]

bro joined today

[wanton spoke]

curious to know the silly questions lmao

I bet nG is yapping with that guy in DMs rn lol about cohomology

[foggy galleon]

If you just want like G(L/K)^ab=C_L/norm C_K then you do need to at least talk about H^(-1), or at least you need something that substitutes it. Neukirch does this in all his books.

The global duality theorem is expressed for i negative when A is just finitely generated, when A is Z-free and finitely generated then you can take all i.

[fierce steeple]

Jk

[foggy galleon]

but I'm not sure to what extent you need negative degrees to prove the more general local duality theorem. It's about knowing cd k=2, H^0, H^1, H^2, and then using Tate's spectral sequence to relate the cohomology of the dualizing module (i.e., mu) to that of the given Galois module. But Neukirch et al. mention this on the proof of Tate's spectral sequence

[foggy galleon]

historically, didn't this terminology arise in algebraic topology where it is clearer why it's a "co"? In any case, you do have this, not sure if it's too clarifying tho

[hallow bone]

It was both a joke and a genuine question lol

Thats a really cool result though

[foggy galleon]

although I guess if you want to define group cohomology in terms of a projective resolution you still have to take Hom, while homology is tensor product

[hallow bone]

the localizations at a prime A_p arently usually f.g. over A right?

ig of course not because k(X) = k[X]_(0) and we have a field extension k(X) / k

[lone jacinth]

Guess it depends what usually means

Sometimes is sometimes isn't

[vague pawn]

why are there 2 enpeaces

[hallow bone]

my account got hacked

lol

[vague pawn]

when

[hallow bone]

week ago

give or take

[vague pawn]

sent you a dm

[limpid horizon]

Apparently + Aren’t = Arent’ly?

[limpid horizon]

hello. Firstly, doesn't p need to be minimal over Ann N for it to be in Supp N? Also I'm assuming Np (x) k(p) -> k(p) is just projection and Np -> Np (x) k(p) is surjective so we get Np -> k(p) surjective? Then, did we want it to be surjective to ensure Np -> k(p) -> Mp is a nonzero map?

[lone jacinth]

It doesn't need to be minimal.

Any nonzero k(p) vector space will have some map onto k(p), you can call it projection if you like.

Np -> Np(x)k(p) is indeed surjective.

And yes you have k(p) a submodule of Mp, so that's why you construct that map.

[limpid horizon]

Ok, so is there a reason we needed Np -> k(p) to be surjective?

[hushed bone]

You just need it to be nonzero

[limpid horizon]

yeah

so there's no reason to state that its an epimorphism?

[hushed bone]

It is tho

[narrow kraken]

I'm not sure how to approach this: Let A be a finite dimensional algebra over a field. Assume that for every simple A-module L the multiplicity of L in the cosocle of A
viewed as a left A-module equals the multiplicity of L* in the socle of A viewed as a right A-module. Show that A is self-injective.

[lone jacinth]

Then injective envelope of a (finite length) module will be the injective envelope of its socle.

Compute the injective envelope of A. What is its dimension?

[narrow kraken]

oh i haven't gotten to injective envelopes yet, let me read about it

[lone jacinth]

Well, the key is basically just that if you have a map M -> E for a finite dim module M that is injective on the socle, then the whole map is injective

Then for L in the cosocle of A you have a good choice of map from L* to an injective module

[narrow kraken]

given R artinian, what's an example of an indecomposable, cyclic R-module M that has more than one maximal submodule (here cyclic means generated by one element, so Rm = M for some m in M)

[lone jacinth]

Consider the path algebra of the square
1 -> 2
v v
3 -> 4

with a commutativity relation, and let M be the ideal generated by the two arrows a:1 -> 2 and b: 1->3

(note M is cyclic generated by a+b)

[narrow kraken]

hm interesting

[worldly zealot]

A while ago I was trying to find a combinatorial rule for the AR translate of a module over certain path algebras, the main one was the quiver for finite product of finite totally ordered sets with commutativity relations

I didn’t try for very long but I couldn’t figure it out in higher than 2 dimensions

It seems like the sort of thing that a smart combinatorialist would figure out pretty easily but I’m neither

[lone jacinth]

I don't know how general you had in mind, but you have these interval modules (sometimes called hook modules) supported on
b < x <= a
for some a and b. Then the AR translate will be the interval module supported on b <= x < a

[worldly zealot]

I guess what would be suitably general would be intervals with finitely many generators and cogenerators

[smoky wasp]

disregard its just base changes

it does make perfect sense

[lone jacinth]

I don't think the AR translate will be intervals then though

So what would be a good description of them

[worldly zealot]

Yeah they wouldn’t - i suppose a good description would be a transformation from the original?

I imagine that computing DTr is the easiest way even if there is something else

By “I imagine” I mean I gave up trying so I just decided that must be true

[lone jacinth]

Can be fun to compute some AR-quivers, but the 2x2x2 cube is already rep infinite, so it gets bad quick

[smoky wasp]

ok I am legitimately losing my mind what

So K field of char p, f functions from K to N with finitely many non zero entries

ive been working on this problem

im trying to construct an isomorphism

(also this is as a Bf algebra in the sense of base change, these are tensor products over K)

(so it just scalar mult in the second component)

ive defines this map which I think ought to be an iso

(or at least induce one)

its K-bilinear

plays nicely with scalar multiplication

and its gonna preserve that (xs x 1) to whatever power is 0, so does whatever it maps to

now the issue is i have no idea how to show its actually an algebra homomorphism???

when I take product

(Second products are just scalar multiplication and those play nicely so who cares)

we get

vs

and im honestly stumped what to do w this

am I just wrong? is this then not the morphism?

[smoky wasp]

hold up I may be stupid

I literally just

defined it on generators

(xy x 1) is not a generator

true!

so I think its fine?

yeah ok its fine im dumb

im terribly sorry I keep posting a question to here and then shortly after figuring it out myself 🥀

[hushed bone]

Damn

I couldn’t even ping

Because it takes so long for the right thing to come up

[limpid horizon]

Are there some typos? In the 4th line should it say Ext^n-1(N, M/x’M) = 0? And should the ses be with M/x’M and M/xM? (Multiplying by xn)

[lone jacinth]

It should say Ext^n-1(N, M/x_1 M)

So the second application of induction would be

Ext^n-1(N, M/x1M) = Ext^n-2(N, M/(x1,x2)M) = ... = Ext^0(N, M/(x1, ..., xn)M)

You could probably rewrite the proof to go the other way though. Maybe that would be more natural

[limpid horizon]

I thought it was Ext^n-1(N, M/x’M) = 0 because Hom(N, M/x’M) = 0

[lone jacinth]

Is that what 1.2.3 says?

[limpid horizon]

Thats 1.2.3

[lone jacinth]

So Hom(N, M/x'M) = 0 by 1.2.3 and Hom(N, M/x'M) = Ext^n-1(N, M)

[limpid horizon]

O

Then its just Ext^n-1(N, M/x1M) = Hom(N, M/xM) by induction hypothesis and then psi being an isomorphism gives Ext^n(N, M) = Hom(N, M/xM)

[abstract patrol]

Can anyone help im so lost 🫩

[dark widget]

give this man postgrad rn

[hallow bone]

i recommend the book "a course in arithmetic"

that might help you :3c

[abstract patrol]

Thats what im saying bro

When we say advanced algebra

We are talking like x AND y right?

[dark widget]

real shit

[abstract patrol]

On god i could solve every problem in this discord

[rose mirage]

c'mon man there's not even a 67 in ts 💔

[abstract patrol]

Exactly

[dark widget]

6+7=13

[abstract patrol]

Okay well i know i am in the wrong section but my homework rlly make no sense i dont even know what im doing anymore

[hallow bone]

woag new pfp

[abstract patrol]

I am 2 months out from my exam and still on an E 🫩

Now TS advanced algebra (not even algebra)

[hallow bone]

#geometry-and-trigonometry

[abstract patrol]

I know i know

I dont acc need help

Well i do

But oh well

[hallow bone]

okay 💔

[abstract patrol]

Got 42/200

Am i passing

[unborn rampart]

I'm a bit confused by this, does B contain 0 or not? I guess it can't if you want it to be an actual basis, but otherwise the multiplication in B is not closed?

[hallow bone]

maybe what they mean is that if you add 0 it will be a semigroup

[unborn rampart]

oh yeah, that's probably the most sensible interpretation

[hallow bone]

sounds like some grading-related stuff

[unborn rampart]

Yep, I think it is. It's Gröbner bases for noncommutative polynomial rings and path algebras, but I think it can be generalized further to graded structures

maybe it's easier if I just go to the generalization 🤔 I don't like thinking about semigroups that are almost semigroups

[signal moon]

Hope this question suits this channel. Let $G$ be a finite group with conjugacy classes $C_1,\dots, C_k$ and irreducible complex representations $U_1,\dots,U_k$. Let $a_{ij}=\chi_{U_i}(C_j)$. Is there a name for the matrix $A=[a_{ij}]$?

[broken turtle]

person2709505

[signal moon]

Oh... it's just the character table isn't it...

[hallow bone]

yes lol

[narrow kraken]

in the proof of (i), how does it follow that lambda is self-injective?

[lone jacinth]

A^* is injective

So if A = A^*, A is injective

[narrow kraken]

but why

[lone jacinth]

Hom(-, A*) = Hom(A, (-)*)

And A is projective and (-)* is exact

[narrow kraken]

oh ic

[ashen echo]

Someone help me

[past cove]

brother in christ

go to #prealg-and-algebra

[ashen echo]

My bad guys

[soft parcel]

Could someone explain this to me. I don't really understand why the alpha-string through mu being unbroken should follow from (7.2) (the classification of sl_2-representations) and Weyl's theorem on complete reducibility. Here S_alpha is the copy of sl_2 correponding to the root alpha.

[hallow bone]

pending postgrad

[limpid horizon]

Lol

[wanton spoke]

lol

when you start seeing numbers bigger than, say, 5, there’s a good chance the level is undergraduate at most

[hallow bone]

unless its some modular forms bullshit

[wary elbow]

The subspace W is a representation of the copy S of $Sl_2$ inside G (I assume the context here is a reductive group G). The positive and negative root spaces of S are spanned by the weight vectors $X_\alpha$ and $Y_\alpha$ in G, so the result follows from the structure of $Sl_2$ reps.

