groups-rings-fields

[coral spindle]

I like "a first course in abstract algebra" by Fraleigh. I just googled "fraleigh first course in abstract algebra pdf" and got a pdf as the first result.

It's that easy

[warm wyvern]

stare at it??

[left estuary]

Use it to understand the definition of a maximal ideal prime ideal etc

[coral spindle]

There's no definition there

[warm wyvern]

then what?

[coral spindle]

You might as well say that a sofa is "some kind of object"

It says nothing

[lavish spoke]

i love how mathematicians just upload whole textbooks onto their faculty pages

[left estuary]

Solve problems with it

[warm wyvern]

I am so confused

[chilly ocean]

l3athalgaming did you take any of the advice you got from that one time?

[warm wyvern]

why do you refuse to learn from a book??

[left estuary]

Yes

[formal ermine]

chixen moment

[chilly ocean]

such as?

iirc most people were telling you to, well, get up to speed on proof stuff and to read books

[left estuary]

Yeah and I tried doing that

[coral spindle]

I'm not really interested in continuing this discussion: two people have given you book recommendations.

[warm wyvern]

do aluffi

it got categories

[chilly ocean]

yeah it's not a month long endeavor. work harder

[formal ermine]

unterstufenkategorientheoretikermoment

[left estuary]

But at the very least compare my knowledge now to then that’s at least something worth noting right?

[chilly ocean]

german language moment

[warm wyvern]

wat

I...

[coral spindle]

Idk why you guys are bothering rn ngl

[warm wyvern]

ig?

[chilly ocean]

no not really to be honest

[warm wyvern]

it's entertaining

[left estuary]

Fair enough

[chilly ocean]

we're not trying to haze you lmao we're just trying to make sure you learn things well

and properly

[left estuary]

Ye I get that

[warm wyvern]

for reals doe, just do everyone (including yourself) a favour and pick up artin or smth

[left estuary]

I have it open rn

[warm wyvern]

nice

[left estuary]

You think there are audio books for this stuff I don’t have much time to sit down and read it but I can listen to it as I go

[coral spindle]

There are not.

[left estuary]

Rip

[formal ermine]

learning math is 50% reading 50% doing exercises

[left estuary]

Man I’m going to have to take more out of my free time then that sucks

[warm wyvern]

...

mofo trying to read math books as if they're light novels lmfao

[formal ermine]

you don't?

[delicate orchid]

devvystation

[warm wyvern]

I do, but I'm built different

[left estuary]

Built with Linux

[warm wyvern]

for the record, this is a joke

[coral spindle]

Maybe I should say something about your approach so far. You seemed to think that "a special subset of a ring" was a definition of an ideal. In fact that is not a definition at all, but a vague description that deliberately obscures detail to get to a different point.

In mathematics, we are not studying objects like in e.g. physics, where the definition of electricity for example must start out with a vague description. In mathematics, we have explicit, precise definitions, that leave nothing up to interpretation or opinion. There is an explicit list of properties that defines what it means to be an ideal.

[chilly ocean]

Ideal? That sounds perfect

[delicate bloom]

what's your goal for learning math?

[coral spindle]

#algebraic-geometry message

[warm urchin]

is there an intuitive way to think about this lemma? [ E/R is the set of orbits]

[coral spindle]

Christ, if you find out an intuitive explanation for Burnside, please tell me

It does become a bit clearer once you've done some rep theory

But it's really not obvious at all

[warm urchin]

ah rip , indeed it did feel a bit random but ile try to find some applications for it somewhere

thanks anw

[delicate orchid]

Wtf is this notation

[warm urchin]

ask my professor

[karmic moat]

why is [0] not in the lattice of subgroups of C12/C2

[delicate orchid]

I’ll try and think of a good way to think about it

It’s definitely less intuitive than orb-stab for sure

[delicate orchid]

[2] gets mapped to [0] by the quotient if I’m reading the notation correctly

Never mind I wasn’t

[coral spindle]

the image of [0] is the same as the image of [6]

[karmic moat]

oh right

[warm wyvern]

LOLLLL

[delicate orchid]

Eh I had the right idea lol

[warm wyvern]

"lemme just learn all of algebraic geometry over the weekend rq"

https://tenor.com/view/gigachad-chad-gif-20773266

[coral spindle]

Oh you're looking for applications! OK

Applications of Burnside's are actually quite easy to find. I'll give one example

Let's say you want to find the number of simple graphs of order 5, let's say

these correspond colourings of the edges of the complete graph K_5 by two colours (think of one colour as meaning 'in the graph' and the other meaning 'not in the graph')

However, if we ignore the symmetry in K_5, we'll be counting a lot of duplicates.

So we act the symmetric group Sym(5) on K_5, and then the number of simple graphs of order 5 is then the number of orbits of this action

Burnside then gives you a really nice way of calculating this number: you look at each element of the group (there are 5! = 120, but you can look at conjugates, so it's much easier) and look at their fixed points

[elder wave]

REAL

[warm urchin]

Oh? So its also a way of counting thats useful when considering symmetry

[formal ermine]

lmaoo

[warm urchin]

Not entirely sure how it does that exactly ile think about it.

