groups-rings-fields

[karmic moat]

eeverything about this quote is so awesome

[thorny cobalt]

wait till bro sees the superbuffed red GM+2 chapter 1 maps with like 10 frame-perfect inputs per room 💔

[karmic moat]

i love that stupid tweet

"frame-perfect inputs" 🤤

[thorny cobalt]

its miserable

i js play celeste for the vibes lwk

adn the music

[karmic moat]

as u should

i like the vibe and music

[thorny cobalt]

haven't touched celeste in a while though

[karmic moat]

tho like in some rooms i kept dying and i would mute the game out of rage

[thorny cobalt]

ah hn

[thorn jay]

https://youtu.be/SFgNEEE3Clg

[thorny cobalt]

what chatper

[thorn jay]

am loving this video

[thorny cobalt]

if its chatper 3 that's understandable

oshiro def pissed me off

ok wait

[karmic moat]

bingo

[thorny cobalt]

if ur a precision lover js play geometry dash

[karmic moat]

is oshiro the one weith the book room

[thorn jay]

tna delrune

[karmic moat]

i hated the book room

[thorny cobalt]

yes 🗣️

[karmic moat]

i love this core whatever it is

[tardy hedge]

last game i liked playing was demons souls remake on ps5

[thorn jay]

RAHHH GEOMETRY DASH MENTIONED

[tardy hedge]

besides that i havent enjoyed a new game in a long time

[thorn jay]

https://tenor.com/view/every-end-geometry-dash-gif-8964709680687702268

[karmic moat]

go figure i used to play a lot of osu

like a lot of osu

like i bought a drawing tablet just to play osu

[tardy hedge]

bro i played geometry dash for years yk

[thorny cobalt]

i see those clips on insta with like "most terrifying gd screasm"

[thorn jay]

same i love it sm

[tardy hedge]

i beat bloodbath years ago hahaha

[thorn jay]

LMAO i know exactly which ones youre talking about

[tardy hedge]

hopefully that level is still relevant

[thorny cobalt]

and its like people screaming the equavlianet of ike an eldrich horror after 97% on unnerfed sakupen circles or some shi

[thorn jay]

i mean yeah its still a generational demon

[karmic moat]

i think i hit 5 digit rankings in osu

[tardy hedge]

yay

[thorny cobalt]

oh damn

[tardy hedge]

beating bloodbath used to be cool ok

[thorn jay]

i feel them tbh

[tardy hedge]

i was like 12th person

to do it

[karmic moat]

i loved osu

[thorn jay]

no way thats crazy

[karmic moat]

i haven't played it in a long long time

[thorny cobalt]

wtf

[tardy hedge]

my video got like 50k views or smth

[thorny cobalt]

are there any btd lovers out there

[thorn jay]

https://tenor.com/view/congregation-jumpscare-gif-15004478797119541730

[karmic moat]

yea btd5 is awesome

[thorny cobalt]

5?

[karmic moat]

wait by btd do you mean bloons tower defense

[tardy hedge]

people who are cracked to that absurd degree in GD remind me of those kind of ppl in math

same energy

[thorny cobalt]

yeah

[karmic moat]

yea bloons td 5

[thorny cobalt]

yeah hella math people ik play gd

[thorn jay]

btd 🔥 🔥 🔥

[thorny cobalt]

the game genuinely js seems frustrating to me i never got it

[tardy hedge]

yea weird thing

[thorny cobalt]

unc /j

[tardy hedge]

bro remember fking bloons world

its deleted off the face of the internet

only I remember it

[karmic moat]

omg bloons td5 was 2011

holy hell

i thought that was the newest one i didn't know they had one in 2017

zamn

[thorny cobalt]

newest one is crazy

[karmic moat]

no :(

[tardy hedge]

https://www.reddit.com/r/bloons/comments/1f1uv82/anyone_remember_bloons_world/

[karmic moat]

oh nah never heard of it

[tardy hedge]

😎

ppl made their own bloons levels lol

[karmic moat]

the big thing when i was a kid was poptropica

[thorn jay]

i love the music i think that helps a lot

[karmic moat]

all my friends played it in the school library

[tardy hedge]

club penguin was my thing

[karmic moat]

well not in the library but during our computer lab

[thorny cobalt]

poptropica was laways big i feel

[karmic moat]

yea probably

[thorny cobalt]

coolmathgames so goated

[karmic moat]

when flash support was taken away from browsers that was so saddening

[thorny cobalt]

ong

[karmic moat]

i think the most recently released game that i really liked was elden ring and its dlc

[crystal vale]

I don't have an idea for the last one, it seems computational

[alpine plank]

Try using the fact that R[X] = R[X - a] for a \in R

[crystal vale]

So R [X,Y, Z] = K[Z], where K = R[X,Y] and K = H[Y], where H = R[X], so K = H[Y-X^2]

So K = R[X][Y-X^2] and R[X,Y,Z] = K [Z] = K [ X^3- Z] so R[X,Y,Z] = R[X][Y-X^2][X^3-Z]

So every polynomial in R[X,Y,Z] is of the form g(X) + ( Y- X^2)( f(X,Y) ) + (Z-X^3)( k(X,Y,Z) ), right?

Thank you @alpine plank

[balmy python]

can you have an infinite group with finite generators with finite order?

don't think so now that i think about it

[tribal moss]

Yes -- we discussed that yesterday; a simple example is that the group presented as < a,b | a²=b²=e > is infinite.

[elfin wraith]

Can anyone give me a push in the right direction to show that a module if projective iff theres another module such that their sum is free? Ive only seen the lifting defintion for projective and in all honesty im not sure how to make use of that for this

[balmy python]

ahh yeah

[tribal moss]

Or, perhaps more concretely, the group of all functions of the form f(x)=a+x or f(x)=a-x where a in Z, under function composition.
It is generated by 1-x and 0-x, which both have order 2.

[quiet pelican]

Hint: you have a surjection from a free module onto your projective module

[quiet pelican]

(This is a general fact for modules, there is a surjection from a free module
And in fact holds for basically any algebraic structure with a concept of "free")

[balmy python]

ah yeah that's also a nice one

so that serves as a contradiction to assuming that if we have an infinite group with a finite number of generators, there is at least one infinite cyclic group generated by a single generator

edited it

[tribal moss]

That's not quite a counterexample because in the example above, 1+x generates an infinite cyclic group.

(Unless you're only speaking of the cyclic groups generated by your one of your finitely many designated generators).

[balmy python]

i get mixed up with terminology so easily in gt lmao

[tribal moss]

The solution to Burnside's problem, which was also discussed yesterday, shows there can be a finitely generated infinite group in which all of the cyclic subgroups, no matter which elements they're generated by, are finite.
(I don't know the particular example, but it took decades to find, so probably too complicated to be explained in a chat anyway).

[balmy python]

ahhh okay

how would I structure my proof around that fact?

[tribal moss]

What are you proving?

[balmy python]

every infinite group has infinitely many subgroups

i did talk about it yesterday hhere but I got quite confused

this is what i wrote

[tribal moss]

Sorry, it looks like we got sidetracked into competing for the nicest proof instead of critiquing yours.

You definitely should not be using Burnside's problem in your proof.

[balmy python]

😂

It's nice to know such a group could exist though

[elfin wraith]

(Sorry for the ghost ping, misclicked) I had gotten this far, and I suspect its staring me in the face and im just being dumb but my current thinking is this.

Take P to be a projective R-module, so we know it satisfies the lifting property, then some other module Q. We then get a surjection between the free module F = R^(PxQ) and P(+)Q. Then using the inclsuion we get the lifting property and whatever, but we dont know that P(+)Q is free, we just have a surjection from F and I dont think this will be an isomorphism in general. I think the issue is choosing Q in the start, im not sure thats the correct way forward, and I dont think that in general I can choose Q such that I do get an isomorphism, so I think this approach is just flawed

[balmy python]

how could I write that proof into a better proof or what's wrong in my proof?

[quiet pelican]

So you have F -> P -> 0
Let K be the kernel, so we get 0 -> K -> F -> P -> 0
Can you prove this SES spilts?

[elfin wraith]

Oh that seems much smarter, yes I should be able to do that

[tribal moss]

Hmm, I think there are three problems here:

1. The way you speak about your case split sounds like you're assuming the group "knows" what its generators are, such that you can ask how many of them there are. But there may be several subsets of the group that each generates it, and some of theses subsets may be finite while others are infinite. In any case you can always declare all the elements of the group to be generators, and then you wouldn't need the finite case at all. So unless your "infinite" case actually works for every group, you need to be more explicit about what the condition you split into cases on is.
2. In the finite case, "at least one of these generators must generate an infinite group" is not true, as demonstrated by my counterexample above.
3. In the infinite case, you're assuming that two different generators generate different cyclic groups. That's not necessarily true -- for example it's generally not true if you've declared every group element to be a generator!
   You could try to fix up these problems one by one, but I suspect you could easily end up with something where the case analysis doesn't actually contribute much to the argument.

[elfin wraith]

Yeah ok this is just split exact by the splitting lemma, since the map F->P is surjective, so its right cancelative. Ive got to run now but ill do the other direction when im back and hopefully stop being dumb lol, thanks for your hints though mico!

[thorn jay]

it is true 🙏 🙏 🙏 🙏

[balmy python]

can you explain what you mean by there being several subsets of the group that each generates it

[tribal moss]

Well, for example, Z is generated by {1}, but it is also generated by {-1}, and by {3,5,42} and by {2,3,5,7,11,13,...}.

