advanced-algebra

[worldly zealot]

im not sure i understand the question

can you clarify what sort of statement you want an analogous version of?

[polar hawk]

Maybe is because my poor English, but also I know the one I ask is an ambiguous question. The analogous statement I asked for is precisely the example I wrote in the first message. I reproduce it here:

From the factor theorem (or first isomorphism theorem) one could conclude that all relevant information about a homomorphism is encoded in the kernel of the morphism given that we know the "ambient object" (group, ring, module, etc). As one would expect from this interpretation, it extends to the dual cokernel as containing the "residual information" of no need, as it could be identified in most cases as the object that annihilates the image of the morphisms.

I wrote it to clarify what I meant for "an interpretation that arises from the common use of a result". In this case, I ask for the an "analogous" for the Snake Lemma.

Well, actually I only ask if someone has thought about it in that way. I know that not every mathematical statement has a deeper meaning or interpretation. Some just assert what they states. Haha.

[neat nexus]

I don’t remember the details. GTM 196 proves snake lemma using a H->H->H in the middle, H I meant homology. so this middle step has something to do with kernel and cokernel since homology group of A-f->B-g->C is kernel of the (g factored by cokernel of f), or equivalently, cokernel of the (f factored by kernel of g)

This I was talking about

[worldly zealot]

so i think rather than the first isomorphism theorem, the relevant notion you want to draw parallels from is that given a homomorphism f: A->B, you have an induced long exact sequence 0 -> ker(f) -> A -> B -> coker(f) -> 0

the snake lemma can be seen as addressing how that interacts with exact sequences. for short exact sequences 0 -> A' -> A -> A'' -> 0, and 0 -> B' -> B -> B'' -> 0, and writing f': A' -> B' and f'': A'' -> B'', all you can conclude is the exactness of 0 -> ker(f') -> ker(f) -> ker(f'') i.e. left exactness. dually for cokernels, you only have right exactness. so in a sense cokernel is the correct way to continue the kernel exact sequence to the right, and kernel is the correct way to continue the cokernel sequence to the left. this is an example of derived functors which you will probably see soon.

[polar hawk]

Well, that makes sense. I completely forgot that very first statement you mention. Morphisms' sequences is a relatively new concept to me so probably is best to keep it up with the material and then come back with these type of questions as one can see wider with the big picture. Thank you for your responses, @worldly zealot and @neat nexus !

[limpid horizon]

What book

GTM 196 ok

[neat nexus]

Yeah. Basic homological algebra Osborne

[limpid horizon]

Ah ya looked familir

Weirdly friendly textbook

I read a bit of the beginning chapters

For Ext

[neat nexus]

I only read chap 7 for abelian categories

[wanton heron]

I need help with 2.6a

The progress so far is B being a Flat A alg allows Ideal \otimes B \cong IdealB

And then I get the obvious inclusion to be true

I.e (intersection I)B \subset eq intersection (IB)

But i haven't used the family being finite yet

So I'm expecting it has to be used somewhere for the other inclusion

On that note

[lone jacinth]

Notice that
0 -> I\capJ -> I(+)J -> A
is exact

[wanton heron]

What is the coker of the map

Ohh

[wanton heron]

Ohkay but tell me this is there some way to look at the map
0->A/\cap I -> \prod A/I -> coker->0

I was thinking of tensoring this with B

The prod being finite means that the tensor distributes

So I was trying to come up with what the cokernel looks like here and if it's same as the coker when
B/\cap IB -> \prod B/IB -> coker

[lone jacinth]

Maybe you want to think about
0 -> \cap I -> A -> prod A/I
instead

[wanton heron]

Right 👍

Yeah that sequence is exact

And the kernel after tensoring with B is also intersection IB

So done ig

[wanton heron]

But what is the canonical homomorphism

Okay nvm

It should be f \otimes b \mapsto f \otimes (1 \mapsto b) i think

But why is something like this canonical

[hushed bone]

Only thing that exist, try writing anything else down

Further point: this is a natural transformation

[wanton heron]

Between what functors

[hushed bone]

Between the bifunctors in M and N

[wanton heron]

Huh

[hushed bone]

But for the purpose of this proof, fixing N, the functor in M

[wanton heron]

Okay but I'm curious now what bifunctors

[hushed bone]

And I mean it’s even a triple functor because there’s functoriality in B

[wanton heron]

HUUUH

[neat nexus]

Two functors (Mod-A)^op times Mod-A times A-Alg ->Ab

[wanton heron]

Oh ok

Thanks

I'll try to work out the details, seems like this excercise is most of it probably

[faint minnow]

I like this guy's videos:
https://www.youtube.com/watch?v=kczJovq3Q9E&list=PLHEKZhtMXuKwNhFOeaFuzqCOvSthG-gbc

Today I started using that playlist to learn homological algebra just for fun.

[echo summit]

I'm trying to figure out this exercise, and I am just very lost. The first part is obvious to me, pulling back a prime p under the inclusion $(S_f)_0\hookrightarrow S_f$ gives $p_0=p\cap (S_f)_0$ which is clearly prime. For the other direction of the bijection, I am trying to just show via brute force that this is prime, i.e. for $ab\in \sqrt{p_0S_f}$ we have $(ab)^k\in p_0S_f$, but this is not getting me anywhere. Does anyone have a hint?

[broken turtle]

Sleepybear

[lone jacinth]

It might help to first show that P is homogeneous

[winged lintel]

the goat 🤯 I haven't seen this one yet

[polar hawk]

Hi. I'm studying semi-simple modules and I started to have some questions about direct sums. In particular, if $\bigoplus_{i\in I} M_i \cong \bigoplus_{j\in J} N_j$ then $|I| = |J|$ and there exists a permutation $\sigma$ such that $M_i \cong N_{\sigma (i)}$. It seems pretty straightforward if all the $M_i$ and $N_i$ are simple, but I can't prove it for modules in general. I'm starting to suspect that this statement is false for general modules. Am I wrong?

[broken turtle]

The Penguin that Drinks Coffe

[polar hawk]

(I forgot to state an additional hypothesis: each N_j and M_i can't be expressed as a direct sum)

[lone jacinth]

It is not true in general. But the Krull-Remack-Schmidt-Azumaya theorem says that it's true when M_i has local endomorphism ring and N_j are indecomposable.

But for the purposes of studying semisimple modules I would assume it is meant that they are simple

[polar hawk]

Thank you!

[broken wren]

Here’s a counter example: let the ring be functions on the 2-sphere. Maybe continuous functions, maybe smooth functions, maybe algebraic functions: A=R[x,y,z]/(x^2+y^2+z^2)^2=1

Then vector bundles are the same as projective modules. The tangent bundle plus the normal bundle in R^3 gives the tangent bundle to R^3, which is trivial. The normal bundle is trivial. But the tangent bundle is not trivial, having no decomposition as a sum of lines
T+A=A+A+A

[wanton spoke]

it’s true whenever you’re in a Krull–Schmidt setting
i mean every module decomposes as a finite direct sum of indecomposables and that decomposition is unique up to permutation (and isomorphism)

[broken wren]

Does Krull-Schmidt imply that indecomposables are simple?

[wanton spoke]

no

[lone jacinth]

I mean, you still need to assume the Mi and Nj are indecomposable

For another nice counter example, R = Z[sqrt(-5)],
M = (2, 1+sqrt(-5)).

Then M^2 = R^2

[desert flax]

I'm not sure which channel to ask this, but I'll ask here:

Let k be a real reductive Lie algebra.
Is Int(k) a normal subgroup of Aut(k)?

[torn harbor]

Here's a simple counterexample: the countable direct sum of Z is isomorphic to the countable direct sum of Z^2

[broken wren]

What is Int(k)?
Did you mean Inn(k), inter automorphisms? That’s normal

[desert flax]

By Int(k) I mean the image of Ad in Aut(k) for any connected Lie group K with Lie algebra k

Int(k) = Ad_k(K) for any connected Lie group K with Lie(K) = k

[broken wren]

What is Int short for?

If the group is connected, isn’t the image of Ad equal to the image of ad?
The image of ad should be normal for fairly formal reasons

[desert flax]

From Knapp's book

The definition is not the same as what I said, but I think it should agree

Oh yeah it is

"Int(g) [...] equals Ad(G) if G is connected"

[quick bear]

Hello, I was wondering how you multiply 2 elements in a Cayley diagram. I’m pretty sure you just follow the arrows from the beginning products vertex to the end product, and the last arrow corresponds the the result. But I may be wrong. My textbook wasn’t very clear

[restive pike]

row first then column

[quick bear]

I’m trying to create a table just from a Cayley diagram though, I don’t have the table

[spice idol]

You choose a sequence of generators to get to the first element and apply that to second element no?

If youve got a cayley graph of a group then the choice of sequence of generators shouldnt matter cuz congruences n shit

[rose mirage]

I like to think of the cayley graph as being "coloured" by the generators, and moving along an edge corresponds to multiplying by the generator corresponding to the colour of that edge

[desert flax]

I love it when a paper asserts something you want to understand without any justification or reference.

Like, yeah, ts is so independently verifiable. Its "obvious".

🙄

[dull cedar]

"We will not insult our reader's intelligence"

[distant harness]

That sounds like a nice way to say "I couldn't find a convincing way to prove this".

[urban granite]

tfw

[formal rock]

this is not as bad

[formal rock]

saying this kind of thing with zero references is inexcusable imo

[golden osprey]

Anyone have a good errata list for Atiyah-Macdonald?

Running a reading group on it this spring and so I'd like a good thing to point people to

https://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald?answertab=modifieddesc#tab-top

This seems not bad

but I'd like to know if there's a more complete list

[wanton heron]

For (a) it doesn't hold when M=0??

[subtle plaza]

0 is always a torsion element. A module is torsion free if 0 is the only torsion element, so the zero module is torsion free

[wanton heron]

Oh.

Yeah then it checks out

[hushed bone]

Thanks I’ll pull this out for my thesis

[hushed bone]

Hi Walter 🙂

[subtle plaza]

Hi Chmonkey 🙂

[summer quest]

What is the complaint here exactly

[desert flax]

Id like to know how its derived

Are these parameterizations supposed to be obvious?

[summer quest]

They kind of are obvious yes

They are not obvious if you do not know any Lie theory

[desert flax]

What is the justification then?