[broken turtle]

Decrements

[soft parcel]

I forgot to add that the context is that V is a Lie algebra module, but I get what you're saying nonetheless

[wary elbow]

Right, so then we're just talking about a reductive lie algebra $\mathfrak g$. The logic is the same.

[broken turtle]

Decrements

[soft parcel]

Yeah, I think I was just confused because I thought that we know strings are unbroken from the theory about root systems, but now that I'm looking at it, the proof of the classificaiton of sl_2 representations also seems to give that strings are unbroken for any such irreducible representation, right?

Which is what we're using here

[wary elbow]

actually when I said "weight vectors" I was accidentally talking about the lie algebra anyhow

[wary elbow]

Yep, that's right. The only extra step is that we're considering a particular copy of sl2 inside the Lie algebra, corresponding to the weight \alpha. It doesn't follow immediately from root systems though, bc now we're talking about weights of an arbitrary representation

[soft parcel]

While you're here, this reminds me of another thing I was confused by. In one part of the book, Humphrey defines the "abstract theory of weights" where he says a weight is the set of all elements lambda in a vector space E for which <lambda, alpha> is an integer for all roots alpha.
Yet later on, he says that to avoid confusion, he will call any element of H^* a weight and a linear function lambda for which all lambda(h_i) are integral to be integral linear functions.

I don't really see the connection between the two definitions

[wary elbow]

What's E in this context? Something on which the Cartan acts?

[soft parcel]

It says that E is a euclidean space with root system Phi and Weyl group W in the intro to the section

[wary elbow]

What book is this -- I probably have it

[soft parcel]

Humphreys - Introduction to Lie algebras and representation theory

Chapter 13 for the first definition, and then now I'm working through chapter 21 where the second definition appeared

[wary elbow]

Ok, I have that. Let me look

[wary elbow]

Ok, so the second definition is just the specific case where E is the dual space of the cartan

[soft parcel]

Yeah. That seems to be in line with what I thought weights were

[wary elbow]

In the development of the subject, you want to start with abstract Weyl groups and then specialize later. That's all he's doing

[soft parcel]

But in allowing weights to be arbitrary linear functionals, you lose any integrality properties for weights then, right?

[wary elbow]

They're not arbitrary. In both definitions, he requires the pairing to take integer values

[soft parcel]

The pairing being?

[wary elbow]

He defines the pairings in both cases. In chapter 13 it's the bracket he defines. In chapter 21 it's the pairing between the Cartan and its dual. The point is that when you take E to be the dual Cartan, these pairings are the same

[soft parcel]

But the pairing condition \mu(h_i) = <\mu, \alpha_i> being in Z is what he defines as a linear function being integral

And then he says that on the other hand, any element of H^* will be called a weight

[wary elbow]

Where does he say this?

[soft parcel]

Bottom of page 112

[wary elbow]

ah ok, gotcha. Right, then the definition in chapter 13 corresponds to integral weights

[soft parcel]

It's just a bit weird to me because before in the book, arbitrary elements of H^* haven't really been of interest from what I can tell

[soft parcel]

Ok, thanks. Good to know I hadn't just misinterpreted something silly

[wary elbow]

Well, when you're dealing with finite-dimensional reps, all weights are integral

However, you do care about non-integral weights, bc they appear in infinite-dimensional irreps

So you don't want to throw them out entirely

[soft parcel]

Do you know where I might find this fact?

[wary elbow]

This follows from the theorem on p.112 since all weights of a finite-dimensional irrep are obtained by starting from the highest weight and then subtracting positive roots, which preserves the integrality

[soft parcel]

Ohhh

I see

That explains it

Thank you!

[wary elbow]

Sure!

[quiet compass]

What would be a good reference for learning about parabolic subgroups (in complex reductive groups)? Assume that I know enough differential geometry. In particular, I am interested in parabolic subgroups associated to one parameter subgroups and so on

[wary elbow]

Any book on linear algebraic groups will cover this. Standard references are Humphreys, Springer, or Borel.

[trim basin]

So I'm looking at an exercise about what we can infer from having an injective/surjective module homomorphism from A^n to A^m

My intuition would be that this is exactly the same thing as with vector spaces

So injective implies n<=m and surjective n>=m

So I (think) I have that with free modules it is possible to define dimension just as in vector spaces

Because I proved that if I have a surjection of a finitely generated module in itself then it is injective. So what that means is that if I have a free module with basis of two different sizes I can just map the bigger base into the smaller one and Id have injectivity so the basis would have to be the dame size

And so basically everything turns into linear algebra and since basis should be mapped to linearly independent sets if my function is injective, then the original base goes to a linearly independent set so the dimension is bigger in the destination module

And for surjective it's just seeing that theres a generating set of the destination module that has size at most the dimension of the origin module

But what's confusing to me is that the book hint is asking me to tensorize by A/m for m maximal

(To prove the surjective part)

And for the injective part it seems to suggest is false

[trim basin]

Well it asks whether its true or not, maybe saying that it seems to suggest it is false is saying too much

[digital parcel]

tensor products preserve surjective mappings (but not necessarily injective). recall that, when m is a maximal ideal, A/m is a field. both A^n and A/m are A-modules, so A^n \otimes_A A/m gives you an A/m-module. but modules over fields are vector spaces, and so this becomes a linear algebra question

[digital parcel]

for commutative rings, if you have an injection A^m -> A^n, then m \le n, though iirc the proof isn't trivial. for non-commutative rings, there should be some counterexample

[trim basin]

Ok so the question is in commutaive rings

[trim basin]

I'm guessing the point is that the tensorization proof shouldnt work here since it doesnt preserve injections

Which would explain why the book asks about it separately

[digital parcel]

right

[trim basin]

But what I dont get is why (in commutative rings) just following the standard proof in vector spaces wouldn't just work for both surjective and injective cases

Because I think it works unless I'm missing something

But I don't trust myself to be able to detect if I made some mistake

[digital parcel]

the vector space argument uses rank-nullity, which im not sure can be carried over

in particular, submodules of free modules aren't always free

[trim basin]

Where would you use rank-nullity?

Oh I see

I mean hmm I guess that would be the standard proof on vector spaces right

[trim basin]

Do you maybe have an example in mind?

[digital parcel]

R = k[x,y] and I = (x,y)

[trim basin]

Ok yes Ive seen that before

Thanks

[trim basin]

Though Im thinking and maybe things like this cant be Kernels of homomorphisms that go to other free modules

[digital parcel]

Wdym? Kernels of maps between free modules aren't necessarily free

Tho tbh I can't think of an example

[trim basin]

Oh in modules you cant necessarily extend a linearly independent set to a basis now that im thinking

Which I guess one would need to prove rank-nullity

Iirc

[digital parcel]

Yeah

also now that i think about it this is probably better suited for #groups-rings-fields

[stoic bone]

is this chat appropriate for asking questions about finite group representation theory?

[stuck dove]

hopefully we have some coxeter group fans here

im having to show that the respective matrix for a Coxeter group of type $\widetilde{D_n}$ for $n \geq 4$ has determinant 0, but because the Coxeter graph branches on both ends, I cant use the recursive formula for calculating this. it really seems like im supposed to use this formula, so im wondering if im missing something

[broken turtle]

hiidostuff

[lone jacinth]

One thing you can do is show that it has a null-vector.

Note you can think of a vector as assigning a number to each node if the graph. Then the action of the matrix is just taking 2 a node minus the sum of its neighbors. Aka it's the discrete Laplace equation.

This you can solve pretty easily, and the solution doesn't really depend on n.

[stuck dove]

Hmm

This might be beyond the scope of what ive learned so far about coxeter graphs

Which is frankly pretty little

[lone jacinth]

What part?

Or what do you mean?

All I'm using is that the entries of the matrix are -1 if there's an edge and 0 if there isn't an edge

(and 2 on the diagonal)

So say you have your diagram D6-tilde:
X X
X - X - X
X X
then start looking for a null vector we just start somewhere:
1 X
X - X - X
X X
now for this to be a null-vector we would need 2*1 - X = 0, alright
1 X
2 - X - X
X X
then we can fill in more
1 X
2 - X - X
1 X
and then more
1 X
2 - 2 - X
1 X
....

Pretty soon you find a null vector

[stuck dove]

To calculate determinants of associated matrices for coxeter groups im only given the recursive formula on the matrices upper left submatrices

And it seems im supposed to use that in particular

[lone jacinth]

What's the formula?

[stuck dove]

det(2A) = 2d_{n-1} - cd_{n-2}

Where the d's represent the determinants of the submatrices

[lone jacinth]

So you don't even define the matrix?

[stuck dove]

And c depends on the number associated to the edge connecting the last node (assuming there's a unique one)

[stuck dove]

A is defined as the matrix st the entry corresponding to simple reflections s1, s2 is -cos(pi/m(s1, s2))

si, sj i should say

[lone jacinth]

Well, you have m(si, sj), so you could calculate this. But okay, something using this recursion formula, 🤔

[stuck dove]

I suppose but I have to do this for all such ~Dn which requires induction regardless

But theres no good subgraphs

[lone jacinth]

It has Dn as a subgraph, so if you have already calculated the determinant for them you could use that

[stuck dove]

Sadly the recursive formula requires the last node to connected to only one other node

[lone jacinth]

But it does

[stuck dove]

Even if I do that though I dont get a determinant of 0 which is impossible for a non positive definite coxeter graphs

[lone jacinth]

Like in
1 6
2 - 3 - 5
3 7
Can't you remove 6 and 7?

[stuck dove]

I have to remove contiguous nodes i think

And the first has to be on the outside

[lone jacinth]

Ok, if you have all those constraints then you just can't apply the formula at all

So then you're definitely not supposed to use it

[stuck dove]

Thats whats so confusing to me though

As the chapter is explicitly about the application of the recursion to positive semidefinite graphs which arent definite

[lone jacinth]

What book is this?