One idea for fixing (1) and (3) that doesn't work would be to say, "okay, take a minimal set of generators" -- but unfortunately some groups don't have minimal set of generators. One familiar example is (Q,+) -- every generating set will contain redundant elements (so there's always a proper subset that also generates the group).

[thorn jay]

how is the set of reciprocals of primes not minimal?

/genq

[rocky cloak]

It's not a generating set

[balmy python]

what even are the generators of Q 😭

[rocky cloak]

I mean it kind of works in that you can fix a finite minimal generating set when the group is finitely generated, and then you can think about an ordered generating set for infinite group.

But I think the case work of finitely vs infinitely generated isn't helpful for this problem anyway.

[thorn jay]

oh right, reciprocals of powera of primes then :P

[rocky cloak]

Well which powers, cause some of them will be redundant

[balmy python]

ah i see something the cyclic groups produced by a generator, I can look at like a group spanned by a bunch of generators?

[thorn jay]

ah yeah i figured it out as soon as i said it lmao, a strange group

[rocky cloak]

Who needs 1/p when you got 1/p^2, amirite

[balmy python]

So like say for some generators a group H is produced, if H is finite, we can extend our group into <H,g_n> where it is a larger group than before

so then <H,g_n> must eventually be infinite after repeating the process

[thorn jay]

just use 1/p^\infty smhsmsh

[balmy python]

hmm but this gives finite subsets

[glad osprey]

Isn't that just Z/2Z x Z/2Z? The wiki page for Burnside's problem says that B(m, 2) is the direct product of m copies of Z/2Z

[tribal moss]

Part of my point is that there isn't any "the" generators for any group.
One possible set of generators for (Q,+) would be {1/1, 1/2, 1/3, 1/4, 1/5, ...}.

[balmy python]

ah but you could have other generators

ahhh

i see

[tribal moss]

This is not B(2,2), because there's no (ab)²=e relation.

[thorn jay]

this is C2 \ast C2

free product

[glad osprey]

I see I don't fully understand group presentation yet

[tribal moss]

In fact unless I'm mistaken, < a,b | a²=b²=e > is isomorphic to my later example of { x -> a+x | a in Z } U { x -> a-x | a in Z }.

[balmy python]

so we want finitely many subgroups of an infinite group somehow contadicting the fact that the group is infinite

hmmm so every element has a cyclic group

the union of all of these must be the group itself

[tribal moss]

Yeah, I would recommend going for "there are infintely many cyclic subgroups" without using a contradiction.

[glad osprey]

lol, for a second I thought that was a counter-example to Burnside's problem, so I was like, why did it take people more than 60 years to come up with that example

[balmy python]

hmmm

without using a contradiction you mean that i shouldn't start off by saying consider the case of finitely many cyclic subgroups?

[tribal moss]

Right.

[balmy python]

woulnd't i want to rule off why there are not finitely many cyclic subgroups?

[tribal moss]

(Mostly I'm trying to hint towards the proof I suggested at #groups-rings-fields message -- so I might be biased :-D ).

[balmy python]

ah

I mean if there are infinitely many cyclic subgroups, wouldn't their union be an infinite group

im bugging out

[tribal moss]

Hmm, it turns out I actually phrased my own proof with a contradiction too.

You can often slice these kinds of things in different ways, so essentially the same proof can look either as a by-contradiction or not.

Another way to state the same idea I wrote yesterday would be:

Consider all the cyclic subgroups.
If all of them are finite, then there must be infinitely many of them because (bla bla bla).
But if one of them is infinite, then there are also infinitely many of them because (other bla bla bla).

[balmy python]

yeah

[rocky cloak]

You can also frame the proof as an induction proof which is fun:

Say there's an infinite group with only n subgroups. If it has an infinite proper subgroup it must have strictly less than n subgroups.

If all the proper subgroups are finite, their union is finite, so there is an element not in a proper subgroup, hence the group is cyclic, contradiction.

So there are strictly less than n subgroups for every integer n, and you can have a negative number of subgroups.

[tough raven]

me omw to get out of subgroup debt by borrowing an infinite group with finitely many subgroups

[tribal moss]

Sounds like a swindle.

[elfin wraith]

Is there any reason to bother with polynomials here rather than just jumping striaght to we can write alpha as a/b in least terms etc? I guess its trying to get at the fact that it would be a root if it was rational, but its just made me doubt myself, it seems to be a pointless extra step to consider g

[rapid cave]

It does sound pointless

[vocal pebble]

I mean it's showing that the min poly has deg. Atleast 2 I guess

[elfin wraith]

Yeah sure I guess, it just seemed like all it did was add a line to the proof so I wanted to double check im not just being dumb

[rapid cave]

Well the proof will be that f is irreducible so the minimal polynomial is of deg 3

[vocal pebble]

I know. I'm just trying to interpret this exercise in a way in which the last line is not pointless

[rocky cloak]

They're not suggesting you add an extra step though.

They're saying in particular you can conclude this fact.

Like proving that alpha is not of the form a/b, so not rational, is a perfect answer to the first part.

And then it's trivial to show that in particular this means it's not a root of ax+b

[elfin wraith]

Yeah I guess, fair enough

[south patrol]

Lol

"Note that the cube root of 2 is a cube root of 2"

[elfin wraith]

This whole problem sheet as been essentially just this, I hope the course picks up a little lol

[noble nexus]

use fermats last theorem

quick corollary

[elfin wraith]

I do love absolutley nuking a problem, but id be amazed if thats not circular lol

[rapid cave]

Eisenstien criterion will be a small nuke

[south patrol]

I prefer the slightly analytic proof that cbrt(2) is irrational

[tribal moss]

What is the slightly analytic proof?

[tribal moss]

I hope there's more to it than that in (b), (c) and so forth.
Otherwise it looks like the exercise is just name-dropping ring-theoretic concepts like Q(alpha) and Q[x] without them really being relevant for what you're being asked to do.

[south patrol]

I learnt of a proof due to Conway (but not sure if it is actually due to him). The proof works to show that $N^{1/m}$ is irrational unless it is an integer. The idea here is this: suppose $N^{1/m}$ is rational and write $N^{1/m} = b+ a$ where $0 < a < 1$ and $b \in \mathbf N$. Now note that the subring $\mathbf Z[N^{1/m}] \subset \mathbf Q$ has a $\mathbf Z$-basis given by $1, N^{1/m}, \dots N^{\frac{m-1}{m}}$ so $\mathbf Z[N^{1/m}] \subset \frac{1}{d} \mathbf Z$ for some $d \ge 1$. But then $a^{k} \in \frac{1}{d} \mathbf Z$ for all $k$, which is impossible since you can find $k$ with $0 < |a^k| < 1/d$

(It is of course easier to state for square roots though)

[elfin wraith]

Yeah later it got us to do polynomial division to show that it’s also not the root of a quadratic, so I guess it’s somewhat relevant but it feels such a trivial thing to even bother mentioning

Then again I probably have a much better background with rings and stuff than most of the people taking this so

[south patrol]

Of course not really too "analytic" but the point is that it never uses prime factorisation or anything and instead uses the order on Q

[tribal moss]

Cute. (But you have two different m's, I think).

[south patrol]

True sorry

I actually had three ms

[cloud walrus]

Prismatic Potato

[south patrol]

that should be fine

[tribal moss]

I'm not convinced that the stated elements are necessarily a Z-basis for the subring, but they clearly span it, which seems to be all we need.

[rocky cloak]

I guess this really proves the stronger thing that Z is integrally closed

[south patrol]

oh yeah

[south patrol]

yeah true, I guess they are indeed a basis by considering the minimal polynomial, but the way i would prove that would make this circular lol. but yeah spanning is all that's needed

[tribal moss]

In fact they're not a basis if we consider for example 4^(1/4).
Except in the sense that they're a basis because we reach a contradiction, and then everything is true. 😆

[tardy hedge]

Hey guys, can i get a small hint on how to solve:

f: M->N R-mod hom,

If f’ : M/IM->N/IN is surjective and I is nilpotent, show that f is surjective

Im not sure what im missing for this problem. Ive been trying for so long

[rocky cloak]

So a good start could be to consider the image of f, let's call it X.

Then the statement that f' is surjective is exactly
X + IN = N.

so if you can prove that IN is contained in X you're done.

It might help to first think about when I^2 = 0

[tardy hedge]

I guess i don’t intuitively understand this that well because that first equation is not immediately clear to me

So ill have to think abt it

[rocky cloak]

A similar approach that might work as too big a hint is to consider what happens if ||you multiply N/X by I||

[rocky cloak]

You should definitely convince yourself of the first equation then

[tardy hedge]

Yeah i did see I^n-1N is contained in im(f) so i was thinking that might be helpful, but i got stuck at that point

[rocky cloak]

Alright, and how did you prove that?

[tardy hedge]

For any n there is m in M st f(m) = n+i1n1, if u just multiply by whatever n-1 i’s then you get that

[rocky cloak]

Nice, this is basically the idea I was hinting at.

Now you just need to go from n-1 to n-2, and so on

[tardy hedge]

Meaning from there show that I^(n-2)N contained in im(f) and so on?

Yeah, i guess thats why im stuck, cause i dont know how to do that yet. Ive been doing this for 3 hours so its probably healthy if i take a break now.

[rocky cloak]

Well the method is pretty much identical to going from n to n-1

The main difference is that instead of multiplying by "whatever n-1" you multiply by "whatever n-2"

[tardy hedge]

I dont see how i went from “n to n-1”

[rocky cloak]

Well you took
n + i1n1 and multiplied by I^n-1 and noticed that I^n-1 * i1n1 was in the image

[tardy hedge]

I noticed I^(n-1)N is in the image

[rocky cloak]

Well that's what you proved

But you used that I^n N is in the image

[tardy hedge]

Man, i guess?

