advanced-algebra

[plucky arch]

[lone jacinth]

What made it click?

[spice idol]

Now now, don't spoil my secrets ;P

[plucky arch]

my past self had some good ideas apparently

[spice idol]

How does someone new to the server already have the postgrad role 🤨

Are you
A dirty hacker?

[digital parcel]

paid mods

[plucky arch]

this is more common than you'd think

[digital parcel]

as usual

[plucky arch]

-# based on the number of cranks i see in #category-theory

[digital parcel]

sorry not "paid"

[lone jacinth]

I know a guy

[digital parcel]

but "made a generous donation to the moderation team"

[spice idol]

A guy named " John Givingmoneytopeople"?

[digital parcel]

oh shit! went to high school with him actually

chill dude

[plucky arch]

kid named finger:

[spice idol]

Do you also perhaps know John Mantle?

[plucky arch]

oh he's one of my opps

[spice idol]

Well yk what they say

The opp of my opp is my friend

[digital parcel]

no actually

[spice idol]

🤝

[digital parcel]

i only went to middle school with him

dont know him that well

[urban granite]

lol this is for real

[hushed bone]

Why does Chmonkey have the honorable role?

Why does honorable have a stupid u in the role name?

[digital parcel]

must've had a nice paying phd

[hushed bone]

Questions science can’t answer

[plucky arch]

isn't it the $\mathfrak{h}\text{onourable}$ role

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

hm i guess not

[digital parcel]

why does mathfrak look so weird in the discord font

[plucky arch]

$\mathcal{h}\text{onourable}$?

[broken turtle]

Pseudo (Cat theory #1 Fan)

[hushed bone]

Lol

[plucky arch]

uh wtf

[digital parcel]

what font is this

[hushed bone]

It’s some weird kind of fraktur

[urban granite]

$\mathfrak{H}\text{onourable}$ i think

[plucky arch]

ah nice

[digital parcel]

ohh

i dont like it im going to file a complaint in meta proposals

[spice idol]

I usually denote my ordered pairs using
$\mathcal{h} a, b \rangle$

[plucky arch]

cursed

[broken turtle]

.enpeace_music

[urban granite]

me when realizing learning math was just learning caligraphies

[plucky arch]

$\mathcal{h} \psi | \chi \rangle$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[spice idol]

Oo I love the inner product of class functions from G to C

[plucky arch]

$\mathcal{h} \psi \mathcal{j} \chi \mathcal{i}$

[broken turtle]

Pseudo (Cat theory #1 Fan)

[plucky arch]

this is so cursed

[spice idol]

13th reason ts

[digital parcel]

wtf is up with $\mathcal p$

[broken turtle]

anamono

[digital parcel]

why is it down there

[spice idol]

Tryna run away

[digital parcel]

yeah prob

[limpid horizon]

Lol

To justify how quotient objects M/N inherit the grading from M and N do we have to use second isomorphism theorem or something?

(M/N)a = Ma/Na

[distant harness]

The glyph is designed such that its "baseline" is where the TeX engine should put the overline part of the complete root symbol.

[lone jacinth]

The degree n part of N should just be the intersection of N with the degree n part of M, so the degree n part of M/N is just the quotient of those two

I guess it comes from the quotient of direct sums being the direct sum of the quotients

[vapid axle]

Wait sorry why is this not always the case

[fierce steeple]

Well it seems you are viewing N (x) M_i' as a submodule of N (x) M, which uses the fact N (x) preserves direct sums (or that functors preserve retracts)

(Because in general like if M is a submodule of M' then the nap N (x) M -> N (x) M' need not be injective)

I would just write this stuff out a bit more explicitly

[vapid axle]

Hmm okay thanks

[spice idol]

Universal algebra and ring theory create horrors beyond comprehension

Literally had to invent a new notation for writing down terms for this shit to fit on the screen

Basically, if you have a term p(x1, ..., xn) then for terms t1, ..., tn, one can write P_i=1^n [t_i] rather than p(t1, ... tn)

This for example turns commutativity of terms into the equation
P_i[Q_i[x_ij]] = Q_j[P_i[x_ij]]

[short vine]

Suppose M, N are finite R modules, f: M -> N a R module homomorphism. Now M' a submodule of M, and it's well known that f induces a R module homomorphism between the quotients M/M' and N/N'. Is it true that f(M)/f(M') is isomorphic to f(M/M')?

[limpid horizon]

Are you sure f induces a map on M/M’?

M’ would need to be in ker f for that

[short vine]

oh yeah forgot to mention that

[limpid horizon]

Then f(M’)=0?

[short vine]

Wait no I meant from M/M’ to N/f(M’)

Which is valid right

[limpid horizon]

Well for f: M->N to induce a map on a quotient M/M’ you need M’ to be in ker f so that its well defined. Then f(M’)=0

Idk exactly the answer to the question yet im just saying stuff

[short vine]

Well I fear that’s only for an endomorphism of M. For general module homomorphism f induces a module homomorphism from M/M’ to N/f(M’) which is still well defined right? f(a+M’) maps to f(a)+f(M’). If a is in M’ then f(a) is in f(M’), so its well defined

[limpid horizon]

Ya u right

[lone jacinth]

Not just isomorphic, but literally equal.

[cobalt sonnet]

How does separability imply non degeneracy of the trace form?

[cobalt sonnet]

I understand the case when the characteristic of k does not divide the order of the extension, since in that case, I can look at the trace of 1 and get the order of the extension

but what about the case when it does

[fierce steeple]

I remember liking this proof https://math.stackexchange.com/a/1303190/932674

In short the idea is that if you have an extension K(a)/K where a has min poly f (not necessarily separable) then f' = 0 iff the trace of all a^k vanish (i.e. trace identically zero)

[broken turtle]

bostock

[ornate atlas]

I think they’re just saying the multiplication is well defined

[willow seal]

i see, just trying to see how you'd show it's well defined

[broken turtle]

bostock

[distant harness]

It is, "regardless of which representative you choose, you get the same element of M when you scale by it".

[willow seal]

ok starting to make sense, thanks

[limpid horizon]

I'm not actually sure how to interpret his question. We are starting with a hom f: M -> N and then what?

[cobalt sonnet]

how do we know the polynomial (x-a_1^k)..(x-a_n^k( has coefficients in k

[cobalt sonnet]

i mean in this proof

[fierce steeple]

Hm now i am confused in the inseparable case

[cobalt sonnet]

what about the separable case?

[cobalt sonnet]

when it's separable we can say something about the coefficients of this polynomial?

[fierce steeple]

When it is separable this follows by Galois theory

Because the poly is symmetric

[cobalt sonnet]

oh right so we are just looking at the splitting field

[fierce steeple]

Yeah like F(a1,... ak)/F is the splitting field of f

But yeah there are of course other proofs of this fact

[lone jacinth]

You get an induced map
f: M/M' -> N/f(M')

Then the image is f(M)/f(M')

[cobalt sonnet]

Also, treating rational polynomials as formal power series makes me uncomfortable

this is the first time i have seen such a proof

[fierce steeple]

Ah fair

To me that is what gave this its charm

But yes the point is to work in F((x)) (= formal power series with x inverted)

[cobalt sonnet]

But yeah, you are right, now that I can justify it, it's elegant

[cobalt sonnet]

But what about the inseparable case?

[lone jacinth]

In the inseperable case it's possible for the trace form to be everywhere 0.

Consider for example L = K(a) with x^p - a^p the minimal polynomial of a and K of characteristic p.

[fierce steeple]

Absolutely — I just meant like how to make the proof work here

The point is that this proof is meant to work in either case to then make a conclusion

[lone jacinth]

Wait, the proof of what?

[fierce steeple]

Sorry the point was to show that the trace form for L/K is non-degenerate iff the extension is separable

And the screenshot was an excerpt of a proof

But I am unsure (or forgot) how to justify a step in general

[lone jacinth]

Well for any element a with minimal polynomial f, the characteristic polynomial will be f^d where d is the degree of the extension divided by the degree of f.

As f is either inseperable or d is a multiple of p, the trace should always be 0.

[fierce steeple]

It should be possible to just deal with the inseparable case separately here and then the proof definitely works in rhe separable case

[fierce steeple]

Yeahh nice

[spice idol]

Is there any literature on "localisation" for commutative monoids, and how it interacts with quotients and stuff?

[ornate kindle]

So I'm reading a bit about Cauchy modules

[broken turtle]

NotABot

[ornate kindle]

And I'm wondering how deeply this connects to the idea of completeness of a space wrt Cauchy sequences

I can see how a Cauchy module would be "complete" related to its dual space, so does the related names go deeper than that?

[indigo lagoon]

I'm currently working through Rotman's HA. At the end of the section on tensor products, he proves that given a torsion group T and a divisible group D, their tensor product will be {0}. Fine by me. But then he remarks that this means we cant make a finite abelian group into a non-zero Q module. How do tensor products change the ring we're taking modules over like that?