[summer quest]

What do you mean the justification is general results in Lie theory

If you go through something like Knapp it gives a bunch of examples like this without much detailed justification after the corresponding general results are covered

The justification is often these standard results and their proofs are often constructive and tel you how to simply read off these answers from the root system and related Weil group combinatorics

[desert flax]

I don't know what standard results you are talking about. Is there a standard/general result for parameterizing K-conjugacy classes of theta stable parabolics?

Its actually surprising to me that such combinatorial parameterizations in terms of ordered partitions even exists

I guess maybe one could try to parameterize K-conjugacy classes of parabolic subalgebras of the complexified Lie algebra in terms of ordered partitions, then then determine which extra conditions on the partitions is equivalent to demanding the parabolic subalgebra is theta-stable.
Maybe for the first part of this, there are some standard/general results available? I will have to look

[summer quest]

Knapp has this spelled out somewhere

That or somewhere else involving the structure theory of real groups

[deft bane]

∫ ln sinx • dx

Can anyone solve this??? It is without limits

[last talon]

Hmmmmm (wrong channel)

[deft bane]

Sorry

[fierce steeple]

#calculus ig

[urban granite]

it's funny that the channel is #advanced-algebra but posting calculus stuffs

lmao

[fierce steeple]

Good question though lol

[digital ocean]

It's also apparently multi variable calculus despite the lack of multiple variables

[urban granite]

,w integrate \intln|sinx|dx

[broken turtle]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by _0mem

[urban granite]

lmao

[wise sedge]

Maybe they meant the derivative (which is just cot(x))

[crisp bone]

Chat what’s a good book for calculating Lie algebras? I have turned to the dark side (theoretical physics) realized I didn’t know the necessary math to do stuff, and I need Lie algebras because that’s what people told me

[urban granite]

If you want to learn specifically about matrix Lie groups and their Lie algebras, see Lie Groups, Lie algebras, and Rerpesentations by Hall

[fierce steeple]

The book by Hall is even better

[past cove]

Knapp has a good book as well

[urban granite]

fixed

[spice idol]

Same Hall as the Hall subgroups?

[past cove]

Nope

[spice idol]

Does anyone have good literature on ringoids (small pre additive categories) and their modules and in particular a notion of ideals

[lone jacinth]

I don't think I have any good literature, but ideals are defined essentially the same as for rings if that's your main question.

[spice idol]

Eh alright yeah

Not really my main question i guess, just in an annoying situation

The situation is that every algebra A gets sent functorially to a ringoid R[V, A], on the set of objects A.

I need some way to map congruences faithfully into some poset of submodules, to have some notion of "sheaf of congruences" and be able to do hom alg with it

[neat nexus]

What is the definition of R[V, A]?

[spice idol]

We first define the category C; this has as objects A, and Hom_C(a, b) = { p(x) \in A[x] | p^A(a) = b }, with composition being composition of polynomials. Formally, here A[x] depends on your ambient variety V, it is the coproduct in V of A with the V-free algebra on one generator.

Then we define \hat{R}[V, A] to be the linearization of C (i.e. take as hom sets the free R-module generated by the original hom sets). At last, for t(x, ..., x) in \hat{R}[V, A](a, b) we impose the relations t(x, ..., x) = t(x, a, ..., a) + t(a, x, ..., a) + ... + t(a, a, ..., x)

Very elaborate construction lol

This arises as the Beck-modules over A in V are naturally equivalent to left modules of Z[V, A]

R just changes the "abelian group object" in the definition to "R-module object"

[neat nexus]

What is Z[V,A] then?

[spice idol]

The above construction with R = Z (the integers)

[neat nexus]

I see. Thank you. What field is this from?

[spice idol]

Universal algebra, i suppose

[neat nexus]

Thank you

[hushed bone]

I’m feeling like I am forgetting some simple fact about local cohomology or something, but what is the injection R/(x1,…,xd) -> H^d_m(R) here? Does this exist for any local ring of dimension d and regular sequence x1,…,xd, or does this require R to be Cohen Macaulay?

[broken wren]

Regular is stronger than CM. Do you mean some other hypothesis?

[hushed bone]

Shit

[wanton spoke]

in a regular local ring you’re automatically Cohen–Macaulay so the map exists and is injective for any regular system of parameters ??

[hushed bone]

I don’t mean regular, just local ring

[hushed bone]

Yeah I didn’t mean regular local whoopsie

[broken wren]

What is n? It is not true for all n. CM means that the local cohomology is all in a single degree

[hushed bone]

There is no n. d is assumed to be the dimension of the ring

I think the answer is just

It’s the span of 1/x1•••xd

[wanton spoke]

it exists in general what fails is injectivity

[hushed bone]

When you write the top degree local cohomology in the obvious way

And regularity of the sequence tells you that the annihilator is exactly (x1,…,xd)

I thought about it some more

Write the Čech complex definition of local cohomology

[wanton spoke]

it’s injective iff x1,…,xd is an R-regular sequence so in particular if R is Cohen–Macaulay then every sop is regular

[hushed bone]

Does showing that first iff just computing colon ideals essentially?

From the Čech resolution

Or is there something else you do

[wanton spoke]

you identify H^d_m(R)=H^d_(x)(R) and use the Čech or direct-limit description to send r mod (x) to the class of r/(x1…xd)

[hushed bone]

Yeah

Okay

Cool

Huehuehuehuehue

Tyty

[wanton spoke]

hueheueheuheue

[wanton spoke]

in your case it s injective

[hushed bone]

Yeah I mean it says so

And I followed what it’s doing

I just wasn’t seeing what element had the right annihilator until I thought about an explicit presentation of the local cohomology group

[spice idol]

https://tenor.com/view/huehuehue-gif-7509268964755959557

[faint minnow]

https://www.youtube.com/watch?v=zgp6irf9w7w

[river junco]

for a lattice L, I_L be a prime ideal iff its lawrence ideal is a prime ideal right?

sorry nvm it's true

[tawny notch]

Haven’t delved into that level yet, I just graduated from undergrad

[spice idol]

Cool lattice or the discrete subspace one

[spice idol]

Congrats on that!

[tawny notch]

Thanks

Gonna start grad school next week in mathematics

With a focus on analysis and algebra

[spice idol]

Stoef sounds very dutch lol

Are you?

[spice idol]

Oo nice, any branch of algebra in particular?

[tawny notch]

Hopefully commutative, like my former professor

He’s still teaching at Clemson

[woven loom]

Why the analysis mixed in

Curious mix

[tawny notch]

I took linear analysis last year

[woven loom]

Not that I’m one to talk with DEs and logic but

[tawny notch]

Fun course

[woven loom]

I see I see

[tawny notch]

I’m on the B2G program

[tawny notch]

At Clemson

[neat nexus]

What book are you reading about this? I want to learn about this Lawrence ideal.

[hushed bone]

LETS GOOOO

[river junco]

https://arxiv.org/abs/math/9912247

[neat nexus]

Thank you

[limpid horizon]

Hey can anyone give me some direction on how to understand this? For reference I havent studied direct limits before

[spice idol]

1. Time to start studying direct limits and inverse limits then! can't go very far in algebra without em :3
2. This is a special example of something called a directed union. This means you have a diagram of embeddings
   A_0 -> A_1 -> A_2 -> A_3 -> ...
   (I assume the induced maps (0 :_M I^m) -> (0 :_M I^n) are nothing more than the inclusion maps)
   and the direct limit of this is in some sense then the "union" of all the A_i, even if they may not be actual subsets of some larger set

as for this case in particular, if m < n, a homomorphism f : R/I^m -> M may be turned into a morphism f* : R/I^n -> M by precomposing with the projection R/I^n -> R/I^m. This is an injective process, as the projection map is surjective.

The isomorphism Hom_R(R/I^n, M) \cong (0 :_M I^n) can be seen "geometrically" in some sense. Elements of Hom_R(R/I^n, M) can be seen as points m on the "affine line" M, satisfying rm = 0 for all r in I^n, i.e. it is the submodule (0 :_M I^n)

(recall that a homomorphism Hom_R(R, M) is the same as choosing an element of M)

[spice idol]

Intuitively, this should also make some sense: if m < n, then I^n is "smaller" than I^m, so the condition that rx = 0 for all r in I^n is much less strict

[trail hinge]

Hi people, I've been messing around a bit with persistent homology and I was thinking given a persistent homology barcode diagram what would be a suitable way to represent said barcode diagram as a function/ scalars/vectors/matrices?

[worldly zealot]

what do you mean

the persistence diagram/bar code is a representation of some homological information, i dont know what it means for you to further represent a represntation of something

[mental escarp]

Do you mean represent a barcode as in “store the information inside a computer”? Like that life and death stuff that I know nothing about

[trail hinge]

Yes so typically a barcode is represented by a sub interval [birth_{barcodei},death_{barcode_i}] I was wondering what would be a useful way to store it inside a computer lately I tried storing them inside a distribution with l_i = abs{birth_i - death_i} but I was wondering if there was maybe a better more representing way to store them maybe some function f(b_i,d_i) -> R^n 1 \leq n

I'm just not sure how would that function look

likle

I'm trying to enhance the meaning of each barcoe

barcode

meaning, I'm trying to find a function that will emphasize the difference between each barcode

I believe the l_i technique doesnt' really do that

I thought about maybe doing a weighted lifespan

where the weights are index dependent

but I haven't tried that yet

[topaz oyster]

I'm pretty sure I've seen them represented by multifunctions

I can't imagine this is a better means of storage

[last talon]

How does this work?
So I’ve tried to follow what they’re saying
In the 0 page we get
$$\begin{matrix} 0 & C’{q+1} & C’’{q+1} & 0\ 0 & C’{q} & C’’{q} & 0\ 0 & C’{q-1} & C’’{q-1} & 0\end{matrix}$$ and so on, with the differentials pointing down
In the 1 page we get
$$\begin{matrix} 0 & H_{q+1}(C’) & H_{q+1}(C’’) & 0\ 0 & H_{q}(C’) & H_{q}(C’’) & 0\ 0 & H_{q-1}(C’) & H_{q-1}(C’’) & 0\end{matrix}$$ with differentials pointing straight to the left, which I will call $\partial$
In the 2 page we get $$\begin{matrix} 0 & H_{q+1}(C’)/\partial H_{q+1}(C’’) & \ker \partial & 0\ 0 & H_{q}(C’) /\partial H_{q}(C’’) & \ker \partial & 0\ 0 & H_{q-1}(C’) /\partial H_{q-1}(C’’) & \ker \partial & 0\end{matrix}$$
After the second page, everything in sight converges because every non-trivial pair of composable differentials is of the form $0 \to X \to 0$
But this (when I then compute the infinity page) seems like I have an off by one error in the $H_{q}(C’) /\partial H_{q}(C’’)$ terms, but I can’t see how?