[delicate orchid]

Fixed points are points that kind of “ignore the symmetry” if you want to think about it like that I suppose

[warm urchin]

Oh , thats true

[left estuary]

There is no goal I’ve already learned the majority of what classes I will take there’s very little that I need to learn in order to pass my math classes for the future I just do this as a hobby

[glossy crag]

Why does this hold? O here is an order of K, i.e. subring of the integers of K that is free of rank [K:Q].

[coral spindle]

In a bit I’ll explain more

I’ll give wnother example

[pallid oracle]

One possible reasoning is that C2 x C2 is a quotient of Q8 (Q8 is a nonsplit extension of C2 by C2 x C2).

So there is a map from Z[Q8] to Z[C2 x C2] coming from the homomorphism Q8 --> C2 x C2. This hom sends -1 to 1.

[coral spindle]

Have you ever done a combinatorics problem like this:

We have 3 red and 3 blue beads. How many circular necklaces can we make that use all of these beads?
The idea is that you cannot simply count the number of ways there are of arranging the beads in a line. Instead, there are symmetries (here rotation and reflection) that mean that we consider certain arrangements of beads in a line as 'the same,' which we might say more precisely as being in the same orbit. This question could use Burnside's lemma with the action of a dihedral group to produce the answer.

[white yoke]

Ah thanks

[warm urchin]

yeah i got the basic idea behind it now , thats a pretty good example ,i also found how its applied to aligning 8 rooks on a chess board so that no 2 attack each other which finally linked the idea for me.

didnt expect this to pop up in so many combinatorial problems lol

thanks boytjie and wew

[median finch]

Anyone know what this is?

[formal ermine]

context?

without any context I'd assume it's the fock space?

[chilly ocean]

what is meant by linearly ordered subset of a group, in definition 3.1

[hidden haven]

I think you can ignore the linearly ordered in that definition. They just mean that the basis should be an ordered set

But in the next theorem they just take a basis to be a set so idk

Probably just think of the analogy with vector spaces

You can require bases to be ordered or unordered

They are requiring them to be ordered here

[chilly ocean]

what's the difference tho? if it's ordered

doesn't seem to change anything

[hidden haven]

Yeah nothing really

In vector spaces the order is useful when you wanna represent maps using matrices

Because then there's a canonical ordering on the rows and columns

Maybe they wanna do something similar here

Probably ignore for now and see if it is used somewhere later

[coral spindle]

An example of what Moldi is talking about re: matrices is the following: If you choose a different ordering, then e.g. your notion of an upper triangular matrix will look very different.

Try computing a change of basis of your favourite upper triangular matrix, where the new basis simply swaps the ordering of the old one

[left estuary]

Hey I get this part now but would the next step be proving by induction?

[chilly ocean]

probably not, no

i don't see how induction gets used anywhere in this

[left estuary]

Just thinking back to a group theory question I did where I proved by induction

[formal ermine]

no

[left estuary]

so this is for proving that 1+x is a unit which is the first part im here but i dont know the next step in proving it

[chilly ocean]

i think my hint was inappropriate

without further detail at least

[left estuary]

Nah I get it cause (1+x)w=1 where w would be 1/(1+x)

[barren sierra]

That's a really good step. Have you tried doing the algebra to see what this simplifies to?

[chilly ocean]

you can do the explaining for me

[barren sierra]

I mean at this point it is literally just do the thing

[left estuary]

It’s an infinite series oh wait is converges right

[barren sierra]

that doesn't matter

in power series rings, convergence is immaterial

[left estuary]

It’s basically just 1-x+x^2…

[chilly ocean]

okay maybe i should explain

[barren sierra]

What is there to explain? It's just do the distribution and it works out. The hard part is coming up with that power series if anything.

[coral spindle]

This is a formal power series. Think of it just as a collection of symbols. There is no meaning to it converging or not, without additional information.

[left estuary]

Yeah I was just confusing it converging at 1 and equaling 1

I get it now

[chilly ocean]

the message immediately following yours

boytjie did it for me

[coral shale]

im confused how we got to formal power series

but alright.

[left estuary]

w = 1/1+x

[barren sierra]

1 / (1 + x) is not a thing in R[[x]] so we use a power series

[chilly ocean]

i was going to say i brought up the power series just as a heuristic tool for finding out what the inverse of 1 + x should be if x is a unit

we are not dealing with formal power series

in R[[x]]

also l3athal please don't ping me for messages that old

[chilly ocean]

i also, frankly, don't think this person is ready to learn what a formal power series is

let's stick to basics

[left estuary]

OH I SEE

When you start distributing it is going to 1

[left estuary]

My bad mate lol

[formal ermine]

yeah

but remember that x is nilpotent

meaning that x^n = 0 for some n

[left estuary]

Ye

That’s where the next part comes in

Going back to the original

r + n = r(1+nr^-1)

nr^-1 stays nilpotent right?