[fierce steeple]

Could you send a screenshot? There are a couple of ways to answer this

[rare walrus]

Well do you remember we take tensor products of bimodules? Q is a left Q, right Z bimodule, so Q (x)_Z A is going to be a left Q module

(I've read Rotman so I remember this)

[fierce steeple]

Idk if this is quite the point

[fierce steeple]

Ah lol

Then yes it is what Boytjie is saying

You can also do the following like

[indigo lagoon]

oh I see

[rare walrus]

I am going to log of for the night so you can pick things up potat

[indigo lagoon]

Like yes I see that from the stuff from earlier but its different with a concrete example, I don't exactly have the intuition for how that's working

[fierce steeple]

If $M$ is an abelian group and $R$ a ring, then an $R$-module structure on $M$ is in particular given by a map $R \otimes_{\mathbf Z} M \to M$ etc

[broken turtle]

Prismatic Potato

[indigo lagoon]

right

[fierce steeple]

Like you need to give a map R -> End_Z(M) of rings, which is in particular a map of abelian groups which corresponds to the above

so this means the only map of abelian groups R -> End_Z(M) is 0, which means it isn't a ring map

But eh there are a couple of ways to interpret this remark lmao. I think mine is a bit closer to what Rotman intends but not sure

[indigo lagoon]

no I think he means via bimodules

[fierce steeple]

I guess the thing is like

[indigo lagoon]

my one gripe with this presentation of the material is how few actual concrete examples there are when it matters

[fierce steeple]

Like there is a difference between giving G the structure of a Q-module and "forcing" it to be one by tensoring but ye

And the former is more interesting here

ig like (1) $M \otimes_{\mathbf Z} \mathbf Q$ has a $\mathbf Q$-module structure as Boytjie says, and like this is one way to turn any abelian group into a $\mathbf Q$-module. But also (2) in fact if you can give $M$ the structure of a $\mathbf Q$-module, then it is unique and $M \simeq M \otimes_{\mathbf Z} \mathbf Q$ as $\mathbf Q$-modules, which means $M \simeq 0$ lol

[broken turtle]

Prismatic Potato

[indigo lagoon]

2 is clearer, I think

thanks

rotman's pacing and order of presentation is pretty bad tbh

[spice idol]

Bump

[limpid horizon]

I'm starting with a quotient of polynomial ring k[x1, ... xn]/I which is Z^n-graded by the grading on k[x1,..xn] and I. We are localizing this at some x=x_{i_1}x_{i_2}...x_{i_n}, and we are putting a grading on (k[x1, ... xn]/I)x by defining deg(r/x^n) = deg(r) - ndeg(x).

Anyway, there is a lemma that says the k-dimension of an a-th graded component (k[x1, ... xn]/I)x is <=1. I'm just wondering how we are viewing that guy as a k-vector space

ok i clearly dont know how to use latex on here great

Idk why i was confused actually I mean there is of course a k scalar action on that thing so its a k-vector space ok fine

[near lantern]

For A an abelian category, consider the equivalence relation ~' on the objects/isomorphism classes of A (ignore universe issues) generated by the relation ~ defined by M ~ N iff there exist objects P, Q and short exact sequences P → M → Q and P → N → Q (0s implicit).

Of course M ~' N iff there exist M = M_0, M_1, ..., M_{n-1}, M_n = N with M_{i-1} ~ M_i for i = 1, ..., n. However, is it possible to bound n (uniformly for all pairs M, N such that M ~' N), perhaps given some conditions on A?

For example, A arbitrary, or A = category of arbitrary/coherent/finite-length/(fg) semisimple/(fg) projective R-modules for some ring/commutative ring/integral domain/division ring/field R? (If necessary to pick one myself, fg modules over a Noetherian commutative ring or finite-length modules over an Artinian commutative ring could be interesting.)

To get the obvious cases out of the way:

• If A is semisimple, n is obviously bounded, since ~' is the same as isomorphism. (The optimal n depends on whether we use objects or isomorphism classes, etc. If we use isomorphism classes and the above definition of ~, n = 0 in this case.) This covers e.g., (fg) semisimple modules over a ring (or more generally, any product or "direct sum" of k-(fd)Vect's for division rings k).
• For A = ∏_i A_i, n(A) = sup_i n(A_i) so that n is bounded for A iff it is bounded for each A_i by a bound independent of i. (Caution: if sup_{i ∈ I \ J} n(A_i) = ∞ for all finite J ⊆ I, then ~' for A is not the product of ~' for the A_i's. But n(A) = sup_i n(A_i) still holds because both sides are ∞.)
• If A is a full subcategory of B closed under sub- and quotient objects and extensions, then A is closed under ~ of B and ~ of A is the restriction, so n(A) ≤ n(B).

[near lantern]

If A = finite-length R-modules, for each indecomposable D in A, let n_D be the minimum number of ~ needed to relate D and its semisimplification (semisimple module with same Jordan--Hölder factors). Then
max_D n_D ≤ n(A) ≤ 2 max_D n_D
since we can relate any object to its semisimplification with up to max_D n_D uses of ~.

For R = k[x]/(x^n) specifically, k a division ring, we can replicate JCF arguments over a field to show the indecomposables are (x^{n-i}) = R/(x^i), 1 ≤ i ≤ n. It seems likely that n_{R/(x^i)} = i-1, but I'm not sure how to prove this ceil(log_2(i)) (max(length(summands))) probably at most halves each time, and for this R we can always halve it using (x^{ceil(i/2)})). So n(A) is of the order of log n.

[hushed bone]

The types of questions people come up with are so weird. Like, I actually totally see why this would be something someone would care about, but in no world would I ever ask this question based on my own interests

It’s very cool

[near lantern]

it happened because I assumed K(product category) = product K(categories), then tried to prove it and failed

[hushed bone]

Lmao

[near lantern]

and now the condition needed to ensure it is here

[hushed bone]

And this shows exactly why I would never ask the question . I don’t care about K(category)

At least for now…..

[near lantern]

K is dimension of a vector space or Hilbert series of a graded vector space but for any abelian category (where it is the "least lossy" analogue in a formally justified sense), so I was thinking about it as a corollary of revising Hilbert series.

[lone jacinth]

So if A is the category of finite length modules, then your question reduces to wether there exists indecomposable modules of arbitrarily large finite length.

This is related to one of the Brauer-Thrall conjectures (now theorem): a finite dimensional algebra has infinite representation type iff it has modules of arbitrarily large finite length.

[near lantern]

At least this tells me that K(graded fin-length R-modules) = (multiplicities of graded pieces) if there are finitely many (finite-length) indecomposables, which is general enough for me to move on if I can't get an answer.

[crude forum]

hey, I have a small dumb question

Given two rings R and S, and a morphism $\alpha:R\to S$ making R into an $S$-algebra, then what's the nuance between this and looking at $S$ as only an abelian group and hence at $S$ as an $R$-module?

[broken turtle]

Goose

[crude forum]

Is it only that $R$-modules are more general as in they include $S$ not needing the pre-existing ring structure ?

[broken turtle]

Goose

[hushed bone]

Yeah

[crude forum]

okay sorry for the low level question 😭 Im just trying to make a mental image of the different algebraic structures

also do you say an $S$-algebra for $S$ the main ring or when $S$ is the domain of $\alpha$

[broken turtle]

Goose

[hushed bone]

S is an R-algebra

R is fixed, S varies

[crude forum]

what do you mean by fixed and varies?

[hushed bone]

R is a base ring

S are the algebras

[distant harness]

A morphism R->S makes S into an R-algebra, not the other way around. This may be contributing to the confusion?

[vapid axle]

I'm trying to show that for $A$-modules $N$ and $M_i$, $N \otimes \bigoplus_{i \in I} M_i \cong \bigoplus_{i \in I} N \otimes M_i$. I've constructed a map $g: N \otimes \bigoplus_{i \in I} M_i \to \bigoplus_{i \in I} N \otimes M_i$ given by $n \otimes (m_i){i \in I} \mapsto (n \otimes m_i){i \in I}$. But I'm having a bit of trouble constructing an inverse for $g$. Any hints?

[broken turtle]

okeyokay

[spice idol]

Maybe you can do it showing that the universal properties are equal?

I.e. N ⊗ ⊕_i M_i satisfies the universal property of direct sum

[limpid horizon]

whats universal property of direct sum? I didnt know it had one

[plucky arch]

it's a coproduct

So morphisms $f : (\bigoplus_{i \in I} M_i) \to Z$ naturally correspond to families of morphisms $f_i : M_i \to Z$, in that you can interconvert between the two

[broken turtle]

Pseudo (Cat theory #1 Fan)

[spice idol]

||Let ι_i : N ⊗ M_i → N ⊗ (⊕_i M_i) = U be the obvious map.

Then consider a family of maps f_i : N ⊗ M_i → Z. These can be extended to a map f : U → Z as follows:
n ⊗ ∑_i m_i → ∑_i f_i(n ⊗ m_i)
Clearly f_i = f ∘ ι_i, and uniqueness follows from the fact that U is generated by the tensors n ⊗ m_i for m_i ∈ M_i, and the f_i decide where these should go. By uniqueness of universal properties we have the desired result.||

[forest turtle]

sniped

[spice idol]

Top of the message

Don't wanna write that whole thing out

The category theoretic proof is imo more elegant and reveals more deep properties though

[plucky arch]

:D

[forest turtle]

which proof?

[digital parcel]

||you can also use the facts that 1. a coproduct is a colimit; 2. tensor product commutes with colimits (this is some exercise in Atiyah-Macdonald)||

[forest turtle]

is the one you gave not category theoretic?

[spice idol]

||Left adjoints preserve colimits, and essentially by definition the tensor product is a left adjoint functor||

[plucky arch]

(strictly speaking - tensor M for a fixed M is a left adjoint)

[forest turtle]

yea, okay, this is what i first suggested in #groups-rings-fields

[spice idol]

Yeah yeah eugh formalism

[plucky arch]

:3

[spice idol]

Who cares about bifunctors anyways /s

[plucky arch]

Hom

[digital parcel]

anyhoo as elegant as category theory proofs may be sometimes we should do ugly ugly proofs

[spice idol]

The humble monk:

[digital parcel]

at least for a first time learning

[plucky arch]

Just don’t be homophobic and we’re good 👍/s

Actually I guess being against bifunctors would be biphobic

[spice idol]

I am very bisexual don't be worried

[plucky arch]

same :P

[spice idol]

Yeah of course!

but I will stop doing them as soon as I can

[digital parcel]

preach

at a certain point even the category theory proofs become ugly

[spice idol]

That is true

Then you gotta go to higher category theory

[digital parcel]

thank god i dont have to 🙏 (yet)

[spice idol]

Category theorists when they have to do a small computation:

[digital parcel]

category theorists when they have to do their job

[spice idol]

Anti work

[forest turtle]

why so many flavors of category theory and not just one encapsulating one

[spice idol]

Isn't that like infinity category theory

[forest turtle]

idk

sounds like another flavor to me

[plucky arch]

Having one encapsulating one would kinda go against the spirit of category theory

[forest turtle]

in what sense?