[broken turtle]

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

[last talon]

Wait bweh
Bweh bweh bweh
There’s a relative shift between the nth and n+1th columns
I hate notation

Yeah that solves it

[wanton spoke]

in this spectral sequence the (p,q) spot comes from degree p+q not degree q

column p=1 comes from C’ in degree q+1 so E1 is H_{q+1}(C’)
column p=2 comes from C’’ in degree q+2 so E1 is H_{q+2}(C’’)

the missing shift

And d1 from column 2 to column 1 is the connecting map delta: H_{q+2}(C’’) -> H_{q+1}(C’) with only two columns it collapses at E2 and reading the induced filtration on H_*(C) recovers the usual long exact sequence for 0 -> C’ -> C -> C’’ -> 0

[last talon]

(We realised)

[limpid horizon]

Can someone please give me some intuition on what a system of parameters is for

One of those terms ive seen so much now but still dont really know what its doing

[hushed bone]

They tell you the dimension of the ring

Them being regular sequences tell you if something is CM

Quotienting by them gets you an Artinian ring

There’s a lot

[limpid horizon]

Yeah i think thats why i dont have a clear picture in my mind of them

Cause they seem to show up everywhere

[supple perch]

does anybody have an intuitive resource to understand eigen values behaviour in graphs and polynomials in depth?
I know they are scaling factors of eigen vectors and have the geometric intuition of what they are with respect to geometric vectors but i can't say the same thing with other forms of vectors like polynomials (polynomials are vectors too you know they satisfy all conditions to be a vector)
I am yet to understand how they are apt ranks for pagerank (website ranking of initial google) and how bivariate graphs have eigen values occuring in + and - pairs and how they are crucial to stability of a system?

[limpid horizon]

Im trying to understand the proof of krull’s principal ideal theorem, but this definition got me stuck. Im not really sure how to interpret it (its a lot going on)

If someone knows an exercise to do in a/m maybe

[ebon cove]

Maybe look up primary ideals in atiyah macdonald; the main point of constructing symbolic powers in that proof is that the symbolic powers of a prime ideal p are p-primary. I don't recall if a/m defines symbolic powers.

[spice idol]

I suppose this is from the view of primary ideals the "correct" notion of powers of a prime

since from the integers, the intuition for primary ideals is that they are of the form (p^n) and so give the prime decomposition of a number

However, in general rings, powers of a prime ideal are not in general primary, and these are the correct notion of "power"

[limpid horizon]

does that result there really need that blurb about m-adic topology?

i havent learned about those topologies

[digital parcel]

the family (x^k) having the m-adic topology means that it is such a family as in this paragraph

you can read about the m-adic topology in, e.g., Atiyah-Macdonald chapter 10

[ebon cove]

But you don't really need to know anything else about the m-adic topology to go from the first displayed equation to the last displayed equation.

[digital parcel]

yeah

oh also i guess a priori it's not obvious that (x^k) gives the m-adic topology, since (x^k) isnt (x)^k as one would take in the definition of the m-adic topology (here m = (x))

but the (x^k) form what A-M call a "stable m-filtration" (because R is noetherian, I assume) and Lemma 10.6 says that it gives the m-adic topology anyway

[limpid horizon]

Thx

[wise sedge]

The answer is no

The 16-dimensional irreducible representations of M11 are a counterexample

[vague pawn]

Let f:A --> B be a morphism in an abelian category. f factors through Im(f) uniquely by a map g:A --> Im(f). How to see that g is epi?

Its obvious to me what this means in Mod_A, but I am a bit struggling to show this using universal properties and in a general abelian category.

[spice idol]

wikipedia has a nice proof showing that in general categories with equalisers, g is an epi

https://en.wikipedia.org/wiki/Image_(category_theory)

[rose mirage]

Freyd-Mitchel embedding, then it's trivial :trollface:

[golden osprey]

Is whatever is in brackets not basically the whole solution or is there something extra I have to check?

[lone jacinth]

g factors as
A -> coimage(f) -> Im(f)

The first is epi because it's a cokernel, the second being an isomorphism is one of the axioms of an abelian category

[lone jacinth]

That's just the solution yeah

[golden osprey]

thanks Atiyah-Macdonald 😵‍💫

[worldly zealot]

lol

[limpid horizon]

Lol

[vague pawn]

the second being an isomorphism is one of the axioms of an abelian category
I don't have this as an axiom

[limpid horizon]

Why is the last implication true

[golden osprey]

oh eh I guess

[plucky arch]

you can prove it being an iso from these axioms

though it takes a little more effort

a good reference for this is borceux handbook of categorical algebra volume 2, chapter 1

[limpid horizon]

If M and N are fin dim then I know M (x) N is fin dim of dimension m*n so if its 0 then either m or n is 0. Thats all i got rn tho

[golden osprey]

no yea that's basically it because you know M and N are finitely generated

I guess I already know that fact from taking linear algebra so I'm not going to bother writing it out

I just thought it was funny that it was basically like [Hint: Answer]

and thought I was missing something

[limpid horizon]

Going from M (x) N = 0 to Mk (x) Nk = 0 really is just from tensoring twice with k?

I thought it was localization at first glance

[golden osprey]

it's dumb notation yea

M_k is defined to be k (x) M

[lone jacinth]

Okay with these axioms:

Say g is not epi, then you have
Im(g) as a subobject of Im(f).

Now prove that the cokernel of
Im(g) -> Im(f) -> B
is cok(f).
Then it follows by (2) that both Im(g) and Im(f) are the kernel of cok(f), i.e. they're equal

[vague pawn]

what do you mean by the cokernel of Im(g) -> Im(f) -> B

cokernel of which morphism

of Im(g) --> B?

[lone jacinth]

Yes, like imagine this diagram
A -----> B -> cok(f)
v v v
Im(g) -> B -> cok(g)
v v v
Im(f) -> B -> cok(f)

[vague pawn]

ahhhhh

where do you get maps cok(g) -> cok(f)?

and B --> cok(g)

[lone jacinth]

I meant to write cok(Im(g))

So everything on the right is the cokernel on the thing on the left

[vague pawn]

Ok

Lemme try this

But I think I might return to this tomorrow

Thanks jagr

[hushed bone]

It literally is the proof yes

[limpid horizon]

Yas

[worldly zealot]

more like atiyah-mcdouble with no pickles

[limpid horizon]

Im trying to understand why that part is true

R is noetherian local. , x1 to xn is a system of parameters. Does this have something to do with E(k) being an essential extension of k?

[low orbit]

I think so. That should mean every element of E(k) is killed by some power of mathfrak m.

any nonzero cyclic submodule Ra subseteq E(k) intersects k nontrivially, hence actually contains k.. this forces Ra to be supported at \mathfrak m and so Ra has finite length (as Ra is finitely generated and northern local)

[digital parcel]

<@&268886789983436800>

other channels too :(

[faint minnow]

I like algebra.

Yesterday I started reading a book about homological algebra to learn the subject just for fun.

[urban granite]

bro started learning on based stuffs just for fun, very much based

[faint minnow]

The book is teaching me about the category Ch(C) for which C is either R-Mod or Mod-R .

[urban granite]

yeah this one is important

in many cases you work with modules

[faint minnow]

homology module

[limpid horizon]

In the next line, some xj is not in p, otherwise (x1, … xn) is in p and that would mean dim R/p <= dim R/(x1, … xn) = 0 contradicting dim R/p neq 0?

[low orbit]

yes, that's basically it

[limpid horizon]

Cool

I am hoping this proof still goes through in the case of R not noetherian local, but *local or whatever it is… has unique homog maximal ideal

I think it probably does though

[blissful adder]

hello! i have something i'm struggling with

here's a decomposition of my semisimple ring A into blocks of simple left ideals

in classic math notes fashion it says "it is clear to see that $B_i=\bigoplus_{j=1}^{\eta_i}{L_{i,j}}$ is a left ideal of A"

[broken turtle]

blutac

[blissful adder]

maybe i'm missing something clear. but this is not clear to me

is it just that a sum of left ideals is a left ideal? is that it?

[lone jacinth]

That's it yeah

[blissful adder]

wow

now i feel stupid

ty :)

[lone jacinth]

I guess the interesting thing is that Bi is also a right ideal (so two sided ideal)

[blissful adder]

thats the harder part :')

[spice idol]

I guess:
Let M be a simple module and N a direct sum of a finite number of M's. Immediately we see we have a decomposition series 0 < M < M (+) M < ... < N where all the factor modules are isomorphic to M.

Suppose now M' < N is a submodule of N. As N is finite length, N/M' is too so there is a decomposition series of N containing M'. By Jordan-Holder it follows then that M' \cong M.

Now let M be a submodule of A isomorphic to V_i. Then M is nontrivial so there must be some B_j such that M \cap B_j is nontrivial, i.e. M < B_j. But B_j is a direct sum of isomorphic modules, so by the above it must follow that M \cong V_j, i.e. B_j = B_i. In other words: B_i is the sum of all submodules of A isomorphic to V_i.

Now the result follows by Schur's lemma, as right-multiplication on a simple module is either zero or produces an isomorphic module.

Is this sound?

[lone jacinth]

I think this works, but you don't need all these theorems:

If M is a simple submodule of A and is supported in Li,j, then the projection onto Li'j is an isomorphism.

So if M is iso to Vi it must be contain in Bi.

Then Vi * a is isomorphic to Vi (or is 0) badaboom

[spice idol]

lmao yeah okay I forgot you could just project ont Li,j

those darn module categories and their biproducts

[spice idol]

how does it follow that whenever M iso to Vi it must be contained in Bi?

I guess it's implicitly known that, no matter what decomposition you choose, the Bi will be the same

[lone jacinth]

Well if it was supported on any Bj not equal to Bi it would be isomorphic to Vj

[spice idol]

that's not necessarily a priori

wait that's the wrong term probably

gah

[lone jacinth]

The projection onto Vj will be an isomorphism...

[spice idol]

oh right

[spice idol]

^

[void plank]

Hey everyone, so I'm trying to follow this post about showing that if f, f' : C --> C' and g,g': D ---> D' are chain maps such that f ~ f' and g ~ g' (where ~ is homotopy), then f (x) g ~ f' (x) g'. https://math.stackexchange.com/questions/148229/tensor-product-of-two-chain-homotopic-maps-are-again-chain-homotopic

We can assume g = g' WLOG. Let F = f - f'. Then F is null-homotopic, so F = dh + hd for some homotopies h.