[formal ermine]

yes

[chilly ocean]

to fully elaborate on my hint: in calculus the power series for $1/(1 + x)$ is obtained by noticing that $$(1 + x)\left(\sum_{k=0}^\infty (-1)^k x^k \right) = \left(\sum_{k=0}^\infty (-1)^k x^k \right)(1 + x) = 1.$$ in a general ring the expression with the sum doesn't make sense. however, if we assume $x$ is nilpotent, then eventually all of the terms will be zero and we get a finite sum like: $$(1 + x)\left(\sum_{k=1}^N (-1)^k x^k \right) = \left(\sum_{k=1}^N (-1)^k x^k \right) (1 + x) = 1,$$ so if $x \in R$ is nilpotent then you can find an inverse for $1 + x$ by this method

[cloud walrus]

TTerra

[chilly ocean]

finite sums make sense in rings

so this is fine

apply this and you are done

[coral spindle]

I love this trick. So cool.

Just wanted to say that

[chilly ocean]

me too

[left estuary]

And because of the proof that 1 + a nilpotent element is a unit then it just becomes a unit * a unit

[chilly ocean]

one of my favorites boytjie

[left estuary]

And the product of 2 elements is an element

[celest furnace]

why does an infinite sum not make sense in an arbitrary ring?

[coral spindle]

Well it's simply not defined

[left estuary]

Therefore r+n is a unit

[coral spindle]

remember that + is literally a binary operation R x R -> R

[celest furnace]

so only accounts for finite sums?

[coral spindle]

there's no meaning to repeating an arbitrary function an infinite number of times

yeah

[formal ermine]

yes

[coral shale]

its undefined

[left estuary]

Yay I’m learning

[coral shale]

by default for ring ops?

[coral spindle]

In general, due to something called the Eilenberg–Mazur swindle, you cannot define infinite sums in all cases

If you want to think about limits and other things, then you can imbue your object (ring, group, module, whatever) with some topological structure

E.g. hilbert spaces have a metric structure

[coral spindle]

Elaborating on this, basically you say that a sum like 1 + 1 + 1 + ... cannot be defined in general in a way that is consistent with the operation '+'

[agile burrow]

eilenberg mazur is neat

[coral spindle]

It's another one of my favourite tricks haha

[chilly ocean]

1 + 1 + 1 + ... = -1/12

[coral spindle]

so true monarch

But yeah I mean

that's a good example, right

[left estuary]

I remember seeing a video on that a while back

[coral spindle]

bc that's not consistent with addition!

[coral shale]

[agile burrow]

I should find other applications of Eilenberg Mazur. I only know the one where if you have a projective module P you can find a free module F such that P + F = F

[delicate bloom]

only reason I ever heard of the swindle was that it could be used to prove that you can't tie a knot in an already knotted string and slide the knots together to "cancel out" to get an unknotted string

[coral spindle]

Someone showed me a very cool application of it to some set theory. Apparently (I forget how, exactly) you can use it to prove that if A injects into B and B injects into A then there is a bijection between the two

It might have been a surjection. I'll try to reproduce the proof sometime.

[left estuary]

Thanks for your help ima keep reading stoof and move on the the 3rd question

[formal ermine]

don't

do the exercises in your book

[agile burrow]

ah yeah, I'm reading on the wiki page that Cantor-Bernstein-Schroeder can be seen as an antecedent of Eilenberg-Mazur

[left estuary]

I’m doing both

Think of it like my current goal vs progress

[chilly ocean]

i am being swindled right now

[left estuary]

Goal is now question 3 progress is exercises in book

[toxic zephyr]

in the dihedral group, what word would you use to refer to the half that has an s and the half that doesn't? cosets?

like what would you call $s^i\langle r\rangle={s^ir^a\mid a\in\bZ}\subset D_n$

[cloud walrus]

nilpotent nix

[coral spindle]

Sure, it's a coset

I don't know of any name for this specifically

Not everything in math has a specific name

[toxic zephyr]

yeah i didn't necessarily mean for the dihedral group specifically, but more just the general term. thank you 🙂

[coral spindle]

Well they certainly are all the cosets of <r> in Dih(2n), so there we are

[pallid oracle]

If you're identifying the elements of D_n with the action they have on the plane while leaving the regular polygon of n sides invariant (this is how the dihedral group is usually introduced), the ones that don't have an s are the rotations (determinant 1), and the ones that do are reflections + rotations (determinant -1).

[next obsidian]

This is actually also algebraic!

D_n is C_n semidirect Z_2

The C_n component tells you how much it’s rotated

And the 0 or the 1 from the Z_2 tells you if you have a reflection or not

So there’s a map D_n -> Z_2 which you can consider as projection to the second component

And that’s the 1 or -1 Monchi is talking about

:D

[toxic zephyr]

oh huh that is really interesting!
yeah the original question i had was about where the cosets get mapped to by a type of function. going to ask my mentor about it tomorrow, but I've been spending the last few days trying to prove some results on my own first. dealing with the cosets has me tripping tho lol.
idk why group theory is harder for me to understand than ring theory. i feel like it should be the other way around lol

thank you both @next obsidian @pallid oracle! ❤️
people in the #groups-rings-fields channel seem really cool haha. i really hope my abstract algebra III course isn't cancelled next semester 🥲

[wooden ember]

im not sure i understand why the prufer group is injective... I'm trying to show it via Baer's criterion but struggling: a morphism from nZ to Z/p^inftyZ is just a morphism from nZ to one of the Z/p^kZ subgroups, so it's the quotient map nZ->nZ/np^kZ. However when n is coprime to p (ie when n is anything but p) I'm struggling to see how this can extend to a morphism from Z to Z/p^inftyZ

[chilly ocean]

hmmm

why is this true?