Im not really thinking straight rn to be fair because im quite flustered

[rocky cloak]

Coolio, maybe go defluster somehow

[twilit wraith]

Ok im looking at my schools algebra prelim from last semester

First question is to show all groups of order 24 have normal subgroup of order 4 or a normal subgroup of order 8

[tardy hedge]

I felt like i used that any n mod IN is in the image

[twilit wraith]

Couldnt this be done using sylow theory or am I missing something

[rocky cloak]

Yes, that's important too I guess

[twilit wraith]

Oh actually the case where |Syl_2| = 3 leaves the problem unfinished I suppose

[rocky cloak]

There is a small extra step after that yeah

[twilit wraith]

Im not immediately seeing why it implies that theres a normal subgroup of order 4

[rocky cloak]

A hint could be that S3 has order 6

[twilit wraith]

Oh I see

So G acts on the sylow 2 subgroups

That action is transitive so the kernel of the action has order 4 and were done

Cool

[tardy hedge]

In the case I^3=0 we have I^2N in im(f), and from that i can show

f(m) = n + i1n1

So f(i2m) = i2n+i1i2n1

f(m’) = i1i2n1 so then that shows i2n is in image sort of thing

So IN is in image(f)

[rocky cloak]

So then you have it in the box yeah?

[tardy hedge]

In the box?

[rocky cloak]

Locked and loaded

Capiched

Squared off

KO'ed

[tardy hedge]

Haha, i mean, this felt kind of anticlimactic, so im probably gonna go back later and reinterpret why i was so stuck on this

[tardy hedge]

I think formulating it like this is not completely intuitive yet

[rocky cloak]

Using more submodule notation and fewer elements makes the argument a little cleaner for my taste. Like
X + IN = N
Multiply by I gives
IX + I^2 N = IN
adding X to both sides gives
X + I^2 N = X + IN = N

Then by induction
X + I^n N = N
for all n.

Then you just need the existence of an n for which I^n N is a submodule of X

[tardy hedge]

Ok interesting, and for us since I^nN = 0 then that last part is automatically true

[elfin wraith]

Ive done the first part but im looking for a hint with the second part. If the sequence is exact, then its clear that $|A| = p^{n+m}$, and then by the FTFGAG, we know $A \cong \mathbb{Z}{p^{a_1}}\oplus\cdots\oplus \mathbb{Z}{p^{a_k}}$ such that $a_1+\dots+a_k = n+m$. Im guessing this is the right way to go based on the first question, but now im a little lost. I imagine I want to say something about the $a_i$ and get some sort of divisibility argument or something? Im not quite seeing it

[cloud walrus]

Nope

[tough raven]

Not an answer, but I think necessarily A is generated by 2 elements (1 in Z/p^m Z and a lift of 1 in Z/p^nZ), so you can take k <= 2.

Also the largest a_i is at least m, so that probably cuts it down further.

Hmm is p^{m+1}, p^{n-1} possible

[elfin wraith]

Oh yeah good spot, but im not sure Im seeing the first thing yet, that its generated by 2 elements

[proper girder]

,,

[thorn jay]

if m = n it seems there is only one such A, Ext^1 vanishes if i havent done anything wrong?

wait no thats wrong lol

ignore me

[rocky cloak]

So I think Raghuram has given you a good hint for an elementary solution.

For a homalg solution: such sequences correspond to elements in Ext^1(Z/p^n, Z/p^m) so by taking the projective resolution
Z -p^n -> Z -> Z/p^n are given by pushouts
Z/p^m <- Z -p^n-> Z

So they are exactly
( Z (+) Z/p^m ) / (p^n, a)
for various values of a

[thorn jay]

god I love hom alg

[tough raven]

Let 0 -> M -> N -> P -> 0. If {ai} generates M, {bj'} generates P and {bj} are lifts of bj' to N,then {ai, bj} generates N.

For n in N, let image in P be pi(n), then pi(n) = ∑_j dj bj'; lift this to ∑j dj bj then pi(n - ∑j dj bj) = 0 so it is in M so n - ∑j dj bj = ∑i ci ai.

[elfin wraith]

God I wish I had time to study it properly and actually be able to make these arguments myself... maybe this summer

[tardy hedge]

i like it too

[tardy hedge]

yeah i see what you mean but doing manipulations on that level is not intuitive to me yet 🙁

[rocky cloak]

Different folks different strokes

[tardy hedge]

I guess youre really just treating it like an equation and manipulating it

[elfin wraith]

Yeah im in the same boat, Ive sat with that solution for a few minutes and I can digest it but no way could I come up with that

[rocky cloak]

Sure, it's an equality of modules, then I do the same thing to both sides

[elfin wraith]

But ive never really spent the proper amount of time with homalg that id like to, I will get there

[tardy hedge]

youre also using a lot how I+J = I iff J is a subset of I right

[rocky cloak]

Yeah

[tardy hedge]

ok well, i will move on from this now. Im hoping i learned something

Thank u

[elfin wraith]

Ok yeah I see this, thank you!

Had to sit on it and write some stuff out but I see whats happening

Ill go try to peice it together now, then its hometime

I once again didnt start ch3 today but im actually going back and filling in the patchy bits of my knowledge so I think its fine

[thorn jay]

whats ch3 of hatcher?

[elfin wraith]

Cohomology

[thorn jay]

ahh ye

hom alg is important haha

[elfin wraith]

Im taking a course on it next sem and trying to just make a start

[elfin wraith]

Yeah I dont know none haha, I know pretty much what I need for hatcher, like the first 3 chapters of Weibel, my issue is just that ive not done many excercises so im pretty bad at using it in general

I tried to do excercises in Weibel when I was looking at it but for issues of time, them being quite concerned with sheafs, and generally being hard it didnt really happen

[thorn jay]

you mean Weibel is concerned with sheafs or your course is

[elfin wraith]

The book was, loads of the questions were about them, and I dont know anything about them, and didnt have time to really learn back then

[thorn jay]

hmm, might be interested in reading Weibel then

[elfin wraith]

Sheaf cohomology was a bit of a tangent for what was already a tangent into homalg on my combinatoral comalg project

[elfin wraith]

He doesnt like introduce them, theres just questions using them

[thorn jay]

lmao, still

[thorn jay]

ah i see, makes sense then

[elfin wraith]

Yeah I kinda just needed to know about projective resolutions and Tor, I didnt have much time to bother with the rest

[elfin wraith]

Yeah ok, pretty sure the final answer is all groups of the form $\mathbb{Z}{p^{m+t}} \oplus \mathbb{Z}{p^{n-t}}$ for $t\in \mathbb{N}$

Theres a third part to the question but I think thats good for now lol, surprisingly simple solution when you put the pieces together but for me at least, damn hard to see

[cloud walrus]

Nope

[tardy hedge]

I learned a bit about Ext but not about some correspondence between its elements and short exact sequences

Thats pretty cool

[thorn jay]

it's one of the main reasons you care about computing it I think

[thorn jay]

right??

[tardy hedge]

Yeah thats fair i havent really seen much where Ext is used. Besides as a way to detect depth

I dont really remember how that argument goes much besides using the LES for ext from a ses using regular elements

[elfin wraith]

My understanding of derived functors is that they’re basically measuring failure of a sequence to be exact, or like I guess, making them exact

[thorn jay]

right-derived functors measure the failure of right-exactness of some left-exact functor

[elfin wraith]

Like the tensor product isn’t exact, so if you want to take an exact sequence to an exact sequence you need to slap some Tor in there

And yeah it’s sided of course

[tardy hedge]

At some point does it go to 0 ?

[thorn jay]

and left-derived functors measure the failure of left-exactness of some right-exact functor

[tardy hedge]

Like R^nF(A) = 0

[elfin wraith]

I don’t know in general tbh, the only derived functors I know are really Tor and Ext and even then only really for categories with enough projectives etc

I know the idea with tor is that you can take projective resolutions and everything works nicely but I’ve no idea how that generalises

[tardy hedge]

Yea i had to learn a bit about that cause local cohomology is also a right derived functor

[elfin wraith]

Where’s jagr when you need him

[tardy hedge]

Haha

There was definitely a lot going on in their constructions

[rocky cloak]

Sounds like you got it under control

[thorn jay]

well the idea with projective resolutions, is that it doesn't matter what resolution you take, the (co)homology will be the same (up to isomorphism of course)

I believe this is done by using projectivity to construct a homomorphism of resolutions and show that, after applying whatever functor you have, it turns into a quasiisomorphism?

[rocky cloak]

In fact all projective resolutions (of a given module) are homotopy equivalent

So taking projective resolution is a functor to the homotopy category

[tardy hedge]

And additive functors or smth preserve homotopies right

[rocky cloak]

Indeed

[thorn jay]

ohh okay i see, that induces a homotopy equivalence between complexes after passing through an additive functor

[elfin wraith]

Damn about as big a compliment as you can get

I suspected that taking resolutions and stuff would work nicely in general as that’s kinda what you spend the first bit of Weibel setting up and I don’t remember it being anything specific, but I’m not confident enough with that to back myself in general

[rocky cloak]

I mean, was there like a specific question? The general definition of derived functor isn't really different from Ext and Tor

[misty meadow]

What are people yapping about just homological algebra?

