groups-rings-fields

[errant shadow]

what does this notation stand for by the way

is it some specific field?

[chilly ocean]

it's the finite field with p elements

[errant shadow]

ahhh, that makes sense

nice, thanks

[chilly ocean]

An isomorphism is a function f , which takes elements of R, \matbb{Z}

[south patrol]

So true

[south patrol]

How was that related to previous discussion anyway

[white oxide]

An isomorphism is a function f , which takes elements of R, \matbb{Z}

[ebon wyvern]

...what

[south patrol]

should be \mathbb

[ebon wyvern]

algebrains in chat what is your favorite isomorphism between R and Z

[chilly ocean]

x -> 0

[south patrol]

Well it's not \mathbb R

it's R

so just take R = Z

:)

Hope this helps!

[chilly ocean]

but those aren't isomorphic!

[sly crescent]

Yes they are

Z is a ring

And rings are usually denoted with R

[chilly ocean]

An isomorphism is an aplication continuos of funtions

[chilly ocean]

whoosh

[sly crescent]

No u

[chilly ocean]

so true

[south patrol]

what about when there isn't a topology

[night onyx]

there's always a topology 😎

[chilly ocean]

the topology is only convex to a subset of Rn

[south patrol]

??

[sonic coral]

i managed to do it by the way he hinted me toward, how would you have proven it?

[uncut girder]

Every sentence Fabian utters is more deranged than the last

[rocky cloak]

So a = b^2^k such that 2^k doesn't divide the order of G.

The order of b divides the order of G, so equals 2^l m for l<k and m odd.

The order of b^2^k is the order of b divided by gcd(|b|, 2^k) which equals m, so a has order m, which is odd.

[frosty pecan]

Their writing looks extremely similar to someone that posts about resolving the Hodge conjecture a lot on Reddit.

[sonic coral]

what if 2^k does divide the order of G though

[rocky cloak]

Then you have to choose k a little bigger

The statement is that for all k we have a = (b_k)^2^k , so in particular it holds for ks such that 2^k doesn't divide the order of G

[topaz solar]

This is such a weirdly specific connection

[delicate orchid]

if you know you know

[topaz solar]

I don’t

[sly crescent]

Christiano de Barros?

[frosty pecan]

No but I don't think I'm going to share the account.

[white oxide]

hmmmm, so I'm tempted to consider the sequence $0 \rightarrow I \rightarrow A \rightarrow B \rightarrow 0$ and show that it splits, but this doesn't even work since $A$ has to be the first nontrivial module in the sequence... I need a map from $B$ to $I$ tho, so it would be natural to consider $0 \rightarrow A \rightarrow B \rightarrow I \rightarrow 0$ but then the splitting doesn't tell me anything i think

[cloud walrus]

okeyokay

[rocky cloak]

So 0 -> A -> B splits, so you have a map B -> A

[delicate orchid]

you recall the proof you did for projective modules? or was that moamen

[white oxide]

that was me, i think that problem was showing that we could always consider a projective module P as a direct summand of a free module tho (assuming that every short exact sequence with P as the last factor splits)

[white oxide]

ah right

oh, so let's call this map $h: B \to A$. could I just let $\lambda: B \to I$ be $fh$?

[cloud walrus]

okeyokay

[white oxide]

oh wait I haven't even thought about commuting yet

[rocky cloak]

Yeah, so that's the map. Now you just show that this commutes

[white oxide]

oh yea

that works

cuz gf = id

yea I gotta learn to not approach things so "visually" I was searching for an explicit map from B to I and not considering composing things

at least when it comes to these problems, idk

but thanks!

[delicate orchid]

if u have a proof for an object...

the proof of the dual is not far behind

[void cosmos]

any examples

of left artinian that is not right artinian

[white oxide]

yea ig so

but like this was a conceptually different problem right

so i couldn't just reverse the arrows to the other problem

or whatever

[rocky cloak]

The ring of matrices of the [[Z, Q], [0, Q]] should work

[void cosmos]

this is right noetherian btu not left noetherian too right?

[rocky cloak]

Yeah

[void cosmos]

cool

tysm

[delicate orchid]

conceptually you were still trying to find a map going backwards

or maybe not but that's usually the quickest way to show things are split exact

[rocky cloak]

Yeah, it wasn't litterly just the dual problem, but the strategy was pretty much the same anyway I guess

[white oxide]

oh okay

well now i have to show that every R-module being injective implies that every R-module is projective, so I guess this will be a better instructive exercise of that notion

[delicate orchid]

yeah ok it's not quite literally the dual unfortunately

oh I don't actually know the proof of this

I'll do it with you

(anything to get out of trying to prove this theorem)

[white oxide]

lol ok cool you're gonna finish wayyyy b4 me

[delicate orchid]

doubt it

I make it look easy because I usually just know how it goes

[white oxide]

so tempted…….

[delicate orchid]

so close! But what if the ||surjective map was out of an indecomposable module??||

[white oxide]

ooh ok

so tempted to click

but must resist

ok if i don't get it in like the next half hour or whatever i'll take a peek

[delicate orchid]

can I smash something together using the horseshoe lemma

that would be funny I think but definitely overkill

[rocky cloak]

A possible hint is that ||you can use the exercise you just did||
Don't know if that's the strategy you want to go for though

[delicate orchid]

||that's what I'm doing lol - it makes sense if the exercises follow each other||

I'm nearly getting like... a homotopy out of this lol

shits wild

[rocky cloak]

I feel like it would be easier to just ||show that every module is a summand of a free module, but then you wouldn't use the previous exercise, so...||

[delicate orchid]

oh actually yes you would

||you'd automatically know the sequence splits by the previous exercise||

[rocky cloak]

||the sequence would split because the kernel is injective. The previous exercise was the converse, i.e. in a semisimple category every object is injective||

[delicate orchid]

||now I'm very confused. It's incredibly trivial that every object is injective in a semisimple category... this exercise is showing that if every object is injective then the category is semisimple the previous exercise was showing that exact fact about the kernel being injective||

wait did I read projective and injective backwards

nope

[rocky cloak]

||the previous exercise was "if every short exact sequence in Mod R splits, then every module is injective"||

[delicate orchid]

we gotta be in two different timelines

[rocky cloak]

||the one okay solved at exactly 26 minutes ago||

[delicate orchid]

yeah this one?

whatever I think I've done this anyway

[rocky cloak]

Indeed

[delicate orchid]

ah wait not quite

one mo

[white oxide]

does the solution use split exactness? i'm experimenting with that and think it might work, idk

considering the sequence $0 \rightarrow P \rightarrow C \rightarrow B \rightarrow 0$ which I know splits since $P$ is injective....

[cloud walrus]

okeyokay

[delicate orchid]

@rocky cloak thoughts? prayers? hopes? dreams? ||the upwards map is induced by the universal property of the coproduct using the injective map from A and the section from B (which exists as the sequence is split) and the downwards map follows from the universal property of free modules using the maps out of A||

wait fuck ||A might not be finitely generated||

ahhhhhhhh so close

I have a feeling I may need a ||second exact sequence||

[rocky cloak]

||I'm a bit confused about what's going on here. Where does the maps from A and B into R^n come from?||

[rocky cloak]

If you just choose the right sequence, it could

[delicate orchid]

yeah ||B needs to be finitely generated too||

I'm too used to that property man

[rocky cloak]

I'm still confused with that assumption, but...

[delicate orchid]

I'm just gonna google it

I've lost interest!

[rocky cloak]

|| you have an exact sequence K -> F -> A, with F free. Since K is injective, the sequence splits, so A is projective||

[delicate orchid]

oh there's a proof using derived functors that's rather nice

not quite what I'm after here

[delicate orchid]

since fucking when is that implication two ways

[rocky cloak]

||that direct summands of free modules are projective?||

[delicate orchid]

||no! That if a SES splits the image has to be projective?!||

||I ran into this exact question in my proof like 10 minutes ago||

[rocky cloak]

||since the sequence splits A is a summand of F||

[delicate orchid]

||yes, if the central module is free, but it could be anything||

||my whole proof strategy was trying to show that the central module HAD to be free||

[rocky cloak]

||I'm confused. You start with A, then you pick some free module that surjects onto A. Like not every short exact sequence has the middle term free...||

[delicate orchid]

I'll try and show it - ||maybe there's something with the universal property of the coproduct+free modules||

[delicate orchid]

||yes, exactly - but inorder for C to be projective we have to show that for any surjective morphism out of ANY module, there's another one such that the triangle does its thing||

[rocky cloak]

||what is C? Wut?||

[delicate orchid]

||sorry, B||

I wrote C on my paper then changed it to B for the diagram

[rocky cloak]

Okay, so your talking about what you wrote before now? I'm lost

[delicate orchid]

the definition of projective is that for each surjective morphism between any two modules it factors through the projective module
||we require the middle term of our SES to be free in order to show that the right term is projective via being a direct summand of the middle term||
||however, by assuming that - we have failed to show that any surjective module between ANY two modules (not always free) factors through our right hand term||
it seems like each of these claims imply each other and not in a good way

[rocky cloak]

||I just wanted to show that every module is a direct summand of a free module. Since that implies it's projective||

[delicate orchid]

I think I'll just accept the ext proof

[rocky cloak]

So you don't accept this?