[stuck dove]

Humphreys's reflection groups and coxeter groups

Chapters 2.4 and 2.5

[lone jacinth]

It doesn't seem to imply that the second vertex needs to only be connected to one thing

So you can just remove the branch to get Dn and Dn-2 union a vertex

The determinant for a disjoint union should just be the product of the determinants

[stuck dove]

Wait why not just Dn-2

Im thinking about removing the two branches at one end

[lone jacinth]

1 5
2 - 4 - 6
3 7

Like with this ordering you get
1 5
2 - 4 - 6
3
and
1 5
2 - 4
3

[stuck dove]

Oh wait I think im seeing what ur saying

Ok now this is making sense

Wait but still, in 2.4 we have that there is a numbering such that the last vertex is connected to only one other vertex

With its edge labeled m = 3 or 4

Wait I suppose we could let the isolated vertex be 1 and the rest are just numbered now Dn would be

In which case we still get 0

[lone jacinth]

I mean that's what's happening above.

7 is only connected to 6 with edge labeled 3

[stuck dove]

Oh I thought it was that the subgraph we end up with is only connected to one

But still this happens

Oh wow no im being super silly

The way it words it make it very obvious that its the initial graph thats numbered

Oops

[quiet compass]

Having thought about it a bit more, I think it will have the correct results, though perhaps from the wrong perspective for what I want. More precisely, in what I am doing, I want to work with "reductive = complexification of compact" and do things differential geometrically

[drowsy niche]

are submodules of free modules over k[F] (k a field and F a free group) free?

related: for commutative rings R (assuming choice) every submodule of a free R-module being free is equivalent to R being a PID

is there a similar characterization for noncommutative rings?

[digital parcel]

I don't think so. Eg k[Z^2] is iso to k[x, x^{-1}, y, y^{-1}]

and that ring has ideals which shouldn't be free

[drowsy niche]

? I'm talking about k[F] where F is a free group

[digital parcel]

oh sorry right it's a free abelian group not a free group

oops

hmm

[lone jacinth]

This is at least true if F is a free monoid. So I would suspect it holds true for groups as well

[drowsy niche]

roughly, what would that proof look like for F a free monoid?

[lone jacinth]

I don't know if I recall much detail offhand, but basically just pick a groebner basis and notice that the submodule is freely generated by it.

Worth noting it's enough to check that left ideals are free, by Kaplanskys theorem

Then you want to notice that two principal ideals either don't intersect at all or one is contained in the other

Hmm, sounds like this should translate to the group algebra pretty immediately

I guess it's a little more annoying since word can get shorter when you multiply them

[drowsy niche]

k[F] is a free ideal ring by corollary 7.11.8 here
https://webhomes.maths.ed.ac.uk/~v1ranick/papers/cohnfr.pdf

so yes, this should be true

[lone jacinth]

For this question you have these results

A path algebra A = kQ/I where I is contained in J^2 for J the arrow ideal, then A is hereditary iff I = 0.

Then for the upgrading from projective to free Q cannot have more then one vertex, so you're left only with the free algebra.

So I guess the result would be an admissible path algebra has this property iff it is a free algebra.

[drowsy niche]

sorry, I'm not familiar with path algebras. Does this in some way address the question for all noncommutative rings? Or is this meant to address why free algebras have this property?

[lone jacinth]

Well, it addresses the question for a large family of noncommutative rings.

Namely, admissible path algebras

[drowsy niche]

does this include all finite dimensional k-algebras?

[lone jacinth]

If k is alg closed it includes all of them up to Morita equivalence

[drowsy niche]

interesting

[lone jacinth]

But it should be easy to see that no finite dimensional algebra has this property (besides fields)

[drowsy niche]

is there a standard reference for path algebras covering this?

I'm not seeing much online

[lone jacinth]

Some standard texts here
https://wiki.math.ntnu.no/ma3203/2024v/pensum

Assem Simson Skowronski is the most approachable.

[unborn rampart]

If a K-algebra R is generated by G, then there's a canonical homomorphism from R to the free algebra K<G>, right?

[fierce steeple]

other way round

to say it is the free K-algebra means this

[unborn rampart]

Yeah, I know there's a projection, but I was hoping there was an inclusion into K<G> too

[silver goblet]

example is if K = Z and R is Z/2 generated by x=1. There is no K-algebra homomorphism Z/2 --> Z<x>

for torsion reasons

[unborn rampart]

Yeah I think I see why you can't have an inclusion into K<G>, like if you try to include k[x, y] into k<x, y> it would force the image in k<x y> to be commutative

[foggy galleon]

is there a condition for a ring R so that it has the following property? If M, N are direct summands of R^n (as R-submodules I mean) then M cap N is also a direct summand. Same question but with M+N.

Actually, I'm interested in the case that M and N are locally free (this is implied by the given assumption if M, N are finitely generated I think?)

[past cove]

this clearly holds PIDs (the first condition)

in the second condition do you mean direct sum or arbitrary sum?

[foggy galleon]

I mean the submodule generated by M and N inside R^n

[lone jacinth]

For the second question:

Consider M = R(1, r) and N = R(1, 0) in R^2. They are both direct summands (and are free of rank 1). Their sum M+N is R(1,0) (+) R(0, r). This is a direct summand iff Rr is a direct summand of R.

So this would imply R needs to be von Neumann regular

I'm thinking this is also sufficient. Not sure.

[drowsy niche]

It should further work for hereditary rings

since then 0 -> M n N -> R^n -> R^n/(M n N) -> 0 splits

[lone jacinth]

Why does that split?

[drowsy niche]

R^n/(M n N) is a submodule of R^n/M (+) R^n/N

which should then be projective

[lone jacinth]

Nice

[foggy galleon]

thanks all, very interesting answers

[lone jacinth]

If R is Noetherian I think the first condition implies hereditary:

Say I is an ideal generated by a1, ..., an. Consider M inside R^n+1 generated by (a1, 1, 0, ...), (a2, 0, 1, 0, ...), ... And N be 0 x R^n. Then their intersection is exactly the kernel of the presentation
R^n -> I. So if this splits I is projective

So I guess this gets you semi hereditary at least

[lone jacinth]

I wrote something I deleted, but property 1 should be equivalent to semi-hereditary. I just argued it implies semi-hereditary. Conversely for R^n = M (+) M' projecting N into M' you get a sequence
0 -> MnN -> N -> M'
the image is a fg submodule of projective so projective, then MnN is a summand of N and hence of R^n

And property 2 should be equivalent to von Neumann regular just because M+N is finitely generated, so it's a summand

[foggy galleon]

I am confused by this lemma. Aren't they saying that M locally free implies M projective? But isn't this false in general if M isn't finitely generated

source: https://stacks.math.columbia.edu/tag/05JN

[past cove]

I don't see them saying locally free anywhere in this lemma

[foggy galleon]

isn't locally projective the same as locally free. In any case, they are saying that M_p projective for all p implies M projective, no?

uhm wait no

[fierce steeple]

Locally projective is not the same as locally free or this would not be a thing they define right

[fierce steeple]

Not quite

[foggy galleon]

so I guess like locally projective is not something you can check on stalks?

[lone jacinth]

Locally free is like this weird thing right? Not just M_p free for all primes p

[fierce steeple]

Locally free means free locally (for Zariski topology)

The Mp thing is equivalent given like fg and Noetherian hypotheses

[foggy galleon]

i c

[potent plaza]

people who read weibel

[subtle smelt]

I am bit confused myself. You seem to be refering to the "In particular part" at the bottom, in which case if M_tilde was locally free, then it admits an open cover on which it is free and hence projective. So M_tilde being locally projective implies M being projective.

[potent plaza]

my purpose is to learn spectral sequences. can i skip and read chapter 5 of spectral sequences immediately?

[foggy galleon]

I think yes

I mean probably be aware of derived functors

[subtle smelt]

We can even show it directly: since the rank is locally constant we can split Spec A into finitely (as Spec A is quasi-compact) many connected components Spec A_i (clopen hence themselves affine) on which the M_tilde is just a free module.

In this case we can realize M as the direct sum of restrictions and pushforwards, where each factor corresponds to the rank on that connected component many copies of A_i viewed as a A-module (by quasicoherence and the equivalence of categories you can note that the pullback and pushforward are identified with tensoring and restriction).

But since A is a direct sum of the A_i, each A_i is a direct summand of the free A-module A and hence projective. So M works out to be a direct sum of projective modules, hence projective.

[hushed bone]

This lemma is a mistake Grothendieck made in EGA. The proof of some things regarding smoothness are incomplete because of this

[limpid horizon]

Why is the characterization of depth/grade using Ext helpful?

Or, I guess I'm trying to look for more intuition on this

Also I wasn't sure why it needed M neq IM. The lemma right before it describing an isomorphism between Hom and Ext didnt need that condition

[hushed bone]

Exts are useful, they’re one of the most studied groups in comm alg

If you want to see a specific application you can look in the last chapter of Peskine Szpiro where they show a result about cohomological depth

At one point they use the fact that you can measure depth via Exts because Exts will vanish on a projective module

[limpid horizon]

Ty

[steady lance]

Also just generally it can be tedious to find maximal regular sequences and translate between rings/modules in that way. Computing Ext kinda lets you zoom out and view the ring as a single object instead of a space with elements.

It can be nice to think of depth as "the first ext that doesn't vanish" and dimension as " the last nonvanishing ext"

Then when you need some result to be true you can just frantically read nlab articles about derived functors and hope for the best

I kinda cheat with this intuition too and think of cohen macaulayness in terms of local cohomology as "the dimension of the simplest hole in the m-adic topology is the same as the dimension of the corresponding affine coordinate ring". So it's somewhat of a condition on different topological notions behaving relatively nicely. I'm not a geometer, though, so this might be very wrong. But it's a nice fairy tale I tell myself

but at least you can say that cohen macaulayness means that there's only one dimension with holes in the m-adic topology (in some sense)

(also if any geometers want to correct me on anything here, please do. I was partially writing this out to see if my intuitions actually do make sense)

[round seal]

finding maximal regular sequences is hard but computing vanishing of exts is easy (at least for M2)

[limpid horizon]

Ty ppl

[unborn rampart]

What does the arrow ideal of the path algebra of this quiver look like? Is it just { a x^n + b y^m | n, m in N, a, b in K }?