[wooden ember]

what do they mean by the notation H^u?

[chilly ocean]

seems like its ${ uhu^{-1} \vert h \in H }$

[cloud walrus]

Gev

[chilly ocean]

These elements h^{u} fall outside of G, but why can't there be such relation? is it because L is a subgroup of X which is a free product?

[wooden ember]

I think it's just because each h_i is an element of G

so that expression is a word in G * <u>

which alternates between elements of G and <u>

so cant be reduced

[chilly ocean]

or is it between elements of G and elements that are neither in G nor in <u> (because uhu^{-1}, h in G, u in <u>)

oh wait it makes it into 3 elements of <u>, G and <u>

kinda makes sense now

[pallid oracle]

Maybe I'm missing something, but I think if in your map from nZ you send n to f(n), you can extend this function to Z by sending 1 to f(n)/n. You can divide by n because it's a divisible group.

I don't remember why it's divisible though

[wooden ember]

isn't it only p-divisible

ahhh wait no that makes sense

[pallid oracle]

Well, being p-divisible is trivial but they're also divisible

[wooden ember]

yeah like if n=p^rm i take some element of Z/p^kZ, i just embed into Z/p^{k+r}Z and in there there's some element such that when multiplied by n gives me our element

or smth

okay so this was in the context of the injective hull of Z/pZ

and i think I kinda see why it is that now by Baer

so like in general if i want the injective hull of M id need some kind of direct limit that mimics this same kind of behavior no?

like id do something kind of like lim R/Ann(M_i)

or smth

cause i want every morphism from I->M to be of the form i->im for some m so i need to make my module grow so that there arent ideals that just annihilate it right?

that's probably not enough though

[pallid oracle]

You don't need to restrict to powers of p. If eg p and n are coprime, then there's a solution for x/p^k · n = 1/p^k, and so you can divide 1/p^k by n, for any natural n coprime to p. Dividing by p is also easy, so you can divide by any natural n.

But I don't know about your other questions

[wooden ember]

yeah but i mean in the context of the factors of the direct limit to allow myself to divide by n i need to go to a high enough prime power that doesnt divide the highest prime power of n no?

or multiplication by n in Z/p^kZ would just be 0

oh sorry p and n are coprime yeah

anyways im getting myself confused over this injective hull business ill get back to it later there are more important things for me to study over right now

[pallid oracle]

If it's not clear I'm just arguing that the group is divisible. So n can be any natural number. You wrote the case where p is a power of a p. I wrote the case where (p, n) = 1. Together, you can divide by any number.

I'm sorry I can't help with the more abstract questions

[wooden ember]

no i was also doing for general n, i was just looking at the divisibility within Z/p^kZ for a high enough power k so that this would hold in the prufer group

[wooden ember]

np thanks for the help

[pallid oracle]

Oh I'm so dumb I read your p^rm as p^(rm). I'm sorry lol

[warm wyvern]

can someone give me an example where this is useful or when it occurs naturally?

examples of R-algebras are also welcome

[wooden ember]

this occurs pretty naturally for algebras

it's easier to see for fields

if im looking at a field extension L\K

what we have is an embedding i:K-> L

so we can consider K as a subfield of L

and now we have a natural K-module structure on L given by your picture

ie a K-vector space

considering fields as vector spaces over their subfields is very fruitful

[lethal dune]

Is this right place to ask about spectral sequences

[coral spindle]

It's the most overloaded channel topic in the whole server 🥲

[lethal dune]

Maybe consider splitting

[warm wyvern]

moving abstract algebra to early uni and turning this into "advanced-algebra" would be gud I think

[warm wyvern]

[warm wyvern]

R as a Q-vector space

[wooden ember]

it's more useful in the setting of algebraic extensions

Namely if you quotient K[x] by an irreducible polynomial, you obtain a field extension K[x]/(f) \ K, and the dimension of L=K[x]/(f) over K is the degree of f

If you let a denote the residue class of x in this quotient we denote this extension by K(a) and call f the minimal polynomial of a over K

To see why it could be useful to consider fields as vector spaces over their subfields, try showing that every finite field has order p^n for some prime p

[warm wyvern]

@coral spindle why ?

[coral spindle]

That is part of the problem: people sometimes come here bc they think it's just advanced algebra

where 'algebra' just means... solving polynomial equations

The issue imo is more like people do group theory in the same place as homological algebra. There's not enough space to talk

[chilly radish]

Yea they don't know that's called algebraic geometry

[coral spindle]

#groups-rings-fields is a fine name

I don't know what would be a perfect splitting-up of this channel, but perhaps having a separate #algebraic-structures channel would be the start of a plan

[warm wyvern]

or maybe we can make a thread for each major area of algebra?

like one for group theory one for ring theory etc

[tender wharf]

feels overkill no?

[coral spindle]

That is simply too many channels

[wooden ember]

honestly the server already has too many channels

[coral spindle]

finite groups, infinite groups, rings, Galois theory, Lie algebras, I mean the list goes on

[wooden ember]

we just have to come to terms with the fact that math is too big to please everyone with just a few discussion channels

like this applies to the other channels too, think about probability

they just have two channels

[rotund aurora]

discussion in here is sometimes very elementary compared with some of the topics you guys often touch

but I guess its fine, idk

[coral spindle]

True, but this is (as the stats somewhat recently showed) is the most overloaded

[wooden ember]

fair enough

honestly im also never sure whether to post a question here or in #algebraic-geometry

[tribal moss]

Tbf, suddenly seeing "homological algebra" at the end of the topic list here feels like quite a gear shift.