[delicate orchid]

I believe you dw

I just don't see why

I know that ||if the third term is projective then it splits - hell I even posted the proof like 30 mins ago lol||

it's the reverse direction

[rocky cloak]

||like you have A, and you want to show that it's projective.

(i) There exist a surjection F -> A from a free module, agree?

(ii) This has a kernel so fits into a short exact sequence K -> F -> A, agree?

(iii) Since every object is injective K is injective so this splits, thus A is a summand of F, agree?

(iv) This implies A is projective.||

[white oxide]

ok so I have the graph [\begin{tikzcd}
&&& I \
0 & A & B & C & 0
\arrow[from=2-1, to=2-2]
\arrow[from=2-2, to=2-3]
\arrow[from=2-3, to=2-4]
\arrow[from=2-4, to=2-5]
\arrow[from=1-4, to=2-4]
\end{tikzcd}]
and want to show that there's a map from $I \to B$, so I'm considering $B$ as an injective module, specifically the diagram [\begin{tikzcd}
& B \
C & I & 0
\arrow[from=2-3, to=2-2]
\arrow[from=2-2, to=1-2]
\arrow[from=2-2, to=2-1]
\arrow[from=2-1, to=1-2]
\end{tikzcd}]
since I want to eventually produce two maps coming out of $I$, was wondering if I'm on the right track or not

[cloud walrus]

okeyokay

[delicate orchid]

yeah that's what I've had in my head for ages

the problem is I just don't believe it - I know I'm being difficult but like

Like I'm supposed to believe that because 0 -> C_2 -> C_2xC_2 -> C_2 -> 0 is a split exact sequence of Z-modules so C_2 projective?

[rocky cloak]

No, you're not supposed to believe that at all

[rocky cloak]

Yeah, or you just want to produce the one map I -> B, but it should fit with the map you already have.

A hint can be to consider ||what do you know about A||

[rocky cloak]

Which point in the proof is making you believe that?

[delicate orchid]

the original claim I think

don't worry about it

[rocky cloak]

Well I have to worry about something

[delicate orchid]

hence my rather direct reaction

cause that sequence immediately came to mind

[rocky cloak]

So what's the theorem you're trying to get out of?

[delicate orchid]

it's basically clifford's theorem for fusion systems

I'm like... 2/3rds of the way there

I think anyway - my supervisor hasn't commented yet

[rocky cloak]

What's Clifford's theorem

[delicate orchid]

it's uhh

when you restrict a character to a normal subgroup you have that the restriction decomposes as a multiple of a sum of conjugate characters

mainly, they're all the same degree and have the same multiplicity

[delicate orchid]

that part is proven!

it's extending that result to "the degree of the constituents of the restriction divide the degree of the character you started with" that's hard

now quick somebody post another question so I don't get scooped lol

[rocky cloak]

Conjugate meaning like galois automorphism or what?

[delicate orchid]

if you have a character \chi of a normal subgroup N <= G then precomposing \chi with c_g (the map sending x -> gxg^{-1}) gives you another character of N

and it sends irreducible characters to irreducible characters

[rocky cloak]

Ah of course, that makes sense

[delicate orchid]

fun exercise! (one I can actually do lol...)

I won't share anymore cause I am paranoid about getting sniped (it kinda happened to me in April lol)

[rocky cloak]

That sucks, what got sniped from you? If you're at liberty to tell?

[chilly ocean]

What is an ring-Chow???

[delicate orchid]

I don't wanna say "scooped" cause that implies some maliciousness on their part lol - it was just unfortunate that we were working on similar stuff - it was determining when the representation ring of a fusion system was freely generated or not and doing some calculations on fusion systems on sylow subgroups of PSL(3,2)

I can send the paper if you want

[rocky cloak]

Sure, don't know how closely I'll read it, but sounds good

[delicate orchid]

https://arxiv.org/pdf/2303.10341.pdf

[rocky cloak]

Hopefully you hadn't sunked too much work into it.

I told my supervisor once that I wanted to look into tilting in exact categories. Then just like a couple of weeks later, I attended a talk called "tilting in exact categories". Pretty funny coincidence, but I can't really say I got scooped

[delicate orchid]

nah - to be honest it probably saved me a lot of work and let me accelerate in other areas way quicker than I otherwise would have. But yeah it sounds a lot like your story

you gotta have the mindset of "oh that's saved me a lot of work! I can't wait to read it"

just don't do it a week before my thesis is due in please ;3

[rocky cloak]

Is your thesis not like a collection of papers you've produced over your degree?

[delicate orchid]

yeah mine will end up like that

cause I'm nowhere near cracked enough to make my thesis be like, it's own contained work, like some people are

[delicate orchid]

and also publishing as you go avoids this very issue

[rocky cloak]

I wonder what Im gonna title my thesis. Like what actually is the common thread between the projects I'm going right now?

"various things about endomorphism rings of nice modules"

[chilly ocean]

Does a ring-Chow have embedded categories?

[rocky cloak]

"things that vaugely involves some generalization of tilting"

[delicate orchid]

"what if we did character theory but on finite categories"

[void cosmos]

what is the endomorphism ring of Z_p^infinity

[rocky cloak]

It's really unfair that rep theory of groups get all those nice character tables. Why can't I get a character table?

[delicate orchid]

working on it boss...

simply study a ring that is isomorphic to a REALLY nice ring

[delicate orchid]

fun question lemme think about it

[void cosmos]

ik the answer

[rocky cloak]

My gut feeling would be p-addic integers

[delicate orchid]

oh yes yes yes

[void cosmos]

yea correct

what a gut

chef

[delicate orchid]

Hom turns colimits into direct limits

very funny

[void cosmos]

wait

whats a colimit

[delicate orchid]

wait there's a few other steps there

[void cosmos]

its like the inverse limit?

[delicate orchid]

no, the direct limit

wait yeah it's not so simple

I had these two things in my head but they don't work out so nicely

[rocky cloak]

Yeah, do abstract nonsense solution would be

$Hom(colim Z/p^k, Z/p^\infty) = lim Hom(Z/p^k, Z/p^\infty) = lim Z/p^k$

[cloud walrus]

jagr2808

[delicate orchid]

call me silly but how do you get that last equality

oh oh

yeah

[rocky cloak]

Hom(Z/p^k, A) = elements of A with order dividing p^k

[delicate orchid]

yurrrrr cause u map the generator to any one of them

[void cosmos]

what type of math is this

this is the coolest shit ever

one of my favs is the like duality with like Hom direct product and sum

[delicate orchid]

Category theory

[void cosmos]

okay category theory arc it is

[delicate orchid]

Left adjoint preserves limits or some nonsense

[void cosmos]

lmfao yea i remembered

the currying one

Hom(M tensor N,P) is Hom(M,Hom(N,P))

lmfao this is dark magic

[delicate orchid]

Oh yeah it is just currying. Based

Take ur little tensor triangle then boom

[void cosmos]

really wanna learn more about Ext and Tor

dk shit baout them

[delicate orchid]

Free-faithful is the easiest one to see imo

Fuckin

Forgetful

Not faithful

[south patrol]

Ext and Tor are fun

[next obsidian]

What

[delicate orchid]

ctorial

[south patrol]

Sure

[void cosmos]

yea

i guess

someone here told me that they measure how bad hom and tensor are not left or right exact

so i just thought why are they so cool arent they just the ker/im

[delicate orchid]

Yeah this is true

For example Tor^i will be 0 if ur tensoring with a flat module

They’re what happens if you take the left/right exact sequence you get from passing an exact sequence through some functor and just keep extending it

[delicate orchid]

And no this is homology (which is related!)