[tardy hedge]

Pre much

[misty meadow]

Nice

[rocky cloak]

Is there more to life really

[tardy hedge]

Lol

[misty meadow]

nope

[elfin wraith]

Specifically about this stuff, I don’t really know how anything works in like non module categories, or what happens if you don’t have enough projectives etc and you can’t always build a resolution

[thorn jay]

derived functors need enough projectives/injectives

though maybe there is some fucky wucky construction that makes it more general who knows

[rocky cloak]

I mean, I guess you can define them as universal delta functors or whatever, but you can't really prove they exist

[elfin wraith]

Ah yes this is true haha

I still don’t really get the deal with universal delta functors, I know they’re somehow a massive deal and everyone loves the Tohoku paper but I don’t really get the big picture

[rocky cloak]

Well, like the main benefit of derived functors is that they give you a long exact sequence.

And a delta functor is just the data of functors that give long exact sequences

[misty meadow]

yup

[misty meadow]

yes

[tardy hedge]

Introducing: dan

[misty meadow]

thank you kiand123

[tardy hedge]

my apologies u been here since 2022. But ur new to me

[thorn jay]

yeah lol, haven't been active much here?

[misty meadow]

I just pop up every now and then and ask questions

[rocky cloak]

Got to boost that blue color in the name

[misty meadow]

and answer some too!

[tardy hedge]

What does the colour in names mean?

[rocky cloak]

Level of activity

[elfin wraith]

How much of your life you waste

[misty meadow]

ONE DAY!

[tardy hedge]

Jagr is yellow cause he Honourable

[thorn jay]

is it really time wasted if its talking about math

[elfin wraith]

If you waste it well it becomes golden

[rocky cloak]

You are "Very active" for example

[tardy hedge]

I sureee am 👑

[vapid vale]

do they just give it at some point

[elfin wraith]

It’s automatic

[vapid vale]

o

[rocky cloak]

A bot measures it somehow

[rapid cave]

I am helpful

[elfin wraith]

There was a fuck up with said bot today though, it’s like 2 weeks behind

[tardy hedge]

Why are u helpful

[rapid cave]

because I am helpful

[tardy hedge]

True

[rapid cave]

I help at the help channels

[misty meadow]

homological algebra stuff I've got a course next term which'll be nice -- had to use a lot for my project last year tho shoutout to my boy kenneth brown his book is goated

[elfin wraith]

Despite all my efforts there’s still never been a homalg course anywhere I’ve been, maybe I’ll get round to it in my PhD…

[misty meadow]

tragic 😔

one day

[rapid cave]

there was last year at my uni

I wasn't that much into algebra at the time

[misty meadow]

I do think it's quite rare

[rapid cave]

why downvote me @thorn jay

[misty meadow]

i've just started my masters (different uni, got a scholarship) and there wasn't one at the uni I did my undergrad

[rapid cave]

am I not allowed to be a normal undergrad

and not know algebra at birth

[thorn jay]

considering in how many disparate fields of math its used yeah it is somewhat strange

[rocky cloak]

I guess people maybe package it into a rep theory / algtop / alggeo course

[thorn jay]

smh

[elfin wraith]

Yeah I think my current uni do a little week or two tangent in the algtop course

[rapid cave]

smh indeed

❤️

[tardy hedge]

WTF

[misty meadow]

you'd think that but don't think so! maybe just like lightly ig not sure

[rocky cloak]

I mean, it's kinda necessary for those courses

[thorn jay]

what kind of courses does your uni mainly offer?

pure logic \ analysis??

[elfin wraith]

But having done my UG and doing my masters at places with so much rep theory AG and K theory you’d think I’d have come across a course in it

I’ve also not managed to even do a rep theory course at either of them 🥀

[misty meadow]

especially even outside of those areas like take group cohomology can go from there to galois cohomology or even like bounded cohomology and stuff if you're more GGT

[rapid cave]

I think my uni is stronger at analysis

but there are some professors who teach algebra

and all of them are amazing

[misty meadow]

my unis quite heavy on algebraic geometry

[karmic moat]

my school does hom alg in a modules course but no dedicated course

[thorn jay]

and no hom alg???

[thorn jay]

I'm gonna be taking rep theory 4th quarter, which means Im gonna be doing group theory and rep theory at the same time

[balmy python]

how is this proof for the question proposed earlier

Take the set of cyclic groups of each element. Since there are finitely many subgroups, our chosen set is finite. The union of these sets cover the group. There are two cases to consider. If we have an infinite cyclic group, consider an isomorphism from the cyclic group to the integers which clearly have infinitely many subsets. If we have only have finite cyclic groups in our set of groups, then their union must also be finite which is a contradiction.

[rapid cave]

aren't you a rep theory master or something

some of you seem like you have an infinite well of knowledge

[rocky cloak]

enpeace is the UA master

[thorn jay]

well I've done some rep theory from Serre's book and also from Erdmann's book on algebras and rep theory

but that's abt it

[rapid cave]

you are the rep theory master

[thorn jay]

which can be seen as the representation theory of clones so good enough

[rapid cave]

and tbh you always suprise me with your knowledge

not just you

again

[rocky cloak]

rep theory of artin algebras maybe.

Someone else can claim groups

[elfin wraith]

Yeah the extent of the rep theory Ive managed to do is enough character theory to define L functions and the like one workshop in my noncom class where we prove schur, maschke and said find some modules of CD_4 or something like that, so like not much

Anyway, I may be doing a thesis on a categorical notion of rep theory in fucked up monoidal categories

[thorn jay]

that sounds fun

[elfin wraith]

It does, but it also sounds possibly too hard

[thorn jay]

maybe there's some nice analogues of character theory at that abstraction

[tribal moss]

The sentence "Since there are finitely many subgroups, our chosen set is finite" looks strange in this context. (Perhaps a leftover from editing?) It would be a better proof simply to remove that sentence.

[elfin wraith]

Then again my other option is some K theory nonsense so uh, gonna be good times

[elfin wraith]

Its not really rep theory per se, but more so, ok we can go to representations, how can we go back. Theres some conjecture about Tanakian reconstruction in these categories that the prof suggested

But I think in 3 months even getting up to speed on the problem will be a challange

[rapid cave]

but are you in any pressure?

[thorn jay]

how it feels to research anything nowadays

[elfin wraith]

I mean Id like to pass my masters ideally

[rapid cave]

how close you are to that

in terms of time

[elfin wraith]

Real. Like I said my other option is K theory stuff so like, im going to be suffering either way lol

[elfin wraith]

Like 9 or 10 months now, the time I have to write the thesis is 3 months though

[rapid cave]

ok yeah thats nothing

[thorn jay]

real talk K-theory research sounds muuchhh harder than this rep theory stuff of monoidal categories, as in to catch up / keep up

[rapid cave]

good luck

[elfin wraith]

Yeah I dont think so tbh, like K theory is fucked, but its not all created equal, it didnt sound like I need to learn infty-catgeories or anything for it

Also the prof who suggested the monoidal stuff I didnt get off on the best foot with so

[thorn jay]

oh right I see

honestly I get scared and cower in a corned everytime I see the word K-theory

infty-categories seem more approachable, somehow

[elfin wraith]

Yeah I mean the stuff he suggested is Milnor K-theory, so like, I just need to deal with tensor algebras I think

[rocky cloak]

I mean, I use K0 all the time. How much harder can it get to just increase that 0 a bit

[thorn jay]

it's weird to me that so much can be said by just a single group

[elfin wraith]

I still genuinely dont really know what it is, and after spending half an hour talking with that prof about it, its not much more clear to me

[thorn jay]

"wow the set of clones on the singleton is so simple! How hard can it be to just add one element.."

The humble post's lattice:

[elfin wraith]

Like its an invarient of a topological space, sure, something about vector bundles, sure, but like WTF do they actually measure and why is there like 6 different branches all called K theory?

[thorn jay]

I love Post's lattice it is so horrendously fucked up

[elfin wraith]

I know its like something something Grothendieck wanted a unified cohomology theory but how does that tie in

[thorn jay]

well if Grothendieck wants something he'll get it

I think that's the most reasonable explanation

[elfin wraith]

Maybe ill take the K theory project just to come at you with an answer

Maybe I can become the resident knower of something

[rapid cave]

Nope (K-theory #1 fan)

[thorn jay]

if you learn UA you can become the second in command 🔥

[rapid cave]

Its for 12am AG session

lets go

[thorn jay]

if you think K-theory is the doable one out of these then I'd say do it!!

[rapid cave]

hope me luck I don't go to sleep randomly

[elfin wraith]

I think I can get something out of each of them, but both profs want me to do novel research which I dont see happening. Ill make my mind up next week, ive still got one more potential advisor to chat to (and possibly a second if he ever replies to my email) but I doubt ill take that since hes a geometer

He is lovely though

[thorn jay]

novel research as master student is goofy

I've heard its not very common

[elfin wraith]

Yeah theres just not really the time for it, maybe if youre at a place with a 2 year masters but ill have like 3 months ish to seriously think about it and work on it, and its seeming like I very barely have the prereqs to even begin to learn how state the problems just now so I dont see myself getting far with research

Especially with the monoidal categories one, bro said people have worked it out up to like char 5 or something then theyre stuck, tf makes you think I can help

[tardy hedge]

Lol

[thorn jay]

I'm sure you'll have an epiphany

[elfin wraith]

Uhh have yall tried induction

[thorn jay]

yeah I solved collatz with it actually

just yesterday

the proof is barely half a page!

I've written it up in google docs, I can send you a google drive link if you want

https://en.m.wikipedia.org/w/index.php?title=Boolean_algebras_canonically_defined
this is a such a strange wikipedia page

it reads a little like an nlab page? in terms of difficulty, but then logic/UA-pilled rather than category theory pilled

[balmy python]

hint?