[plucky arch]

To me category theory is actually a rejection of the notion of a “theory of everything”

So having a theory of everything for category theory would feel silly

[digital parcel]

i heard some quote that was like "math is the language of the world and category theory is the language of math"

[forest turtle]

you mean like how it was originally developed to study alg top?

[plucky arch]

Yeah I disagree with both of these

[forest turtle]

and that it’s original intent wasn’t to be a new foundation for math

[plucky arch]

Yeah to me the categorical perspective is like

It’s ok to not have a theory of everything

And to have and consider multiple perspectives on the world

So long as you have ways to translate between them

[spice idol]

It feels like category theory rather underlies everything than incapsulates everything

If that makes sense?

[forest turtle]

ye

[plucky arch]

I wouldn’t even go that far necessarily

I just view it as a perspective one can take with regards to maths

[forest turtle]

what, so it’s a framework to reason about math?

[plucky arch]

Yeah

[spice idol]

Right, "everything" is a strong word

Lol

[plucky arch]

Mhm, that’s really my point

[forest turtle]

i knew what u meant, enpeace lol

[spice idol]

Yes, but I feel the need to clarify because there are certainly people who take everything literally

[plucky arch]

In general I’m pretty biased against theories of everything

[plucky arch]

Sorry

[spice idol]

Uh, 'everything' literally

[plucky arch]

Tone can be hard to read over text

[spice idol]

Nah nah not you

[digital parcel]

i literally dont take anything literal at all

like literally

[plucky arch]

[spice idol]

And certainly not meant in a derogatory way

[plucky arch]

coughs string theory coughs

[forest turtle]

not even myself fr

[spice idol]

Its kind of what UA was for algebra; not a grand unifying theory but rather a general framework that algebra sits inside

Of course, UA is much more than that now

[digital parcel]

not a unifying theory
has universal in the name

[plucky arch]

Mhm

[spice idol]

Grrr

[plucky arch]

Like I don’t want to oversell category theory

And treat it as some mystical secret of the universe or whatever

I just find it neat

It’s been a useful perspective for me

[digital parcel]

shhh you're gonna upset the HSCTs

[spice idol]

Pff, imagine me unironically saying "yeah guys UA will soon replace ALL of algebra, so just stop what you're doing and join us, cuz it'll all be irrelevant anyways"

[plucky arch]

But there’s no axiom of mathematics that says category theory always has to be useful

[spice idol]

But this is the view that some people have about category theory

[plucky arch]

Yeah…

[forest turtle]

i think that is most certainly flawed

[plucky arch]

For me category theory is very against the idea of there being One True Perspective on anything

So the idea that we should take category theory as the One True Perspective on maths feels… not in the spirit of the subject

[forest turtle]

what are alternatives that you like to consider?

[plucky arch]

I mean set theory is cool

[forest turtle]

recently i have been trying to learn type theory, and universal algebra is creeping in somehow. also, yea, set theory

[plucky arch]

There’s lots of maths (e.g. combinatorics) which doesn’t feel categorical

Or analysis

[spice idol]

Muhahah

Some parts of UA also have little to no resemblance to category theory

[forest turtle]

that’s cool

[spice idol]

For example, commutator theory and tame congruence theory are almost purely lattice-theoretical in nature

[plucky arch]

That is interesting!

[spice idol]

And they have proven incredibly fruitful. So much so, that the latter even basically gives us all the knowledge about finite algebraic structures that we need

[plucky arch]

that's pretty neat :>

and see this is the thing

to me the world would feel a lot smaller and less pretty if there was just One True Perspective on everything

i actively enjoy having multiple perspectives and switching between them as appropriate

[spice idol]

Yis

I do like seeing category theorists have to reinvent lattice theory sometimes lol

[plucky arch]

:P

[crude forum]

yes yes it makes sense now thanks 🙂

[plucky arch]

i know some people don't feel the same way and like the idea of there being one ultimate way to view the world, or maths, or physics or whatever

see e.g. the search for the grand unified theory in physics

but to me this is just... fundamentally unappealing

[crude forum]

is it possible to generalize the notion of morphisms of $R$-modules where the base rings change

[broken turtle]

Goose

[crude forum]

I was thinking of saying: $\phi:M\to N$ would be such a morphism of respective $R$ and $S$-modules, if endowed with a $\alpha:R\to S$:
$\phi(r\cdot m)=\alpha(r)\cdot \phi(m)$

[broken turtle]

Goose

[spice idol]

Yeah, that's just a morphism of R-modules after restriction of scalars

@plucky arch look it's a heteromorphism I think

(sorry for ping)

[limpid horizon]

What do u mean here? M and N are S-modules ?

[spice idol]

Taking the functor taking S modules to R modules

[limpid horizon]

Idk if i understand the setup

[digital parcel]

if you have a ring hom a : R -> S and an S-module M, then you can realize M as an R-module

that's the setup

[limpid horizon]

Yeah and then if you instead started with an R-module M and you had R->S then the S-module S(x)R M is “extension” of scalars to S right

Just verifying the stuffs

[digital parcel]

yes because you can regard S as an R-module

and then take the tensor S \otimes_R M

see atiyah macdonald chapter 2

[limpid horizon]

Yis

[plucky arch]

yeah quite possibly

[spice idol]

Yis

[forest turtle]

Explicitly, there is a "heterogenous" category of modules whose objects are pairs (R,M) where R is a ring and M is a module over that ring.
A morphism (R,M) -> (R',M') is a ring homomorphism f : R -> R' and an abelian group homomorphism h : M -> M' such that f(r) h(m) = h(r m) for all r in R and m in M.
Equivalently, this is a ring homomorphism f : R -> R' and an R-module homomorphism h : M -> f*M'.
The R-module f*M' is obtained by "restricting scalars", i.e., f*M' is the R-module obtained by defining r m' = f(r) m' for r in R and m' in M'

-# courtesy of dilligentClerk for explaining this to me

[spice idol]

Is that category useful?

Or is it just "there" for completeness

[forest turtle]

im not sure. i think that depends on what you care about?
dilligentClerk says he is interested in this catgeory, at least, in one of our conversations (Grothendieck constructions were mentioned... but i am still uncomfortable with them)

i suppose for now, it's there for completeness until somebody gives reason as to why one should care about this category

i think its nice that it forms a category tho

[spice idol]

Yeah!

Also cool is fixing some abelian group A and looking at all the possibilities for A to be a module over rings R

That can be formed into a category; the comma category (Ring ↓ End(A))

What was most surprising to me is that this category has a notion of the first isomorphism theorem

In my opinion that is not obvious; it isn't a variety in the sense of universal algebra

[forest turtle]

do varieties in general have a first isomorphism theorem?

[spice idol]

While the comma category (A ↓ V), where V is some category of algebraic structures like groups rings or modules can be interpreted as some category of algebraic structures

[spice idol]

Very much

A variety in the sense of universal algebra, by the way

[forest turtle]

did some scanning of the wiki page

[spice idol]

Given a suitable interpretation, every isomorphism theorem can be generalised to universal algebra

[forest turtle]

making me want to study ts now

[spice idol]

(in fact, given any diagram J of algebraic structures in V (say, groups), the category of cocones of J in V is naturally again a variety)

The equational theory is very messy but it proves that every colimit exists

[spice idol]

It's super neat I think

[forest turtle]

any references that you recommend?

[spice idol]

Burris' and Sankappanavar's "A course in universal algebra" is good

But it does not mention any connections with category theory, for that you can use the older Universal Algebra by Grätzer

[plucky arch]

grothendieck constructions are quite cool

[spice idol]

An R-module structure on an abelian group A can be seen as a generalised element of End(A)

Lol

[weak lodge]

Chapter 3 of Borceux, "Handbook of categorical algebra, volume 2: Categories and structures" is a good crash course on universal algebra from a categorical perspective. I used it for my undergrad thesis

[plucky arch]

yep~

[spice idol]

This is algebraic geometry coded

Mfw you can generalise algebraic geometry until it becomes the study of classes of algebras closed under embeddings

That's somewhat unrelated but it's still funny

[forest turtle]

they seem cool! a lot to digest for me atm tho

[plucky arch]

I would recommend getting familiar with bundles/fibrations

[forest turtle]

yea, yesterday i kind of started a crash course on fiber bundles

proving some basic properties

hence the talk about locally trivial bundles yesterday

[spice idol]

Do bundles and fibrations only pop up in AT/pure category theory?

Or rather, where do they pop up

[plucky arch]

My perspective on this is

The notion of “indexed family” pops up all over math

Even just ordinary functions f : A -> B can be viewed as an A-indexed family of elements of B

But you also have indexed families of sets, of open sets, of functions, of vector spaces, of G-torsors, …

Bundles/fibrations are one perspective on how to encode such indexed families formally

[spice idol]

Mhm

I see

Okay, I've been playing around with the (V ↓ T) thing where T ∈ V

And it's very weird, because there is a notion of quotients, but if (A, f) is an object in the category, then you can only quotient by congruences contained in the kernel of f?

Strange because usually it's quotients containing ker f

But oh well

Hm

[crude forum]

so in a sense it's a bit like for bundle maps

[cobalt sonnet]

Is there a classification of finite-dimensional k-algebras?

The closest i have come across is this theorem but when is B/J(B) separable?

[forest turtle]

i am actually not too familiar with bundle maps yet. maybe somebody else has something more insightful/correct to say.
but if i had to answer, maybe you mean in the sense that bundle maps are fiber-preserving, heterogenous module maps should be multiplication preserving with respect to f?