[void cosmos]

hahaha wait why isnt it just ker/im

[south patrol]

What do you mean by that

[delicate orchid]

I dunno what you mean by that. If you have some exact sequence A -> B -> C and then tensor it you get a long exact sequence … -> Tor^i(B) -> Tor^i(C) -> A -> B -> C -> 0

Replace the is with 1

And add tensors in

I’m in mobile I cannot be bothered

You get the picture

[void cosmos]

yeahh

[delicate orchid]

Please explain what u meant by that lol

I think you mean homology but I’m not sure

[void cosmos]

i meant like

like i just thought okay how much a sequence is not right exact

well then just mod out

ker / im

like

okay let me try to think of an example

if like u have some exact sequence 0-->A-->B-->C-->0

take some functor call it F

for F(A)-->F(B)-->F(C)-->0 to be exact we have to like see at each point the ker and im

so if u define the other functor to be just exactly that

then wont u get that its right exact by definiton

[chilly ocean]

What does n_rt(a) mean
Positive real one? or any complex numbers which x^n=a?

[void cosmos]

like ker(F(B)-->F(C))/im(F(A)-->F(B))

[delicate orchid]

Yeah that’s homology

And it does measure how non-exact a complex/sequence is

Tor/Ext are what are called derived functors and are specifically for tensor/Hom

[void cosmos]

yea so whats Tor

like

how do u define i

[delicate orchid]

As the derived functor to Tensor lol that is the definition - I think the Wikipedia definition is alright (or nlab if ur brave)

[white oxide]

is this ideal not just {0}??

[delicate orchid]

(2) in Z/4Z

Wait which ideal

Doesn’t matter actually

[white oxide]

because 0 is in the set and {0} is a prime ideal right

wait

WAIT NO

that's only in an integral domain

tricky tricky stuff

what are some examples of rings with zero divisors that are not Z_n for n not prime?

[delicate orchid]

M_n(R)

[white oxide]

oh ya

[delicate orchid]

R[x]/(x^2)

[white oxide]

oh ya

[white oxide]

wait.... this "set" would be a subring right?

[delicate orchid]

C[G] with G a finite group

[delicate orchid]

Hmmm

I mean yes it’s a ring

Depends if you need your sub rings to have the same identity or not

*multiplicative identity

[white oxide]

right

well what if this set with zero divisors and 0 WAS the prime ideal

hmmm

i'm looking at Z8 rn lol

[delicate orchid]

Then you’re in an integral domain and the result is trivial

Wait

Sorry misread

[white oxide]

wait huh

ur good

[delicate orchid]

Entering mild divine orthogonality but other than that I’m as right as rain old chum

[white oxide]

that wasn't sane

[dense void]

I would like group like structure for multiplication: Fₚ[√1] (but this does not make sense)

[delicate orchid]

Sqrt of 1 is 1 so this is just F_p

[white oxide]

i feel like contradiction is the way to go here... assume that given any ideal of this ideal it's not prime and deduce a contradiction from the fact that we're working with zero divisors and 0

[dense void]

I would like that √1 is treated as an extesion

so √1 √1 = 1, but √1 ≠ 1

[delicate orchid]

So you want like, F_p[x]/(x^2-1)?

[dense void]

x² -1 is (x+1)(x-1)

[delicate orchid]

This isn’t a field, it’s isomorphic to (F_p)^2 by the Chinese remainder theorem

[delicate orchid]

That’s how you’d formally define a new square root of 1

*adjoin, not define

[dense void]

that was also my idea

[delicate orchid]

Well there you go

As for the multiplicative structure uhhh I’m not sure how the unit functor plays with products lemme think

Right it should just be C_{p-1}^2 for the multiplicative group

[white oxide]

if R is a commutative ring then R is immediately isomorphic to its opposite ring R^op right?

[next obsidian]

Via the identity

[white oxide]

yessir

abstract algebra god?

is it true that if $\varphi: R \to R^{op}$ is an isomorphism, then $\varphi = id$? it seems like that's the case and I'm going to try to prove it

[cloud walrus]

okeyokay

[delicate orchid]

for commutative rings yes lol

it is NOT true for non-commutative rings which is why rings are a bad object

[next obsidian]

Wew no it isn’t lmfao

[delicate orchid]

can't embed as a category? lame!

[next obsidian]

This is saying isomorphism => id

That’s not true for like, Z

[delicate orchid]

burrpppp

prove it

[next obsidian]

-1

[delicate orchid]

what about it, loser

[delicate orchid]

https://tenor.com/view/stickydacat-gif-26502121

[next obsidian]

@obsidian sleet

[summer path]

what is even happening

[next obsidian]

The invasion

[open sluice]

algebra

[delicate orchid]

my divine connection is glowing like autumnal dayspring

[delicate orchid]

anyway I uh suggest ceasing this exercise now that I've actually read it lol

[white oxide]

well I'm trying to give an example of a ring R that's not isomorphic to it's opposite ring

[delicate orchid]

if the identity is indeed a ring isomorphism from R to R^{op} then precomposition with the map multiplying by any unit gives you another non identity isomorphism

[white oxide]

and I found that any non commutative ring works i think

because we must have $\varphi(ab) = \varphi(a) \circ \varphi(b) = \varphi(b)\varphi(a) = \varphi(ba)$ and so if $ab \neq ba \in R$ then our map would not be injective

[cloud walrus]

okeyokay

[delicate orchid]

uhhh lemme thing :glowingmind:

we have to map U to U^op which is natural so like w/e but then this doesn't play nice with addition so yeah

[white oxide]

bro said addition

[next obsidian]

????

[delicate orchid]

you have to extend it to a ring homomorphism you muppet

[white oxide]

nah

[delicate orchid]

alternatively T_2(F_2) is not isomorphic to it's opposite

[white oxide]

addition = multiplication

[delicate orchid]

le sigma algebra has arrived or something

[night onyx]

where does the last equality come from?

[next obsidian]

This doesn’t make sense

[delicate orchid]

hell, where does the first equality come from

[white oxide]

oh wait ur right

OOPS

[delicate orchid]

it's insane!!! it''s crazzyy

[next obsidian]

Definition of homomorphism

[delicate orchid]

nah

[next obsidian]

They used circ to mean the product in R^op

[delicate orchid]

think about it real hard

[night onyx]

well the first is the fact that it's a homomorphism with the opposite ring

the last part doesn't follow from anything

[white oxide]

yea that's true

hard to think when the people in front of you are making out in the library

time to move

[delicate orchid]

sorry I'll stop

[night onyx]

lol wow, throw a book at them

[chilly ocean]

join them

[next obsidian]

Babe no…

[delicate orchid]

nooo not the wholesome bookerino

[delicate orchid]

sorry bby.... okey''s grind comes first...

[next obsidian]

TTeppAllCrabsAreBlue

[open sluice]

flex on them with your knowledge of advanced algebrum until they vacate the premises

[chilly ocean]

i just had that stink with max and the mods come to mind recently and i wanted to honour it

[next obsidian]

I forgot about that

Lol

[night onyx]

first thought is no, any automorphism of a commutative ring is an isomorphism with the opposite ring, so just look for non-trivial automorphisms of some commutative ring

[next obsidian]

I said that already

-1

[night onyx]

oh sorry I missed that

[next obsidian]

🤯

[delicate orchid]

what about it, loser

[next obsidian]

Can you believe wew said it was true

Lol

What a dummy

[night onyx]

lol wouldn't have said that if I saw that, sry

[delicate orchid]

ngl I saw "id" and stopped reading

[next obsidian]

I should start implement similar practices

[ivory trail]

Tor and Ext end up being cohomology actually, the long exact sequences are the long exact sequences in cohomology

[next obsidian]

See “Wew Lads Tbh” and stop reading

[delicate orchid]

I penanced for my sins

[summer path]

i see name griefed lmao

where do i keep reading

[delicate orchid]

oops! What's that? You don't have a yummy can of unopened pringles on your desk?... yikes... sucks to suck ig....

[white oxide]

oh shit that's facts

[delicate orchid]

when I say it I get ignored I'm going to cry

[summer path]

pringles are salty

[open sluice]

hyperbolic paraboloid

[summer path]

saddle

[old spire]

Two questions: i) Are root systems a special case of Coxeter systems ii) Have the finite Coxeter systems been classified fully?

[white oxide]

okay i gave up on the exercise i'm sure it has something to do with considering the ring R generated by some chosen matrices or something smart but i'm no lin alg expert sadly!

[delicate orchid]

wait what

we said it wasn't true like 19232 times

[white oxide]

wait what

[delicate orchid]

yes, and the finite coexter groups have been classified fully and their corresponding diagrams are known

if you want to read about coexter groups I recommend Humphrey's book on them

[white oxide]

what's the best way to show that I prime => I maximal in Z again? quotient it out and show it's a field?