Now define D(h) = dh + hd. The post has a claim that D(h (x) g) = d(h (x) g) + (h (x) g)d = Dh (x) g - h (x) Dg.

But I'm confused how I'm supposed to interpret the compositions d(h (x) g) and (h (x) g)d. From what complex is the differential d acting on h (x) g coming from?

[limpid horizon]

Lol nah fr i feel like this would be tricky to check

Ill get to that at some point …

A chain map between complexes induced a map on homology, i forgot what a chain homotopy is tbh

Two different chain maps f,g are said to be homotopic if there exists some h where fh+gh = something?

[void plank]

f - g = dh + hd

[limpid horizon]

Is the point kind of like, if we have two different chain maps f,g we wouldnt necessarily expect them to induce the same map on homology, but if f,g are homotopic then they really do induce the same map

[void plank]

Yes, homotopic chain maps induce the same map in homology

[spice idol]

yes

[lone jacinth]

Presumably d is the differential on C(x)D and C'(x)D' respectively

Or maybe I don't understand what you're asking...

Like d_[C(x)D] = d_C (x) 1 + (-1)^sgn 1 (x) d_D

Where the sign is determined by the koszul rule

I guess they also have some weird sign convention since they say
df + fd = 0 instead of df = fd for chain map f

But you can just flip the sign to whatever fits your convention

[void plank]

I suppose they might be talking about the differential on the Hom complex? Df = df - fd

But yeah, the signs are weird. And no, you understood my question correctly (as always 🙂 )

[limpid horizon]

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

the wiki page for this is so bad lol

[storm basin]

Is there a textbook out there that attempts to take a purely categorical approach to algebra (ring theory module theory etc?)

For example of the type of discussion im looking for: a group is a 1 object category where all morphisms are isomorphisms. Group homomorphisms are functors between groups. A subgroup of a group G is an equivalence class of monomorphisms in grp with codomain G, and equivalence relation such that two monomorphisms are equivalent if there is an isomorphism between them commuting with the monomorphisms. Then define a ring somehow categorically (?) and define an ideal as a subobject of the underlying additive group with special properties (?)

Part of my motivation is because I feel there is a lot of things in algebra where in the classical textbooks you either have to be willing to handwave or willing to dig into the exact set theoretic definition of constructions to verify eg that various maps are well defined or that various maps are actually equalities vs isomorphisms etc. (For example: the construction of the field of fractions of a ring doesn't actually contain the original ring as a set theoretic subset)

Another thing that annoys me is that a lot of algebraists (and mathematicians in general) seem to write down isomorphisms without actually specifying what the maps in question are. In many cases it's somewhat obvious (eg the map $(M \otimes N) \otimes W \rightarrow M \otimes (N \otimes W)... although again, my understanding of this map is more based on a particular set theoretic construction of the tensor product than it is from any universal property) but in many cases it's quite unclear what the map actually is and recovering it to me seems like it often requires thinking hard about constructions and elements. I wonder if someone has done the trouble of taking a more categorical approach where the maps are defined via universal properties or something.

[spice idol]

plenty of these issues aren't with the fact that these things are defined using set theory, rather that you're implicitly using universal properties

[storm basin]

another example of things where you either handwave or think too hard abt elements: the field of fractions doesn't even contain the localizations as set theoretic subsets. At least with the standard construction

[spice idol]

to that end I'd recommend Algebra Chapter 0

[storm basin]

I know you can say to yourself ok retroactively just declare the subset of the field of fractions to be the localization and pretend we were working with that the whole time but these approaches aren't very satisfying to me

similar thoughts abt various quotients and stuff too

[spice idol]

well the reason is that these embeddings are incredibly canonical (i.e. all diagrams commute due to the universal properties) so you might as well treat them as subrings lol

[storm basin]

yes true

[spice idol]

the "correct" way of thinking about this (and this what algebraists do) is indeed categorically in terms of universal properties

[storm basin]

but it is quite annoying because it feels a little bit like "canonicity creep" where you get many of these things are very canonical but then as you start stacking these isomorphisms on top of each other to get slightly more complicated identities it all of the sudden is no longer obvious what the maps are

I struggle with this a lot with tensor product stuff and homological algebra

[spice idol]

at some point category theory is needed

for me the construction of a tensor product is almost never used

maybe for basic properties but at some point universal properties are much more elegant

[spice idol]

in any case:

[storm basin]

Thanks

[storm basin]

I find it's like useful for showing the maps exist/are well defined but i find it quite hard to check they're injective or surjective or whatever with just the universal properties

for example let's say M_i is a directed system

and you want to prove that (lim M_i) otimes N cong lim (M_i otimes N)

the universal property gives you the map but i struggle with showing that it's an iso

[spice idol]

lim here is the direct limit yes?

[storm basin]

yeah

[spice idol]

in that case it is rather easy: tensor products are left adjoints and hence commute with colimits

[storm basin]

yeah that's the purely categorical way to do it but that doesn't give you any intuition really for what the maps are then

whereas if you try to construct them directly using universal property of direct limit and tensor product you get the maps quite clearly but it's no longer clear they're isos

so yeah it's just kinda annoying to me

[spice idol]

well, colimits have to do with "addition" (direct sum) and "quotients" (coequalisers), and tensor products kind of, trivially commute with those

[spice idol]

I mean if you write them down I'm sure that it's not too hard to show they're inverses

[storm basin]

yeah but then we're working with representatives like when you write them down. i guess im trying to have my cake (understand what the maps look like) and eat it too (do it in a construction independent way)

[spice idol]

in general direct limits are a little annoying to work with because you're constantly working with representatives

there's no real going around that using direct limits

unlesss you use category theory :>

[steady lance]

I love the term "canonicity creep" and I'm gonna steal this in the future lol

[fallow ibex]

Can anyone help me understand the concept called Cartan Subalgebra? Ive recently been researching donuts/torus and this has come across my research paper im reading. Thank you

[plucky arch]

This is exactly what yoneda helps with

[storm basin]

Hmm im not following how Yoneda is relevant here.

[plucky arch]

It’s what you use in tandem with universal properties to show isomorphisms like this

[storm basin]

Been a while since I looked at Yoneda but im confused as to how just constructing the map with universal properties + applying yoneda implies it's an isomorphism without actually doing any module theoretic work or any work specific to the specific universal property in question. Unless you're saying we use Yoneda to prove that left adjoints preserve colimits which true ive seen that proof (though as I mentioned i dont find it very satisfying)

[plucky arch]

The latter is true to an extent but you don’t even really need that

[lone jacinth]

Isn't that what you get though?

Like you get a construction of the map and by abstract nonsense that map must be an isomorphism

It's not like you have to choose one or the other

[storm basin]

If you prove the other object satisfies the universal property too then there’s no issue.

That’s not exactly what I was doing though

For example if you want to prove (lim M_i) otimes N cong lim (M_i otimes N) one way to do it is to construct via the universal property of the direct limit a map from the right side to the left. Then you construct a map in the other direction via the universal property of the tensor product. Then you check that the composition is the identity. The last step is hard to do without working on the level of a construction.

Anyway another part of the point is that doing a proof via saying for example lim (M_I otimes N) satisfies the universal property of the tensor product is also not completely trivial

Especially without working with a construction

[lone jacinth]

Well, you have a natural map
lim(M(x)N) -> limM (x) N
coming from the universal property.

Then -(x)N being left adjoint tells you this map is an isomorphism

And you can describe this natural map explicitly using whatever construction of direct limits you have

Like if what you want to do it so understand what the maps look like explicitly and have a construction independent proof, then just do that.

Prove it construction independently and construct the map.

Unless you're saying you want a construction independent proof that depends on a construction there isn't really a problem

[plucky arch]

Yes that’s why it’s easier to instead show that maps out of them are naturally iso

[plucky arch]

$\text{Hom}((\lim M_i) \otimes N, Z) \cong \text{Hom}(\lim M_i, \text{Hom}(N, Z)) \cong \int_i \text{Hom}(M_i, \text{Hom}(N, Z)) \cong \int_i \text{Hom}(M_i \otimes N, Z) \cong \text{Hom}(\lim (M_i \otimes N), Z)$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[fierce steeple]

Where here I guess lim should be colim or varinjlim

[spice idol]

yeah

it's supposed to be a direct limit

[plucky arch]

This is why I prefer the coend notation

[woeful crane]

The Wikipedia entry on Cartan subalgebras does a decent enough job and has examples, you should check there. If you want more detail, there's an appendix in Fulton-Harris on Cartan subalgebras.

What's the paper you're reading?

[void plank]

if f, f': C ---> C' and g, g': D ---> D' are pairs of homotopic maps, is there an explicit formula out there for the homotopy between f (x) g and f' (x) g'? I know what it is stenographically, but the indices are a massive pain to figure out explicitly

[plucky arch]

presumably you do the homotopy componentwise

[void plank]

Yes, but I guess I was worried about the signs

Then again, I suppose I can just tack on the signs component wise according to the Koszul rule

[plucky arch]

the signs?

[void plank]

The homtopy map at the tensor product level will pick up some minus or plus signs according to the Koszul sign rule right?

[plucky arch]

idk what that is sorry

[lone jacinth]

I mean, this is the exercise you had before right?

If the signs don't come out right you can just change them.

Like if h(x)g gives the wrong sign then just use the map with component
(-1)^i (h(x)g)_i
in degree i or whatever

[void plank]

Yeah, that's what I eventually figured. Thanks!

[storm basin]

Yeah I realized this later. If you want construction independent then by definition you don’t really know what the maps look like since you don’t know what the elements look like. If you want to know what the maps look like do it based on a construction. It’s fine though

[broken turtle]

Onkos

[solar turret]

So I started using Homological algebra by Weibel

And i got stuck at definition

So the first question is why they denote all d_n as d only?

And what does it mean by the kernel of d_n is the module of n-cycles of C ?

[solar turret]

I think they are just definition

[undone idol]

very often, the boundary operators / differentials of a chain complex arise from the same (topological) operation performed at different "levels"

[solar turret]

okay, they define the category of chain complexs, so if C and D are two chain complex and u is a morphism C to D, so author said u maps boundary to boundary and cycles to cycles, so i shown that u(Z_n) \subset Z'_n and u(B_n) \subset B'_n, do i have to show equality there?