[glad osprey]

Weird so this means that the main article on Boolean algebras define them non-canonically

[thorn jay]

well they're a lot more beginner-friendly

[thorn jay]

this for example almost explicitly defines Boolean algebras as the representation theory of the maximal clone on the set {0, 1}, and derives all the rest from there

[glad osprey]

I see the point of wanting to cover this content in wikipedia, but I feel like at a certain point it turns wikipedia into just an online textbook, or kind of a community edited blog

Whenever I read wikipedia I'm constantly reminded of the expression "a camel is a horse designed by committee"

[thorn jay]

in that sense its "canonical", and I do agree that is the correct way to view Boolean algebras (I believe you can extend this to give an algebraic logic for any logic system)

[thorn jay]

that's fair, but nlab wouldn't have this stuff which I think is a little sad

[glad osprey]

Why not?

[thorn jay]

it's too logic-y

at least I skimmed the boolalg page on nlab and it didnt mention nearly half of the stuff covered in this wikipedia page

[glad osprey]

I guess the wiki article is decent, it covers a lot I just wish it didn't have that weird title and intro

[thorn jay]

https://en.m.wikipedia.org/wiki/Analysis_of_Boolean_functions
what the fuck 😭

[thorn jay]

I agree with the title, fwiw

defining it as the representation theory of the clone of boolean functions

[balmy python]

If an automorphism takes generators to generators and we can have infinitely many generating sets for Z, can we not have far more automorphisms?

[thorn jay]

well if there are infinitely many generating sets, then there are infinitely many ways to permute them no?

but Z has not many automorphisms - this is precisely because there are only a very small amount of singleton generating sets

an automorphism must map a generating set to a generating set of the same size

[glad osprey]

An automorphism sends generators to generators, but the converse isn't always true: you can't necessarily send generators to generators and expect an automorphism

[rapid cave]

actually I had a small research project on my second year about this

[tardy hedge]

Good question tho tbh

[thorn jay]

oo no way

[rapid cave]

it was about fourier transforms of boolean functions like this

It was wacky and we didn't really achieve much

[thorn jay]

is there some nice rep theory perspective of this?

[rocky cloak]

A more appropriate formulation might be to say that an automorphism takes a generating set to a generating set.

Then it might be more clear why an automorphism can't map {2, 3} to {1} even though both are generating sets.

[woven trout]

Im trying to show PSL_2(F_q), where q=p^n, p prime not equal 2, is a simple subgroup of SL2(F_q). I have that SL2(F_q) is generated by all transvections (upper and lower unit triangular matrices in the 2x2 case), and now i want to show if K is a normal subgroup of SL2(F_q) which contains a transvection, then it contains all of the transvections

so first i suppose K contains the matrix [[1,a],[0,1]] for a not equal 0.

Then because we are working in SL2(F_q), we have an easy formula for inverses

we want to show if we take these elements (w,x,y,z) in F_q, then we can produce any element in the top right while producing 1 in the top left and bottom right, and 0 in the bottom left

not sure how to do that easily though

ah wait

lemme cook

hm yea

so we get az^2 in the top right

which def does not generate the units of F_q?

[rocky cloak]

Notice that
[1, a]
[0, 1]
raised to the nth power is
[1, na]
[0, 1]

[misty meadow]

we do offer hom alg as a course! that's what i was saying!

[rocky cloak]

So that gives all the upper triangular unipotent matrices

[thorn jay]

oh i think it was someone else who said that then lol? my memory is garbage

[misty meadow]

no worries! :)

[rocky cloak]

Okay only p of them, I see q not prime

[woven trout]

right

[balmy python]

wdym by this

[thorn jay]

I think I misinterpreted what you said lol

ignore that

[balmy python]

ahhh

[rocky cloak]

But you can multiply by squares and you can add stuff.

Seems reasonable that every element of Fq would be the sum of some number of squares

[balmy python]

so say if i were to find the automorphism group of the integers, how would i do so if there are many different generating sets to choose from?

[rocky cloak]

Just pick whichever you like

[balmy python]

but how do i know if they give all automorphisms

[rocky cloak]

Generating sets don't give automorphism...

[tribal moss]

If you pick a nice generating set -- such as {1} rather than {3,5} -- then it's easier to see what the ways to send that set to some other generating set is, and then you can check which of those actually produces an automorphism.

[balmy python]

understood

[tribal moss]

(In this case, there are two ways to send {1} to some generating set, and it turns out both of them extend to automorphisms).

[balmy python]

automorphism -> generating sets mapped to generating set of the same size but not the converse

[balmy python]

yeah

[tribal moss]

(In contrast, most of the infinitely many ways to send {3,5} to a two-element generating set don't extend to automorphisms).

[woven trout]

let A={na: n in {1,...,p}}. Is it true that F_q={bz^2: b in A and z in F_q}?

[balmy python]

how do we know that

[tribal moss]

So it's easier to classify the automorphisms based on what happens to {1} than based on what happens to {3,5}.

[rocky cloak]

Probably not. Would at least depend what n is

[tribal moss]

I happen to know it because I already know that the only automorphisms of Z are the identity and negation. So {3,5} has to map as either f(3)=3, f(5)=5 or as f(3)=-3, f(5)=-5. There are a myriad possibilites such as f(3)=17, f(5)=2 that send {3,5} to some generating set, but won't match any of the two possible automorphisms.

[woven trout]

I mean A is the set of all things in the additive group generated by a

[balmy python]

what if we didn't know the automorphisms of Z

[rocky cloak]

Yes, I understood that

[woven trout]

oh

sorry

second n

should be 2

[tribal moss]

The easiest would probably be to find them first.

Nobody is (I hope) saying anything like "if you pick some generating set, here's a method for finding the automorphisms by staring at that generating set, which always works".
At most, it's a case of "it can often help with identifying automorphisms to pick some generators and consider what an automorphism has to do to those generators". (And it is much more likely to help if you pick a small set of generators than if you pick a larger one).

[rocky cloak]

Okay yeah actually it will be.

If b is a square then bz^2 ranges over all the squares and when b is nonsquare bz^2 ranges over all nonsquares

[woven trout]

why does the second assertion hold?

[rocky cloak]

I mean bz^2 are all nonsquare and they make up half of the elements, so that's all of the nonsquare elements

[woven trout]

ah ok, I didnt know the number of non-zero squares in a field of order q, q not even, was 1/2(q-1)

[tribal moss]

It has to be, because each nonzero square has two square roots, and every nonzero element is the square root of something.

So there's one square for each pair of {x,-x}.

[woven trout]

why does the square for the pair {-x,x} and {-y,y} need be distinct for x not equal y?

[tribal moss]

Because if x²=y²=a and x isn't ±y, then the polynomial X²-a would have four different roots, which is not possible in a field.

[woven trout]

ah true

[tribal moss]

(It's different in characteristic 2, because then X²-a² factors as (X-a)(X-a), so then an element can have only one square root. So in a finite field of characteristic 2, every element must be a square; otherwise there wouldn't be enough squares to go around).

(Both of these alternatively follow from the (slightly deeper) fact that the multiplicative group of a finite field is cyclic).

[ivory ore]

If F is a free abelian group of rank n, then Aut(F) is isomorphic to the multiplicative group of all n x n matrices over \Bbb{Z}. with determinant = ±1

got the forward the direction with the argument that the inverse has to exist and taking determinant of phi phi^-=1.

anyclue of the converse?

[quiet pelican]

What should the morphism related to an n x n integer matrix look like?

[quiet pelican]

Also have you proven that two different autos can't have the same matrix?

[ivory ore]

you mean the End or the map from End(F) \to M_n(Z)

[quiet pelican]

I mean, given A \in M_n(Z), what 'should' the element of Aut(F_n) related to it look like

[ivory ore]

i mean not all of MnZ gonna correspond to an Aut, but i haven't done the explicit sort of construction

for the x_i \mapsto x_j it will be permutation of identity

[rocky cloak]

Do you know about the adjugate matrix?

[ivory ore]

looked it up, yeah i have some naive idea about adjoint

[rocky cloak]

Alright, well from it it follows that invertible determinant implies the matrix is invertible

[ivory ore]

how convenient, just the units of Z

[rocky cloak]

Yeah, the same argument works for any commutative ring

[ivory ore]

yeah, which i just noticed and thought of a nice tool to have in pocket

[tardy hedge]

Most of them wont even be homomorphisms right?

[tribal moss]

Correct.

[tardy hedge]

I need to review these basic ideas heh…

[tribal moss]

What exactly is the converse you want?
That if G is any group (abelian or not) and its automorphism group is isomorphic to GL(n,Z), then G is free abelian of rank n?

[rocky cloak]

They prove automorphism implies determinant ±1, so presumably the converse they wanted was determinant ±1 implies automorphism

[tribal moss]

Ah, makes sense.

(Turns out what I proposed wouldn't even be true: Z/3Z and Z/4Z both have the same automorphism group as Z).

((And a bit of googling even found me a torsion-free example: the additive group of rationals with square-free denominator also only has has {1,-1} as automorphisms.))

[tribal moss]

(((And tensoring any nontrivial free abelian group with that gives a non-free abelian group with the same endomorphisms and thus the same automorphism group.)))

[tardy hedge]

do we just need to make sure where we send the generators satisfy the relations the generators had among themselves?