[fierce steeple]

Yes it appears a lot

[forest turtle]

just to clarify, you mean the hetergenous category of modules?

[fierce steeple]

One reason is like you want there to be a functor {rings} ->{(large) categories} which sends R -> LMod_R and like

[forest turtle]

-# what is LMod_R?

[fierce steeple]

Sorry lol maybe this is not ideal notation — i mean left R modules

For some reason this is common notation in my area lol

[forest turtle]

i use R-Mod for left R-modules and Mod-R for right R-modules

[fierce steeple]

But anyway uh the corresponding Grothendieck opfibration is the projection from the category you gave to rings

Like under straightening/unstraightening

Anyway the important thing being that this is a nice way to even set up the functor

[forest turtle]

yea, this is how he explained it to me, but i think there is a lot of data to sift through for me rn. it hasn't settled in yet

[fierce steeple]

Cause i guess really it is a pseudofunctor and things are more involved if you want to write down stuff

This category appears a lot (for the same reason) if you want to do deformation theory

Like say you have an Z/p- module M and you want to study ways in which it comes from reducing a Z/p^2 module

In these sorts of situations you want to consider modules over different rings

You can also get a similae thing with vector bundles for the same reason

[forest turtle]

why is this a natural thing that you would want to consider?

[fierce steeple]

Well maybe this is not very interesting in itself, but it can be "globalised" in geometry (e.g. deforming vector bundles or things like that) and there it is quite interesting / more subtle

Like if you "deform" some geometric object then you may wonder whether you can deform other things you attach to it

[forest turtle]

that does seem more natural to consider

thanks

[near lantern]

Not a proper answer, but see https://math.mit.edu/~poonen/papers/dimension6.pdf for a classification when commutative and dim ≤ 6, k algebraically closed.

[lone jacinth]

Every basic finite-dimensional algebra is the tensor algebra of a species with admissible relations.

And every finite dimensional algebra is Morita equivalent to a basic one, so I guess can be described by the extra information of how many summands of each projective the regular module has.

I think that's the closest you can get to a classification.

[fierce steeple]

Silly q but what does basic mean here

[lone jacinth]

That the decomposition of A into indecomposable projectives don't have any repeated summands

[fierce steeple]

Thanks

[vapid axle]

Going back to this, is the best way to invoke the universal property for direct sums and not construct inverses for "canonical" maps (which haven't made themselves obvious yet, at least in one direction...)

[spice idol]

I think so, yes

[vapid axle]

Okay, I'll try that

30 mins left to solve this exercise then I'm moving on

Have already spent 3 hours on it 🤦‍♂️

[spice idol]

At least the proof I gave with the universal property really didn't use any trickery at all

[plucky arch]

-# universal properties :D

[vapid axle]

Hmm I'm thinking about defining the extension $\varphi: N \otimes \bigoplus_{i \in I} M_i \to G$ by $\varphi\bigl(n \otimes (m_i){i \in I}\bigl) = \sum{i \in I} h_i(n \otimes m_i)$

[broken turtle]

okeyokay

[vapid axle]

Given arbitrary maps h_i: N (x) M_i -> G

[spice idol]

That's it!

This is well defined as by the direct sum only finitely many m_i are nonzero

[vapid axle]

Oh sweet yeah that was part of the reason for why I thought of the map

And then hopefully the canonical maps are just $\iota_j: n \otimes m_i \mapsto n \otimes (0, 0, \dots, 0, m_i, 0, \dots, 0)$ (by abuse of notation)

[broken turtle]

okeyokay

[spice idol]

If it's obvious then it's probably true

(the motto of concrete instances of universal properties)

[vapid axle]

lmao facts

[spice idol]

Pseudo is this fax

[plucky arch]

To a good approximation, yeah

-# also why are you asking me lol

[spice idol]

If you have (cat theory #1 fan) in your name you're obv an expert on it

[vapid axle]

Wait I'm actually stupid I've shown that the tensor product distributes over direct sums now I'm struggling to show that if M is flat then each M_i is flat

I know if we have a map N (x) M_i -> N' (x) M_i we want to factor through the direct sum somehow then compose with the isomorphism we just proved

fuckkk this shit pmo 🥀

Okay maybe it's easier to prove that each M_i is flat implies M is flat first

[spice idol]

Flat means tensoring with it is exact right

[vapid axle]

ye

[plucky arch]

There’s a difference between being a fan and being an expert

[vapid axle]

equivalently, it suffices to show that tensoring preserves injectivity

anyways I might be doing too much rn

[spice idol]

:P real reason is cuz ur cool and I wanna talk w people who I think are cool

[spice idol]

Mhm

[plucky arch]

🥺

[spice idol]

Getting lost in sound synthesis atm but after that I can see if I can write a quick proof and give you sum hints

Hehe

"sum"

[vapid axle]

[
\begin{tikzcd}
{\bigoplus_{i \in I} N \otimes M_i} && {\bigoplus_{i \in I} N' \otimes M_i} \
\
{N \otimes \bigoplus_{i \in I} M_i} && {N' \otimes \bigoplus_{i \in I} M_i}
\arrow["{(f \otimes 1_{M_i}){i \in I}}", from=1-1, to=1-3]
\arrow["{\psi{N'}^{-1}}", from=1-3, to=3-3]
\arrow["{\psi_N}", from=3-1, to=1-1]
\arrow["{f \otimes 1_M}"', from=3-1, to=3-3]
\end{tikzcd}
]
I'm thinking about something like this where the $\psi_N$ are isomorphisms

[broken turtle]

okeyokay

[vapid axle]

Just have to show commutativity because everything else is injective(?)

[forest turtle]

yea, assuming each component module is flat first

[vapid axle]

Yea

so that's that direction done

the other direction seems a bit more annoying

[forest turtle]

pretty sure you can just flip the vertical arrows here.

from the top left corner, following down right and up will be injective if M is flat

[vapid axle]

Oh and that implies that each f (x) 1_{M_i} is injective right

[forest turtle]

i feel like yes. trying to work it out now

yes, because you have natural inclusions into the direct sum. and those are injective

[forest turtle]

so you can extend this square up with two nodes

N ⨂ M_j and N’ ⨂ M_j for fixed j

connected to the original square with downward directed edges

𝜄_j : N ⨂ M_j —> ⨁_i (N ⨂ M_i)

𝜄’_j: N’ ⨂ M_j —> ⨁_i (N’ ⨂ M_i)

and f ⨂ M_j : N ⨂ M_j —> N’ ⨂ M_j

will be injective

[spare solstice]

I'm reading Gaschutz theorem and it seems like it has some redundancy

Supposing G has order not coprime to l then by Cauchy's theorem it has an element x of order l.
The element (x-1) is nilpotent of order l in F_lG

So it should be in the Jacobson radical, no?

Oh wait

xM is not necessarily a left module

[lone jacinth]

x isn't necessarily in the center, so the ideal generated by (x-1) need not be nilpotent

[cobalt sonnet]

Is there a general classification of commutative Frobenius algebras over any field?

[lone jacinth]

I think there's probably too many to hope for a useful classification

[cobalt sonnet]

But can we always say that it is the direct sum of fields and indecomposable annihilator algebras?

Independent of the field characteristic or closure?

[cobalt sonnet]

This is what I found when I tried to find this online but the author doesn't mention any assumptions on the field

[lone jacinth]

Do they give a proof?

Anyway, your algebra should split into a product of indecomposable ones.

Then for each of those A/J (algebra modulo radical) should be simple and the socle will be simple so any ideal is indecomposable. Including the annihilator of A/J (unless J is 0, in which case it's a field)

The argument doesn't depend on the field

[cobalt sonnet]

This is from Abrams PhD thesis but the proof is not explicitly given

[lone jacinth]

I mean, either N(A) = 0 or it doesn't

[cobalt sonnet]

and this decomposition is as algebras not just vector spaces?

[lone jacinth]

Yeah, I mean a decomposition of vector spaces would be nothing. All vector spaces break into 1d spaces

[spice quiver]

Oh yeah? Find such a decomposition of {0}

[rose mirage]

k \ominus k

[spice idol]

{0} is not a real vector space the feds have been lying to you

[distant harness]

{0} is the direct sum of no copies of k.

[fierce steeple]

exactly

[cobalt sonnet]

Yeah, makes sense. I was thinking of Wedderburn's theorem but it's different. Like the decomposition is as a linear space only because the radical doesn't have a unit right? otherwise both are closed under multiplication

[vapid axle]

Can I have a hint for the this problem? I've wrote A[x] as a direct sum of Ax^i for i >= 0, and I want to show that each Ax^i is flat

[spice idol]

Each Ax^i is isomorphic to A as an A-module

[vapid axle]

Ye but how do we know that A is flat

Oh wait

It's because tensoring with any injective map of A-modules N -> N' gives N (x) A -> N' (x) A

which is just N -> N' again

[spice idol]

A (x)_A M ≈ M

As the identity functor is exact, tensoring by A must be exact

Lol

[vapid axle]

idgi

ur the goat btw

[fierce steeple]

Like point is - (x)_A A is the identity functor

[vapid axle]

oooo

and the functors preserve monomorphisms

[spice idol]

(as good as)

[fierce steeple]

Up to iso

[fierce steeple]

Well it preserves everything

[limpid horizon]

Functors in general dont preserve monomorphisms tho right

They do if they r left exact?

[spice idol]

It's more that the identity functor is exact, which (outside the context of specifically right exact functors) is a lot stronger

[fierce steeple]

Ydah lol

[spice idol]

ye, that's basically what left exact means

But the language of exact sequences is frankly easier

[lone jacinth]

Preserving monomorphisms is quite a bit weaker, but being left exact means preserving finite limits, and preserving pullbacks implies preserving monomorphisms.

[limpid horizon]

Oh yea i guess i only know the defn of exactness in abelian categories ?