[undone idol]

equality is not necessarily guaranteed

for instance like any other (abelian) category we want to have an initial object, the zero chain 0 -> 0 -> 0 -> ... -> 0 -> 0

and you basically never get equality with this thing

[solar turret]

okay

[lone jacinth]

So the proof of exactness at M(x)N should use that psi' is epi, because otherwise it's not true.

I'm not sure I quite follow what you're saying next. Are you saying because you don't use that psi' is epi you can now prove that Tor(M, N) is 0? That would be correct, but you do need that psi' is epi, so that's the flaw

@polar hawk

[solar turret]

Here I am confused on which ring I have to show { (Hom(A, C_n) ) } is a chain complex over Z or R?,

I don't see any problem in both cases

[last talon]

"Chain complex of abelian groups" so over Z

[solar turret]

I thought the same

Okay thanks

And i think they are not assuming R has to be commutative

[lone jacinth]

If R is commutative then you in fact get a chain complex of R-modules

If R is not commutative it's only abelian groups

[solar turret]

Yes

[solar turret]

I got stuck at second part

Do I have to find under A = Z/nZ, what will be my B_n and Z_n ?

[lone jacinth]

A = Z_n, not Z/n

[solar turret]

Ah

My bad

[solar turret]

I got it

Man, lots of diagrams, lots of objects and morphisms , how do you guys manage them?

[urban granite]

well there isn't other thing than not trying to write down everything lol

[lone jacinth]

Buy a bigger black board

[urban granite]

tbh chain complexes doesnt feel to have a lot of objects in it after knowing what is a mixed hodge complex lmfao

[forest turtle]

you just start getting used to it. knowing the purpose of the structure you’re working with is helpful, as well as memorizing a few key examples. this is what i have done while learning about internal categories, double categories, bicategories, monoidal categories, enriched categories, etc.

all of these have a lot of structure to remember, but there are a few examples to remember for each type of category, along with what each one is supposed to accomplish.

[wanton spoke]

chain complexes: just the skeleton ☠️
mixed Hodge complexes: skeleton + organs + nervous system 🧠🫀🩻

[solar turret]

Stuck at the converse part

[lone jacinth]

One thing to think about, say:
The inclusion Z_n -> C_n is an element of Hom(Z_n, C) that gets mapped to by the differential. What needs to happen for this inclusion to also be in the image of the differential

[solar turret]

i used this fact in second part, but here if i can define morphism Z_n -> C_(n+1) such that compositing with d(n+1) gives inclusion mapping, one thing i am thinking that we know Z_n lie in im (d_(n+1) ), so for each z in Z_n, i can get x in C_(n+1) such that d_(n+1) (x) = z.

[lone jacinth]

What you're saying here is correct, but I ask what would it mean for the inclusion to be in the image of the differential of Hom(Z_n, C)?

[solar turret]

i don't know i am saying correct or not, its mean there exists a mapping Z_n to C_n+1 such that composing with d_n+1 gives inclusion

don't give answer, i am trying to come up with your right answer

[lone jacinth]

So then the way for Hn(C) to be 0 is for Cn+1 to surject onto Zn, but for Hn(Hom(Zn, C)) to be 0 you at least need such a splitting map Zn -> Cn+1.

Now are these conditions equivalent or not?

[solar turret]

i don't think, but let me think more

[solar turret]

so i am thinking in this way, let's say there G1 -> G2 a group homomorphism and it is surjective but it is not necessarily that there should be G2 -> G1 homomorphism such that their composition G2 -> G2 is identity on G2, we can take abelian group and G1->G2 any non injective surjective mapping, Z/4Z -> Z/2Z, 1 -> 1

[prisma gull]

That idea is critical and really important in homological algebra. when do we have maps between complex with same homology?

In sense of category, we do not have inverse maps even we have same homology and induced maps are isomorphism

Keep in mind, first chap of Weibel is trying to say when they have such map or not.
When they have same homology

[plucky arch]

Quasi isomorphism ?

[prisma gull]

Yes

Quasi-isomorphism and chain homotopy

Chain homotopic equivalent gives quasi-isomorphism, but sometimes even their homologies are isomorphic, they don't have such opposite direction maps.

First chapter of Weibel is story of chain homotopy and why it cannot cover all quasi isomorphism

Which means Homotopical equivalents is not enough to reflect, equivalent of homology

[fierce steeple]

Idk what the question is

[solar turret]

@fierce steeple maybe that's the question, i was asked for converse part, which i got it.

[solar turret]

i don't get it this, what is homology

[fierce steeple]

Ah yeah this was a weird question lol

[prisma gull]

Oh you will see that in the first chap of Weibel

In my memory, it's on 1.5

[solar turret]

okay

[wanton spoke]

homology is homology

it is a way to measure the shape of a space by detecting its holes in different dimensions

[prisma gull]

Ah I thought he's talking about homotopy

Homology is just homology
Quotient, ker/image

[prisma gull]

In the topological sense, yes

[solar turret]

can i get some intution why we are interested in chain complex?

why these chains are important? How its comes into picture?

[plucky arch]

Hm

[plucky arch]

Do you have intuition for exact sequences

[solar turret]

tbh no, i just know they are chain with some property

[plucky arch]

Ok I’ll try to release an article on this soon

I’ve given this explanation a lot of times and it’d be better to write it down than to repeat it

The gist is that you can think of exact sequences as “indicator functions for algebraic structures”

[solar turret]

okay

[prisma gull]

My direction is quite different

In category theory, you can think about functors with preserving kernel and cokernel or not

Homology is technically the way to measure how much kernel and cokernel acts on does maps

So you can measure, how much your functor respect kernel and cokernel
If you calculate the homology of functors

That's algebraic intuition

On the other hand, homology can reflect shape of space or differential structure of spaces

[plucky arch]

Interesting, how

[prisma gull]

Dervied functor will answer this

[plucky arch]

Oh those

I’ve heard of them but I don’t quite understand how they link to homology of chain complexes

[solar turret]

so we are in category, where notion of kernel is well define

[prisma gull]

You can think exact functor as a functor between abelian category, which preserves kernel and cokernel
Right?

In this sense, when you have right exact or left exact, which is fail to preserve kernel or cokernel

Then you will see homology is none-trivial after we apply thoes functor to exact sequences

This show how much your functor respect kernel or cokernel

[solar turret]

actually i don't sure about notion of kernel in category, i know equaliser thing

[prisma gull]

(Kernel is equaliser) in abelian category

[solar turret]

my category knowledge is just very little, trying to build

[lone jacinth]

One motivation is that understanding modules is generally hard. So you might understand them by looking at generators and relations. This gives rise to projective resolutions.

Then when you apply some sort of functor (like Hom(A, -)) to a projective resolution it might no longer be exact.

This then gives you the definition of a chain complex

[plucky arch]

A kernel is a special kind of equaliser

An equaliser with the zero map

[urban granite]

well not a big thing because in most cases you will work with R-modules

or at least you can convert to this case

[prisma gull]

By Freyd theorems I guess

[solar turret]

and now i encounter abelian category first time in Weibel, so that's mean for each set Hom(A, B) there is notion of addition of morphism such that Hom(A,B) is an abelian group? and does that addition notion concide with other addition notion in Hom(C,D)?

[solar turret]

yes

so that's mean we need 0 map

[prisma gull]

Abelian category need more than that

[plucky arch]

This is only an Ab-enriched category

An abelian category has some further exactness properties (meaning, properties to do with limits, colimits and how they interact)

[prisma gull]

And also finite biproduct

Which mean product and coproduct will agree for finite index

[plucky arch]

Yes, I’d classify that as an exactness property

[solar turret]

so do i have to stop here, and go to read about abelian category?

[prisma gull]

Yeah you are right, I just give more explanation

[plucky arch]

Mhm

[prisma gull]

Weibel has chapter of abelian category

Just keep read it

And it also has a appendix

[plucky arch]

It’s something like - you need finite biproducts, kernels, cokernels, every monomorphism is normal, every epimorphism is normal

[solar turret]

okay

[prisma gull]

Cleary, if you know what is functor,limit,colimit,adjoint
You have enough knowledge to read chapter 1-2

[plucky arch]

I found the definition easier to understand once I’d seen the various flavours of monomorphism and epimorphism you can define

Ordinary, extremal, strong, regular, effective, split

[prisma gull]

Good to know

Pretty sure good to know, but not necessary for now

[prisma gull]

Maybe my explanation of this would not cleary tell whole story.
In addition, you can think category of chain complexes and construct derived category by localization with their model structure.
Then you may have embedding of each object in abelian category (with good axioms, to construct resolutions and so on) to the derived category.
What actually shows each resolutions are in same category with chain complexes, in sense of derived category.

I think it's still not enough to tell why it's connected but...

It's beginnings of story

[polar hawk]

Thank you for take the time to answer. I finally realized that after some reflection on my proofs. I could see the necessity of psi' being an epimorphism after reading some other proofs of the statement about _\otimes N I. 😄

[urban granite]

<@&268886789983436800>

[wanton magnet]

In an abelian category, given two maps $f:C\rightarrow B$ and $g:B\rightarrow A$. I want to argue that the map $im(f:C\rightarrow B)\rightarrow im(g\circ f:C\rightarrow A)$ is an epimorphism? It is not hard to prove this map is surjective if A, B, C are sets, How can I prove this using only arrows?

[broken turtle]

Dong_Valentino

[wanton magnet]

I was using the definition of epimorphism and want to argue a=b if $a\circ k=b\circ k$. I do not know how to proceed.

[broken turtle]

Dong_Valentino

[wanton magnet]

Is there any hint?

[lone jacinth]

So far you still haven't used all the power of abelian categories.

In particular ker(cok f)) = cok(ker f) would be useful

I think this is even true for preabelian categories, though a little unsure: ||all you really need is for the canonical map C -> im(f) to be an epimorphism. I believe this is true also in preabelian categories, but it would be much harder to show.||

[wanton magnet]

Thank you. Now I see that.

[wanton magnet]

I may need a bit more help with this. Suppose additionally right vertical arrow is an epimorphism. I want to show that the middle arrow g is also an epimorphism.

I want to prove by definition. D is any object and I want to prove a=b. The right upper vertical arrow is epi tells me the induced map a'=b'. How can I argue that a=b from a'=b'?

[lone jacinth]

I often find it simpler to try to show that a-b = 0.

So naming a-b for c, notice that cg = 0 implies cgf = 0. Then another arrow can be added to your diagram

[limpid horizon]

Is im some kind of categorical image?