[wooden ember]

yeah what motivated that decision?

[formal ermine]

it has algebra in its name

[wooden ember]

I usually ask homological algebra questions in #algebraic-geometry

since it's closer to commutative algebra i guess

[south patrol]

homological algebra should just go in alg top jk

[tribal moss]

Is there any use of homology in algebraic geometry? I sort of assumed it had been put here because it was considered a kind of common ground.

[wooden ember]

Oh I mean y’all would know better I just used to ask there for lack of anyone telling me to do otherwise

[tribal moss]

I'm serious, actually -- what little I know about homology connects just to alg top, and I'm unsure what else there is.

[warm wyvern]

is it really?

I wouldn't know tbh

but aluffi literally does homological algebra two sections after introducing modules lmao

[wooden ember]

Yeah idk it’s just cause I’m learning homological algebra during a comm alg class. I think it has some uses in algebraic geometry with étale cohomology or whatever

Allufi based

I imagine that’s pretty standard honestly yeah

That’s more or less the order I learnt stuff

[delicate bloom]

I guess in a simple way, genus is used to categorize algebraic varieties, like elliptic curves are genus 1 cubic curves

idk how in depth it goes past that either

[warm wyvern]

question about terminology

[tribal moss]

Pethaps I feel it's a gear shift because it's the one thing in the list where I don't even know any basic definitions ...

[wooden ember]

I mean honestly at this point isn’t there a cohomology theory just about everywhere

[tribal moss]

Don't ask me; I've never learned what a "cohomology theory" IS.

[wooden ember]

What field do you study?

[tribal moss]

None. Computer science back when I did study.

[wind steeple]

Yes of course ! It's used to define sheaf cohomology for instance

This gives you a cohomology theory for algebraic varietied

It's used to define the etale cohomology as well which is used to prove the weil conjectures for instance

[coral spindle]

Although I don't have a good understanding of it, you can rederive Riemann–Roch entirely via cohomology of sheaves

[next obsidian]

Homologvial algebra isn’t just present in algebraic geometry, it’s very present in commutative algebra

[coral spindle]

Present in noncommutative algebra too

[warm wyvern]

[chilly radish]

I'm offended by this

[elder wave]

Good

[chilly radish]

Try defining the derived category without noncomm algebra, fucking geometers have no respect smhsmh

[next obsidian]

Okay, I use a model category structure

[chilly radish]

Prove the localisation exists and has nice properties

Derived category is just an ore localisation

[solar glacier]

when i tell my non math friends im taking abstract algebra they just think im sitting there factoring polynomials all day long lol they think its the same as college alg and alg II lol so i gave up and call it group/ring theory now lol

like they go "oh i took algebra in high school" ............ lol

[tribal moss]

I've wondered if we could reduce misunderstanding by reviving the (somewhat antiquated-sounding) term "higher algebra" instead of "abstract".

[south patrol]

I imagine that would lead to some people asking about lurie lol but good point (I imagine such questions would be relatively rare anyway)

[solar glacier]

or just #group-ring-field-theory lol

[tribal moss]

Oh, looks like nLab is squatting on "higher algebra" for some weird n-categorical stuff. 😵‍💫

[formal ermine]

lurie?

[prisma ibex]

Homological methods are absolutely central in modern algebra. It's hard to do much of anything without them when studying CA, AG, AT, etc

[gleaming fable]

Hello
Let's assume the relation R in which :
∀x,y ∈ N* , xRy⇔∃k∈N s.t x = y^k
(So sorry again if my terms are not correct idk most of them in english)
Now Let's assume that a subset A⊂N* is A={2,4,8}
The question says find the intervals of minorants(aka elements m such that mRx) and I fount that they are {64^n | n∈N}
Now the next question sais find out if there was an inferior, which I believe is 1 cuz
∀x∈A, 1Rx
But the solution says that the inferior doesn't exist, do you guys know why? Or if the solution is correct from the first place?

[tribal moss]

Perhaps the real problem is that "groups, rings, fields, vector spaces, modules, etc" really ought to have a channel in "early university".

[prisma ibex]

Ehh probably not

[next obsidian]

There’s no analysis channel in early uni

[tribal moss]

Which also feels strange to me.

[next obsidian]

If you compare it to every other channel other than #proofs-and-logic analysis and algebra feel qualitatively different

Even part of the “proofs” part feels different than the other early uni channels

Also since this is the internet it invariably becomes 🇺🇸-centric

Where analysis and algebra are like, 3rd year courses

[south patrol]

Okay that explains it lol

[formal ermine]

how long is a bachelor in the us

[delicate bloom]

about 4 years

[formal ermine]

what do you do in the first 2 years

[delicate bloom]

party

[formal ermine]

lol

[chilly ocean]

first one was easy but I'm stuck on the second one

[formal ermine]

what's the serious answer

[coral shale]

in the first semester at least, thats serious

early university section encompasses a lot of first yr i suppose.