[night onyx]

yeah Z_p being a field is enough

[summer path]

Z_p

[old spire]

I know the classification of the groups; I'm interested more in the systems

[dense void]

I figured it out. I am ret*****, thanks.

[white oxide]

i feel like i should prove (i) implies (iii) first lol

but fuck it we going 1 --> 2 --> 3 --> 1

[old spire]

I'm also wondering about a possible analogy:
root system : Coxeter system :: compact simple Lie group : compact Riemannian symmetric space

I'm speculating this analogy exists. But forget the speculation. How are Coxeter systems classified?

[night onyx]

lol trew! this I was assuming

[white oxide]

ugh this problem is annoying

[night onyx]

lol give it some time! not all questions are 5 minutes ones

[white oxide]

yea ig

my midterm's tomorrow i'm so fuckeddd

[delicate orchid]

you can speed this up by first showing that ||Z is a PID|| if you're not allowed to use that as just a standard fact

[white oxide]

oh yea i'm using that fact

i feel like i'm close, i let a + I in Z/I and am trying to show that ar - 1 is in I for some nonzero r in Z

[delicate orchid]

which implication is this for

[white oxide]

uhh prime implies maximal

so i quotiented about by it and trying to show it's a field

[delicate orchid]

I strongly suggest going the other way

ii => i is immediate for a start

and as you said so is i => iii (cause PID)

[white oxide]

ok i'll try (ii) => (i) => (iii) => (ii) then

i feel like i'm getting somewhere with (i) => (ii) tho but idk might be hard to do

[delicate orchid]

iii => ii is way easier to apply your quotienting idea to

[crystal turtle]

That's the order my first thought would've been too

[white oxide]

oh so (iii) => (ii) is just saying show Z_p is a field

[delicate orchid]

exactly

[white oxide]

which i'll try to reprove

cuz i forgot that proof

[delicate orchid]

which should be fairly quick via ||bezouts|| you probably don't need it but I don't care it's easy to remember

[white oxide]

oh yea (a, p) = 1 => ar + sp = 1 => ar \equiv 1 mod p

[white oxide]

here since a PID is a Bezout domain i could just say ar + sp = 1 is in (a, p) right

but that's like kinda cheating

[crystal turtle]

Why would that be cheating

[white oxide]

cuz i dont' wanna do it with that assumption

it makes it too easy

[night onyx]

lol no, these are facts about Z, you're free to use them

[crystal turtle]

We have machinery for a reason: to make our lives and proofs easier

[night onyx]

same with prime factorization, that's used in a few of the implications too

[crystal turtle]

It would be a waste to learn the machinery and then not use it

[white oxide]

alright

then i guess i'll prove PID => Bezout domain then

i had a solution for this question before tho and it didn't use the fact that PID was a bezout domain

wait

oops

this is trivial isn't it

cuz I principal

J principal

I + J principal since we'er working in a PID

an I + J is an ideal

[white oxide]

yo i'm actually a fucking moron. i'm trying to prove that in a PID irreducible => prime, and am assuming a divides bc and a does not divide b. trying to show that a divides c

holy shit i'm an idiot i've been sitting here for 20 minutes and have been able to make no progress

[crystal turtle]

don't call yourself a moron

[white oxide]

bro i literally can't think of a single thing 💀

[crystal turtle]

Well

Do you have PID ==> UFD?

[white oxide]

no

[crystal turtle]

I think i've seen the proof for UFDs, and then just combined it with PID ==> UFD

[crystal turtle]

Well okay then. Since a is irreducible, it should generate a prime ideal, no? Combining that with this should give a start

[white oxide]

well i'm assuming i don't know that

[crystal turtle]

why

Like are you expected to be able to prove literally everything from scratch for your midterm?

If not, then you should feel free to use results, assuming you've seen them in class or in assigned reading

[night onyx]

yeah you're proving too much, like imagine having to prove the uniqueness of prime factorization whenever you want to use it, this question is specifically about Z, you should be using what you know about Z

[crystal turtle]

I mean he might be on a different question now

But still. PID ==> UFD is a standard fact that you should be able to use.

[white oxide]

yeah, i just want to get practice with using the definition of prime and irreducible and stuff

[white oxide]

i guess not

how do people get insights

or how are ppl so creative

when it comes to proofs

like my dumbass just writes down definitions and tries to derive things directly from them

idk how you do it man

[open sluice]

think more and be really, really patient

[crystal turtle]

The developments you learn in a class took place often over like 50 years lmao

With some things being even further apart than that in earlier classes.

[open sluice]

if you aren't getting enough sleep then go do that because sleep is the source of most of your creativity

[crystal turtle]

i need to sleep more

[white oxide]

Ig that’s true. I’m aboutta burn out lol midterms got me studying 7-9 hours per day

[night onyx]

you're doing it how you should be doing it, it's tough to see the forest for the trees but seriously, you're not asking stupid questions or approaching things badly

learning math deeply is like a 5-10 year journey just to not be considered a "student" anymore, don't underestimate how hard it really is

[white oxide]

thank you, yeah sometimes I need to really remember that math is hard for everyone ig

[night onyx]

Yeah, I mean even when I started my masters I felt insecure compared to everyone else about my proof writing skills, being a basketcase around exam time like I was going to forget everything, then you hear phd students talk and you feel like a complete moron, everyone feels like that

[safe meteor]

why is this true what is the intuition behind it?

i got given this defintion?

[mighty spade]

You just need to prove that if f=gh, and neither g or h is a unit, g and h have strictly lower degree

In fields

In fact, this holds for entire rings, since deg(fg)=deg f + deg g for polynomials over an entire ring

[tender wharf]

it follows almost immediately from the definition

so what you have is that the degree of a unit is 0

[mighty spade]

fields are entire rings by definition, so it follows immediately

[safe meteor]

ah ic ty

i knew all degree 0 were units but i kinda forgot that all units are degree 0 -_-

for polynomials

[terse crystal]

This is a general thing. Any vector space of finite dimension n, which is F^n, over a infinite field F, can’t be equal to finite union of proper subspaces. Hint: consider vectors (1,x,x^2,…,x^(n-1)) : x from F

[mighty spade]

You also need to know that both deg f and deg g are less than deg fg, which isn’t necessarily true for rings that aren’t entire.

[safe meteor]

entire?

[mighty spade]

Integral

[safe meteor]

i dont follow

oh ic

[mighty spade]

integral domains

[safe meteor]

i was actually wondering why integral domains were used with warning for the whole polynomial/factorisation topic i guess thats a good way to justify it

[mighty spade]

Because if you have f*g equals 0 for f and g non-zero

then obviously the degree is lower

[mighty spade]

yeah

[chilly ocean]

a space vector by definition is an isomorphism of the characteristic field $F_{n}$, but it can also be quadratic in $F_{2}$ since being "finite" it is composed of quadratic unions.

This is not true for a Vector subspace, in general it is a technical contradiction (since it is not finite), you can prove that F_{2}=0 "for example" at each point of a Vector-subspace (where the latter does not have finite unions)

[coral spindle]

But the only isomorphism of the field F_p is the trivial one

Wat

[tribal moss]

(I'm not properly awake, ignore me).

[delicate orchid]

Damn it tropo. I thought we’d finally found a space vector…

[errant shadow]

cool, thanks! I'll write it out.

[delicate orchid]

le vandermonde determinant has arrived or some such

[terse crystal]

Exactly

[open sluice]

hey i did my linear algebra final project on this

[chilly ocean]

@coral spindle but why is it trivial?

[delicate orchid]

a finite field of size p^k has exactly k automorphisms - specifically powers of the frobenius automorphism, here k = 1 so the only automorphism must be trivial

[chilly ocean]

I am not considering k, only p. Here the isomorphism is ordinary. Now in $k$ the isomorphism is not so trivial

[delicate orchid]

what

I'm going to presume there's some sort of language barrier here

[south patrol]

Nice method

There's also a nice geometric ish method I know uh

[terse crystal]

What is it?

[south patrol]

So suppose we have V our finite dim vector space and proper subspaces U_1,...,U_k. Let's assume that there are no containments between the U_i or we can just chuck one out lol. Then pick v in V \ U_1 and w in U_1 \ (U_2 u ... u U_n).