[thorn jay]

yes

[tardy hedge]

thx bruhski

[thorn jay]

because what you actually do is construct a homomorphism from the free structure on those generators and proves it factors through the quotient, which is isomorphic to the group/module/whatever

more generally, suppose a presentation A = <X | R>, where R is some set of equations with variables in X (i.e. relations), then a function f : X → B extends to a homomorphism A → B if and only if the equations in R hold in B, when each x is substituted with f(x)

so you only have to prove that "enough" relations hold

[tardy hedge]

i think im trying to understand what maps are factoring through

so we have M->N we construct the free structure on generators of M, so F(M)/ker = M

[thorn jay]

Say S generates M. Then there is a quotient map p : F(S) → M, and an inclusion function i : S → F(S).

Any function f : S → N extends to a homomorphism f* : F(S) → N. This map "descends down" to a homomorphism f' : M → N if and only if ker p is contained in ker f*

hold on, I can draw a diagram that may explain this (and the terminology) better

[tardy hedge]

im not even seeing where we getting a M->N

[thorn jay]

okay, so S ⊆ M is a generating set

this is expressed using this diagram commuting (p is the unique surjective homomorphism making this diagram commute, and the other two arrows are the usual inclusion functions)

say now, we want to construct some homomorphism f : M → N, by specifying some function f|_S : S → N

that induces a unique map f* : F(S) → N like so

this is by evaluating the expressions in F(S) at f|_S

[tardy hedge]

Ok

[thorn jay]

cleaning up a bit, we now want to factor this map f* via p, to get the homomorphism f : M → N

this is what i mean by "descending down", you want to "descend" the induced map f* down to the quotient

[tardy hedge]

Ok i kinda see what u mean better now

[thorn jay]

in a sense f* is the "naive" homomorphism constructed from f|_S, not taking into account the structure of M at all, just that it is generated by S

[tardy hedge]

Brb but im listening

[thorn jay]

factoring through p shows that the function f|_S actually respects the structure of M (that is, all the relations that the generators satisfy).

If we let the image of s ∈ S in F(S) be denoted by |s|, youve got this

So indeed, ker p ⊆ ker f* if and only if each relation satisfied by S in M is also true in N. But by properties of quotients, ker p ⊆ ker f* is exactly the condition for f* to descend down to M!

a presentation M = <S | R> is telling you that ker p is generated by R, so then we get my original statement that a function f : S → N extends to f' : M → N if and only if every relation in R is satisfied after substituting each s ∈ S with f(s)

this line of reasoning exactly translates to any algebraic structure you can think of

[tardy hedge]

Thats good stuff ill read this

Gtg rn for calc tutorial tho

They are doing product rule 🥺

[near gazelle]

Sorry, if I interrupt. But I have a pretty quick question I think. My book restricts the notions of prime/irreducible elements to integral domains. Is this strictly necessary? Why can't we talk about irreducible elements in $(\mathbb{Z}/4\mathbb{Z})[X]$ for instance?

[cloud walrus]

Ante0417

[thorn jay]

good luck 🙏 🙏 🙏

[tardy hedge]

I dont think integral domain is necessary to talk about them. I guess you dont have the prime implies irreducible thing if youre not in a domain though

[thorn jay]

if R is not an integral domain, then there are a couple different ways to define irreducibility, and it behaves nowhere near as nice

[near gazelle]

Why would the definition of irreducibility change though.

[thorn jay]

not change, there are just multiple ones which are useful, but only equivalent in integral domains

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

the wiki article covers this

[near gazelle]

$p$ is called irreducible if a factorization of type $p = xy$ with $x,y \in R$ implies $x \in R^\ast$ or $y \in R^\ast$.

[thorn jay]

lmao

x, y ∈ R implies x ∈ R or y ∈ R

[near gazelle]

no wait

[tardy hedge]

True

[near gazelle]

I cant do the latex thing 😂

[thorn jay]

use \ast

[tardy hedge]

Ngl i dont know what u put in that p map

[thorn jay]

anyways, thats called strongly irreducible

[cloud walrus]

Ante0417

[thorn jay]

its the evaluation map

[near gazelle]

aight ty

[thorn jay]

sends |s| to s in M

[tardy hedge]

Why are u evaluating it at some sum and = 0?

[thorn jay]

you want to know the elements of the kernel of p

[tardy hedge]

Because those are dem relations

[thorn jay]

and the computation shows that ker p is exactly the set of expressions in elements of S that evaluate to 0 in M

[thorn jay]

thats what you show, yes

[tardy hedge]

So are you trying to understand what properties p needs to have so that f o p equals that f* map?

[thorn jay]

we want to know when such an f exists

[tardy hedge]

Oh ok i thought we already had f

[thorn jay]

no, thats the whole thing we want to construct

[tardy hedge]

Ok that makes more sense

[thorn jay]

we begin with a function f|_S : S → N, want to construct a morphism f : M → N such that f restricted to S is indeed f|_S

[thorn jay]

we are trying to look for a condition on f* such that we can construct such and f.

[rapid cave]

Yoooo enpeace handwriting

[tardy hedge]

I know it blew my mind first time i saw it too

[tardy hedge]

Now im trying to remember why the f* appeared and why its relevant ig

[thorn jay]

universal property of free modules

[rocky cloak]

Like the definition parses, but I don't think it's super useful.

Like in Z/6 you have that 2 is prime, but not irreducible

[tardy hedge]

Meaning we can have any F(S)->N morphism based on saying where we want S to go in N

[thorn jay]

exactly

[thorn jay]

and the desired morphism f existing is equivalent to f* factoring via p in this diagram

thats what the uniqueness part of the universal property says

if there is an f : M → N with f(s) = f|_S(s), then f* must equal to f ∘ p, because f*(|s|) = f|_S(s) = f(s) = f ∘ p(|s|)

[tardy hedge]

Ok i kinda see what ur saying there but need to think it through a bit

[thorn jay]

take your time

this isnt super easy, haha, but very important to internalise imo

[tardy hedge]

Agreed, which is why i do want to understand this one lol

Whats the best way to think about the uni property of free modules, is it the fact free modules have the property that sending its basis anywhere, to any module (like as a set function) always extends to a unique map of modules?

I think dummit and foote talks about it as something like any set S can give a free module with basis S, and then it shows that property with the maps in the diagram

@thorn jay ill come back to this but ngl i probably should leave it for later

Ive been sooo distracted by this discord lately

[thorn jay]

yes the above is correct

[tardy hedge]

ok, i feel like dummit and foote were emphasizing the fact that a set S can have a free module F(S) with basis S

idek

[thorn jay]

i mean that is a very important property

lol, but not the most important

[tardy hedge]

is it part of the universal property though

[thorn jay]

no, the universal property gives a characterisation of the free module

[tardy hedge]

Ahh

Thats what uni properties do

So a module in the wild that has that property is a free module

[thorn jay]

yes

but that property is always with respect to some subset

[tardy hedge]

In one minute im talking abt product rule

[thorn jay]

if a module M has the universal property of free modules wrt a function f: X → M, then it must be naturally isomorphic to F(X)

[knotty badger]

What do you mean by naturally isomorphic in this case

[thorn jay]

i think canonical map be better. An isomorphism φ: F(X) → M such that f = φ ∘ i, with i: X → F(X) being the inclusion

[knotty badger]

I see

[tardy hedge]

Ok i see what u mean ig, not sure i see the significance of it though

Like im not so sure why you said its “with respect to some subset” anyway

[thorn jay]

the universal property is not necessarily something of the module itself, but rather a property of a module with respect to some subset of (or more generally a function to) the module

because you have to specify some generating set for the universal property

and the statement "there exists a basis B ⊆ M" is equivalent to "there is a subset B ⊆ M such that M has the universal property of free modules wrt to the inclusion i : B → M"

[thorn jay]

notice how in the top diagram here, we are required to know the inclusion map S → F(S)

[tardy hedge]

Ok i think i get it

[vapid vale]

free module they did nothing wrong

[proven hearth]

can someone help me with this ? #1428824151077945344

[delicate orchid]

You might want to look into algorithms that solve the word problem for finite groups, or algorithms that reduce group elements into irreducible words in some fixed set of generators

[proven hearth]

okay so if this is a finite groups i can search for similitudes with others finite groups that already have an algorithm to solve them ?

this is a finite group

thx i'm going to search for this

[tardy hedge]

i just want module back bro

[marble hinge]

Hello! I am trying to prove one of the problems from Herstein's "Topics in Algebra". I've got a proof, but it feels a little bit hacky. I couldn't do anything better yet though. I just wanted to discuss if you think that the proof is good and maybe think about some alternative proofs.

Here is my proof (used Overleaf for the first time in a while!)

First I wanted to prove it more directly, i.e. try to capture the essence of this N, write its definition without intersections

I don't know what normal is 🙂 that's in the next chapter

I guess this just prepares me for the next chapter -- I think this is good approach 🙂

I like doing those problem-oriented books, that asks you to prove many intermediate results and just guide you by splitting it into small chunks and sequencing appropriately. So I don't mind

OK, then that's that 🙂 It also looked to me like it's also N(N) = G, i.e. normalizer of N is whole G

I wondered if there is a name for such thing, and it turns out there is 🙂

[rocky cloak]

You learn about normalizers before normal subgroups?

Wacky stuff

[marble hinge]

anyway, so what do you think about the proof?

[rocky cloak]

Nice proof

Pretty much the canonical one I'd say

[marble hinge]

Herstein introduces normalizers, centralizers and center in the exercises (maybe he doesn't use them much later, not sure)

[marble hinge]

cool! Glad to hear that. I've spent like an hour before I came to this idea of just plugging in the definition (quite an obvious idea in retrospect, I guess I just didn't want to deal with hairy combinations of intersections)

I tried characterise the resulting elements of N instead (i.e. the ones that are present in every single xHx^{-1} -- tried to get some properties of them, but that didn't help much with proving the second part...