[chilly ocean]

oh wait these relations force y to be of order 20 at most

I'm dumb

[celest furnace]

A lot of General education classes (where I am, it’s at least 1/2 of your degree)

[formal ermine]

general education classes?

[elder wave]

you should swap to one of the discussion channels for that

[chilly ocean]

what is $SL(2,\mathbb{Z})$?

[cloud walrus]

Gev

[chilly ocean]

2 x 2 matrices integer entries determinant 1

thx

[agile burrow]

that's a cool group

[coral shale]

hmmm

[chilly ocean]

SL means determinant 1, 2 means 2 x 2, and Z means entries in Z

[coral shale]

thats not a group right

[chilly ocean]

to unpack the notation

[coral shale]

or is it

nvm it is

[chilly ocean]

wait

how is it a group?

the inverses

prove it

just prove it

what's the inverse of a 2 x 2 matrix

[toxic zephyr]

general Ed stuff mostly. so various history, English, social studies, art, etc. stuff alongside basic major classes like calculus.

[chilly ocean]

what if the determinant is 1

just write it down lol

[chilly ocean]

okay the determinant is 1 so its still integers

cool group indeed

[coral shale]

isnt it det +-1

or nah

[chilly ocean]

1

[coral shale]

nvm ok

[chilly ocean]

Im about to attack an exercise of presenting SL(2, Z) as an amalgamated free product

[agile burrow]

that's very cool

maybe my favorite amalgamated free product

[warm wyvern]

it's a Z-module

[chilly ocean]

hmmm

seems to work

[agile burrow]

Yeah, maybe this isn't the most illuminating proof

Still neat though

[chilly ocean]

what could I do to improve it tho?

[agile burrow]

Well from this perspective, not much. It is just rote computation

But there are lots of cool interactions between free products and actions of groups on trees and graphs

That's called Bass-Serre theory and it's worth looking into if you're interested in this stuff

Serre's book on trees is a good reference (or so I've heard, I haven't actually read through it myself)

[chilly ocean]

is it combinatorial group theory or does it diverge from it?

trees and stuff like that

[agile burrow]

I'd say it leans toward combinatorial group theory, but I'm sure there are other directions to explore

Like intersection of combinatorial and geometric group theory

[chilly ocean]

my advisor gave me these, to prepare for my undergrad thesis which will probably be in combinatorial group theory

[agile burrow]

Fun stuff, could be worth talking to your advisor about. They'd probably give better advice than me lol

[chilly ocean]

he also said there's representation theory if I'm interested but his main research is in combinatorial group theory

[agile burrow]

Rep theory is fun too, yeah. I used to know more than I do now but it always takes me like 2 or 3 runs before I start to actually understand something

[chilly ocean]

it's really confusing since I'm gonna spend so much time on my thesis, it will determine my future

[agile burrow]

I'm not too sure about interactions between rep theory and combinatorial group theory but I don't doubt that they exist

Best of luck with your thesis, I'm sure it will go well if you put in the time

[chilly ocean]

thx

I have more than a year, I hope I won't fail

what are the most active areas though?

I know I still have a lot of fthe basics to learn

[agile burrow]

I'm not the right person to ask. These are probably questions better suited for your advisor

[formal ermine]

I don't understand why every z-module is an abelian group

[tribal moss]

Every module is an abelian group.

... together with a scalar multiplication, but when scalars are from Z, there's only one way the multiplication can work.

[formal ermine]

there's only one way the multiplication can work.
why?

[tribal moss]

Because it needs to distribute, so for example 5·m = (1+1+1+1+1)·m = 1·m + 1·m + 1·m + 1·m + 1·m = m+m+m+m+m.

[warm wyvern]

I can't parse this lel

what's it saying

[warm wyvern]

Z is initial in Ring

[tribal moss]

It says that every subobject is the kernel of some morphism.

In Grp, only normal subgroups can be kernels; in Ring, kernels don't even need to be subrings.

[warm wyvern]

what does "every monomorphism in R-Mod is a kernal" mean tho

[tribal moss]

That's just the catheaded way to talk about subobjects without pointing to "subsets of elements". A subobject is identified with its injection morphism (which is always mono).

[formal ermine]

ah

[tribal moss]

For example, Q is a subgroup of R, viewed as abelian groups. In the category-theoretic phrasing we can't speak about "subset" directly, so instead we declare that it is the identity function Q -> R (which is a good homomorphism -- and indeed a monomorphism too) that "is" the subobject.

[warm wyvern]

I see

everyday passes and I hate category theory more and more

[formal ermine]

"Every field extension L/K of degree 2 is normal. Let f(x) in K[x] be irreducible with a root a in L. if a is in K then f(x) = (x - a) and the statement holds. assume a is in L\K. after making it monic f is the minimal polynomial of a over K and we get 2 = [L : K] >= [K(a) : K]= deg(f) >= 2. therefore L = K(a) and deg(f) = 2. thus f splits over L."

I don't quite understand the last part. what does "f splits over L" mean?