[lone jacinth]

Well sums of those things

So x + x^2 is also in there for example

[unborn rampart]

Oh, right 👍 does it make sense to talk about the subalgebra generated by the edges? How would it differ from the arrow ideal?

[astral ginkgo]

What do you mean by algebra generated by the edges exactly?

[lone jacinth]

I mean, it makes sense I guess.

It would look like k + J for J the arrow ideal

So in this case it would be k[x, y]/(xy)

Unless your subalgebras aren't unital, then it would just be the arrow ideal

[unborn rampart]

I see, thanks 🙏

[unborn rampart]

Just the smallest algebra containing the edges (or the non-trivial paths or whatever)

[fallen tree]

Hey everyone is Aluffi's notes from the underground any good?

[sacred sentinel]

I really liked it for self study

[soft parcel]

Does anyone know how to vertify that R^theta is closed under reflections? I've attached a picture of where I'm stuck

[limpid horizon]

I'm not sure why you need IM neq M. Is it something with nakayama lemma?

ok its just to make sure M/xM nonzero

[hushed bone]

An M-sequence means that M/xM ≠ 0

If x is a maximal M-sequence but IM = M, you could have a non-zero divisor in I, because it could be that x,yM = M

So it doesn’t contradict that fact that x is a maximal M-sequence

[limpid horizon]

You could have a nonzero divisor of M/\mathbf{x}M in I?

[hushed bone]

Yes

Because that non-zero divisor could then make M =( x,y)M

So there’s no contradiction

[vague pawn]

What does it mean for a ring to be integrally dependent on a subring

[lone jacinth]

I would guess that it's an integral extension. Maybe you have some context...

[vague pawn]

context:

[lone jacinth]

Then yes, R is an integral extension of k[x1, ..., xn] is the statement of Noether normalization

[limpid horizon]

idk i think im missing something here cause if we're already saying x is a maximal M-sequence then doesnt that already mean there wouldnt be any nonzerodivisors in I of M/xM

[hushed bone]

Dude I literally just explained why you need IM ≠ M for this to be true

[limpid horizon]

Are you saying that if x,yM = M then y is a nonzero divisor of M/xM?

[hushed bone]

No

Okay

Exercise

[hushed bone]

Write a proof of this

And I will tell you where it’s wrong

[lone jacinth]

The definition of M-sequence includes M/xM being nonzero right?

[limpid horizon]

yeah

ok ill be back

[lone jacinth]

So then it could be that x isn't maximal if you drop this requirement

Which would give you trouble

[soft parcel]

bumping this, in case

[hushed bone]

@limpid horizon did you figure out where it’s necessary

[limpid horizon]

Yeah you need M neq IM to rule out the possibility of there still being a nonzero divisor of M/xM in I just with M/(x,y)M = 0

[hushed bone]

And why is that necessary?

[limpid horizon]

If M/(x,y)M = 0 then M = IM. Or were you asking something else

[hushed bone]

Like, what’s wrong with M/(x,y)M = 0?

[limpid horizon]

Well if we didnt have M neq IM then we wouldn’t be able to say that Ext^n is really nonzero? Cause there still could be a nonzero divisor

[hushed bone]

Okay what I mean is what’s the relevance of having a non-zero divisor but then having M/(x,y)M = 0

[limpid horizon]

Its not M/xM-regular .?

[hushed bone]

Right!

[limpid horizon]

Yes I was kinda forgetting that part of the defn of M-sequence

[hushed bone]

There’s the other condition on a regular sequence which you often can forget because you’ll often work in a local setting where Nakayama just ensures you never worry about it

But without that you have to actually worry about that condition, and this leads to depth having some weird ass properties

Like if you aren’t Cohen Macaulay you can have the depth like do weird shit when you localize further

This is why S_n isn’t as simple as just “all maximal ideals are depth >= n” or something

[limpid horizon]

What’s S_n here?

[hushed bone]

Serre’s condition

For all primes p, depth M_p >= min{ht p, n}

[limpid horizon]

Thats neat

[hushed bone]

Pretend I just said M instead of R, you can actually define it for modules but you pretty much never will

So it says that for primes up to height n, the ring is Cohen Macaulay

But also for higher primes past that the depth is always at least n

There’s also R_n which asks that for up to height n primes the ring is regular

Serre’s famous result on normality says that normal = S_2 + R_1

[limpid horizon]

What is n?

[hushed bone]

The subscript

[limpid horizon]

Any integer?

[hushed bone]

I mean I guess

But you should not pick a negative one

[digital parcel]

S_2 and R_1 my beloved

[steady lance]

Normie pick tbh

[digital parcel]

why

[hoary tree]

Find The Sum
2 0 2 7
1 4 + 8 3

[steady lance]

Very bad pun on “normal”

[hexed tangle]

Is there a way to show that the differentials of a totalization of a double chain complex wihtout a massive computation compose to zero?

[lone jacinth]

I mean, it's not really much computation.

The differential is just the sum of a bunch of differentials most of which compose to 0.

The only thing that doesn't vanish immediately are the commutative squares of the double complex

Which just cancel eachother by definition of a square commuting

[hexed tangle]

what I get dosen't vanish 🙁

[hexed tangle]

Can I send you a screenshot of there I am?

[digital parcel]

Ohh

[hexed tangle]

dosen't it go from p+q = n to p+q = n-1?

[lone jacinth]

Depends on your convention, but it doesn't matter

You can replace all n+1 with n-1 if you want

[hexed tangle]

fair

[spare fractal]

In a proof by contradiction: G is a minimal simple group (minimal simple meaning simple and every proper subgroup is solvable), P a Sylow p-subgroup of G.

The following lemma has been proven:
Lemma 4.7: Let 1 \neq T \leq N_G(P). Then N_G(T) \leq N_G(P).

The author now concludes that N_G(P) must be a Frobenius complement in G (whence it is not simple due to the existence of the normal Frobenius kernel).

The only characterisation of Frobenius complements for general groups known to me is one from a separate source (Theorem 7.7 in Gorenstein's Finite Groups) where it is easy to verify that 1 =/= N_G(P) <= N_G(P), so N_G(N_G(P)) = N_G(P). However I can't really seem to get any hold on how this implies that N_G(P) \cap N_G(P)^g = {1} for every g \notin N_G(P). Fiddling around with N_G(P) \cap N_G(P)^g =/= {1} for some g \notin G and applying the hypothesis seems to not really lead anywhere.

Is there some simple argument I am overlooking?

[limpid horizon]

depth M/xM = depth M - n (for x an M-sequence of length n) just follows from every maximal M-sequence having the same length right?

[past cove]

yes

[lone jacinth]

Don't know if this is the argument they have in mind, but:

Let T be the intersection of P and P^g.

Then N(T) <= N(P) and N(T) <= N(P^g).

Now as P is a p-group N_P(T) strictly contains T. Let h be an element in it, not in T.

Then h is in N(T) so in N(P^g) which would mean h in P^g. But then h is in T contradiction. Hence we must have T = 1.

Then G satisfies the definition of a Frobenius group acting on P by conjugation.

[spare fractal]

Using growing normalisers seems right on track actually.

[spare fractal]

Oh actually, the frobenius complement is supposed to be N_G(P), so we'd need to verify that N_G(P) \cap N_G(P)^g is trivial for g not in N_G(P)

[lone jacinth]

Well, maybe it's not the argument they had in mind then, but it works

[spare fractal]

Oh that's what you meant!

[lone jacinth]

Wait, no N_G(P) is still the Frobenius complement in my argument

It's the subgruopthat fixes P under conjugation

[lone jacinth]

Like you only need this if you insist on using thm 7.7

Which they don't claim they do

[spare fractal]

Uh do you have any link or something to somewhere where a frobenius complete is defined pretty much like this

[lone jacinth]

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

[spare fractal]

I am sorry I may be a bit slow here, but I don't see the full step of how this fits the permutation group description or the disjoint conjugates description

[lone jacinth]

G acts on the set of p-sylow subgroups by conjugation.

As the N(P)s don't intersect no element fixes more than one sylow subgroup

Then N(P) is the subgroup fixing P hence a Frobenius complement

[spare fractal]

but how do we see the N(P)s don't intersect? this only assumes the Ps intersect

[lone jacinth]

Hmm, oh yeah. I guess that's an issue

[spare fractal]

this argument could be fixed I think if we had some way to force N_G(P) to be nilpotent but i am not sure this is true?

[lone jacinth]

Hmm, how about this:

Let Q be a q-sylow subgroup of N(P). Then the N(Q) <= N(P) condition gives us that Q must also be q-sylow subgroup of G (or trivial). If N(Q) = N(P) for all the sylow subgroups, then N(P) is nilpotent.

If N(Q) < N(P), then you can just replace P by Q and continue by induction

[spare fractal]

So the idea is to induct downwards using potentially different primes until one of them has a sylow subgroup with a nilpotent normaliser?

[lone jacinth]

Yeah, or I guess simplifying a little: pick p so that |N(P)| is minimal.

If N(P) and N(P)^g intersect. Then it will contain an element h of order q for some prime. Then N(N(...(h)..)) will eventually contain N(Q) for some sylow subgroup. But Then N(Q) <= N(P) n N(P)^g contradiction

[spare fractal]

So the total argument becomes:

Pick sylow subgroup P of G with |N_G(P)| minimal.

Claim: N_G(P) is nilpotent.
Proof: Assume false, then there exists a nontrivial sylow q-subgroup Q <= N_G(P) with N_G(Q) < N_G(P) since N_G(Q) <= N_G(P) by the lemma. Since |N_G(Q)| < |N_G|, Q must fail to be sylow in G by the choice of P. However, N_G(Q) must contain a larger q-subgroup (every q-group is contained in a sylow q-group, q-groups are nilpotent), a contradiction!

Let K = N_G(P).
By the lemma, N_G(K) = K.