Cuz idk about limits in category theory yet

[spice idol]

(right, everything I've said is implicitly about functors between abelian categories)

[fierce steeple]

Left exact more generally does not require additivity but for abelian cats etc one usually assumes that and then it is true

[vapid axle]

exactly that's what i meant!

[lone jacinth]

If we assume all functors are additive, then being left exact means preserving kernels.

Just preserving monomorphisms should still be weaker though I think

For example the functor from Ab to Ab taking a group A to nA the subgroup of multiples of n is additive and preserves monomorphisms, but is not left exact

(n>1)

[spice idol]

Lmao

[vapid axle]

I'm a little bit confused as to what they mean here. Localizing at p restricts to all prime ideals except those which are contained in p (including q). But then what's the next step after? Quotienting by A_q, so that we get A_p/A_q? But the remark says that we only restrict to prime ideals containing q if we quotient by q and not A_q

[spice idol]

Quotienting mod q

Meaning that they quotient by qA_p

[vapid axle]

Sorry what's qA_p

[spice idol]

And it's a pretty nice theorems that says (A/q)_{p/q} ≈ A_p / qA_p

[spice idol]

Do you know localisation?

[vapid axle]

Yeah I'm just not familiar with that notation

[lone jacinth]

It's the ideal generated by q

[spice idol]

It's the ideal generated by { a/1 | a ∈ q }

[vapid axle]

Where q = (q/1) | q in q?

[spice idol]

Yeah

[vapid axle]

Also by lie in between p and q do they mean contains p but contained in q

Bc doesn't quotiening by q restrict to ideals which contain q

[spice idol]

Other way around

q is already contained in p

[vapid axle]

Okay good

I was just used to numbers between x and y meaning all those z such that x <= z <= y lol

[spice idol]

Ig but they do denote it p ⊃ q, so it is the same the ordering is just reversed

[lone jacinth]

I mean, between just means between

Before one after another

[vapid axle]

So if a prime ideal in A_p "contains q", they mean it contains the ideal generated by (q/1) right

[spice idol]

Yes

[vapid axle]

fascinating, fascinating

If p = q do we have A_p/(pA_p) \cong (A/p)_p

[spice idol]

Yes

Gives a field

Because you're quotienting by the unique maximal ideal of the local ring A_p

(which is pA_p)

[vapid axle]

capital!

[fierce steeple]

You may have seen this but this is called the residue field $\kappa(p)$

[broken turtle]

Prismatic Potato

[fierce steeple]

Very cool

[spice idol]

The fairly straightforward fact that Spec R is quasi-compact takes some pretty cool universal algebra to prove in its generality

Namely, the fact that for a congruence θ generated by X, one can prove that a,b is in θ if and only there is a chain of unary polynomials (a term t(x1, ..., xn) where all but one variable is replaced by an element of the algebra) p_0, ..., p_n and (x_1, y_1), ..., (x_n, y_n) ∈ X such that:

1. a = p_0(x_0)
2. p_i(x_i) = p_i+1(y_i+1)
3. p_n(y_n) = b

Then, under some conditions and if the maximal congruence is finite generated, if there is an open cover Spec^F A = ∪_i D^F(a_i, b_i), we have that the congruence generated by X = { (a_i, b_i) | i ∈ I } is maximal, and so by the fact that the maximal congruence is f.g. we can use the above described Malcev chains to show that there must exist a finite subset X° ⊂ X giving a finite subcover

[spice idol]

Oh lmao never mind I just came up with a much nicer proof that works more generally

[spice idol]

The idea is that, for any algebraic closure operator, no closed set can both be finitely generated and have an infinite minimal generating set. Then, we can translate an open cover into a generating set wrt the radical of the top congruence, which must have a finite subset also generating it, translating back into a finite subcover

[atomic crow]

Hey, can someone give me an intuition on the Baer sum for abelian group extension. I don’t understand why this should be natural

Also, I want to prove that for two morphism f,g : A -> A and E in Ext(A,C), we have f•E + g•E = (f+g) •E and I don’t see an easy way to do it

[lone jacinth]

So we know the elements of Ext(A, C) correspond to short exact sequences
0 -> C -> E -> A -> 0
and that Ext is functorial with the action of morphisms corresponding to taking pullbacks and pushouts of these sequences.

Now we have two such elements
0 -> C -> E -> A -> 0
and
0 -> C -> F -> A -> 0
what would be a natural way to add them together?

Well, the most natural thing might just be to take their direct sum
0 -> C^2 -> E(+)F -> A^2 -> 0.

But of course those gives us something in Ext(A^2, C^2) not Ext(A, C) :(. Luckily we have canonical maps A -> A^2 and C^2 -> C, so if we just apply these we get something in Ext(A, C) :).

And that's exactly the Baer sum

As for your second question you can think about the map
f(+)g : A^2 -> A^2
and how you can apply this in the middle of this process

[atomic crow]

Thank’s that’s so clear

[limpid horizon]

hey i just want to verify, what would be the map d0: C^0->C^1?

lets say x = x1,x2,x3 so d0: R -> Rx1 (+) Rx2 (+) Rx3

im trying to verify that this is a cochain complex

[spice idol]

Still on the local cohomology

[limpid horizon]

are there going to be alternating signs?

[limpid horizon]

idk what u mean

Oh, yeah

theres just been a lot on these couple pages that is difficult for me to understand

[digital parcel]

d0(r) = (r/1, r/1, r/1)

it's just the image of R in Rx1, Rx2, and Rx3

[limpid horizon]

ok, does that follow from the defn of the map there

[lone jacinth]

The sign depends on s (which variable you remove). In this case s=1 for all the maps, so no signs

[limpid horizon]

am i getting messed up on my example because we're not just labelling x we are labelling the labels ?

[lone jacinth]

We are labeling the labels yes.

So for example if it was Rx2,x4 then x2 would be the first variable and x4 the second

[digital parcel]

yes

[limpid horizon]

thanks

have not been feeling motivated the last little while

not sure if its worth it

[digital parcel]

let me write out the d1 differential

hopefully it'll help

give me a sec to tikz it

[limpid horizon]

Sure you can if you want thanks, i think i could get it now that i know i was getting the labelling probably wrong

Was just wanting to test d2 o d1 = 0

I was trying an example x =x1,x2,x3 but for something like localizing at x2x3 i didnt treat the label on x2 as s=1, i treated it as s=2

Which probably why my signs were messed up

[digital parcel]

okay. here's a computation of C1 -> C2, which you should work out first and then check with the below

from this it'll be clear that C0 -> C1 -> C2 is zero

[indigo lagoon]

Just to make sure I'm not going crazy... this is a mistake, right? I think this should say $\text{By induction we have}\ P_{n+1}=X_{0} \bigoplus ... \bigoplus X_{n}\ \text{(since}\ P_{0}={0}, P_{1}=X_{0})$

[broken turtle]

ost

[lone jacinth]

Yeah X_0 got lost in the mix

[indigo lagoon]

my guess is he went back and changed the indexing after the fact without double checking

[limpid horizon]

If we have a chain complex of R-modules, and we put a grading on the modules, can we say we have a graded complex?

I guess im not sure what the definition is for graded complexes. Is there a condition needed to be placed on the maps that respect the grading too?

I assume so

[digital parcel]

i don't think "graded complex" is a term

[limpid horizon]

Ok, they use it here in bruns herzog

[digital parcel]

there's differential graded modules which is a different thing

[limpid horizon]

“We extens this grading on the components to Ci …” etc

I will look more closely at it but im assuming the grading we put on the localized Rx has it so that the maps dk: Ck->Ck+1 has (Ck)a map into (Ck+1)a. Is this true?

[lone jacinth]

If the homology is supposed to inherit the grading then the differentials should be homogenous

[limpid horizon]

Yeah. I guess they just say it tho not justify it

I guess i could try writing it out

[fierce steeple]

Also known as chain complexes

[fierce steeple]

One notion is just a graded object in complexes, i.e. a tuple (M_n)_n of complexes

You can then take a direct sum if you want to get a complex with a homogeneous differential

[limpid horizon]

The (M_n)_n is like a direct sum of complexes Mn?

[fierce steeple]

See next message

[limpid horizon]

i dont think i get the point

[edgy pond]

In the direct sum decomposition of CG in proposition 3.29, does each End(Wi) occur in the direct sum (dim Wi) many times? Or is End(Wi) isomorphic to the direct sum Wi with itself dim Wi many times already?

[cerulean cove]

how to start adv alg from zero

[swift cove]

well, you gotta add the identity into the mix

[agile cypress]

i need x and y

[scarlet ermine]

it's the second option. as a vector space End(Wi) is isomorphic to the direct sum of Wi with itself dim Wi many times. Think of this as matrices---if Wi is the vector space C^n, then End(Wi) is the set of matrices on C^n, i.e., the vector space C^{n^2}

one thing also is to pay attention to what you mean by "isomorphism"! the decomposition
R = sum (Wi)^{dim Wi}
is an isomorphism of representations (or equivalently as CG-modules). Both the left and right side are vector spaces, equipped with a linear group action.

The second statment is an algebra isomorphism. Both have multiplication---in CG you multiply by declaring e_g * e_h = e_{g*h}, and then extending by linearity. On the right hand side, you multiply by composing endomorphisms in each summand

[silver goblet]

there's probably a better channel for this like #1021175428326633542 btw

[edgy pond]

Ah this makes so much sense if i think about it that way

Thank you!

[scarlet ermine]

yeah no problem!

[edgy pond]

I have a couple more questions from page 37 of Fulton-Harris.

2. Why are the left ideals, which are isomorphic to one irrep of CG, generated by idempotents?

3. If a given irrep W itself corresponds to a minimal ideal generated by an idempotent, then how can the given projection operator be an idempotent? Yes, since it is a projection it becomes an idempotent too but if it projects onto multiple copies of W, this big idempotent itself should separate into multiple idempotents each of which projects onto a single W. Is this a correct interpretation?