[storm basin]

IMO they’re just useful computational tools and you see why we use them after seeing them come up naturally in a ton of different areas of math

Have you done any differential form stuff

Or like combinatorial topology

Simplicial complexes etc

[wanton magnet]

The image here is defined in an additive category. The image of a map f:A-->B is the ker(coker(f:A-->B)).

[wanton magnet]

I will give it a try tomorrow when I wake up. Thank you.

[storm basin]

Or from a more algebraic sense you have some map of modules and you want to study what the kernel is when you tensor some sequence that was originally exact with something

Or many more examples of useful chain complexes throughout math

[digital parcel]

read the intro to the chapters of hatcher's algebraic topology book

he has some good stuff motivating chain complexes

probably like chapters 1 and 2

[solar turret]

No, none of them

[solar turret]

so direct product of R-modules is product in catgeory of R modules

[swift cove]

As well as the coproduct

[lone jacinth]

Finite coproduct anyway

[solar turret]

i can show by constructing explicit maps but can i get this by using some property of H_n functor ?

[lone jacinth]

It's not true in a general abelian category, so you'll have to use something.

Follows from sum/product comuting with kernels/ cokernels though

[solar turret]

i shown \sum (Z_n / B_n ) isomorphic to \sum Z_n / \sum B_n

[lone jacinth]

This part is true in general.

Something that might fail is
Zn(Sum C) = Sum Zn

[solar turret]

what is mean by Z_n(Sum C)?

[lone jacinth]

Like the direct sum of complexes is Sum C, and then you take the kernel of the differential at degree n

[solar turret]

i think it is true Z_n (sum C) = sum Z_n

[lone jacinth]

In mod-R sure

[solar turret]

yes

yeah it will may problem in other abelian cat

[lone jacinth]

So my point is that using the explicit construction is probably the best solution, since it doesn't hold in general

[solar turret]

okay

thank you

[solar turret]

Is there any difference between monic and monomorphism in any category? And in abelian category kernel are equaliser of f and 0 so they need to be monic

So like Ch(R mod) , can we define chain complex of any abelian category?

[plucky arch]

yes

chain complexes can be viewed as additive functors from some fixed category J

[solar turret]

explain

[solar turret]

got it, in R mod it is easy to see, but in other category i don't think it will be easy to see even it is true there

today i get to know that what does it mean of Ch in "Ch-monkey", it is chain complex, right?

[compact summit]

I thought monic and monomorphism meant the same thing? I'm just trying to check my understanding as well haha

[plucky arch]

you use the category $\mathcal{Z}$ defined in the screenshot

[broken turtle]

Pseudo (Cat theory #1 Fan)

[solar turret]

yes

[compact summit]

Nice!

[solar turret]

so object of \mathcal Z are? Z , \mathcalZ(n,m) ?

[plucky arch]

the objects are integers, the morphisms are given in the def

$\mathcal{Z}(n, m)$ is the abelian group of morphisms from $n$ to $m$

[solar turret]

okay

[broken turtle]

Pseudo (Cat theory #1 Fan)

[solar turret]

i see

they have to write Hom(n,m)

and composition of n and m is ?

[plucky arch]

"composition is defined by multiplication whenever it takes values in a nontrivial group"

[solar turret]

yeah, i don't get it, what does it mean

[rose mirage]

I think it's terribly worded as well. I think they mean that the composition map Z(m, n) x Z(k, m) -> Z(k, n) maps (f, g) to fg if n-1 <= k <= n and to 0 otherwise

mimicking how d^2 = 0 in chain complexes

[solar turret]

if f is monic and f is kernel of map g: B->C, how do i show g is co ker f?

[prisma gull]

Because

Let's m be monic of morphism f

then fm=0 gives universal property to cokernel

which maen cokerm m has unique maps to cod(f)

keep in mind this diagram, and consider zero arrow from left hand sides

mono gives it's directly

this is what I'm saying

this unique maps allow you to prove for arbitrary f, since it's universal property replaced it

So what do you want to fill out on left side of this diagram?

[solar turret]

m be monic of f?

[prisma gull]

m is monic

just monomorphism

[solar turret]

and f?

[prisma gull]

f is any maps

[solar turret]

okay

[prisma gull]

some morphism

what you said

[solar turret]

i want to show this

so here you assumed m is given monic polynomial and m is kernel of f

and we already have existence of co ker m

so we get the unique map from C to D

so B -> C your co-ker m?

[prisma gull]

yes

B->C is coker

Let's assume you have another maps to B
composition to coker(m) is zero

then you will see what I meant

[solar turret]

okay

but why i need that ?

why i need 0 map A to C?

[prisma gull]

what do you want to prove is m is kernel of coker(m)

isn't it?

[solar turret]

yes

but i want to show it is enough to show m is kernel of some morphism

[prisma gull]

yes

so assume you have some morphism like that

and filled out left hand side with arrows

[solar turret]

you mean 0: A->B

[prisma gull]

Showing m is kernel of some maps is enough...

means\

(m is kernel of some maps) => (m is kernel of it's cokernel)

is what you want to show

[solar turret]

yes

[prisma gull]

so you assumed (m is kernel of some maps) is true

[solar turret]

okay i see what you said

[prisma gull]

so you have such diagram

good

[solar turret]

ah no

m is kernel of some maps, okay

[prisma gull]

so f is some maps

what I collected

[solar turret]

yes

yeah

[prisma gull]

then cokernel has universal property of such dash arrow

right?

[solar turret]

yes

[prisma gull]

then

when you make new arrow like

E->B

such that composition with cokernel is zero

what happen?

It will directly prove your statement, if you can figure out.

[solar turret]

then i think it gives a unique map E -> A such that the triangle commute

[prisma gull]

yes

since it's kernel of f

you got it

then it's same as having universal property of cokernel as well

[solar turret]

yes

[prisma gull]

good

[solar turret]

ah i see

oh great

thank you

it is something which comes natural after sometime spending time with this stuff, or am i dumb?

[prisma gull]

it's natural because you are not friendly with category theory

That's why I recommended you to read category theory book in the same time

[solar turret]

Yes I am reading

[spice idol]

yeah, it's just a lot of practice with category theory

[neat nexus]

I am confused. Should we define Hom(n,n)=Z=Hom(n,n-1) and a functor F: Z->A mapping 1 in Hom(n,n) to 1_F(n), mapping 1 in Hom(n,n-1) to d_n: F(n)->F(n-1)

[limpid horizon]

Why is Ass(R/I) the same as the primes associated to I (in I’s minimal primary decomposition)

I know that 1) each prime in Ass(R/I) contains the annihilator of R/I, which is I. And, every prime associated to I does contain I.

[rose mirage]

that's what I had in my head yeah

[limpid horizon]

Its ok i see the theorem in eisenbud

[solar turret]

why Rotman defined hom(A,B) to be a set?

[rose mirage]

does it matter?

[solar turret]

I don't know, maybe because we will work only with local small category therefore there is no problem

[spice idol]

yeah usually you do want that

[rose mirage]

well if he's defined all categories as locally small I doubt you'll be working with locally large ones

[subtle smelt]

Hallucinatory messages popping in and out of existence 😔

[spice idol]

you better take your meds

[subtle smelt]

Silence shadow

What's the most colloquial example of a locally large category, besides Cat ig

[rose mirage]

is Pos locally large

Grpoid has to be lol

[rose mirage]

it isn't

[subtle smelt]

Could be, but size issues are for nerds raaah

[spice idol]

functor categories

like the yoneda embedding stricly doesn't exist for nonsmall categories if you only accept locally small categories, as [C^op, Set] may not be locally small

I'm pretty sure?

[prisma gull]

Because, usually beginner of category theory doesn't have enought knowledge to treat class and logics.
Basically they wants to ignore size problem, by assuming everything they treat is small enough.

Also, even though they have some additive structure(which mean it's pre-additive category),
it's not necessary to realize in category of some algebra

we can still use internal strucutre with some maps
to realize it has algebraic structure.

For example, Lets G is a group

We can think two different way

1. object of category of groups
2. object of sets with some morphism define operations, inversions, etcs

That's why we have good approach for Hom_C(A,B) as a sets.
Even though they can have some structure.

If you wants to know some approach to construct category in rigorous way(still not enough for logician tho).
I will recommend you to read beginnings of "Categories and sheaves" from Kashiwara

This book follows some old fashioned(or can be seen as approach without 2-category sense).
But still have good construction of category.

I will say, just read beginnings(u-sets, and small sets) and abandon this book.

Okay, I think I didn't mention why it's hard to treat them?
If they are class, we can not directly quotient all hom sets and make new category.
For example, homotopical category K(A) of ch(A).

[limpid horizon]

if there exists a "contracting homotopy" of a complex (so that, the identity and zero chain maps are homotopic), it implies the complex is exact. is that just because, if id: H^i -> H^i with H^i nonzero then it sends something nonzero to something nonzero, so obviously that cant be the same map as just the zero map

[limpid horizon]

Also, can someone pls give me some intuition on why people care about the cohen-macaulay property? It's been hard for me to get a good understanding there, because it seems like its relevant to so many areas of math

so its hard to see the big picture reason of what is going on

it feels like the CM property implies 5 million things lol

[worldly zealot]

other people can answer much better than me but i can try to say something in the meanwhile

if you know about the quasicoherent sheaf attached to a module, then a module sort of looks like a vector bundle over the spectrum of the ring. if this ring locally has dimension n, then a regular sequence is a a hyperplane (so reducing the dimension), and then another hyperplane (which intersects the first one to reduce the dimension again) etc etc. then being cohen-macaulay means that you can do this hyperplane reduction n times, which is some sort of rigidity property of the module – it never happens that these intersections of hyperplanes reduces the dimension too fast.

for why people care about it, it has some useful applications, which i dont know fully but i can think of a few

the dualizing complex is concentrated in one degree, meaning you dont need to work with complexes to use duality; "miracle flatness" states that a morphism from a finite type cohen-macaulay scheme to a finite type regular scheme with equidimesional fibers is flat; theres also a sense in which the cohen macaulay locus of a scheme retain all the homological information about a scheme

but to be honest its not clear to me why this rigidity property means all these other great things philosophically

[limpid horizon]

that sounds quite interesting

[digital parcel]

CM rings are common and you have already seen some useful properties of CM rings (that is, those relating to local cohomology)

if you can convince yourself that depth is useful, then it is also natural to ask "when is depth = dim?" which is the CM property

[ebon cove]

The Auslander-Buchsbaum formula is quite useful, in connection with depth.