[warm wyvern]

thx tho

[tribal moss]

There's something to be said for this view in everyday mathematics too. Say, if you have constructed R via Dedekind cuts, and you want to say Q is a subset of R, then you end up implicitly claiming that 1/2 and { q in Q | q < 1/2 } are the same thing, which is absurd. So you end up either pretending this isn't a problem, or doing a lot of tedious footwork in your definition of R such that the rationals can still represent themselves. OR you decide that it's enough that there's an injection from Q to R which preserves all of the operations (arithmetic, ordering) we're interested in.

[warm wyvern]

I find that as nothing more than pointless pedantry tbh

but it doesn't bother me too much

coz it makes sense, I have to admit

[coral shale]

f splits over L means you can factorize f into linear factors in L[x]

so like all its roots are in L, essentially

f(x) = \prod (x-a), a in L

[formal ermine]

why does it split over L

[coral shale]

cus ur like

chopping that polynomial

up

into linear factors

?

remember polynomials are 'over' a ring

polynomials over K are polynomials in K[x]

[formal ermine]

yeah but like

how do we know that its roots lie in K

[coral shale]

oh you meant in context

[formal ermine]

yee

[coral shale]

I'm not sure i follow the argument you've copied

Let f in K[x] irreducible with root a in L.
You've shown L = K(a) ?
And L/K is degree 2?

If the above, then a must be the root of a degree 2 polynomial. Hence f is degree 2, and +-a are its roots (nvm dont think thats necessarily true)

[rotund aurora]

If f has degree two and has one root a in the field, then (x-a) divides f, therefore we can write f(x)=(x-a)(x-b), where the coefficients are in the field.

[coral shale]

yh im handwaving, thats better

[rotund aurora]

idk why the proof that rectangle cube copied uses so many words

[formal ermine]

this is the original one

[formal ermine]

yeah but if a is not in K

[coral shale]

(x-a) divides f in L[x]

then f(x) / (x-a) must also be in L[x]

[formal ermine]

ah

[rotund aurora]

What you want to prove is that if a polynomial of degree 2* f has a root in the field, then the other root will also be in the field

[formal ermine]

yeah I get it now

thank you

we want our extensions to be galois because otherwise we'd permutate some roots to themselves, right?

[coral shale]

well there are 2 things

[formal ermine]

normal cuz of fundamental theorem

[coral shale]

not normal means you're missing some roots you could've permuted to

not separable means that what would've been distinct roots are the same, and you lose some permutations as a result

[formal ermine]

yeah ok

[rotund aurora]

Galois extensions K<L allow the most freedom in terms of automorphisms. It turns out that this will allow a very rich relation between intermediate fields of the extension K<L and the subgroups of Aut(L/K)

[formal ermine]

the fundamental theorem only requires a finite normal extension though?

[rotund aurora]

Can you share what you mean by "fundamental theorem"?

[tribal furnace]

Wouldn't that be the Galois correspondence?

[next obsidian]

You need the extension to be Galois, you need separability to get almost anything

[formal ermine]

correspondance

in my lecture script it says finite normal extension

[next obsidian]

Are you only working over Q?

[formal ermine]

no

[next obsidian]

Only in char 0?

[formal ermine]

no

[next obsidian]

Well then it’s wrong lol

[coral shale]

if its not separable or not normal your galois group is smaller

[next obsidian]

You can have finite normal extensions in char p with infinitely many intermediate sub fields

[rotund aurora]

non separable stuff seems pretty interesting tbh, many pathologies

[flint crater]

So how do I calculate all possible ringhomomorphisms of Z[\sqrt{27}] -> Z/119Z? I haven't a clue where to even start. I was thinking perhaps the first homomorphism theorem somehow?

[rotund aurora]

Mmh idk but are there even any? The image of 1 will generate the whole ring, and then you have no room for sqrt 27

No?

Oh wait nvm thats irrelevant

[solar glacier]

Gaussian integers w rad 27?

[rotund aurora]

Like you send 1 to 1 and sqrt 27 to one of the roots of x^2-27 in the field Z/119Z

[flint crater]

In case you're asking the definition it's the set {a+b\sqrt{27} | a,b ∈ ℤ}

[solar glacier]

Yes ok

Gaussian integers

I’m tryna think how first iso can be applied

[rotund aurora]

So there are obviously at most 2. And I think those are indeed homomorphisms, because sqrt 27 doesnt interact with addition

[solar glacier]

Cause what are the elements of ker of the homomorphism

Just 1,-1

[rotund aurora]

This is all you need to know

[solar glacier]

Since the norm on Gaussians w rad 27 is a^2+27b^2

Forcing b to be zero

[solar glacier]

That actually makes somewhat sense

Is we want this to be in the kernel it leaves for a to be plus or minus 1

[tribal moss]

Are we sure it isn't a problem here that 27 is not square-free?

[rotund aurora]

Was gonna mention that. For some reason I thought 27 was a quadratic residue lol

[tribal moss]

3 and 119 are coprime, so 27 is a quadratic residue iff 3 is -- but 3 is not a quadratic residue (fairly quick to check using the CRT even if one doesn't remember the quadratic reciprocity law, since 119 = 7·17).

[delicate bloom]

Since $\phi(1)=1$ we also have $\phi(27)=27$, which has order $16$ in $(\bZ/119\bZ)^\times$. This forces $\phi(\sqrt{27})=\alpha$ by squaring to have $27=\phi(\sqrt{27})^2=\alpha^2$ which makes $\alpha$ have order $32$. But CRT forces order of the multiplicative group to be lcm(6, 16) = 48, and since $32 \nmid 48$, there are no ring homomorphisms.