Claim: K \cap K^g is trivial for all g not in K
Proof: Assume false, then let T = K \cap K^g. T is a proper subgroup of the nilpotent K, so T < N_K(T). But now N_K(T) <= N_G(T) = N_G(K) \cap N_G(K)^g = K \cap K^g = T, contradiction!

Thus K is a frobenius complement in G.

Seeing this I am most certain the author likely had something else in mind but thank you!

I realise just now that there was a hidden oddness assumption required on the prime p behind P to work so this still doesn't fully work as is 😭

[drowsy niche]

idk if this is the right channel for such a request, but does anyone know where I might find a copy of lecture notes for Kontsevich's 1998 lectures Triangulated categories and geometry at ENS?

(found: https://www.lamfa.u-picardie.fr/marin/docs/coursKont.pdf)

[wanton spoke]

which ENS are you talking about?

[drowsy niche]

Paris

[wanton spoke]

I think you can just check in the library or go to Jussieu's library

you'll find something

[drowsy niche]

oh tbc I meant the lectures were at ENS paris lol

I don't mean where to find the lecture notes at ENS

[wanton spoke]

ah my bad

[drowsy niche]

I unfortunately am not at ENS

[lavish gull]

Hello I've got a homework question about Noetherian rings and Noetherian modules. Is it better to ask here or in one of the math help channels?

[limpid horizon]

Here

[lavish gull]

I have been asked by my Module Theory lecturer to flesh out the proof of part 1 of the following theorem in Steven Roman's Advance Linear Algebra

Theorem 5.8 (part 1)
Let R be a commutative ring with identity.

1. R is Noetherian if and only if every finitely generated R-module is Noetherian.
   Roman then goes on to say
   For part 1), one direction is evident. Assume that R is Noetherian and let M = <<u_1, ..., u_n>> be a finitely generated R-module. ...
   I've gotten so turned around, I'm not even sure which direction is supposed to be "evident" let alone how to flesh it out.
   It may be helpful to state the preceding theorem
   Theorem 5.7
2. An R-module M is Noetherian if and only if every submodule of M is finitely generated.
3. In particular, a ring R is Noetherian if and only if every ideal of R is finitely generated.
   I apologize for the fairly technical question, but Google has been of no help to me

[hallow bone]

well, R is a finitely generated R-module

so that gives you one direction

[lavish gull]

Oh for goodness sake

Thank you!

[hallow bone]

for the other, maybe begin by proving that the product/direct sum of two Noetherian R-modules is Noetherian, and then use the correspondence theorem

I'm not sure what the intended method is, but this is how I would do it

(more generally any extension of a Noetherian module by a Noetherian module is Noetherian)

[hallow bone]

that happens, lol, sometimes you just don't think about one little thing that makes it trivial

[fierce steeple]

Yeah I think this is the best path, like prove that given a short exact sequence 0 -> M -> N -> P -> 0 of modules, TFAE:

1. N is Noetherian
2. M and P are Noetherian

Fun times

[hallow bone]

pretty diagrams

like, to prove that N is Noetherian is essentially using the fact that the congruence lattice is modular

fun times

[fierce steeple]

exposing my ignorance but wdym by this

[fierce steeple]

or it's automatic if you do the other direction

Actually lol I find this a funny and trivial application of five lemma lol

[hallow bone]

Let
A0 < A1 < A2 < ...
be an ascending chain of submodules in N, and consider the following two:
A0 \cap M < A1 \cap M < A2 \cap M < ...
A0 + M < A1 + M < A2 + M < ...
one will be an ascending chain in M and the other will be an ascending chain in P (because of the correspondence theorem), so they both stabilize at some n. Thus we have An < An+1 with An \cap M = An+1 \cap M and An + M = An+1 + M. If An =/= An+1, then we would have an N5-sublattice which is impossible because of modularity, so An = An+1 and we are done.

[fierce steeple]

[hallow bone]

:3

this is what N5 looks like

[hallow bone]

well no one knows lattice theory in this day and age

[past cove]

I hate the nerve construction man

[hallow bone]

😔 you hate hasse diagrams?

[hallow bone]

I guess the five lemma with the first and last modules 0 is essentially the modularity property

lmao

in some interpretation

[fierce steeple]

Lol fair enuff

[worldly zealot]

what is this for you? this is a very familiar diagram to me but i imagine in a completely different context

[hallow bone]

this is the hasse diagram for a poset

this happens to also be a lattice (meaning that every pair of elements has both a join and a meet)

important about this one specifically, is that a lattice is modular (a relaxation of distributivity) iff it doesn't have this specific lattice as a sublattice

[hallow bone]

it happens that the submodule lattice is always modular, hence I can conclude that An = An+1, as else it would produce an N5 sublattice

[worldly zealot]

i c

this is the lattice of torsion classes for A_2 and i don't really know general lattice theory so i was curious

interesting

[hallow bone]

well then you've got an easy proof that this lattice is not modular :P

what are torsion classes of A2?

[worldly zealot]

A2 is a quiver (path algebra), torsion classes are certain subcategories of the module category

they have a torsion free counterpart and vaguely generalize the notion of the torsion, torsion free pair of subcategories of Z-mod, the lattice of torsion classes of an algebra (under inclusion) is studied a bunch

[fierce steeple]

Quiver in ecstacy

[hallow bone]

that's interesting that N5 shows up there

is the lattice structure of torsion classes super important?

[worldly zealot]

yeah but it wasn't entirely relevant to me so i don't know it too well

https://arxiv.org/pdf/1711.01785

this is a very important paper in the field though, jagr probably knows this stuff better

[hallow bone]

oo I might look into this, thanks

it's fun to see lattices pop up in math

I'm gonna be honest, I try to avoid pure lattice theory as much I can because I don't know it well lol

all the pure UA people used to do everything mainly lattice-theoretically, I'm glad Malcev conditions exist

[worldly zealot]

i think this stuff is pretty cool, the lattice of torsion classes for the kronecker quiver is extremely intriguing

https://arxiv.org/pdf/2102.08527 good exposition

[limpid horizon]

[lone jacinth]

The hasse diagram of the torsion classes for An is an associahedron, so that's fun

[hallow bone]

that is cool

[cloud karma]

Trying to find the minimum value k such that any element R^m (x) R^n can be written as a sum of k pure tensors. My guess would be k is the minimum of m and n but I'm not sure how to show this in a clean way. Using the isomorphism R^m (x) R^n = M_(mxn)(R) we can take a pure tensor x (x) y and map it to a rank 1 matrix x y^T but then how do you go from there?

R here is real btw

[lone jacinth]

So then the question is just how to write a matrix as the sum of rank 1 matrices.

You just pick a basis for the image and compose with the various projections

[cloud karma]

I think I figured out the lower bound but not the upper one. Suppose that A is a sum of rank 1 matrices then rank(A) <= k. There exists matrices of rank min(m,n) for example we can embed I_min(m,n) into M_mxn so representing such a matrix requires k >= min(m,n)

Oh and the upper bound is obtained by SVD

god why did this take me so long

[grave rain]

-# hi enpeace

[near lantern]

So true

Although, why it the generalisation to UA taken to be congruences and not subalgebras?

[hallow bone]

the subalgebra lattice actually doesnt contain much information about the algebra

[near lantern]

Oooh, what are the five torsion classes?

[hallow bone]

congruences in a sense carry the "equational logic" of an algebra, so e.g. identities satisfied by the congruence lattice of every algebra in a variety can be used to deduce the existence of terms with certain properties

[near lantern]

Surely in an Artinian category, Serre subcategories are just sets of simple modules?

[near lantern]

Wait that's too cool

And why that

You'd think a permutahedron

What do other types give then

[near lantern]

OK actually Noetherian_cong(A) = Noetherian_sub(A^2)

so I really don't get it

[near lantern]

I see.

[hallow bone]

not every subalgebra of A^2 will be a congruence

[near lantern]

Um

Oh equiv rel

Right

[hallow bone]

yes

[near lantern]

What's an example of such an identity? Isn't bot(cong lattice) the diagonal, which has no non-trivial identities?

[hallow bone]

well, congruence modularity for one

call an algebra congruence-modular if its lattice is modular (that is, doesnt have N5 as a sublattice)

[near lantern]

Oh identities of the lattice

[hallow bone]

a variety is congruence-modular, iff there exist 4-ary terms m0, ..., mn such that the variety satisfies:
m0(x, y, z, w) = x
mn(x, y, z, w) = w
mi(x, y, y, x) = x for all i
mi(x, x, y, y) = mi+1(x, x, y, y) for even i
mi(x, y, y, z) = mi+1(x, y, y, z) for odd i

[near lantern]

Why does this capture equational logic of A though? I can see why individual congruences (especially on free algebras) do.

[near lantern]

So for example, why should I expect a result of this type to be true?

[hallow bone]

a result of this type can be intuited as a "canonical proof"

so, because modularity holds for every algebra (and in particular the free algebra), you can expect there to be some "canonical proof" for the fact that the congruence lattice is modular, and this will be in the form of some terms that satisfy certain equations

[near lantern]

Hmm. I don't quite see how to read it...

[hallow bone]

modularity is quite a bad first example to intuit this kind of stuff i admit

you could write whole books about the intricacies of congruence modular varieties though lol, its a deep subject

[near lantern]

IG the universal pair to check is (<x = y> + <z1 = w1>) ∩ <z1 = w1, z2 = w2> ⊆ (<x = y> ∩ <z1 = w1, z2 = w2>) + <z1 = w1> or something.

[hallow bone]

yes exactly, something along those lines

[lone jacinth]

I'm not sure it's a particularly good answer, but you have this correspondence between torsion classes and support tau tilting modules where edges in the hasse diagram correspond to support tau tilting modules that only differ in a single summand.

Indecomposable modules of An are just determined by their support so intervals in [1, n]. These you can think of as diagonals in an n+2-gon, and two are tau-rigid if the diagonals don't cross.

So support tau tilting modules are triangulations of an n+2-gon, which you can translate to bracketings and get the associahedron

[near lantern]

But this looks like it should need 6-ary terms

[near lantern]

Ooooh

IG diagonals could be positive roots in the root system

[lone jacinth]

Let P2 be the simple projective, P1 the projective injective and S1 the last simples.