Consider the affine line L through v in the direction of w

Now U_1 doesn't intersect L (since if v + λw is in U_1, so is w)

[delicate orchid]

this was the proof strategy I was trying to guide them down I think

[south patrol]

oh lol

and then you show intersection with U_i has cardinality <= 1 by similar arguments

What's nice about this is uh i think it shows that this is impossible even with finite fields, provided like

k <= size of field

[delicate orchid]

true

[chilly ocean]

@delicate orchid I will try to explain, the only trivial or ordinary isomorphism is true for F_{p}\otimes{}F_{2} (where F_{2} is a quadratic field and finite in F_{p}) here is true the trivial-isomorphism under a finite-dimensional vector-space V.
In k the Isomorphism is not trivial, here for example the Abel-Jacobi map emerges.

[delicate orchid]

I don't know what you mean "in k". k is an integer

[chilly ocean]

Yes it can be integer, but under the isomorphism of an Abel-Jacobi map, that is why I affirm that in k the isomorphism is not trivial

[delicate orchid]

you used the phrase "in k" again. That is nonsense. k is just a number. You asked why the finite field F_p has no non-trivial field automorphisms (well, that's what I presume you meant by "is an isomorphism of the characteristic field", because a priori that is nonsense), and I explained using a fairly elementary result in Galois theory.

I do not and will never care about whatever the Abel-Jacobi map is

[chilly ocean]

This proof has been used in various contexts of the algebraic geometry

[delicate orchid]

I don't care. A vector space is not "by definition is an isomorphism of the characteristic field", it is by definition a module over a field.

Being generous and assuming by "an isomorphism of the field" you meant "an automorphism", then this is obviously false. As F_p has no non-trivial automorphisms and yet has many vector spaces over it

[rotund aurora]

Wasnt the Abel Jacobi map sending sums of points of an elliptic curve to the divisor class or something

[delicate orchid]

is it relevant to the discussion of vector spaces over finite fields and is there an integer associated with it

well

relevant to the definition of vector spaces over finite fields

[rotund aurora]

I dont really know about it, but when I breafly read about it, it was all over C

a formal sum is principal if the p_i vanish of order n_i (if the n_i are negative then it will be a pole of the respective order, or the other way around idk) for some meromorphic function on X

[chilly ocean]

This does not make sense, a characteristic field is by definition a vector field. Since the character is a polynomial.

@rotund aurora defines the Abel-Jacobi map (there are many people with new proofs like Voisin-Sacca for example)?

@delicate orchid What university are you from? ha ha ha

[delicate orchid]

could you link me a source which defines a vector field

[south patrol]

a characteristic field is a section of the tangent bundle of a manifold

[chilly ocean]

@south patrol

exactly

[delicate orchid]

right, ok then I can see that they're vector fields

now i must reiterate: what does this have to do with the automorphisms of F_p

[rocky cloak]

Are there any more properties? If not isn't that just a vector field?

[chilly ocean]

[south patrol]

Bruh lol I was just joking

To me char field is meaningless or would refer to the prime subfield

[rocky cloak]

Bamboozled again

High level trolling

[delicate orchid]

that teaches me to simply go "oh yeah I guess vector fields do kind of look like that lol"

not making the connection that we're talking about a completely different kind of field like the moron I am

[rocky cloak]

English really shot themselves in the foot with that one

The chat could have been called groups-rings-bodys

[chilly ocean]

@rocky cloak I don't understand???

[elder wave]

german moment

[rocky cloak]

Neither do I man, neither do I

[open sluice]

🇯🇵 does this too, I think because it's literally translated from 🇩🇪
群 (gun) for group
環 (wa) for ring
体 (tai) for body

[teal vessel]

"prove that ZxZ is not cyclic"

can't I just say that ZxZ is all lattice points on the cartesian plane, and since the identity of the group is the origin, and any generator I wish to employ must be a single lattice point, each potential generator can only generate a particular "line" of elements since for any two points on the cartesian plane, they must be colinear, and I need something that is non-linear in order to describe the plane?

or, I suppose another way to say it, the dimension of the associated vector space is 2, and I don't have enough linearly independent basis vectors to span the space if I only get to define one generator

[lusty marlin]

Yeah, this argument makes sense(in an 'intuitive', hand-wavy manner)

[lusty marlin]

What vector space?

[rocky cloak]

A lot of languages take their terminology from German. 🇳🇴 is among them for example

[crystal turtle]

Not super precise

[crystal turtle]

Also not a vector space

[lusty marlin]

Exactly

[delicate orchid]

I kinda like this argument lol

[crystal turtle]

Intuitively I have no issues with it, that's a good way to think about it

[delicate orchid]

I mean not even intuitively, it shows that Z^2 must be 2-generated as a Z-module

[lusty marlin]

It hasn't been 'fleshed out' properly, that's all

[delicate orchid]

I ain't got no flesh it's all robotics on this mofo

[teal vessel]

sorry, don't have the word for "vector construction made with a scalar ring instead of scalar field" if it exists

[delicate orchid]

free module

[coral spindle]

Lattice also works in this case.

[teal vessel]

like, this description makes perfect sense to me, and it's good I'm not barking up the wrong tree, but it's frustrating when I don't have the language to formalize the concept lol

[terse crystal]

Ring is かん not わ btw…

[open sluice]

fuck I forgot

i did actually find out it was kan a few days ago because i was watching some videos

[terse crystal]

Yeah we use 音読み here

[open sluice]

o

[coral spindle]

Never seen someone specifically write that out just to say on'yomi

[terse crystal]

Didn’t know it’s from German. I guess a lot are from German? Like ideal of a ring comes from number theory by a famous German mathematician ? I forgot his name

[coral spindle]

absolute nerd

lmao

[south patrol]

German

[elder wave]

I was gonna type some patriotic germany nr1 thing but ended up having second thoughts

[delicate orchid]

ich liebe deutschland jawohl!!!

[rocky cloak]

Dedekind I think

[south patrol]

Hm quick q so like uh so like

[coral spindle]

Wikipedia says it was Kummer

https://en.wikipedia.org/wiki/Ideal_(ring_theory)#History

[rocky cloak]

Could call that a combined effort I guess. Though I guess Kummer came up with the name

[south patrol]

Let $D_i$ be divison rings over $\mathbb R$ and $n_i \ge 1$ some pos integers and $A := \prod_i M_{n_i}(D_i)}$ I know any f.g. module $V$ over $A$ is isomorphic to a direct sum of $D_i^{n_i}$ where basically $M_{n_i}(D_i)$ acts on $D_i^{n_i}$ by matrix multiplication and everything else acts trivially. Is there any simpler way of proving this than what I currently have: $1$) show that modules over a product of rings are in correspondence with tuples of modules over each ring (i.e. $(R \times S)$-Mod $\simeq $R$Mod $\times S$Mod) and then $2$) show that $M_n(D)$Mod $\simeq $D$Mod where $\mathbb D^{n}$ corresponds to $D$, so that every module over this matrix ring can be decomposed as a direct sum of copies of $D^n$

[cloud walrus]

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

[delicate orchid]

https://www.youtube.com/shorts/YxR7xOUKt9s

[south patrol]

I suppose the first step is unavoidable, but for the second we could just write M_n(D) as a direct sum of copies of D^n and show M_n(D) is semisimple directly instead

[delicate orchid]

this is kind of like a converse to Artin-Wedderburn?

[south patrol]

But thenidk the fastest way to show that M_n(D) is semisimple with only simple module D^n other than using the correspondence I just wrote down

[delicate orchid]

as you've said you know that A is semi simple

[south patrol]

So I may just keep what I've got lol

Basically I am trying to write some intro to rep theory thing and trying to keep stuff as self-contained / easy as possible lol

[teal vessel]

Suppose there exists some arbitrary generator of ZxZ (a,b). because (a,b) as a generator means that all elements will be of the form (na, nb) the ratio of any elements generated by (a,b) will always be a/b. However, for all elements of ZxZ (a,b) there also is an element (b,a), and for any a != b, a/b != b/a which means that (b,a) cannot be generated by (a,b). ergo ZxZ cannot be generated by a single arbitrary generator (a,b). (also if a=b you can only ever have elements where a=b, but that last element of the proof is trivial)

[south patrol]

Sort of. It's a bit weird, like rather than tyring to work out the A-W decomposition using reps or whatever, I'm findg representations given the A-W decomposition

[delicate orchid]

yeah

[south patrol]

Yeah, nice

So actually you have basically like

[delicate orchid]

so you know that the regular rep decomposes nicely

hmm

[south patrol]

It's essentially the same proof as showing that Q^2 is not one dimensional

as a Q-vector space

and in fact you can use that fact for Q to deduce it for Z

[delicate orchid]

with slight retooling we could do some kind of idempotent decomposition to find them

[south patrol]