[rocky cloak]

If you want more words, the N in this exercise is called the core of H

[marble hinge]

Ah, cool, thanks, that's helpful!

so now I have lots of new terms: core, centre, centraliser, normaliser and normal groups 🙂

so I suspect that there are examples of H and G so that core, C(H) and N(H) are all different, right?

[rocky cloak]

Sure

[marble hinge]

I already figured that in S_3 C(H) != N(H) when H = (e, (123), (321))

Another meta-question: is it better to discuss such questions/proofs here or in a separate help-channel?

(I am new to this Discord)

[rocky cloak]

If the questions are related to groups, rings or fields, then here is good

[thorn jay]

proof looks good, very detailed and explained well to be honest. I wouldn't be surprised to find this in an introductory textbook

[thorn jay]

(dont tell mods but this channel is a cooler #discussion too sometimes)

[marble hinge]

great, thank you!

will keep discussing (relatively) tricky problems here then 🙂

[rocky cloak]

I have to keep my streak of staying away from #discussion

[thorn jay]

the healthy choice

i regret to inform that i have a couple messages in there when i am summoned

(usually something UA related)

[tardy hedge]

I never go there ever

I like my algebros

[rocky cloak]

That dark shadowy place.
That's beyond our borders. We must never go there, Kian

[thorn jay]

lion king reference?

https://tenor.com/view/simba-lion-king-shadowy-place-shadow-place-simba-shadowy-place-gif-26625263

[tribal moss]

That's Opportunity's half of the planet. We must never go there.

[thorn jay]

it is infested by people who dont know abstract algebra...

truly the worst kind

[rocky cloak]

The best thing Disney ever did

[thorn jay]

they were good but now theyre bad

[karmic moat]

I have studying role to keep me away from there

[rocky cloak]

Makes me think of
https://youtu.be/VfCYZ3pks48?si=D0k6QLtVInulmGp4?t=40s

[thorn jay]

exactly what i was referencing

https://tenor.com/view/link-counsciousness-gif-13187339

[thorn jay]

OH NO POTATO IS A MOD

guess the mods found out ...

[south patrol]

Wdym

[thorn jay]

.

[south patrol]

Lol

[velvet hull]

it matches the profile now

[thorn jay]

wdym

[south patrol]

The pink

[thorn jay]

oh yeah true

[karmic moat]

One who cannot reach the grapes says they must be sour

[thorn jay]

how poetic

[elfin wraith]

It used to be the piss role, new colour is for sure better

[thorn jay]

kaki

[karmic moat]

Shoutout aesops fables

I didnt even notice a change in the color tbh

[thorn jay]

everytime i see something poetic in a silly place im reminded of that one stupid tweet

[karmic moat]

the cathedral one?

[thorn jay]

yes

[elfin wraith]

There are cathedrals everywhere for those with eyes to see them or something?

[karmic moat]

yea

and it's a picture of an evian water bottle by jordan b petersen

[tardy hedge]

Ok i get it up to (and including) here, but the fact that there was all that hidden background involved when u said “descend down to quotient” is like 😂

Its interesting so far though

[thorn jay]

im sorry lmao this stuff is engrained in my system cuz its so fundamental to UAG

[tardy hedge]

Nah this has been super helpful for me so far

[thorn jay]

yippee

[crystal vale]

Q[\sqrt{2} ] has only 2 ring automorphism, right?

a+b\sqrt{2} -> a + b\sqrt{2}, and a + b\sqrt{2} -> a - b\sqrt{2}

[rotund aurora]

you can try proving this if you are unsure

[crystal vale]

i proved, just to be confirm that i did it correctly

[rocky cloak]

Free implies torsion free over an integral domain, if that's what you're asking

[rapid cave]

torsion (in modules) is when ax = 0 for some x in your module and a in the ring the module is over

[rocky cloak]

What's your definition of free module?

Because free modules are just direct sums of copies of R

{1} is a basis for R

Anyway, ca = 0 implying c=0 is just the definition of R being a domain

[rapid cave]

there is no meaning to infinite sums

sure but that depends on some metric

so you need extra structure to say that

how would you define an infinite sums of polynomials?

but what is x + x^2 + x^3 + ... +?

this won't be a polynomial

yeah

[tardy hedge]

but then theres that whole thing with the definition of torsion being weird when R not a domain vs a domain etc

[karmic moat]

There are cases where you can have a “metric” and your elements can be infinite sums, eg the power series ring R[[x]] and Laurent series ring R((x))

But you need to take some care in defining those

[rapid cave]

but really there its a formal sum

[karmic moat]

That’s why i said elements and you have to take some care

[rapid cave]

is there reason to talk about torsion in non-domains?

maybe that was not the best way to phrase it

[thorn jay]

yeah you just take non-zero divisors iirc

[rapid cave]

ok

this makes me think, is there like a way to turn a non-domain into a domain?

like a functor or something.... idk cat theory

[thorn jay]

prime ideals 🔥

[rapid cave]

probably taking all of the non-zero divisors of the ring or something

[thorn jay]

that doesnt form a ring

[rapid cave]

the small problem with this is that you need to choose a prime ideal

[thorn jay]

yes exactly

[rapid cave]

do you have an idea for something more canonical?

[thorn jay]

i think theres a problem with that lol

hmm ive got an idea on how to prove that

i think

[rapid cave]

would you like to share?

[thorn jay]

let me eat first 🔥 🔥

[rapid cave]

sure

no pressure

[rocky cloak]

I guess the simplest examples would be say you have a ring like
KxF
where K and F are two fields.

What would be the integral domain you associate it with?

[rapid cave]

I don't have a good answer to that

probably the trivial ring

because every nonzero element is a zero-divisor

[rocky cloak]

Well, by the most common definition the trivial ring is not an integral domain, but if we allow that then I guess you can just mod out the ideal generated by all zero-divisors an repeat that until you get an integral domain

Will not be functorial though if that was a goal

[thorn jay]

okay, notice that a universal morphism f : R → S with S an integral domain would induce a surjective map f* : Spec S → Spec R. Then we see that f*^-1(V(ker f)) = V(f(ker f)) = V(0) = Spec S. Thus, as f* is surjective, this must mean that V(ker f) = Spec R, and so ker f must be the nilradical. But ker f is a prime ideal, so this means that the adjunction cannot exist in general. @rapid cave

This argument nicely generalises to coherent conditions

[rapid cave]

wdym by coherent?

[thorn jay]

its a thingy im working on

paper published, maybe sometime

[rocky cloak]

The fact that there can't a universal map from R to an integral domain I guess is easy to see by just considering something like R = Z/6

I'm guessing a functor from commutative rings to integral domains that restricts to the identity on integral domains is not possible either

[thorn jay]

alternatively: let p be prime in R
=> p = f^-1(q) for q prime in S
=> ker f = f^-1(0) < f^-1(q) = p
so ker f must be the unique minimal prime hence the nilradical

[thorn jay]

the gist is that they provide a nice framework for general classifications of congruences that are preserved under preimages and quotients. Regardless, they hold surprisingly much power imo

[rapid cave]

why does kerf being the nilradical is bad?

[rocky cloak]

There are rings where the nilradical isn't prime

[rapid cave]

oh its that simple

[thorn jay]

yup

[white oxide]

Could I please have a hint for showing that if x is in p, then x is a zero divisor

[white oxide]

Wait never mind I just realized I proved => wrong

For => do I need Zorn's lemma

wait no that's for constructing maximal elements uhhh

Oh wait, maybe I could just say that if ax = 0 for some nonzero a, then x in (0:a) \subseteq r(0:a), which is prime

and is the smallest prime ideal containing (0:a)

I think that works

[queen quarry]

Wordle 1583 4/6*

⬜🟨🟨🟨⬜
⬜🟨⬜🟨🟩
🟨⬜⬜🟩🟩
🟩🟩🟩🟩🟩

[thorn jay]

so true

[amber burrow]

General question

Do you actually write while solving a problem

I’ve been finding more and more that I just do it in my head

Like I’ll just try different things in my head

[south patrol]

I would say this depends a lot on the problem, how hard it is etc. Personally (as a phd student) I find the conceptual stuff often gets done when I am just mulling stuff over, but then often to get to some understanding you need to look at examples and things, and try to pursue half-formed ideas etc

[elfin wraith]

Yeah I can certainly sometimes just do stuff in my head, and I’d say that’s generally where the bulk of the work happens, but there’s no getting around the fact that sometimes you’ve just got to try some stuff when you’ve only got half the idea, or maybe there’s just a long computation to do

[tribal moss]

Ideally, I think, as you get more experienced, the complexity of problems you tackle should increase simultaneously so it keeps hovering just around the point where you need to start writing things down to keep everything straight.

[karmic moat]

If I don't write down stuff I find that I often miss important details or get small subtelties wrong

[south patrol]

lol the number of times i thought i had finished a problem i'd been working on for weeks/months only to notice i swapped two things by accident or smth in my head

[karmic moat]

have u read/heart of Alg Top from a Homotopical Viewpoint by Aguilar-Gitler-Prieto?

If so what do u think

[south patrol]

Seen a bit and iirc looked p good but haven't read it too properly sorry

But iirc one of the better books for this stuff

[karmic moat]

Nice I’ll add it to the list then

[alpine plank]

dont we need F to be algebraically closed for jordan normal form to exist

[quiet pelican]

Take a field extension K of F such that the JNFs you care about exist
G and GL_2(F) are subgroups of GL_2(K) so we can compute in GL_2(K), then restrict back

[languid trellis]

Am I wrong in thinking that the nilradical and jacobson radical of this ring are both (y^2+1) (in R = C[x,y]/I)

wait no

give me a moment

As we know prime/max ideals of R <-> prime/max ideals of C[x,y] containing I. I = (X(Y-1)(Y+1)) + (Y(Y-1)(Y+1)) = (Y-1)(Y+1)((X) + (Y)). So, I \subset (x-a,y-b). One way of expressing this inclusion is in terms of the evaluation map. if x -> a, y -> b, then if I is mapped to 0 under evaluation, then I is in (x-a,y-b).