Then the torsion classes are
mod A,
0,
add P2
add P1(+)S1
add S1

[near lantern]

And a tilting module could be a chain from a simple root to the longest root where you add one simple root each time

Is this the order complex of the root poset under dominance?

One more topic added to my reading list 😔

[lone jacinth]

I'm not sure I know what these words mean

[hallow bone]

there is a particularly pleasing equivalent condition to modularity, called the shifting lemma
however, i dont yet see any intuitive way why these should be equivalent.

there's also things about higher dimensional equivalence relations, which i want to look into some time, and they perhaps give a more direct reason to care about modularity specifically

[near lantern]

Let a <= b, a, b positive roots, if b-a is a positive combination of simple roots. This makes positive roots into a poset.

The order complex bit is very wrong actually

But I'm suggesting the torsion classes are given by maximal chains in this poset.

[lone jacinth]

Hmm, so this seems to only depend on the composition factors on the modules, while torsion classes generally has to do with quotients.

Though for An it shouldn't make a difference

[near lantern]

Aren't Serre subcategories closed under submodules and quotients? In particular, if it has P1 doesn't it have P2?

[lone jacinth]

They're not Serre subcategories

[near lantern]

Oh my bad

I hallucinated that

[lone jacinth]

Torsion classes are closed under quotients and extensions

That's their defining property

[near lantern]

Usual torsion modules are though, aren't they?

[lone jacinth]

Yes

[near lantern]

Maybe there's a correspondence but not as direct as chain S ↦ add {indecomposbles corresponding to S}?

That certainly doesn't work for the A2 list you gave, since then add P2 (+) P1 would be torsion instead of mod A.

[near lantern]

Maybe it's this but you close S under quotients (which makes the answer depend on the direction of arrows in the quiver, as it should).

[lone jacinth]

Funnily enough the shape of the hasse diagram does not depend on the direction of the arrows

[near lantern]

Hmm

Believable by some reflection functor business

[lone jacinth]

Yeah, or you can move the entire thing to the derived category

[near lantern]

But the specific indecomposables that make up the torsion class represented by a vertex could change with the arrow directions

[lone jacinth]

Yeah

[mystic path]

A general question - I have some commutative unital semigroup, and I'm interested in some ideal of this semigroup. What are the go-to directions in this scenario?

[worldly zealot]

this has an interesting connection to the cluster algebras of type A_n (which are isomorphic to the Z-form of the coordinate ring of Gr(2, n+3)) where clusters are noncrossing triangulations of an n+3 gon and mutations are changing the diagonal of a square inside the triangulation. then the bridge between this and support tau-tilting modules is that the cluster category includes exactly the shifted projectives as the additional indecomposables

(not directed to you but this was the most relevant message to reply to)

[soft parcel]

Could someone who knows about root systems of Lie algebras could take a look at the following stackexchange answer?
https://math.stackexchange.com/a/1331730/1747796

I don't feel like the finishing argument on root strings is correct since I at least don't know of a reason why the pairings <alpha, (beta + i alpha)^v> at the upper end of the root string should be positive

But if someone who knows more about the subject can correct me, that would be nice

Ah, nevermind

It works if one just takes gamma instead of the coroot of gamma

[quick warren]

idk if this is the place to say it but

how do i express a permutation as a product of cycles again?

i forgot lmaooo

OH WAIT nevermind

i rember!

[waxen fractal]

because you already found the answer, i can offer the brainrot answer instead

brainrot: a permutation is a group action by the group (Z, +)

every group action decomposes into orbits

further brainrot: this is the coyoneda lemma

[near lantern]

Hmm but isn't the main content in writing a G-set as a disjoint union of G/H's for H a subgroup of G than in writing it as a colimit of G's?

[waxen fractal]

yeah actually you're right, i suppose ill have to take it back. it's not obvious that a quotient of disjoint unions would actually always be a disjoint union of quotients (this is very untrue in general i believe)

further brainrot redacted. only pay attention to the mid tier brainrot

[hallow bone]

case in point: the trivial top congruence lol

[maiden wadi]

Keith I like ur brainrot

[waxen fractal]

that's one of the best compliments you could ever give me

[fierce steeple]

I'm so confused lol

[waxen fractal]

ok, the details is that, given a permutation f on X, you have that n \in Z acts on x \in X by sending x to f^n(x)

and, given a Z-action on X, you can look at how 1 acts on X to recover the permutation

[fierce steeple]

Oh wait yeah just a single things yes sure lol

So this is jusy saying a map Z -> Aut(X) of groups is just a choice of a point lol

[waxen fractal]

yeah, this generalizes to all free groups

action by a free group on k generators is a choice of k permutations on a set

[fierce steeple]

By definition lol

Fair lol

[waxen fractal]

do u wanna hear more brainrot stuff related to this

actually ill say it in case captainsnake reads it too

there are two distinct ways to freely turn a self-map into a permutation, and they relate to the monoid homomorphism from (N, +) to (Z, +), and applying either base change or cobase change, but for just general actions and not modules

actually i gtg so ill leave it at that, but it's one of my favorite pieces of "brainrot"

[hallow bone]

universal property

[fierce steeple]

Yeah I think like it is trivial in a way I did not think about lol

[waxen fractal]

the walking permutation

✍️

[hallow bone]

also #1203471755449073774

[waxen fractal]

is this a thread for shitposting about algebra

[hallow bone]

its a better discussy

[waxen fractal]

the walking discussion

[digital parcel]

no it's a very serious thread

[hallow bone]

serious topics

like what kind of object Rome is

[waxen fractal]

btw this chat name throws me off

who tf calls it advanced algebra

it sounds like a bizzaro name

[hallow bone]

many a pre uni student comes here

[waxen fractal]

it makes me wonder if im advanced enough

[hallow bone]

somehow having given themselves the undergrad role

[waxen fractal]

like will i measure up

to the advanced ness

[hallow bone]

youll be funee

[waxen fractal]

this server gives me more imposter syndrome than TAU tbh

ppl say it's the opposite but nah

[hallow bone]

how so?

[waxen fractal]

vibes

[lone jacinth]

What should it be called?

[fierce steeple]

Algebra

[waxen fractal]

abstract algebra

[fierce steeple]

Ig abstract algebra was what it was called once

[lone jacinth]

So delete the #groups-rings-fields channel?

[fierce steeple]

Though abstract algebra is kind of redundant as a name lol

[lone jacinth]

Before it was split into two channels

[waxen fractal]

i mean that's groups rings and fields not abstract algebra

i dont see the contradiction

[digital parcel]

at least in the US, i've seen courses called "abstract algebra" that cover groups/rings/fields

so it might cause confusion

[fierce steeple]

Yee this is what I mean

[waxen fractal]

well that's an introduction to abstract algebra

[fierce steeple]

I would call those things abstract algebra if needed lol

[waxen fractal]

they're abstract algebra too

i mean to say groups rings and fields are a subset

so it makes sense as a channel name

[fierce steeple]

And advanced is the more advanced stuff

[lone jacinth]

Almost as if there was a channel for the intro stuff and a channel for the more advanced stuff

[waxen fractal]

im confused

they're both in the advanced category

and advanced is not a word anyone uses anyway

[fierce steeple]

Yeah I think the advanced vs early uni distinction is just like not that real

Like in the UK here many of the advanced channels here are stuff you start immediately anyway like real analysis

[waxen fractal]

anyway i dont rlly have a license to complain

cuz i have done jack shit

once i get a proper license then i will tho

[last talon]

Yeah I could complain for hours about how the division into topics here (especially in the early/pre uni sections, but also in the existence of early uni altogether) is just
US-centric to the point of being impossible to intuit if you’re from any other system

[fierce steeple]

Yeah eg idk what discrete maths refers to that isn't absorbed by other channels

n choose k maybe idk

[lone jacinth]

Graph theory and generating functions...(?)

[waxen fractal]

anyway i got a question for this chat

[fierce steeple]

True but isn't graph theory more like combi structures

[lone jacinth]

Damn, too many channels

[fierce steeple]

To me this is not a standard topic to cover until later or smth

[last talon]

I think it’s like
Graph theory (A level decision maths)
Not
Graph theory (3rd year uni maths)

[digital parcel]

i guess it's like "basic graph theory" vs "more complicated graph theory", kind of like probability-and-statistics vs advanced-probability

[fierce steeple]

Ah ok

Lol

[digital parcel]

i suppose anything that's in Rosen's discrete math textbook falls under discrete math

[fierce steeple]

Idk that book but sure

[digital parcel]

it's the holy grail of discrete math classes in the US

i think

[lone jacinth]

At least my discrete math class covered number theory, first order logic, graph theory and complexity theory

[fierce steeple]

Yee sure

I never did any graph theory or complexity theory at uni lol

[waxen fractal]

nvm ok id be trolling if i asked it in here, ill ask it somewhere better

[lone jacinth]

We got a tease over here

[fierce steeple]

Étale

[hallow bone]

keeping us at the edge of our seats

[fierce steeple]

Can someone pls proof check my work

Oh didn't realise this isn't #1203471755449073774 lol

Ig it morphed into that de facto

[astral ginkgo]

One of these is an advanced channel and the other is an early uni channel

[fierce steeple]

Sure ye ig as above it was just I didn't rly view graph theory as an early uni thing but fair if it is the basics

[wanton spoke]

? where can I see it?