[digital parcel]

geometrically, cohen-macaulay-ness is like having nice singularities though I don't know enough about that to elaborate. someone else def can tho

[worldly zealot]

i mean "elaboration" could just entail that regular local rings are CM. and so the various properties that a CM ring (like those listed above) has are certainly true for nonsingular rings

[digital parcel]

(side note: since you're applying for phd programs these might also be good questions to email your potential advisors and get your foot in the door lowkey)

[digital parcel]

fair enough

somehow i got the impression that being CM was a lot deeper than that but maybe i was just overthinking it

[worldly zealot]

im sure it is but as far as the claim of nice singularities, that at least gets there

[limpid horizon]

Yeah thats a good point, CM stuff is interesting to me so i def would be interested to study it further

[spice idol]

kianthecmmaster

[limpid horizon]

Lol

[worldly zealot]

as a side note, CM modules are hugely important in the math that im interested in – the McKay correspondence can be stated in terms of CM modules, and the category of maximal CM modules admits a "stable" category which is triangulated and a very important category of study – its equivalent to something called the singularity category which spiritually measures how singular a scheme is

[ebon cove]

there's a connection with singularities of the minimal model program as well: klt implies cm in characteristic 0 iirc

[digital parcel]

it's kind of a shame you don't want to study in the midwest since there's lots of good faculty in the midwest who work in CM settings

but at the same time i get it. kansas isn't very exciting

[limpid horizon]

yeah just kinda hard. I did look into lawrence, kansas and columbia missouri

lawrence is like one street

[worldly zealot]

i do think its funny that grothendieck was just living in Lawrence, Kansas

[limpid horizon]

Lol i didnt know that

[worldly zealot]

he was doing homalg there

for like a year

[digital parcel]

yesss

his kansas paper

[subtle smelt]

Oh hey, my class will cover this paper soon
https://arxiv.org/abs/math/9908027
Can you yap more about it 👉 👈

[worldly zealot]

xddd

do you know the classical mckay correspondence?

[subtle smelt]

Nope

It is a derived cats (🐈) class, so the angle is from the other side

[worldly zealot]

so the finite subgroups of SL(2,C) are the cyclic groups, the binary dihedral groups, and binary tetrahedral, octahedral, and icosahedral groups

two infinite families and 3 exceptionals – this is already interesting because it is the same classification as simply laced dynkin diagrams

[ebon cove]

what's a crepant resolution, again?

[worldly zealot]

one preserving the canonical class

[ebon cove]

ahh thanks

[worldly zealot]

their action on C^2 extends to an action on C[x,y], so you can consider the ring of invariants

and it turns out for all of them, the ring of invairants can be written as the form C[u,v,w]/(f), where f is singular at the origin, so you can resolve this singularity

and if you compute the minimal resolution in each case, you always get a bunch of P1's which intersect transversally in some places. if you make a graph where each node is a P1 and intersections are an edge, you get the dynkin diagrams back - cyclic groups becoem A_n, dihedrals D_n, and the polyhedral groups go to E6,7,8

this was known in the 1930s

McKay observed that for these groups, if you take their irreducible representations as vertices of a graph and add edges when tensoring an irrep with the standard 2 dim rep has another irrep as a summand, you get the same dynkin diagram with an extra node (the affine diagram) corresponding to the trivial representation

[spice idol]

did bro just name his papers after the place he wrote them

[worldly zealot]

then it was shown that the reason the geometry and representation theory are talking to each other is because you can define something called the G-Hilbert scheme, which yields a crepant resolution of the singularity, and this is sort of the end of the story

in dimension2, a crepant resolution gives the unique minimal resolution

[digital parcel]

oh sorry actually i think it was his Tohoku paper that has lots material from his work at kansas

but i swear there was some "kansas paper"

[limpid horizon]

Lol

[worldly zealot]

so naturally one wants to extend this to higher dimensional complex varieties and some other finite groups of automorphisms, but crepant resolutions need not be unique, so it gets a bit more complicated

[digital parcel]

https://webusers.imj-prg.fr/~leila.schneps/grothendieckcircle/Kansasnotes.pdf

[worldly zealot]

but these different crepant resolutions will be related by codimension 2 operations called flops, and it was shown that flops yield equivalences of derived categories of coherent sheaves

so to lift the correspondence, BKR argue that you have to look at derived categories, and this proves to be incredibly powerful

[subtle smelt]

BKR?

[worldly zealot]

bridgeland king reid, that paper

[subtle smelt]

Oh damn

You said stable earlier, is that in the infinity🐈 sense or something simpler?

[worldly zealot]

something simpler, you essentially kill off the projective objects

by turning every morphism factoring through a projective into a zero morphism

[subtle smelt]

Can you describe what's the idea there? If my original cat had enough projectives to begin with, then I could form the appropriate bounded derived cat as just the homotopy cat of projective complexes. Hard time seeing why you'd want to kill them off in any case

[worldly zealot]

this isnt forming a derived category from the category; it boils down to the MCM category being something called a Frobenius category, and the stable category of a frobenius category is triangulated, where the shift functor is the 0th syzygy functor

[subtle smelt]

I know but in one scenario we get all information from projective objects (provided there are enough of them) and in the stable case we throw them away. That's a bit confusing

[limpid horizon]

Sorry random question in the middle: I was checking why if the chain maps f,g are homotopic then they induce the same map on homology. Its really just as straightforward as, checking f^n-g^n maps cycles to boundaries?

and it obviously does by just checking the equation

[spice idol]

reading this rn and pretty fond of the coordinate transformation idea lol
gluing of schemes ahh

[worldly zealot]

as for some motivation, there is a very fundamental thing called the auslander-reiten translate \tau, and \tau P of any projective is 0, and \tau^{-1} I of a injective is zero. so in that setting its natural to describe it as an equivalence between stable and costable categories

i am not too knowledgeable on the intuition, jagr could explain it much much better than me so hopefully he spawns in

i mean the fact that the stable category of the MCM category is equivalent to the singularity category (which i havent defined but is very useful in like, string theory) is maybe motivation enough?

and the fact that we have a very natural triangulated category means people can use techniques like tilting theory to talk about triangle equivalences

[subtle smelt]

This kind?
https://share.google/ge9k3ckhXc2mksnha

[worldly zealot]

ya

i have never seen a share.google link

[subtle smelt]

What Google gave me on "copy link" 🤷

[worldly zealot]

a conspiracy is afoot

[limpid horizon]

"ahh" ahh

[spice idol]

""ahh" ahh" ahh

[limpid horizon]

true

[fierce steeple]

Lol this terminology is funny to me as like a htpy theorist

[worldly zealot]

i dont really know in what sense its stable

[wise sedge]

It is well-known that PSL(2,F) is simple for any field F with at least four elements. Does this result depend on associativity of multiplication, i.e., is PSL(2,F) simple for any commutative semifield F with at least four elements?

[steady lance]

https://tenor.com/view/mmmkay-mr-mackey-south-park-alright-okay-gif-19580399

[edgy pond]

a problem from my course asks us to classify irreducible Q-representations of Z/pZ for prime p. How should I go about solving this?
I know there are exactly p one dimensional C-representations
but over Q i don't know how things change.

[lone jacinth]

Many ways to approach this

You can note that representations of Z/p correspond to modules over the ring Q[x]/(x^p - 1). You can determine the modules just from the structure of this ring

Another approach could be to think about when certain matrices are conjugate to rational matrices. This you can determine from the eigenvalues, which can then tell you exactly which C-representations are Q-reps

[limpid horizon]

im not really sure if i understand the significance of those being actually equal versus just isomorphic

[ebon cove]

The goal is to show that multiplication by xj is an isomorphism. It has already been established that xj is injective, so xj E(R/p) is a submodule of E(R/p) that is isomorphic to its domain (which happens to be E(R/p) lol). That is different from saying that xj is surjective (which is proved by the contents on p.132).

IOW you want the underlying equality not just the underlined isomorphism if your objective is to show that this map xj: E(R/p) -> E(R/p) is an isomorphism,

underlined*

[neat nexus]

What book is this from?

[limpid horizon]

Bruns and Herzog

Title: Cohen Macaulay Rings

[neat nexus]

Thank you

[limpid horizon]

Thanks. It looks like that is needed for the homotopy to be well defined, its “undoing” localizations eg. Ex1x2 -> Ex2

[ebon cove]

Yes

[limpid horizon]

does that kind of thing show up in other places or is this just some trick that works for this proof?

Conceptually this proof is quite hard for me because of all the indices and notation (and alternating sums ugh)

[hushed bone]

Think about Z vs the set {2x| x in Z} < Z

call the latter set E. Then Z/E = Z/2Z

But E ≈ Z

And Z/Z = {0}

So what matters in this quotient isn’t just the submodule up to isomorphism, it’s what it literally is set theoretically

[empty stump]

What is a good math quote ?

Oh for love

I got the equation

Be back later 🤍

[spice idol]

so true

[rose mirage]

I got a bit of an open ended question. Does anyone know of any sources that discuss statistical/probabilistic results on the proportion of atomic monoids that admit unique factorisation?

[limpid horizon]

How does that imply there are li such that xi^li a = 0?

I think this may be related to why I dont understand the first part of the proof too, its saying if xi^li x = 0 then somehow m^k x = 0 for some k

[limpid horizon]

It does follow from this blurb but i was hoping i didnt have to go learn about I-adic topology

[golden osprey]

the I-adic topology isn't that bad to learn about

granted idk any good sources, I learned it in a commutative algebra course in my UG

[digital parcel]

atiyah-macdonald

[limpid horizon]

Ok yeah ill do it

[limpid horizon]

Im struggling with trying to show that (x^k) satisfies the conditions above for it to define the m-adic topology

Can someone give me a hint pls?

A few things i know: (x1,…xn) subset m, some m^k subset (x1, … xn)

And i think i need to use the fact R is local somewhere

I dunno what im missing seems like smth obvious

[lone jacinth]

So the notation is a little weird, but it seems that
(x^k) means
(x1^k, x2^k , ...)

So then you can just for example notice that (x)^kn is contained in (x^k)

[wanton spoke]

what do you want to prove?

[limpid horizon]

That the family (x^k) satisfies the condition above wrt m so that the m-torsion is the same as (x^k)-torsion i suppose

[lone jacinth]

The idea is basically just that if an expression in x1, ..., xn has large degree, then it has large degree in one of the xi

[wanton spoke]

I = (x1,…,xn) is m-primary you have rad(I)=m and taking powers doesn’t change radicals so rad((x1^k,…,xn^k))=m for every k in a Noetherian local ring any m-primary ideal contains a power of m so for each k there exists j with m^j contained in (x1^k,…,xn^k)
conversely, since each xi lies in m each xi^k lies in m^k so (x1^k,…,xn^k) is contained in m^k

[limpid horizon]

Taking powers of generators doesnt change radicals?