So, certainly I \subset (x,y), (x-a,y-1), (x-b,y+1), a, b \in C.

so now we take the intersection?

(x-a,y-1) \cap (x-b,y-1) = (x-a,y-1) \cdot (x-b,y-1) due to coprimality

hm

Wait actually

for C[x], we know the jacobson radical is trivial, as otherwise it would imply the existence of a polynomial with unbounded degree

in particular

\bigcap (x-a) = (0)

So, \bigcap (x-a,y-1) = (y-1)

\bigcap (x-b,y+1) = (y+1)

so, we are left to do (x,y) \cap (y-1) \cap (y+1). These ideals are coprime, so we can simply multiply

to get that the jacobson radical is (x,y) \cdot (y-1) \cdot (y-1)

and I believe the nilradical is the same in this ring

actually we know the nilradical is a prime ideal

hmmmmm

[vapid vale]

since these are coprime, use chinese remainder theorem

[languid trellis]

ohhhhhhh

and this is I ?

(X, Y)(Y^2-1) = (XY^2 - X, Y^3 - Y) = I

so N(R) = 0?

and as spec \cup (0) = maxspec in C[X,Y], J(R) = 0

Actually, I've just thought of another way to show this

[vapid vale]

youve shown I = (x,y) \cap (y-1) \cap (y+1), so C[x,y]/I \cong C x C[x] x C[x]

[languid trellis]

By CRT, we have the iso

$$\mathbb{C}[X,Y]/I \rightarrow \mathbb{C}[X,Y]/(Y-1) \times \mathbb{C}[X,Y]/(Y+1) \times \mathbb{C}[X,Y]/(X,Y) \cong \mathbb{C}[X] \times \mathbb{C}[X] \times \mathbb{C}$$

[cloud walrus]

swifteeee

[vapid vale]

yeah

[languid trellis]

and C[X], C have N(R) = 0 each

we know by a previous exercise that N(R x S) = N(R) x N(S)

so N(C[x,y]/I) = 0

same for the jacobson radical

[vapid vale]

yez

[languid trellis]

thank you!

working with polynomial rings is still hard : (

[vapid vale]

i agree

although theres a nice geometric payoff to it i think

[languid trellis]

i am unaware of that at this moment

though i am waiting excitedly for the day we introduce varieties

[vapid vale]

well i think the better commutative algebra intuition you have the more exciting and meaningful algebraic geometry is

[proven hearth]

visualization is a really good tools

[elfin wraith]

As an R module yeah

[karmic moat]

in general Hom_R(R, M) \cong M for an R-module M

[south patrol]

Freddy Fazbear?

[karmic moat]

W

[south patrol]

Yeah then u need the op

Function composition is one hell of a drug

[elfin wraith]

It is the kinda thing that’s obvious once you’ve seen it, but not before then

They probably should say something but I’m guessing your prof just kinda glossed over it without thinking

[south patrol]

Videogame character lol

[thorn jay]

how are they doing the multiplication here

[south patrol]

But sounds like R R R

[thorn jay]

in my opinion for f, g, endomorphisms it should be (f*g)(x) = f(g(x)), but ig that's another convention

[rocky cloak]

The trick is to work with right modules, then you get End(R) = R, no op needed

[karmic moat]

the trick is to work with everything commutative, then none of this is needed

[south patrol]

No that is the convention here I believe lol

[karmic moat]

🤤

[thorn jay]

huh?

[rocky cloak]

fg(x) = f(g(x)) is indeed the usual convention

[thorn jay]

but then for f_r : x |-> rx we have f_r(f_s(x)) = r(s(x)) = (rs)(x) = f_rs(x). How does that turn into R^op

[rocky cloak]

Well f_r isn't linear.

The correct map should be
x |-> xr

[thorn jay]

oh right

noncommutativity affects that

[south patrol]

f_r is right R-linear however ig

Ascend

[rocky cloak]

Right modules also makes it more natural to think of M as an
End(M)-R-bimodule

So you don't need to take any ops or consider bimodules on the same side

[elfin wraith]

Btw jagr do you know anything about Verlinde categories or Tannakian reconstruction/duality? I don’t know if that’s so much the rep theory you do but you also seem to be a never ending source of algebra knowledge

[thorn jay]

just work with commutative rings

[karmic moat]

that's what i said and i got downvoted

[elfin wraith]

Someone find the image

[thorn jay]

I have it but like

I'm not gonna react to myself

[karmic moat]

the jesus one?

[thorn jay]

😭

[karmic moat]

is this the right jesus one

[thorn jay]

nO

it's "what's the matter commutative boy, afraid you might get an interesting theory?"

[karmic moat]

oh

embarassing

[rocky cloak]

I will say my knowledge about that is zero, but I might just look it up

[elfin wraith]

I’ve reached out to a prof who suggested a masters thesis to me in some intersection of these ideas, and it sounds very cool but also very difficult and I was just wondering if you had any thoughts at all (if you knew about them)

I think potentially me getting to the point of understanding the problem he wants me to work on is feasible, but actually progressing the research probably less so

It’s that or I do some Milnor K theory stuff so I think I’m just suffering in either case

[karmic moat]

milnor goes to a restaurant and says "waiter waiter one more arrow in my diagram please"

[rocky cloak]

You're not really supposed to progress research during a master thesis I guess

Getting to the point where you then can start research in a phd is already good I think

[elfin wraith]

This prof seemed like he really wanted me to, but he also seemed to assume a lot of knowledge from me generally (he kinda grilled me about alggeo for a while prior to this because I mentioned potentially doing something with D modules)

But yeah, idk, I have about enough understanding to almost digest what those definitions are but I don’t know if perhaps you had anything insightful to say about them lol

But what I’ve done with monoidal categories so far has been cool, and this seems to be rep theory related in a special monoidal category which seems like a mix of words I enjoy

[rocky cloak]

Yeah could be fun.

Don't know how fun it is if your advisor is a hardass though...

Good to have an advisor you click with

[elfin wraith]

Yeah this is my hang up, it’s why I’m not fully committed to it

[rocky cloak]

Not that someone who pushes you is necessarily bad, but it needs to be enjoyable first and foremost

[elfin wraith]

I’m meeting with someone who seems lovely this week but it’ll be for something alggeo related and I’m not overly fussed about alggeo. The other option is the K theory project which seems fine and the advisor seems fine, but none of that is hugely exciting

[karmic moat]

is it better to work with a hardass advisor in a field you love or with an awesome advisor in a field you dont care for

[elfin wraith]

Yeah it is a tough one, I’ll kinda have all the information by the end of the week but I’m not sure where I’m going to go yet.

[thorn jay]

it seems like kiand ended up in the latter

[karmic moat]

hmmmm

[thorn jay]

yeah

same construction as for groups/monoids/posets

[tardy hedge]

This is how im feeling right now (im in a masters program) but i dont want to do combinatorial algebra

I kinda want to do more fancy stuff lowkey but i still need to study a lot more to get there 🤷‍♂️

[earnest delta]

I wanted to understand concept of conjugate in simple terms?

[kind temple]

what are you finding confusing about it?

[cedar roost]

New here on the server (thanks!), stop me if this is not appropriate, but is it OK to share links to papers about Rings here?

If that's OK, here's one I bumped into that I like. Not very recent, but I had previously not considered the application of rings to deep learning etc. (I'm working on a category-theory-inspired deep learning model at the moment, so I've been looking for alternative approaches to compare to.)

https://arxiv.org/abs/1409.2544

I've been trying to use Topological Data Analysis + Decision Field Theory to accomplish what this paper is about, but maybe their approach is easier, still trying to figure that out.

[vapid vale]

what in TDA

[thorn jay]

holy persisten homology

[raw anvil]

cool algebra

[tardy hedge]

tda moment

[elfin wraith]

so bar coded

[tardy hedge]

you could probably start learning it if u wanted to

the other part is just topology rudiments ig

at least to get started on understanding it

I started learning about it last year and I didnt know alg top

yea!

cuz its like overlap with statistics too yk

sort of thing

i do want to learn more stats tbh

[rocky cloak]

Simplicial homology I guess. But could learn as you go probably

[tardy hedge]

what u get

nice thats cool

id read that

it would be kinda nice to learn stuff that isnt so abstract

lol

def not i do not work with my hands lmao

im so bad at that

ig i wouldnt be opposed to doing like data science type of stuff so id be down with learning more stats

i was gonna do a project with some bio guy on tda but i didnt go thru with it

last year

i got kinda uninterested ig

Was probably also because i saw persistence modules and whatever and i didnt have enough algebra background to understand the theory easily

[elfin wraith]

You don’t need a crazy amount of algtop but you will need to understand simplical homology

[rocky cloak]

Representation theory of quivers is also useful, sure

One of us! One of us! One of us!

[tardy hedge]

measure theory poo

[quiet pelican]

Take it!

[kind temple]

that’s measurement theory

[quiet pelican]

A first course in representation theory (ie a course on finite group representations) is the worst algebra course imo
But also like eat your vegetables it’s good for you

First and last are not mutually exclusive 🙃

[rapid cave]

Real... I never understood the purpose of rep theory while doing the course