[fierce steeple]

Lol I was joking

It is not in a good state to share rn

[wanton spoke]

okay lol

[fierce steeple]

Lol

But thank lol

Hru radu

[wanton spoke]

fine and u it was a long time lol

[fierce steeple]

Yee

I'm okay just headachey

[wanton spoke]

same

[fierce steeple]

sorry to hear ;-;

[astral ginkgo]

Please do keep it on topic

[tough lynx]

tips on proving zariski's lemma? just began studying Commutative Algebra a few days ago

[scarlet ermine]

is this more-or-less your Zariski's lemma? that if K is a field and is finitely generated as a k-algebra, then it is actually a finite field extension of k (so K is not just fg k-alg, but fg k-mod).

the proof I remember is this:
(1) use Noether normalization to realize the ring extension k < K as going through (something isomorphic to) a polynomial ring k < k[x1,...,xn] < K, where k[x1,...,xn] < K is finite (i.e., K is a fg k[x1,...,xn]-module, rather than just algebra).
(2) use dimension/integral ring extension theory to say dim k[x1,...,xn] = dim K, and then dim K = 0 b/c field, so you actually need n=0 (so there are no variables).

so the tools you'll need is Noether normalization and some mild dimension theory (i.e., how dimension interacts with integral extensions, and the calculation of the dimension of a polynomial ring). I think Noether normalization is very cute, but somehow I have trouble remembering dimension theory.

For learning this material, I personally liked Gathmann's notes (here: https://agag-gathmann.math.rptu.de/class/commalg-2013/commalg-2013.pdf) because they feed you some geometric intuition without being very long, but they also really do not have enough exercises. When I went through these I tried to prove most of the smaller theorems/lemmas before reading them as exercises, and I thought that was helpful. But you can find this material just about anywhere, it's probably all fine

[lone jacinth]

Don't think you need any dimension theory, just show that for example 1/x1 isn't integral over k[x1, ..., xn]

[fierce steeple]

The other common proof is via this Artin–Tate lemma

[scarlet prairie]

doing a course where we went through roughly 1-6 of weibel's homological algebra, skipping the parts related to algebraic topology. i have to give a 2 hour presentation on some chosen topic 3 weeks from now. Any suggestions? I am looking for something interesting but hopefully not difficult, im still not particularly comfortable working with category theory.

for reference, the other students are doing presentations on derived categories and projective representations

[lone jacinth]

2 hours is a pretty long presentation.

Maybe something like the proof that the three different ways to define Ext are the same (equivalence classes of extensions, derived functor of Hom(A, -), derived functor of Hom(-, B))

And I guess that probably doesn't take 2 hours, but you can spin it into a presentation on Ext in general I guess

[fierce steeple]

Then turn it into a presentation on spectral sequences

[worldly zealot]

2 hours wow

[tough lynx]

Thank youuu

Im also using his notes as well

[scarlet prairie]

we have balanced tor in the class before so might be abit too similar.

[scarlet prairie]

its more of an informal presentation, probably not as intimidating as it sounds

[astral ginkgo]

Maybe you can do something about simplicial sets and dold kan?

That's also in weibel

[ornate atlas]

I found these notes on Dold-Kan yesterday which is a pretty nice exposition, if you do want to consider that route https://math.uchicago.edu/~amathew/doldkan.pdf

[limber wharf]

Are there any alternatives to serge langs book that roughly covers the same topics with rigor?

[foggy galleon]

lang's books are rigorous and tend to be good

if you are asking about algebra, dummit and foote is also nice

[hushed sandal]

irr representation of SL_2(R) obtained as symmetric tensor prod of (R^2) why.

[warped jolt]

Lang’s books suck (fact checked by real american patriots)

Dummit and Foote is good for algebra though I agree

[hushed sandal]

anyone

[woeful crane]

Probably the cleanest way to see this is through the general theory of "highest weight vectors". This is covered for complex Lie groups in Fulton Harris, or for linear algebraic groups more generally in Humphreys. I'm not sure, but I would guess the strategy for SL_2(R) is to use the classification of SL_2(C) first and then relate it to SL_2(R) via complexification/restriction.

There's also other approaches through, say, the Langlands classification theorem, but I think it's probably not as easy to see the symmetric powers that route.

Here's a nice note from Casselman with a bunch of these details:
https://personal.math.ubc.ca/~cass/research/pdf/Irr.pdf

[wary elbow]

🤎 Dummit and Foote. Very friendly book that's also sufficiently rigorous.

[hushed sandal]

i need lie algebra approach so would like to see from there

[hushed sandal]

i cant see how it comes, does it uses , enveloping algebra?

[signal frost]

hi guys

[woeful crane]

No, no enveloping algebra. As I said, you need the theory of highest weight vectors. Did you look at the sources I suggested?

[hushed sandal]

i know that theory from hymphres lie algebras

[hushed sandal]

the specific proof i couldnt find

[hushed sandal]

can u mention the page number of fulton

[woeful crane]

Chapter 11, which starts on page 146 covers the classification of irreps of sl(2, C).

Chapter 26, which starts on page 430 covers the passage between real and complex Lie algebras, and more specifically page 439 discuss how to relate their representations.

From this second part, you can see that the representation theory of sl(2, R) is essentially "the same" as sl(2, C).

I suppose all you need to do to complete the verification in terms of reps of SL(2, R) is just differentiate symmetric power reps, and check the weights.

[hushed sandal]

those symmet power confuse me as hell, i thought they came from universal enveloping hole

[woeful crane]

If that's the source of confusion, then you should probably focus on understanding that. Are you comfortable with tensor products?

[hushed sandal]

yeah

everything is fine for me

i was like why sym power of R^2 ONLY

[woeful crane]

🤷 that's just the way it is for SL(2, R)/SL(2, C).

[hushed sandal]

okay will be back after some pondring on this

[woeful crane]

I mean, it's like asking why all irreps of an Abelian group are 1-dimensonal. I don't know what other reasons to say other than the proof.

[wary elbow]

This is because the full tensor algebra is "too big" -- it's not irreducible. However, it's straightforward to check that homogeneous polynomials of a fixed degree form an irrep.

[hushed sandal]

for all n

[wary elbow]

Correct. Intuitively, this is because the action of sl_n on degree-m polynomials in n variables (for any m) just permutes the variables.

[hushed sandal]

can we generalizer it to other semi simple guyz

[wary elbow]

You can always take symmetric powers of standard reps, but I don't recall whether they're still irreducible if you take an arbitrary semisimple Lie algebra that isn't type A.

[hushed sandal]

ok thanks

[wary elbow]

It should be easy to find answers -- just Google something like "symmetric powers of representations of semisimple Lie algebras"

[near lantern]

(EDIT: I do have a concrete question that can be asked without this context. See the last message in this contiguous sequence.)

Let A be a (unital associative) algebra over a commutative ring k. Recall (or learn) that A is separable iff the A-linear action map A (⨯) V → V splits (A-linearly) for all A-modules V naturally. (More concretely being separable is actually a structure, not a property, and can be defined as a p = ∑_i pi (⨯) pi' ∈ A (⨯) A such that (1) ∑_i pi pi' = 1 and (2) ∑_i a pi (⨯) pi' = ∑_i pi (⨯) pi' a for all a ∈ A. This is equivalent to a (A,A)-bilinear splitting of multiplication: A (⨯) A → A, or a natural A-linear splitting as above.)

Then if A is separable, any ses of A-modules which splits k-linearly also splits A-linearly.
I know one proof of this but I would like to see another one, which a priori only works for k a field (or semisimple), generalised.

The first proof is in a "computational" spirit following the classical proof of Maschke's theorem:

• For V an A-module, End_k(V) is an A-bimodule.
• A summand U of V along with a complementary W is specified by an idempotent e ∈ End(V).
• U is A-stable iff eae = ae for all a ∈ A. (Indeed, this says (1-e)Ae = 0, i.e., if we start with U, apply A, and take W-component, we get 0.)
  Similarly, W is A-stable iff eae = ea for all a ∈ A, and both are stable iff ea = ae for all a ∈ A.
  Now if we pick A-stable U with some k-linear complement, we have the corresponding e ∈ End_k(V), an idempotent such that eae = ae for all a ∈ A.
  Then pe := ∑_i pi e pi' ∈ End_k(V) is

1. an idempotent (pepe = ∑_{i,j} pi e pi' pj e pj' = ∑_{i,j} pi pi' pj e pj' = ∑_j pj e pj' = pe by (*), (1))
2. commutes with A (omitted, just use (2)) and
3. satisfies (pe)e = e, e(pe) = pe.
   The first two points mean pe is projection onto some U' along W', both A-stable, and the last two identities say U ⊆ U', U' ⊆ U.

The second proof is module-theoretic. The argument is simply that any V which is projective as a k-module is projective as an A-module (since A (⨯) V is then projective over A and V is a direct summand). Thus if k is a field, then A is semisimple. (This is how the theorem is usually stated.)

Is it true that if k → A is a homomorphism of rings and any A-module which is k-projective is A-projective, then any ses of A-modules which splits k-linearly also splits A-linearly?

[near lantern]

I should have probably just led with this.

[lone jacinth]

I guess not an answer to your question, but a different proof:

If
0 -> U -> V -> W -> 0
is a sequence of A-modules that split as k-modules, then
0 -> A(x)U -> A(x)V -> A(x)W -> 0
splits, but the first is a direct summand of the latter so it must also split.

[near lantern]

Ah

OK direct summand as ses

That's obvious

Thanks

This is what I wanted

[lone jacinth]

So key that the splitting of A(x)V -> V is natural

[lone jacinth]

Okay stupid example: let k=Z and A = F2[x]. Then you have things like
A -x-> A -> F2

[near lantern]

Right

[lone jacinth]

Idk if there are some reasonable extra condition to make this true

[near lantern]

I would be interested in the relation between reflection (by the forgetful functor) of projective, injective, semisimple, flat, and ses splitting.

ses splitting is probably the correct one to call A "relatively semisimple" over k.

I wonder if there are any classifications for relatively semisimple algebras over non-fields.

[verbal panther]

can an operation be idempotent associative and invertible?

non trivially at least?

[hidden shore]

i dont think so

$$x * x = x$$
$$(x * x) * x^{-1} = x * x^{-1} = 1$$
$$x * (x * x^{-1}) = 1$$
$$x * 1 = 1 \implies x = 1$$

[lyric rapids]

Can any one give me like a good list of concrete example of split complexes and it's application? I know some in algebraic topology but there should be more, right?

[rugged gull]

Anyone got a good pitch for why someone should take abstract algebra? I need it to graduate but I need to find 1-2 more people to take it, and by that I mean find and convince them. Otherwise they won't even offer the class.