4. the young symmetrizer c_lambda is not an idempotent yet it generates an irrep of S_d. Is that an exception to the statement in 2?

I’d appreciate it a lot if someone could answer a few of these questions.

[lone jacinth]

2. the irreps appear as direct summands of CG, so they will be generated by idempotents. (The projection CG -> W is given by multiplication by an element, and since it's a projection we can choose that element to be idempotent)

3. I'm not sure I follow your question. W appears as a summand in CG in several different ways, each way will give you an idempotent.

4. it's not an exception. For example the ideal (-1) = (1) is generated by an idempotent even though -1 is not an idempotent.

[ornate atlas]

Start by working through either Dummit and Fotte or Artin, they’re pretty comprehensive introductions to algebra

From there just pick a direction, commutative algebra, homological algebra, representation theory, etc

[edgy pond]

Firstly thank you for all the answers.

2. why is the projection CG -> W has to be given by multiplication by and element?

3. now that I’ve slept on it i see that the big operator given in the text that starts with dim W is an idempotent that generates End(W_i) is that correct?

4. so you’re saying we can find an idempotent element d that generates the same ideal c lambda generates

[lone jacinth]

2. Any CG-linear map from CG to itself is given by multiplication by an element. (Just consider where 1 is mapped)

3. yes that seems correct to me

4. yes

[edgy pond]

I see. Okay everything makes more sense to me now. Just one more clarification i need.
3) the projection operator given is an idempotent that generates End(W_i). Moreover, end(W_i) is isomorphic to direct sum of dim W_i many W_i. We can find another idempotent in CG that projects onto a single W_i, rather than End(W_i) itself. In conclusion, an idempotent does not necessarily generate an irrep all the time. However an idempotent always projects onto a subrepresentation. Is this correct?

[lone jacinth]

Yes, any idempotent gives a projection onto a subreptesentation of CG, this is true very generally.

An since CG is semisimple it is in fact also true that any subreptesentation of CG is given by an idempotent.

And for example the idempotent corresponding to W(+)V will be the sum of the idempotents for W and V

[edgy pond]

ah i see.

thank you so much for clarifying things

rep theory is hard :/

can we say that V_lambda is then generated by the idempotent (1/n_lambda) . c_lambda?

and is there a nice intuition for young symmetrizers? What do a_lambda and b_lambda represent in the abstract context? They are projections onto a subrepresentation of C[S_d]. and if we apply them back to back, we end up projecting way down onto an irrep of C[S_d]... so if i write the regular representation as R = + (V_lambda)^(dim V_lambda) then it is as if a_lambda is projecting onto a direct sum of the same V_lambda and b_lambda is extracting the irrep V_lambda out of this repeated direct sum

is this a close interpretation of what is really happening with young symmetrizers? or am i just dreaming?

[vapid axle]

Is the second assertion true since
[A[x] \otimes M \cong \bigl(\bigoplus_{i \geq 0} Ax^i\bigl) \otimes M
\cong \bigoplus_{i \geq 0} Ax^i \otimes M
\cong \bigoplus_{i \geq 0} A \otimes M
\cong \bigoplus_{i \geq 0} M
\cong \bigoplus_{i \geq 0} Mx^i
\cong M[x]]

[broken turtle]

okeyokay

[spice idol]

That's on the levels of A modules yee

[digital parcel]

Alternatively you can just construct an explicit map

[spice idol]

But A[x] ⊗_A M carries a natural A[x]-module structure (extension of scalars) so I'm guessing it's asking for an isomorphism on the level of A[x]-modules

[vapid axle]

Yeah I came up with one from A[x] x M -> M[x] that was pretty natural

but the other direction was annoying

[digital parcel]

True

[vapid axle]

Is using the definition of prime the best way to show this, I tried to show that A[x]/p[x] \cong A/p but that didn't seem to work

[fierce steeple]

It isnt quite this

(A/p)[x]

[vapid axle]

Ah that works though right

[fierce steeple]

But then i would just prove that you have that iso and ur done

[vapid axle]

Okay bet

[spice idol]

I love ideals defined by I has property X iff R/I has property X'

They usually make coherent conditions

[fierce steeple]

These r fun

[hard kite]

For (c) why could phi not be a sum of 4 irreducible representations of degree 1?

Then the norm^2 would also be 4

Also the abelianisation of Q_8 has order 4 so there are 4 distinct irreducible representations of Q_8 of degree 1

[spice idol]

If V = W1 ⊕ ... ⊕ Wn is a decomposition of some representation of G, and gWi = 0 for all i, then gV = 0. Hence, if φ is a direct sum of degree 1 irreducibles, then the kernel of that representation must either be or include [Q8, Q8], and therefore cannot be faithful

φ as described, however, is faithful, and therefore cannot be a direct sum of degree 1 reps

[limpid horizon]

im not sure how the grading on Rx extends to a grading on C^i. I am trying to write an element of C^i as a sum of homogenous components but im not sure how that is working

[hard kite]

I do not understand the first part but I do understand understand the conclusion I think. Any degree 1 rep of Q_8 factors over the abelianisation of Q_8 so a direct sum of degree 1 reps factors of the abelianisation so it can’t be faithfull

Is this reasoning correct ?

[spice idol]

Yes that's basically what the first part said

[hard kite]

Ah ok good

[spice idol]

If g is in the kernel of the representations of all Wi then it must be in the kernel of the representation of V

[hard kite]

Ah yea

I see now

You’d take g to be in the commutator subgroup

[spice idol]

Exactly, I probably should've said that haha

[viscid glacier]

hello

i'm trying to prove that given P irreducible in Q[X] with exactly one real root x, if z is a complex root of P other than x then its real part is irrational

[hard kite]

Ah np it’s pretty obv looking back

Thanks

[spice idol]

No

Np*

[limpid horizon]

nvm its ok. A lot of the graded components are 0 which makes things work

[bitter kayak]

Just curious what are people’s opinions on whether the differentials in double complexes should commute or anti commute

[lone jacinth]

I mean, doesn't really matter right.

It's easy to change between them by just adding some signs.

If you're taking the total complex you would need anticommutativity, if you're doing some kind of diagram chase you might prefer commutativity.

I don't think it runs deeper than that

[fierce steeple]

and of course you can appropriately change the definition of total complex anyway if need be

[bitter kayak]

I see, I was just wondering if there is a choice that tends to be easier in the long run, but it sounds like homological algebra is just the study of objects which have multiple sign conventions lol

I think the direction of the shift functor is genuinely very confusing though

[lone jacinth]

Like you would expect it to go the other way, or it's just unclear which way the shifting should go?

[bitter kayak]

There are multiple conventions afaik

At least for cohomology

Sometimes A -> B[1] is a degree 1 map but sometimes its degree -1

[lone jacinth]

There are?

[bitter kayak]

Unless I’m an idiot

Idk which convention do you use

[lone jacinth]

Well, the shift is always in the opposite direction of the differential.

Whether you call a map A -> B[1] degree 1 or degree -1 seems more like a convention for how you grade homomorphisms

[bitter kayak]

Lol don’t tell me there are multiple conventions for this too

Degree 1 for me means that it goes up in degrees, like a map C* -> C^{* + 1}

[lone jacinth]

Well like with anything graded there are two conventions, homological and cohomological grading

[bitter kayak]

This is for cohomology i guess

[lone jacinth]

Yes, this is always the meaning of degree 1 as far as I'm aware

[bitter kayak]

Okay i guess it’s for both

[fierce steeple]

i prefer not to think about non-zero-degree maps lol

[bitter kayak]

Chain homotopies are not degree 0 though

[lone jacinth]

Then making chain complex is very hard

[fierce steeple]

fortunately this is one of the things that is actually consistent under cohomological/homological conventions

[fierce steeple]

Well i mean you can say a map C_n -> C_{n-1} etc

what i mean is maps of non-zero degree between complexes

[bitter kayak]

Okay just to clarify, the differential is degree -1 for homology and degree +1 for cohomology, right?

[fierce steeple]

in the sense that shifting commutes with change of convention lol

[bitter kayak]

Okay I’m pretty sure the shift conventions in weibel and the stacks project differ for cohomology lol

[fierce steeple]

oh lol

now i am confused tbh

[lone jacinth]

I never really think about homological grading, but for cohomology
A -> B[1] looks like maps A_n -> B_n+1
and
the differential on the Hom complex is
d: Hom(A, B) -> Hom(A, B[1])
so I would want Hom(A, B[1]) to have degree 1

For homology
A -> B[1] looks like A_n -> B_n-1 and
the differential is the same
So sticking with homological grading I guess you'de want Hom(A, B[1]) to have degree -1

[fierce steeple]

stacks project's convention ensures that V[n] (V in degree 0) to be concentrated in homological degree n or cohomological degree -n, which is a nice compatibility

Like if you want "H_n(C) = H^{-n}(C)" to hold then this convention is forced upon you

[lone jacinth]

But are there any conventions where that doesn't hold?

[fierce steeple]

Weibel's conventions violate this

Which is bad to me lol

[lone jacinth]

Wait, so does Weibel consider triangles to be
A -> B -> C -> A[-1] ?

I can't make sense of this convention

[fierce steeple]

It seems so

[lone jacinth]

Seems they actually do. I really didn't think anyone used [-1] for the shift

[fierce steeple]

Yeah me neither lol

Bruh

[lone jacinth]

I mean, it's consistent, just weird that [-1] would be the thing that takes center stage over [1]

[fierce steeple]

In derived AG stuff i regularly have to switch between conventions but i did not realise some use a convention like thos

Everyone i am aware of uses our one

[fierce steeple]

Well it also means you would gave to write [1] for homologically-graded derived cat right

Which feels funny

But i guess it means like you just have to change convention in a different way

Like D_<=0 is equal to D^<=0 now

[spice idol]

Pretty sure Hilton, Stammbach's book Course In Homological Algebra uses it too lmao

[lone jacinth]

Wait, I see what's going on now.