Yup

So really I just need to show M_n(D) is semisimple

I suppose you could just show the jacobson radical = 0 etc

[delicate orchid]

that's easy - submodules of a semisimple module are semi simple right

or am I smoking WEED

[delicate orchid]

using the ol $e_{\chi} = \frac1{|G|} \sum_{g \in G} \chi(g)g^{-1}$

[cloud walrus]

Wew Lads Tbh

[south patrol]

Hm sure

[south patrol]

I do wonder what the fastest way to do this is lol

would you go for Jacobson radical

Tbf, I quite like the idea of using this correspondence since it's more structural

[teal vessel]

petition to redefine the object declension of weed as used ("we" is a subject pronoun with "us" as the associated object pronoun)

[delicate orchid]

these project your big module A onto the M_n(D_i)s, via a -> e_{\chi}a
then J(A) = 0 => J(e_{\chi}A) = 0 right

[south patrol]

So maybe I'll just keep that

[south patrol]

Yup yeah

[delicate orchid]

yeah sorry I'm getting distracted trying to prove this using idempotents when you've explicitly said you want it self contained

use the jacobson

[south patrol]

Yeah and then ig i'd have to explain a bit about the Jacobson radical but that's probably nto hard

[night onyx]

I like it, you can also check that (1, 0) and (0, 1) both being multiples of (a, b) leads to a contradiction (just a slightly different approach)

[teal vessel]

"prove that Z2xZ is not isomorphic to Z" uh, Z is cyclic, Z2xZ is not.

[delicate orchid]

Z2xZ is TWICE AS BIG!!

also Z2xZ has torsion but Z does not

[night onyx]

lol also Z2 x Z contains a copy of Z2 in it, but Z doesn't

[teal vessel]

yeah, that too, I've been getting drilled with cyclic groups for this chapter

I had just proven that ZxZ2 wasn't cyclic, so I just went "a pairwise cartesian product is isomorphic to its commuted product, see earlier example"

[south patrol]

Oh yeah this is that funny thing from stack exchange I remember

Somebody was like uh

To show $\mathbb Z \times \mathbb Z \not \cong \mathbb Z$, note that $\mathrm{End}{\mathbb Z}(\mathbb Z) \cong \mathbb Z$, whilst $\mathrm{End}{\mathbb Z}(\mathbb Z^2) \cong M_2(\mathbb Z)$. The former is commutative; the latter is not.

Lol

[delicate orchid]

End_Z(Z) = M_2(Z)
HOW

[south patrol]

Anyway wew lads

Oh it's just a 2x2 matrix

you can send the generator boys anywhere

[delicate orchid]

I think you missed a ^2

[south patrol]

don't think so

2x2 matrix over Z should be a map Z^2 -> Z^2

oh sorry

[delicate orchid]

and there it is

[south patrol]

I'm blind

Lol

[cloud walrus]

potato

[south patrol]

Lmao

Anyway

Wew I have found a better way to do the matrix thing i said lol

I just forgot some stuff from my rep theory course

Lol cause we learnt a lot of in the specific case of group ring

[teal vessel]

the best thing about mathematicians is that we all know what we're talking about, but talking about math is hard so we all sound like idiots, so we can each individually feel like the smartest person in the room.

[delicate orchid]

no I was going to go on a rant about the Automorphism groups of metacyclic groups how could you do this to me

[south patrol]

Just show that M_n(D) is semisimple (since it is a sum of D^n)

and then like

f.g. module over ss ring is ss

gg

Lol

[delicate orchid]

it's semisimple because it's a direct summand of a semisimple module surely

[south patrol]

Wdym by "it"

[delicate orchid]

M_n(D)

mb

[south patrol]

Which semisimple module is it a direct summand of

[delicate orchid]

the group ring

by decomp of regular rep

[south patrol]

Of which group?

[delicate orchid]

I don't quite frankely care

any finite group

[south patrol]

Lol

Okay idk that feels kinda wrong way round though like finding a group

What I'm dealing with isn't like group reps

[delicate orchid]

oh are we not assuming A is a group ring?

right then yes it's semi simple because of what you said

[teal vessel]

so in the middle of this section on cyclic groups, I get "show that QxZ2 is not isomorphic to Q" and I'm now mildly confused as to why it's here. Q is already non-cyclic, isn't it? it's equivalent to ZxZ*. so I can just go with "they're different sizes" but still

[delicate orchid]

and I doubt there's a quicker way to show it unless J(M_n(D)) = 0 like SUPER obviously

[delicate orchid]

it is not equivalent to ZxZ*, ZxZ* is isomorphic to Z2xZ

it's the field of fractions of Z which is what I think you wanted to get at but nvm we chill

and yes Q is not cyclic

[chilly ocean]

@delicate orchid read this paper https://arxiv.org/abs/1502.04040

[delicate orchid]

I think you were getting vector spaces and vector fields confused

[teal vessel]

like I said, I always sound like an idiot.

[delicate orchid]

we all do dw

[teal vessel]

but the heart of my question still remains: why is this asked here when this is the exercises for cyclic groups??

just a general confusion

[delicate orchid]

Z2 is cyclic lol

[teal vessel]

well yeah, but QxZ2 isn't

[delicate orchid]

same torsion argument works

[teal vessel]

haven't hit torsion yet

[delicate orchid]

that's fine, I'll let you prove it

[teal vessel]

I may have accidentally used torsion in my proof for the integer variation, but I aint't got a definition to work with, so I won't find out for like 3 months at this rate

[delicate orchid]

did you use the fact that there is an element in Z2 x Z that squares to the identity but there isn't one in Z

[chilly ocean]

read the paiper,

[teal vessel]

no, just the fact that regardless of which "parity" of generator you choose, you can't hit everything, and there's no way to generate one from the other.

[delicate orchid]

oh right yeah, chapter on cyclic groups

you either get {(0, 2k-1), (1, 2k)} or {(0, 2k), (1, 2k-1)}

[teal vessel]

interesting next question to self-study: is it possible to create a cyclic group through careful cartesian products of non-cyclic groups? My intuition says no, so it's time to go back and reread the chapter and prove it.

[chilly ocean]

@delicate orchid , no I don't confuse vector space with field,

[delicate orchid]

I think you can prove it right here and right now

I believe in you

[teal vessel]

I mean, if we have two non-cyclic groups, then for any subgroup with form (0,a) there's no singular element "a" to generate the entire group. This leads to entire swaths of inaccessible members of the product according to what element is in the first spot, mutatis mutandis.

[delicate orchid]

latin... cringe!

but yeah, if you can't generate just H with an element of H how u gonna generate G x H with a single element

[teal vessel]

I'd do it in old Norse but it's not as catchy

[delicate orchid]

it's more like I just have absolutely no idea what it means because I'm not a lawyer or a doctor

[teal vessel]

"changing what must be changed.". In other words, extend this basic argument to the rest of the idea, making adjustments as needed

[delicate orchid]

if (a,b) generates GxH then (0,b) must generate H, contradiction.

[teal vessel]

Same vein as "the remainder of the proof is left as an exercise for the reader"

[delicate orchid]

now find the cases where GxH is cyclic

[teal vessel]

Well, it is a necessity that both G and H are cyclic, to start with (from the above argument and its symmetric variant). I'm making lunch so I'll be figuring it out as I go.

[oblique matrix]

I'm having trouble seeing why this is true rigorously

assuming commutativity

(p1, p2)=1

I can see that ord_p(x1x2)<=p1^a1 p2^a2 clearly

but not the other direction

ignore the subsript p

I'm assuming this is an abelian group

[delicate orchid]

I'm going to just say ord(x) = a and ord(y) = b
if there was some smaller number, say d, such that (xy)^d = x^dy^d = 1, then d would have to be a multiple of both a and b, and since they're coprime, the smallest number that is both a multiple of a and b is just ab

one moment, slight technicality I missed

[oblique matrix]

I'm having trouble seeing why x^dy^d=1 implies that d=ka and d= lb

[delicate orchid]

that's the technicality I missed

[oblique matrix]

certainly if x^d=1 and y^d=1 individually

[delicate orchid]

yeah cause this doesn't hold actually

[oblique matrix]

oh what

why does it hold in Z_p?