And the only other campuses in my university system that offer it are either super far or only offer it in spring, which would push back my graduation by an entire year

[last talon]

Is that pushing it from 3 to 4 years, 4 to 5 years, or something else?

[waxen fractal]

hey, can you sign up for this class, you can drop it after it starts

[last talon]

If you want to do any pure maths in the future, you should absolutely know it
Depending on how far the course will go you might be able to talk about insolvability of the quintic

[ornate atlas]

Are you not on a maths or maths related degree? If they’re in maths physics or CS it’s just kinda of a no brainer that they should take algebra (I mean I guess not otherwise algebra would be running)

[hallow bone]

abstract algebra is like needed the moment you do anything besides pure analysis

[waxen fractal]

tmk it's not unusual for classes to run with 1 student in some scenarios

so im confused why this is even a thing

[hallow bone]

so unless they want to be stuck doing inequalities for the rest of time...

[waxen fractal]

assuming it's 100% arbitrary red tape, this might be optimal

[hallow bone]

im confused how there's seemingly only one person wanting to do abstract algebra lol

[novel nimbus]

isnt it required for a degree ?

[waxen fractal]

if you do need to convince them algebra is cool, i recommend asking them what kind of math they like, and applying the concept of a symmetry group to that area of math somehow, because the convenient thing is that most areas of math have some associated symmetry groups, so you can say, "we basically study your area of math", to anyone

[rugged gull]

I'm a 31 year old part time student. So depending on how you look at it... to 4 years for the back half of a Bachelor's?

I had an associates in paramedicine before this and now i'm getting a bachelor's in applied mathematics

My concentraton is in cryptography. The other track is Data Analysis/Data Science

Which is VASTLY more popular in our already tiny/less than 10 year old major program

[novel nimbus]

what course is this where the onus to graduate is on the student for the course

[hallow bone]

oh it's super easy to apply abstract algebra to cryptography lol

[rugged gull]

I think there may be about 10 total upperclassmen cryptography track applied math majors

[novel nimbus]

i mean you can literally complain

[rugged gull]

We can go to an alternate campus and do what's called ePermit it

[hallow bone]

what country is this

[rugged gull]

But they don't run the classes there often either, and somtimes it's a bit weird trying to get things to mesh. Push comes to shove I can do independent study if I can find a professor who's down

USA

[hallow bone]

figured tbh

[rugged gull]

Can't you tell by the lack of math majors?

[drowsy niche]

could you graduate if that's the case?

if so that would be a preferable option

[rugged gull]

Also, I go to a school that started as a criminal justice school. the applied math degree program was born from the cybersecurity department

[novel nimbus]

this is honestly slightly insane

so you have to find people for a course they are offering

[rugged gull]

Five years ago they created the applied math

major*

They aren't offering it

[novel nimbus]

W ____T____F

[rugged gull]

It's supposed to be every other fall but they've never run it

Because it's offered and then not enouh people sign up (we need four)

[drowsy niche]

I was/am in a somewhat similar situation at a not great uni and doing reading courses is really the only way I was able to get anywhere with math

[rugged gull]

So this year it wasn't offered but I need it to graduate. So they said if I can find three more they'll make it a course and run it

[drowsy niche]

if you can talk to your department head or any professors that you like, there's probably a way to get around this

[rugged gull]

there is. I just don't want to go to another campus for one class when I'll be in two others at my main campus

And the other campus is at least 40 mins away

[novel nimbus]

okay i was a grad too in my years and we had very obscure courses but not for mandatory ones

talk to the dept

[rugged gull]

The dept head is being very helpful/gracious and trying to help me find a solution.He just sent every cryptography major an email to poll us for interest

But he said to talk to current classmates to see if I can convince any of them as well

[novel nimbus]

hope ya figure it out @rugged gull : )

[rugged gull]

thank you! I hope so too

I want to graduate before my layoff health insurance extension expires

So I can hopefully work on getting a job during my final year that kicks in after I graduate (and would allow me to retain health insurance).

[soft parcel]

Am I missing something or is the => direction not true (Z(L) = [L,L]^perp => b nondegenerate)?

I mean, can't we just take b trivial and L abelian?

[verbal panther]

is there an algorithm that can determine if a set of axioms and identities is trivial/inconsistent?

[hallow bone]

yes, recursively generate every deduction

i.e. every substitution and replacement

it's not a particularly good algorithm

but you're not gonna get much better

[charred knot]

part b of this question in (the version of) atiyah macdonald (that i have) seems to have a typo, since X is a space and A_1, A_2 are rings... should the exercise be to show X is Spec(A_1 x A_2) or that A is A_1 x A_2?

i would assume it's the latter since that makes more sense with being related to part c, but i just wanted to make sure i wasn't proving something false lol

[hallow bone]

latter lol

[astral ginkgo]

Those are the same thing

But they probably meant as rings

[waxen fractal]

let $A$ be an abelian group. let $\varphi \colon A^\omega \rightarrow A$ be a linear map such that, for all $x \in A^{\omega \times \omega}$, we have $\varphi(i \mapsto \varphi(j \mapsto x_{ij})) = \varphi(j \mapsto \varphi(i \mapsto x_{ij}))$, that is, it doesn't matter if you apply $\varphi$ on rows or columns first. is there a common name for this structure / property?

[broken turtle]

keith

[waxen fractal]

technically, this question could go in #groups-rings-fields, because all you need to understand it is knowledge of what an abelian group is, but i assume that is not what the channel is intended for, despite its name

i should add, my motivation here is to generalize the concept of an R-module for a commutative ring R. also, I've proven some nice things about them on my own

[torn harbor]

seems quite restrictive to be honest, do you have any examples of nonconstant functions with this property?

[waxen fractal]

yes, consider the p-adics Z_p and the map that sends (x_i) to x_0 + p * x_1 + p^2 * x_2 + ...

(this is actually not my own example, I've asked about this a while before and jems9 gave me this example)

but you're right tbh, what has sparked my interest again recently is that i found a nice impossibility result. specifically, there must exist a cofinite subset S of ω such that varphi restricted to A^S is not surjective onto A

[astral ginkgo]

How does what you wrote above assist you in generalising modules?

[waxen fractal]

modules over a commutative ring can be thought of as this for finite arity, satisfying some equations (although for non-f.g. rings, you need to introduce more operations of the same kind)

[astral ginkgo]

But you can have non f.g. modules, so I'm not sure how general modules are represented by this set-up (I'm not sure I understand how f.g. ones are either).

Also what do you mean by f.g. ring? Finitely generated as an abelian group?

[waxen fractal]

by f.g. ring, i mean that it's a quotient of Z[x_0, ..., x_n] for some finite n

[astral ginkgo]

I see

[waxen fractal]

ok to elaborate,

we can view an R-module as an abelian group A with an action by R

we can view elements of R, in a specific R-action, as linear maps A -> A

and in fact, we can axiomitize R-modules by choosing some generating set of R (as a ring) and writing down what is essentially a presentation of R, combined with the ring action axioms

the reason i say finite arity is because, if you try to go to higher arities, because linear maps A × A -> A are just two independent maps A -> A, we get essentially the same data

the commutativity condition i gave is actually a general condition for algebraic structures to be enriched over themselves. this is why R-module homs have an R-module structure only when R is commutative

[astral ginkgo]

So how are you collating these maps in a way where maps A^\omega -> A generalises it?

[waxen fractal]

(any pair of operations must be homomorphisms for each other, in the correct sense)

also, something nice is that if two rings R, S are commutative, then R-mod and S-mod are equivalent iff R and S are isomorphic. so we actually lose no information viewing a commutative ring as its category of modules, and thus as a category of a specific kind of algebraic structures

[astral ginkgo]

It seems to me you need to view this as a collection of maps

[waxen fractal]

so, a linear map A^n -> A can be thought of as sending a tuple (a_i) to r_1 * a_1 + r_2 * a_2 + ... + r_n * a_n

[astral ginkgo]

Oh sure in the finite case you can recover the original maps

[waxen fractal]

so this is essentially asking, what if rings had some notion of infinite weighted sum

[astral ginkgo]

So what do you do for non fg rings?

[waxen fractal]

you add more operations

[astral ginkgo]

Wdym

[waxen fractal]

actually, the unbiased way to do it, would be to add an operation for every r ∈ R

the generating set perspective is purely for computational convenience

[astral ginkgo]

What do you mean exactly when you say operation. Are you defining additional algebraic operations so that this is no longer a ring?

[waxen fractal]

ah i should be more clear

im talking about operations on abelian groups

i should give a concrete example

define the following kind of algebraic structure: (A, +, 0, -, f). it consists of an abelian group A, and a unary operation f such that f is linear, and f(x + x) = x

these are precisely Z[1/2]-modules, because 1/2 "generates" the ring Z[1/2]

in particular, we can present Z[1/2] as Z[f]/(2f-1)

[astral ginkgo]

How does this allow you to present modules over infinitely generated rings with finite information?

[waxen fractal]

consider the ring Q. we can present it as Z[1/2, 1/3, 1/5, ...] with the appropriate relations, i.e. p * 1/p = 1

thus, we need an infinite signature for our algebraic structure (a signature is a list of operations that is part of its data)

but, each operation itself is unary

[astral ginkgo]

So you are expanding the algebraic structure

[waxen fractal]

yeah

[astral ginkgo]

Ok

But then doesn't the information of the ring get stored in these operations, so what would the map A^n->A even be in this case?

If you're defining a symbol for each r in R it seems to me the map A^n->A is superfluous

[waxen fractal]

yeah, to be clear, arities higher than 1 are superfluous in this context

i think what is more notable is that they are in fact superfluous

in general, you cannot describe an algebraic structure with only unary operations

abelian groups are nice enough that binary operations break apart

actually, i think it's true of groups too, but they don't behave nice enough to be well-studied, but there is the concept of "groups with operators" for those curious

[astral ginkgo]

So in this case I don't really see how the above generalises the case of modules over non f.g. rings

[waxen fractal]

to be clear that is not my goal

for all rings, not just commutative ones, the category of R-modules can be viewed as the category of abelian groups with some unary operations that satisfy some equations