Weibel defines [1] the same for homological and cohomological grading. And then the minus sign just comes from them using homological grading

Nope, that's not it.

Nvm, I'm still confused why they do this

[fierce steeple]

Lol

[spice idol]

Hm, nevermind, they don't define it using a shift, rather they define it using homomorphisms of degree -1 for homology and 1 for cohomology

[vapid axle]

What is the point of homological algebra

To find information on groups from other groups via arrows and exactness?

[digital parcel]

sure, or more general algebraic objects

hom alg also gives you nice invariants about things

it's pretty natural to want exactness, eg exact forms in differential geometry (see the de Rham complex) and other stuff in algebraic topology

(i should say though that the hom alg you do over groups is diff from, say modules, because Grp is not an abelian category)

[fierce steeple]

I take the POV that "cohomology"-type things are often interesting invariants of geometric or algebraic objects in maths — think of singular (co)homology, de Rham cohomology, or things for algebraic objects. If you want to systematicallg study these things, you get homological algebra

Like i dont think "measuring exactness" is really enough of the story

[bitter kayak]

Lol okay so the conclusion is weibel bad nice

Well i guess weibel just thinks of shifting as always going in one direction and uses the same direction for both

[lone jacinth]

Surely "hom alg you do over groups" means hom alg of ZG or RG modules. Not some generalization of hom alg to Grp

E.g. group cohomology

[digital parcel]

right

true

[bitter kayak]

This is not true right? In the stacks convention I think V[n] is always concentrated in degree -n whether it’s homology or cohomology

[bitter kayak]

Idk for me it’s extremely motivated by homotopy theory, and from my understanding that’s historically where it comes from

[fierce steeple]

Yeah i think i am wrong

Now i am confused lol

[bitter kayak]

I think stacks is consistent with A -> B[1] always being a degree 1 morphism. So triangles in homological convention would be A -> B -> C -> A[-1]

And A -> B -> C -> A[1] with the cohomological convention

[lone jacinth]

Yeah, so they agree with Weibel for homological grading and disagree for cohomological.

Which I'm fine with, because who cares what conventions you use for homological grading (just never use homological grading, problem solved)

[fierce steeple]

My life now feels a lie

[bitter kayak]

Lol

[fierce steeple]

Maybe i am in a weird bubble or smth

Or have been wrong for a year lmao

[bitter kayak]

Wait so are you used to it always being +1 for triangles?

Cuz ig that’s a third convention lmao

Lol I think Lurie uses the potato convention

This is ridiculously confusing

[fierce steeple]

Or more like i use the Lurie convention

And that is presumably the bubble to which i am referring aha

[bitter kayak]

Makes sense lol

[fierce steeple]

But basically i want [1] to be suspension

[bitter kayak]

Uh but that fails for cohomology in your convention right

[fierce steeple]

Wdym

[bitter kayak]

V[n] in cohomology would be concentrated in degree -n right? In lurie convention?

[fierce steeple]

Ye but that is fine

Basically we use H_n = H^-n and demand that C -> 0 -> C[1] be a triangle (regardless of convention)

[bitter kayak]

I see makes sense

[slender kestrel]

I'm trying to read up on what in synthetic differential geometry is known as Weil algebras, but the nlab article on them seems at least partially incorrect - can anybody here help me make sense of it?

specifically, the article I mean is this: https://ncatlab.org/nlab/show/Artinian+local+ring

[fierce steeple]

Wassup

[slender kestrel]

it reads to me like it says that "local Artinian ring whose residue field is R" and "R-algebra of the form R ⨁ V where V is finite-dimensional and everything in it is nilpotent" is the same thing

but... isn't every field containing R an Artinian local ring and an R-algebra?

in particular, C would be an Artinian local R-algebra but not of that form R ⨁ V

[fierce steeple]

They are distinct in general yes

If you you have an Artinian local k-algebra so that you have a map k -> A and the residue field is k and k -> A -> k is an iso (or better, identity, wlog), then you can split A = k (+) V as you have

[slender kestrel]

right, that rules out the C-example

[fierce steeple]

The C example feels a little irrelrvant to me as the residue field isnt R anyway

[lone jacinth]

It has the wrong residue field though

[fierce steeple]

I think an issue here is when people say "Artinian local k-algebra" they may implicitly mean augmentation or smth. But in general the ring you are an algebra over can be distinct from the residue field

[slender kestrel]

yep, that's what confused me probably

because C is a local Artinian R-algebra in the reductionist sense - but I take it by "local Artinian K-algebra" they probably also meant it being compatible with the residue field in the way you explained

[potent plaza]

what does bigraded mean

[fierce steeple]

It is worth noting that in general Artin local rings with residue field k dont even admit a map from k

e.g. Z/p^2

[fierce steeple]

Yeah

[fierce steeple]

Having two gradings

What they mean precisely depends on context

[slender kestrel]

oh, huh. I somehow thought every local ring was an algebra over its residue field
must've misremembered

[potent plaza]

so like as if two multiplications so that R_i . R_j = R_i+j?

two "." operations?

sorry if im interrupting

[fierce steeple]

Well lol this means you omitted context

[potent plaza]

spectral sequences

[fierce steeple]

Do you want a bigraded ring

[potent plaza]

bidegree

[fierce steeple]

Or bigraded module etc

[potent plaza]

ring

[digital parcel]

Example is k[x, y] with deg x = (1,0) and deg y = (0, 1)

This is a Z2 grading

[potent plaza]

what

[fierce steeple]

Then you can view this as a ring R with a decompoition into pieces R(i,j) as an abelian group such that the multiplication takes R(i,j) x R(a,b) into R(a+i, b+j)

[potent plaza]

a grading for me is multiplication..

[fierce steeple]

Huh

[slender kestrel]

one last question maybe - given a local Artin k-algebra with that compatibility condition and hence splitting k ⨁ V, how do I know that V is nilpotent and finite-dimensional? you don't have to spell everything out, but a few hints would be nice

[potent plaza]

like yeah i get that u get a decopmosition with R_i . R_j = R_i+j

so for example polynomaisl and polynomial decomposition

[potent plaza]

how does translate tho

[lone jacinth]

I feel like I recall proving that if A is a local k-algebra and A/m = K, then A is also a K-algebra.

I.e. there exists a section of A->K.

[potent plaza]

k[x,y] = k[x] (+) k[y]?

[lone jacinth]

The only issue comes from A not having prime characteristic (which is guaranteed by it being a K-algebra)

[slender kestrel]

I'm not super well-versed in ring theory so I found the nlab article hard to follow, and the only other sources I've found on Weil algebra just introduce them as algebras of that form k ⨁ V directly

[potent plaza]

sorry bits been a while

[rare walrus]

k[x,y] = k[x] (x)_k k[y]

it's not a direct sum

[potent plaza]

yeah so thats not a "grading"?

[rare walrus]

Nope

[fierce steeple]

I can explain lol

[fierce steeple]

This is not even really true - there need only be an inclusion

[digital parcel]

How is what I gave not Z2-grading on k[x, y]?

[potent plaza]

what

[fierce steeple]

I will write smth up when i am done with dinner

[potent plaza]

okay im confused as fuck so i will just shut up and hopefully someone just defines it properly

[digital parcel]

Unless by bigrading you mean something other than a Z2 grading for which I apologize

[fierce steeple]

But it seems you are a lil confused about the notion of a grading

[potent plaza]

yes

i am

[fierce steeple]

But i can write it up dw

[rose mirage]

https://tenor.com/view/zelda-cdi-kingharkinian-dinner-wandofgamelon-gif-25539234

[potent plaza]

probably

lmao a grading is just

some way of decomposition

so that the multiplication of the ring gives R_i*R_j is in R_i+j

thats it

[lone jacinth]

So a bigrading is this, but you have R_(i, a) * R_(j, b) in R_(i+j, a+b)

[digital parcel]

more or less yes

this is a Z-grading (assuming i and j are integers)

[potent plaza]

so two gradings

sub a and sub i

so that what uw rote is true

right

[lone jacinth]

Yup

[potent plaza]

so for example

singular cohomology is a graded ring with grading being the dimension of the cocycle

right

[lone jacinth]

yes, but that's not bigraded

[potent plaza]

yeah

whats a "Z-grading"

like whats Z here

[lone jacinth]

k[x, y] is bigraded with x^i y^a having degree (i, a)

[digital parcel]

integers

but you can do this over any monoid

[potent plaza]

i know

[potent plaza]

right

[lone jacinth]

Z is the symbols used for the grading

In R_i, i is an integer

[potent plaza]

right yeah mb

got it guys

tysm all

[fierce steeple]

Note one issue is that there is a general notion of a graded object but a graded ring is not a graded object in rings

Lol

[fierce steeple]

Sorry lol had to go. But note (1) no element of V can be a unit (2) any element of an artinian ring is either a unit or nilpotent

this gives you that V is at least locally nilpotent (i.e. every element nilpotent)

Now I guess you wanna know why V is f.d.

Note that in general a local Artin k-algebra need not be of finite dimension over k (try to come up with an example if you want)

[slender kestrel]

without the compatibility condition any infinite-degree field extension should be a counterexample

[fierce steeple]

exactly

[slender kestrel]

but you're saying for local Artin k-algebras with the compatibility condition it still doesn't hold?