[delicate orchid]

maybe I'm smoking weed

[delicate orchid]

take a divisor d of ab and assume (xy)^r = x^ry^r = 1
then 1 = 1^a = x^(ra)y^(rb) = y^rb
this implies that a divides rb, but since a and b are coprime, we have to have that a divides r. Identically we obtain that b must divide r as well, so ab must divide r

thus r = ab

there we go

[oblique matrix]

from wikipedia on order (group theory)

[delicate orchid]

we're in an abelian group

[oblique matrix]

yes

[delicate orchid]

of course it's completely arbitary in a non-abelian group

<x, y | x^a = y^b = (xy)^k = 1> gives you a group where x,y, xy are whatever order you want

end of the third line is the result you're basically showing here

alternatively we could do something really funny and embed everything into some symmetric group and then the result follows from cycle length results

[delicate orchid]

altenratively again, consider the orders of the groups C_a = <x>, C_b = <y>, C_a x C_b, because (a,b) = 1 there is an isomorphism from C_a x C_b to C_ab = <z> sending xy to z

[oblique matrix]

shouldn't it be y^ra and b divides ra? where do we use that x and y are coprime?

this looks like it works as long as the gcds are relatively prime

oh never mind we don't need x and y coprime

[delicate orchid]

we definitely do

the conclusion that a divides rb => a divides r is only true when (a,b) = 1

[oblique matrix]

(a,b)=1 but not (x,y)=1

[delicate orchid]

x,y are elements of a group

not a gcd domain

[oblique matrix]

right I was thinking of Z_p as an example

and then I realized (x,y)=1 wasn't necessary for Z_p

so it works in any abelian group

[delicate orchid]

Z_p is a field so gcd is pretty boring on it

[sly crescent]

Z_p isn’t a field

[delicate orchid]

Z/pZ is a field

I understand the notation is overloaded but we were clearly talking about cyclic groups

and besides local rings are basically fields ;3

[hidden haven]

What about completions

[delicate orchid]

slow your roll...

[teal vessel]

some general musings I haven't worked precisely through yet, but an example of a potential cyclic group is if the orders of the "factor" groups are finite and coprime. both halves of the element would cycle through their respective cycles until they eventually meet up, and since they're coprime, until you reach the LCM (which is the product of the respective orders) you wouldn't have any repeat elements. I think that should be enough elements to satisfy AxB

[delicate orchid]

yeah this is correct

and you know that each of the factor groups is cyclic so

[teal vessel]

I'm not sure if that's the only way it can happen (because I haven't written it down to double check) but this seems to be the best setup for it. If they weren't coprime you wouldn't have enough members. I think having infinite order would cause problems as well.

[delicate orchid]

you yourself have shown that Z/2Z x Z isn't cyclic

[teal vessel]

I have, which shows that it needn't be cyclic, but I'm just making sure in my mind that it cannot be cyclic if one or both members has infinite order.

[delicate orchid]

I don't understand how the proof would change if you swapped out 2 for any n

[teal vessel]

it shouldn't. I'm also overthinking and conveniently forgetting that all cyclic groups are isomorphic by order

[chilly ocean]

wow, nice to meet you phoenix from plato

[teal vessel]

I'm notoriously bad at going "this is a group, I'm still learning groups, therefore there may be possibilities I haven't yet considered" instead of "this is a group, if it satisfies the conditions, it's a group of X variety, otherwise, it isn't, all other properties are either irrelevant or derivable from the defining conditions"

[chilly ocean]

just comprehend basic properties of isomorphism lol, and it wont require theorem proofs for your attempts at making exceptions

[white oxide]

yo, if I'm considering $M = \mathbb{R}[x]/\bigl((x - 1)^2(x^2 + 2)\bigl)$ and I want to show that $M(x^2 + 2) \simeq \mathbb{R}[x]/(x^2 + 2)$, it would suffice to construct an isomorphism $\varphi: \mathbb{R}[x] \to \bigl((x - 1)^2\bigl)$ (since $\bigl((x - 1)^2\bigl) = M(x^2 + 2)$ by $f(x) \mapsto f(x)(x - 1)^2$ right?

[cloud walrus]

okeyokay

[white oxide]

and show that it has kernel $(x^2 + 2)$

[cloud walrus]

okeyokay

[south patrol]

What does your notation M(a) mean , a-torsion?

[white oxide]

${m \in M \mid (x^2 + 2)^i m = 0 \text{ for some $i \geq 0$}}$

[cloud walrus]

okeyokay

[white oxide]

i'm a fucking moron tho, assuming that that was the right map I didn't even show that it had kernel (x^2 + 2) on the exam lmfao

but hey at least i showed it was an epimorphism

hopefully the grader knows where i was headed and knew i goofed

[solemn dew]

sucks when you find out your mistakes after the exam haha

[white oxide]

yeah so ANNOYING

also there was a question about showing that if G is a nonabelian group of order 10, then there's only one Sylow 5-group, so I assumed for contradiction that there was another, then they must be conjugate (gHg^{-1} = K) and somehow deduced that either g was in H or in K, so H = K, was wondering if that's the right strategy?

[delicate orchid]

this group is isomorphic to D_10

but

number of sylow 5-subgroups is 1 mod 5 and divides 10

[white oxide]

oh

fuck

[delicate orchid]

so there's one sylow 5-subgroup, and the number of sylow subgroups is equal to the index of the normaliser

so it's normal

[rocky cloak]

I mean if it worked it worked

[delicate orchid]

^

a proof is a proof

[white oxide]

true but i kinda vaguely said that HK = G without justifying it, then deduced that either g is in H or g is in K without justifying it as well, so...

[rocky cloak]

If H and K are distinct, they make up 9 out of 10 elements

[delicate orchid]

9 out of 10 elements recommend HK

[rocky cloak]

So I guess there is one pesky last element that needs to be argued for

[white oxide]

ok cool i said they must be distinct lol

[rocky cloak]

But yeah, seems like this should be a fine argument. How detailed you need to explain it might depend on the grader though

Either way, I say good job

[white oxide]

cool! thanks :)

[delicate orchid]

the groups of order 10 are C_10.... and D_10.... the result follows....

100 marks

[fossil shore]

Any books or resources to learn how to work with group presentations? E.g. how to find all subgroups of a given group presentation?

[delicate orchid]

pretty sure that's undecidable in general

if you have finitely presented you might be able to do Todd-Coexter

[fossil shore]

I see. Thank you

[delicate orchid]

I don't have any book recommendations unfortunately

[barren sierra]

is there a reason you want to do this?

like is there a larger goal?

because like Wew mentioned, it's a really really hard problem

[safe meteor]

why does showing N(2) does not divide N(1+sqrt(-5)) and N(1-sqrt(-5))

imply that 2 does not divide 1+sqrt-5 and 1-sqrt-5

this is the norm function for reference

oh ic i assume im missing some basic number theory is it cuz if a|b then a^2|b^2? and just use the contrapositive of that? or is there some other reason

[lusty marlin]

We know that Sₙ is generated by the transpositions (i i+1), where 1 ≤ i ≤ n-1.
But how do we actually go about writing some arbitrary element of Sₙ as a product of such transpositions?

Context: This came up when I needed to write an element of Sₙ (when examining it as a finite reflection group) as a product of simple reflections (i i+1).

[topaz solar]

Well, what’s (1 2)(2 3)

And the (2 3)(1 2), idc which is which because both are useful here

[lusty marlin]

(1 2 3)

[lusty marlin]

(3 2 1)

[rocky cloak]

The first thing you probably wanna do is write an arbitrary transposition as a product of these, hint ||what's (i, i+1)(i+1, i+2)(i, i+1)||

After that you can sort of think about how you'd sort a full bookshelf. Like maybe you'd start by moving the book you want at position 1 there, then you'd move the book you want at position 2, etc until the bookshelf is sorted

[rocky cloak]

The norm function is multiplicative, so N(ab) = N(a)N(b). So if 2 was on the form a(1 + sqrt-5) then N(2) would be off the form N(a)N(1+sqrt-5)

[safe meteor]

i follow what your saying but i still

dont quite see how that implies 2 does not divide 1+sqrt-5

im clearly missing something obvious i just dont know what...

[rocky cloak]

So imagine 2 = a(1+sqrt-5).

Taking the norm in both sides gives

N(2) = N(a)N(1+sqrt-5)

So

4 = N(a)6

So if 2 was a multiple of 1+sqrt-5, then 4 would be a multiple of 6

[chilly ocean]

Why does a Chow-ring associate an arithmetic measure for a subscheme?

This is producing linear-bundle of class $\mathbb{P}^{1}-$ at each point of the ring-Chow ?