[wanton spoke]

i think yes

are you okay with this @lone jacinth ?

[lone jacinth]

Sure

[wanton spoke]

if a is in rad(I) then a^N is in I so (a^N)^k = a^{Nk} is in I^k hence a is in rad(I^k) conversely if a is in rad(I^k) then a^N is in I^k ⊂ I so a is in rad(I)
for generators: (a1^k,…,an^k) ⊂ (a1,…,an) gives one inclusion of radicals for the other each ai is nilpotent mod (a1^k,…,an^k) because ai^k=0 there so every element of (a1,…,an) is nilpotent mod that ideal meaning it lies in the radical

[limpid horizon]

Is this a hard exercise?

I should do it

O

[lone jacinth]

I would just say if
m^t in (x1, ..., xn) then
m^tnk < (x1, ...,)^nk < (x1^k, ...) though

[lone jacinth]

I would say it's immediate from the definition

[limpid horizon]

I was thinking a power (x1, … xn)^n like that would have generators x1x2 etc that wouldnt be in (x1^k, … xn^k)

But yes i guess its one of those things where if u choose high enough every generator would have a high enough exponent

Ty

[ebon cove]

It's essentially the Pigeonhole principle: https://en.wikipedia.org/wiki/Pigeonhole_principle#Strong_form

[soft parcel]

Could someone explain why (S2) holds in the Serre relations? I get that alpha_i - alpha_j is not a root, but I don't see why that implies that alpha_i + alpha_j cannot be a root.
This seems to me to be equivalent to saying that the root string of alpha_i through alpha_j has length 1, consisting only of alpha_i. But the root length in this case should be <alpha_i, alpha_j>. Is there a reason why this should be 0?

[lone jacinth]

[[xi, yj], h] = [xi, [yj, h]] - [yj, [xi, h]]
= -alpha_j(h)[xi, yj] - alpha_i(h)[yj, xi] =
(alpha_i(h) - alpha_j(h))[xi, yj]

as alpha_i - alpha_j is not a root it must be that [xi, yj] is 0

I'm not sure the relevance of the sum being a root to you

[soft parcel]

I the class notes, in the proof it says [x_i y_j] is in L_{alpha_i + alpha_j}, which they write to be 0

But I guess that's for some other reason then

Not because it's not a root

[lone jacinth]

Probably just a typo of a - becoming a + then

[soft parcel]

This is great though

Oh I see

Ok, thank you!

[limpid horizon]

Originally I thought Ra contains k because k has no proper submodules, but thats only if k is viewed as a k-module. the actually contains k part is using field structure right? if ra = k1 then r'ra = 1 so Ra contains the generator of k as an R-module

[limpid horizon]

i could also just use Ass E(R/m) = {m} (thm in BH). Ra subset E(R/m) so Ass Ra = {m}, so m is in Supp Ra. There also cannot be anything else in support because minimal elements of support and ass coincide, and m is maximal. Is that ok?

Then since Ra = R/ann(a), rad(ann(a)) = m so m^t subset ann(a), so m^t a = 0

[low orbit]

Yes. I like your Ass route..

perhaps worth noting that from Ra subset E(R/m) you only get Ass(Ra) subseteq Ass(E(R/m))= {m}, Ass(Ra) is either Ø or {m}. Since Ra ≠ 0, Ass(Ra) ≠ Ø if R is Noetherian..

[low orbit]

I believe this doesn’t rely on k being a field except insofar as maximal ideal <=> simple quotient

[thick oriole]

Ass route

[limpid horizon]

How can i see that cohomology here commutes with direct sums?

[lone jacinth]

Just directly from the construction of direct sum / cohomology I guess

[limpid horizon]

Yeah homology always commutes with direct sums basically?

[lone jacinth]

It does for module categories / AB4 categories

(it's just the definition of AB4)

[limpid horizon]

Ok, and the way im thinking of it is if you have two complexes you can form another one where each chain is the direct sum of the first two. Then its probably easy to check that cohomology of that is the direct sum of the cohomology of the originals

[lone jacinth]

If you're just taking the direct sum of two things it just follows from additivity

It's infinite direct sums where things are more subtle

[limpid horizon]

Ok, in the proof there its using that theorem about decomposing an injective module into indecomposable summands

Im not sure if there are only finitely many

[lone jacinth]

Not necessarily finite no

[limpid horizon]

Do you know why they can assume I is indecomposable there then?

[ebon cove]

It can be checked directly that $\Gamma_{\mathfrak m}$ preserves arbitrary direct sums, and that is all you need for that step.

[broken turtle]

not-affine-mathematician

[ebon cove]

You have to use the fact that only finitely many components of an element of a direct sum are nonzero for that.

[ebon cove]

Good question. I am almost certain I glazed over that step when I read this initially.

[limpid horizon]

Yeah, im wondering why we are allowed to argue with that gamma functor tho

Is that not something like assuming what we want to prove is true or smth

[lone jacinth]

Well an infinite direct sum is still 0 iff all summands are 0

[limpid horizon]

But cohomology still commutes? Like H^i( direct sum E(R/p) (x) M) = direct sum H^i(E(R/p) (x) M)

[rose mirage]

ah but 0+0 is 8 if you wedge sum them so how can you be so sure!?

[digital parcel]

it could even be infinity...

an infinite wedge sum of 0's is the hawaiian earring

[lone jacinth]

It's not actually

Fun fact for you

[digital parcel]

oh wait you're right

they're not homeomorphic

[ebon cove]

You use the definition of $\Gamma_{\mathfrak m}$. The more difficult direction goes as follows If $(x_i){i\in \mathcal I}\in \bigoplus{i\in\mathcal I} M_i$, and for each $i$ there exists $n_{i}$ such that $\mathfrak m^{n_i}x_i=0$, then use $n=\max{n_i}$ and note that $\mathfrak m^{n}(x_i)_{i\in\mathcal I} = 0$. However, $n$ may not be an integer. This is handled by setting $n_i=1$ for all $x_i = 0$. There are only finitely many $x_i\ne 0$.

[digital parcel]

this is so embarrassing... wow...

[lone jacinth]

Topologists earing are compact and infinite bouquet is not

[broken turtle]

not-affine-mathematician

[digital parcel]

right yea

i forgot about that

[limpid horizon]

I guess i meant we are proving H^i(- (x) C) are right derived functors of Gamma, so i wasnt sure how you can argue some things using Gamma if thats what we are trying to prove

[ebon cove]

Oops lol

Tensoring with C_i preserves arbitrary direct sums, and so does cohomology of a complex of R-modules. So I guess I don't understand where an issue would arise.

Am I wrong about the cohomology statement?

[limpid horizon]

where did we use M being finite?

im just needing (a)

i dont think finite is needed for a

[lone jacinth]

Proof doesn't seem to use it no

Not that R is Noetherian either (for (a) anyway)

[limpid horizon]

maybe noetherian for existence of injective hulls?

[lone jacinth]

Nope

[limpid horizon]

Ok

[lone jacinth]

I mean, I don't know how you've proven it, but it's not necessary

[real sinew]

How to show $\mathbb{C} \otimes \mathbb{R}^{2n} \cong \mathbb{C}^{2n}?$ (as R modules). I tried showing $\mathbb{C}^{2n}$ satisfies the universal property, but I'm not sure what map to use. I thought maybe $f(z, (a1,....,a_{2n})) = (za1, ..., za_{2n})$ but I'm not sure

[broken turtle]

Tiessie

[limpid horizon]

Why not just tensor product commutes with finite direct sums

And C (x)R R = C as R modules

[real sinew]

I guess it is immediate from that haha

I didn't know about those properties yet when trying this exercise this morning so I hadn't considered that

[limpid horizon]

Yeah thats fair they are used all the time with tensor product stuff

But maybe ur explicit map works too

[real sinew]

It feels like a natural choice but I had trouble coming up with the map making the diagram commute

[limpid horizon]

The universal property of tensor products is wrt bilinear maps

So idk how u can say C^2n satisfies the universal property

[forest turtle]

it’s just that R-bilinear maps out of C x R^2n correspond to R-linear maps out of C^2n

[real sinew]

What do you mean? It's a tuple of a module and a bilinear map right? So C^2n and my f

[limpid horizon]

Ah ok yeah

[forest turtle]

i think it’s worth figuring out how to show this, for example, if you are going to be making the identification and working concretely with it

[limpid horizon]

How do we show that the tensor product is the unique abelian group with this property up to iso? I remember dummit and foote proved this with an extra condition that so and so generates D as an abelian group i dont remember the details

[plucky arch]

Yoneda

[limpid horizon]

Wowza

[plucky arch]

It guarantees anything satisfying a universal property is unique up to iso

[limpid horizon]

Cool

[plucky arch]

There’s a reason why it’s such a fundamental result

[limpid horizon]

Yeah i hope we cover it in my class. My class is focusing on double categories

[worldly zealot]

its funny how easy yoneda is to prove

its like a math prank but its real

[plucky arch]

Yoneda is one of those statements where the proof is really easy, the hard part is understanding what you’ve just proved

[worldly zealot]

exactly

[plucky arch]

There are lots of interesting interpretations

Representables as free presheaves on a single generator is a common one

I like the is-does duality perspective

As well as the orthogonal complement one

[spice idol]

rrahhhhhhhhh

that's contravariant yoneda at least

[plucky arch]

It’s nice but i found it difficult to fully understand until i learned how to construct more general free presheaves

It would be like having the only example of a free group you know being Z

[spice idol]

yus

[plucky arch]

And i think the cleanest way to do that is with coends

But probably not something you can get away with when you first learn yoneda

That’s why i prefer the other perspectives

[spice idol]

I like yoneda in the ways it arises in the context i care about

which happens to be the representable functors

[plucky arch]

What other kind of yoneda is there

[spice idol]

I meant interpretation lol

[limpid horizon]

I just learned about defn of epi and monomorphism today lol

[spice idol]

epis are kinda yucky

they dont behave well

[plucky arch]

These are pretty important

There are also lots of flavours of epis and monos

Split, effective, regular, strong, extremal

[limpid horizon]

The definition didnt immediately make sense to me but thats probably all of cat theory

[plucky arch]

Monomorphism means postcomposition is injective