[fierce steeple]

it does hold, just the proof i know is slightly more complicated

One thing I like (which was emphasised to me by jagr lol) is that if A is a local artin k-algebra with residue field k and M an A-module then you can consider the restriction of scalars along A -> M and this will preserve lengths

[lone jacinth]

If A is artinian it has finite length so
A/V (+) V/V^2 (+) ... is finite dimensional

But this has the same dimension as A

[fierce steeple]

so length of A over A is the same as the length of A over k, i.e. dim

[slender kestrel]

I have no idea what length is, I'll have to look that up

[fierce steeple]

which is what jagr has just said aha

[digital parcel]

jordan holder

[lone jacinth]

Length = number of composition factors in composition series

[fierce steeple]

If $M$ is an $A$-module then the length is the sup of $n$ such that there is a sequence $0 = M_0 \subset M_1 \subset \dots \subset M_n = M$

or equivalently what jagr said

[broken turtle]

Prismatic Potato

[lone jacinth]

georgia dropper

[fierce steeple]

it is easy to see that if A is a field then, length coincides with dimension

but in general, length is different to the notion of rank - for example A might not even have finite length over itself

[digital parcel]

who is georgia and why would they drop her

[fierce steeple]

You can show that a commutative ring is artin if and only if it has finite length over itself

[slender kestrel]

length still meaning chains of submodules here, not chains of ideals?

[fierce steeple]

Remember a left ideal of A is precisely an A-submodule of A

[slender kestrel]

oh, right

[lone jacinth]

Left ideals anyway

[fierce steeple]

sure sorry I mean to say commutative here lol

https://stacks.math.columbia.edu/tag/00IU and the succeeding section are great for this btw lol

(artin rings + length)

jagr do you remember how to prove uh

(Maybe I am missing hypotheses)

If A is a local artin ring with residue field k of char 0, then A -> k admits a (central?) section

i think this is hard and uses theory of cohen rings or smth

maybe it is only true if k is algebraically closed but i think it works in general

and may also need to assume commutative

[lone jacinth]

Yes, definitely is hard.

But if k=Q for example it is easy

[fierce steeple]

I guess you just need to check that for all n, n is invertible on A, which can be checked modulo the maximal ideal, gg?

[fierce steeple]

but cool thanks

and k in characteristic p and perfect the same thing is true but with W(k) instead

i believe i reproved this a month or two ago lol

[near lantern]

This might be in Serre's Local Fields chapter 3 IIRC

[fierce steeple]

oh nice

[lone jacinth]

So let's say k is an algebraic extension of Q(x1, ...) as all rings are.

First we have a splitting for Q, then we can lift the xi arbitrarily.

[near lantern]

Then you have separable extensions which lift by Hensel's lemma(?)

[fierce steeple]

oh nice

[lone jacinth]

Yeah you might need seperability for it to be true...

[fierce steeple]

i think for W(k) the argument I had was that you prove the stronger statement that if A is local artin with residue field k, then the hom set Hom_{/k} (W(k), A) is a singleton. Then you decompose A -> k as an iterated square zero extension and do some induction

[slender kestrel]

I'm afraid I still can't parse this sentence, even though I think I know all the words

restriction of scalars is something that you do along a ring morphism - so what is M here? an A-algebra?

[near lantern]

There is actually an adjunction Hom (W(k), A) = Hom(k, res_field(A)) or something like that.

[fierce steeple]

Oh yes, true. I guess this is part of the usual "universal strict p-rings" story

i now feel silly aha

[near lantern]

If M is a finitely generated module over a commutative ring R, and I = (x1, ..., xn) is a finitely generated ideal such that IM = M, then the Koszul complex K(x1, ..., xn) (⨯) M is exact, right?

[fierce steeple]

M is an A-module, but you can still restrict scalars

[lone jacinth]

M is an A-module that becomes a k-vector space by
k->A

[fierce steeple]

Like if $\varphi \colon A \to B$ is a ring map and $N$ is a $B$-module then you can define $a.n \coloneq \varphi(a).n$

[slender kestrel]

ah, so restriction of scalars along k -> A, not A -> M

[fierce steeple]

yeah

[broken turtle]

Prismatic Potato

[fierce steeple]

oh sorry, typo lmfao

Very bad typo

[slender kestrel]

I was wondering where you even got a map A -> M from

thanks, that makes more sense now

thank y'all for all the help

[fierce steeple]

But yes the point was if you have an $A$-module $M$ and you are in the like $k \to A \to k$ situation, then $\mathrm{length}_A M = \mathrm{length}_k M$. This follows by [Stacks project, Lemma 02M0] applied to $A \to k$

[broken turtle]

Prismatic Potato

[fierce steeple]

Then in our situation, take M = A to conclude that A has finite k-length (and hence k-dimension)

[lone jacinth]

Then the idea is that you break M into simple modules which are A/V modules.

But k -> A -> A/V is an isomorphism, so the dimension is the same whether you think of it over k or A/V

[near lantern]

(My argument: because (using ideas and some notation from Eisenbud) there exists i0 acting as 1 on M by Nakayama's Lemma and then K(x1, ..., xn) (⨯) M (⨯) ∧^{n+1} R = K(x1, ..., xn, i0, ..., i0) (⨯) M = K(i0, ..., i0, x1, ..., xn) (⨯) M is exact up to degree n because (i0, ..., i0) is "regular" on M.)

Nvm Eisenbud does this later

[near lantern]

Let I be an ideal and M a fg module over a commutative Noetherian ring R; assume IM ≠ M. Let x1, ..., xr be a maximal sequence in I which is regular on M. Does the ideal J = (x1, ..., xr) depend on x1, ..., xr or only on I and M?

[calm trellis]

Is there anything in particular that differential galois theory "achieves nicely"?

Most of my course consisted of learning prerequisites and only at the end we got to Picard Vessiot extensions, differential galois groups etc

I understand the galois correspondence etc but i haven't seen a case where the tools from algebraic groups actually achieve something here

it just feels like forcibly translating diff eqs into an algebraic setting for now

[atomic crow]

I got the idea (I think you can take the pushforward of f(+)g), but idk how to write the proof. Btw I struggle with proofs with Ext because I always come back to explicit definition of objects (for exemple push and pull). Do you have any technic that can make these type of proof easy ?

[vapid axle]

Am I stupid? Why is the last sentence true? If $x_i \in M_i$, then $(\mu_j \circ mu_{i, j})(x_i) = \mu_j(x_i)$ and I'm stuck

[broken turtle]

okeyokay

[fierce steeple]

Follows from condition (2)

Uhh

Like mu(x_i) = mu(mu_ij(x_i)) by construction

And then uhh

[vapid axle]

I'm stuck on the tutorial bro

Anyone know?

Oh wait I see, \mu_{i, j}(x_i) is not in the quotient yet. So we project it to the quotient via \mu_j, and since \mu_i(x_i) = x_i, these two are equal in the quotient

More precisely, we have $(\mu_j \circ \mu_{ij})(x_i) = \overline{\mu_{ij}(x_i)} = \overline{x_i} = \mu_i(x_i)$

[broken turtle]

okeyokay

[rose mirage]

ew colimits EW EW EW EW

you're essentially just composing two inclusion maps

[vapid axle]

Actually, they're called direct limits 🤓☝️

[rose mirage]

they're colimits mate

lets not use more words than needed

[fierce steeple]

ye

[digital parcel]

Why are they called colimits anyway

Why cant colimit be limit and limit be colimit

[rose mirage]

because they're the opposite of limits

[digital parcel]

I can never remember which way to draw the arrow

[rose mirage]

colimits are forcing things to be equal and limits are finding things that are equal - the latter is a lot more "limit-y" in the analytical sense to me than the former

[digital parcel]

Hmm okay I see it

Yeah I buy it

[spice idol]

I like limits more because going against the arrows feels cooler than going with them

[bronze surge]

at minimum, it has very useful applications in computer algebra for symbolic calculus (analogous to solvability problems for polynomials in ordinary Galois theory). I don't know a great reference to be honest, I only know the basics (and moreso just differential field theory as opposed to full-on Galois aspects) from Chapter 12 of Geddes, Czapor, Labahn, which is not a good book imo

[slender kestrel]

very dumb explanation, but: colimits are limit-y things with a co in front of them, limits are limit-y things without

so products are limits, coproducts are colimits

and for products, the projections go from the product to the factors... so for general limits they should go from the cone point to the objects in the diagram as well

[limpid horizon]

for graded objects, does 0 have a degree? 0 would be in every graded component right?

[digital parcel]

Correct, 0 doesn’t really have a degree

[lone jacinth]

Alright, so the ideas is the composition
A -> A^2 -f(+)g-> A^2 -> A
it's equal to f+g.

So the pullback of
C -> E -> A
along f+g can be broken into 3 steps.

After the first step you get the same thing as pushing out
C^2 -> E^2 -> A^2
along C^2 -> C, so changing the order of pushing and pulling you can apply f(+)g to
C^2 -> E^2 -> A^2
In which case you should see that this is the direct sum of the pullback along f and g respectively.

[hidden locust]

Doesn’t it prove e^(-x^2) has no elementary antiderivative

The proof is analogous to the unsolvability of the quintic

Seems pretty cool to me

Idk if that’s an answer to your question

[crude lotus]

https://www.3blue1brown.com/blog/exact-sequence-picturebook am reading this thing about exact sequences by Ravi Vakil and i don’t really understand what the image im supposed to be having here is, does anyone understand this lol

what is the bulge for

what is the “water flow” respresenting

oh ok i read further and it seems those things were not that important (?)

[lone jacinth]

It's just to distinguish which is the quotient object and which is the subject.

[crude lotus]

right

[calm trellis]

yeah i am somewhat content with the solvability part

maybe there's more theory on solvable algebraic groups

[limpid horizon]

if C(delta) represents a simplicial chain complex, what would a notation like C(delta)[j-1] mean?

some shifting indices of some sort i think but I dont really get it

shifting "homological" dimension or something?

[calm trellis]

is that the part of weibel with the mapping cone

[limpid horizon]

what is this term C_-1? The basis is just [] so I dont get it is that still like a rank 1 module?

[digital parcel]

If C is a complex and k an integer, C[k]_n is C_{n+k}

Basically your degree k terms become degree 0 in the shifted complex

[limpid horizon]

context