[karmic moat]

ok so i totally forgot to finish this problem lol, is this right?

we have $ra + 1 = bs$ implies $ra - bs = 1$, multiplying by both sides gives
[ \begin{aligned}
x + y &= (ar-bs)(x+y)\
&= arx + ary - bsx - bsy\
&= ary - bsx
\end{aligned} ]

[cloud walrus]

ana(functor)mono(morphism)

[karmic moat]

ig i could change the negative to positive since -bs = 0 mod b too

and then you can map bsx + ary to (bsx, ary) and x + y to (x, y) giving a bijection between <x> (+) <y> and <x+y>?

ig a shorter way is since R is a PID we have bezout's

so a and b coprime implies there exists r, s such that ar+bs = 1

then you get x + y = bsx + ary

[slate forge]

How to prove that P in A[X] is irreducible with A integral domain, c an invertible element, d whatever element iff P(cX+d) is irreducible

I would love a hint plz

[karmic moat]

ig alternatively you have that Ann(<x+y>) = Ann(x) cap Ann(y) = abR = Ann(M) so <x+y> cong M?

but idk how to show equality

as opposed to isomorphism

[karmic moat]

this is true bc <x> (+) <y> and <x+y> are cyclic

[rocky cloak]

Yeah, so what I was getting at was that

bs(x + y) = bsx = (ra+1)x = x

So the module generated by x+y contains x (and therefore also y)

[karmic moat]

ohhhhh ok

[safe meteor]

ah ic ty

[drifting plover]

Is there a formula for the number of matrices in SL(2, Fq) with any given trace? This seems kinda random to me but the numbers seem to be q(q-1) or q^2 or q(q+1):

[coral spindle]

Cool question

[delicate orchid]

agreed

[coral spindle]

I have to say that the answer seems to be related to the order of SL_2(F_q) which is q(q+1)(q-1)

[drifting plover]

Yeah well, the sum of these numbers is q(q+1)(q-1) 😆

[delicate orchid]

almost like trace is constant on conjugacy classes or something

[coral spindle]

Those multiplicands are to do with the maximal tori of SL_2

[drifting plover]

Dammit, I'm bad at this stuff

[coral spindle]

q+1 is the nonsplit torus, and q-1 is the split torus

q relates to the unipotent radical of the borel, all having trace 2

[delicate orchid]

it's ok homedawg I nearly said "constant trace matrices form a subgroup" which is so false it's unreal

[coral spindle]

I think

[delicate orchid]

the relation to parabolic subgroups is cool though

[coral spindle]

There's gotta be a way to use the fact that every element has a unipotent and semisimple part aghghahf

I can't see it yet

[delicate orchid]

well we know by orbit stabiliser that the number of matrices with blah blah trace have to divide the order of the group

[coral spindle]

Nice observation

[delicate orchid]

I agree with the torus nonsense. There's a copy of p_-^{1+2} in the non-split case and the split case is obvious

wait is that true. Probably

[coral spindle]

Yeah I've studied the nonsplit case in this particular case

Related to the calculations I've been doing, funnily enough

Le epic McKay

anyway

Hm I don't know what the trace would actually be in the nonsplit torus. Any idea wew?

[drifting plover]

I find it interesting that these terminologies (Borel, Cartan, split, nonsplit) are used in things like SL(n,p)

Bruh, they're finite groups

[coral spindle]

Well yes, it's a finite group of Lie type

You can use all the junk from SL(n, \bar F_p) in this finite group

[drifting plover]

Yeah I haven't had the chance to learn these things properly unfortunately :/

[coral spindle]

Join the club

[drifting plover]

Split = diagonalisable, I'm guessing?

[coral spindle]

Yeah that's pretty much it. There's some details.

[drifting plover]

So like Cartan or something

[coral spindle]

No so we're talking about a torus

a torus is split, not an element

[drifting plover]

Right... so somehow the q(q-1) elements form a torus?

[coral spindle]

No it's just q-1 elements

q elements form upper unitriangular elements. That's the unipotent radical of the Borel

there are q-1 elements in the split torus. Those are just diagonal.

Then there are q+1 elements that remain. They're harder to spot.

Give me five minutes. Bonnafé has a book on representations of SL_2(q) where he has an exercise where you describe the nonsplit torus.

[delicate orchid]

if it's what I think it is it's in James-Liebeck as an exercise

[coral spindle]

Yeah this gives us all we need!

[delicate orchid]

one mo

[delicate orchid]

the exercise is describing reps of SL(2, q) and you lift from subgroups of order q+1

[coral spindle]

The trace should be surjective from F_q^2 onto F_q right?

[delicate orchid]

which sounds a lot like what we're doing

[coral spindle]

It sounds like Harish-Chandra induction lol

[delicate orchid]

sure

[coral spindle]

You inflate from a torus to a borel, then induce to the whole group

that's Harish-Chandra

right?

[delicate orchid]

the elements of order q+1 are of the form
\begin{pmatrix}
0 & 1 \ 1 & r+r^{-1}
\end{pmatrix}
with r \in F_q^2-F^q
does this look like the torus to you?

[cloud walrus]

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

[delicate orchid]

fuck off my latex is divine

[coral spindle]

Wtf D:

[slate forge]

Up

[drifting plover]

Suppose P is reducible, so P = QR for some Q and R, then P(cX + d) = Q(cX + d)R(cX + d), so P(cX + d) is reducible, and vice versa?

[slate forge]

Wow im incredibly dumb, thank u

[drifting plover]

Oh I guess my code is wrong for q not prime lol

[last spoke]

Is it important to define lattices

[coral spindle]

nothing is important; we are all going to become dust

Please elaborate on what your question means.

[last spoke]

Doing number theory shenanigans, lecturer has decided to skip defining lattices

And modules but that's old news

[slate forge]

Define dust.

[coral spindle]

So it sounds like it's not important for that lecturer

[last spoke]

Yea but like, idk if I'm gonna be fucked later lol

[coral spindle]

[last spoke]

I mean I'm lost now lol

[karmic beacon]

Assuming that a finite group (ℤ/nℤ)* has a subgroup (generated by α) of large prime order p, is it possible that discrete logarithm with base α is still easy (for some n = k p + 1)?

[last spoke]

Well he's using them now so idk

[drifting plover]

Which exercise is this?

I guess I still don't quite get it: you have a bunch of tori that are subgroups, but these constant trace sets are not subgroups at all, and I don't see how they're related

I have a copy of Bonnafe, and from what I'm reading I can determine the sizes of the conjugacy classes based on whether they're in the image of the d and d' maps, but that still doesn't tell me which conjugacy classes have the same trace so I can add them up

[drifting plover]

How did you get this? I can't find seem to find any such element e.g. in SL(2,5) :/

[terse crystal]

Can you guys give me some examples of normal subgroups of GLn(Z) other than SLn(Z)

[rotund aurora]

The center

[drifting plover]

Is SLn(Fq) the unique index q-1 subgroup of GLn(Fq)?

[terse crystal]

Thanks. I can’t think of others. Is this classified? Like all normal subgroups of GLn(Z)

[rotund aurora]

no idea, but definitely an interesting question

[terse crystal]

Do you have recommendations of books about general / special linear groups? I am interested in both general case, over a domain or something, and over finite field particularly.

[coral shale]

Best way to hunt is probably kernel of actions.
Eg. action on the x-coordinates of all points

[delicate orchid]

The trivial group

[terse crystal]

I see, thank you

[coral shale]

i was tryna think of non-obvious ones but i cant

[delicate orchid]

Yeah there aren’t many normal subgroups in GL over a finite field so makes sense that there aren’t many over Z too

[errant shadow]

y'all, is it necessary for the left inverse of a function to be surjective?

[coral spindle]

Yes

Why is this in #groups-rings-fields ?

[errant shadow]

uhh, I got this question while doing something else in abstract algebra so I accidentally ended up putting it here. My bad.

[coral spindle]

I assumed there was gonna be more to it

[coral shale]

dont u get it after writing the defns down

[drifting plover]

I figured this out: turns out Tr(k) occurs q^2 times for k = 2 and k = q - 2; it occurs q(q+1) times precisely if k^2-4 is a quadratic residue mod p; it occurs q(q-1) times for everything else

Thanks for the help!

[coral spindle]

Nicely done

[delicate orchid]

me staring at the lists of numbers until God tells me what they mean

[coral spindle]

inshallah the funny numbers have a funny pattern

[delicate orchid]

funny sequence I'm looking at atm

[drifting plover]

Yeah I had that intuition that it has something to do with quadratic residues, but I wasn’t sure exactly what until I proved it

[delicate orchid]

found it.

[coral spindle]

Bruh

[delicate orchid]

yeah that's the one

[coral spindle]

[delicate orchid]

index of the normaliser of some subgroup of S_{p^2}

[coral spindle]

Gigabrain wtf

[delicate orchid]

that's not the right expression lemme think

[crystal turtle]

That would just be (p-1)^2