[cloud walrus]

Merosity

[formal ermine]

the one in my lecture establishes a correspondence between separable intermediate fields and subgroups

if the extension field is separable then the requirement for intermediate fields to be separable can be dropped

that's where my confusion came from

[stuck cosmos]

Is there an analogue to polar decomposition of GL(n) using SO(1, n-1) instead of SO(n)?

[flint crater]

Wait why exactly does 27 map to 27?

[tender wharf]

phi(27) = phi(1) + phi(1) + ... + phi(1) 27 times

[flint crater]

Oh yes right right

[median pawn]

How would you check that T is a ring homomorphism?

[formal ermine]

check that it fulfills all requirements

• well defined
• T(ab) = T(a)T(b)
• T(a + b) = T(a) + T(b)

[median pawn]

T(x+y) = (x+y mod m_1,...,x+y mod m_r)

we need to show x+y mod m_j = x mod m_j + y mod m_j?

@formal ermine

[formal ermine]

yes

[median pawn]

yeah cool this is obvious

that's how addition is defined for equivalence classes, and it's well-defined

[formal ermine]

I know that if $M$ is a finitely generated torsion free $R$ module then $M$ is free and $M \cong R^n$

[formal ermine]

is the other direction true as well?

[delicate orchid]

The other direction might be even easier to show lol

[formal ermine]

ah just what I thought

thanks

[wooden ember]

So this is a super old message so dont ask me how i came back across it but now im actually grown up enough to understand this. I think i kinda get what you meant now. So like I have an injective map A-> B which gives me a bunch of injective maps on the finitely generated submodules. When I tensor I get a bunch of maps on the finitely generated submodules, and if I can show each of them is injective then the induced map on the direct limit is too?

[nocturne bone]

So I'm trying to come up with a counter-example to disprove this, but I can't find one. I think that the statement is true as n_5 = the number of Sylow 5-groups is st n_5 divides p, and p is prime, so its only divisors are 1, p

[formal ermine]

I'm having a hard time reading that

would you mind writing that as latex?

[nocturne bone]

I'm not good enough with Latex but I can definitely write it out neater

[formal ermine]

thanks

[cloud walrus]

Joe 1

[formal ermine]

you need to find n_p not n_5

[nocturne bone]

Thanks!

[formal ermine]

5 = [G : H] = |G/H| = |G|/|H| <=> |H| = p^n

so H is a p-sylow subgroup rather than a 5-sylow subgroup

[nocturne bone]

That makes sense. So then isn't |H| = p^n by definition of it being a sylow p-group?

[formal ermine]

|H| = p^n means it has index 5

but also that it's a p-sylow subgroup (by definition of those)

so you gotta find n_p and show that it's 1

[nocturne bone]

Ye that makes sense

Thank you!

[formal ermine]

np

[agile burrow]

This forward direction is only true if R is a PID. Take the ideal (2, x) in Z[x] for a counterexample.

[formal ermine]

yeah only talking about pids

mb forgor to specify that

[agile burrow]

Oh ok

[glossy crag]

I realise that by the basis theorem there is a basis e_i of L such that c_ie_i is a basis of T(L), but why should the c_ie_i then be the image of some basis e'_i of L?

[wooden ember]

Isn't this just smith normal form?

[next obsidian]

it's not that simple

because submodules of M do not correspond to submodules of M (x) N, this fails because tensor product isn't left exact

you either have to show that if an element sum m_i (x) n_i is 0 in M (x) N that there's finitely generated submodules of M and N, M' and N', such that Sum m_i (x) n_i is in M' (x) N', and is zero

which you do by adding in all the elements you need in order to write out that its zero

left adjoint commute with colimits

the easier way

[next obsidian]

it's true over a valuation ring

Fg ideals are principal (and you're an ID) => flat iff torsion-free

Combine with flat and fg over a local ring => free

hi walter :)

[agile burrow]

Oh cool, I wasn't sure about more general settings

Hi chmonkey :)

[next obsidian]

actually you can like

use a lot of theory to get the result for PIDS sort of in this vein

torsion free => flat => projective cuz finitely presented

then use projective => free

but obviously this is really dumb :)

[agile burrow]

Perhaps

[wooden ember]

right okay so once i tensor i can't consider that im taking a direct limit over submodules, but the direct limit of the tensored modules will still be the tensor of the direct limit because left adjoints commute with colimits is what youre saying?

[glossy crag]

Sure, but using just the theorem for free modules (N\subset M both finite free => exist bases x_i and c_jx_j), how do I get this?

[formal ermine]

what's an example use case of a differential ring that is not a polynomial ring

[next obsidian]

Yes this is exactly right!

And direct limits are exact

So it suffices to show that the corresponding diagram on the fg sub modules are exact (it isn’t necessarily equivalent though, but because flatness is about every tensor product in this case it is equivalent!)

[wooden ember]

unless im mistaken, T takes the free module L to a free submodule of L of the same rank so picking a basis of L and looking at the image under T there's some choice of basis in L that makes T(L) be a multiple of each basis vector

this choice corresponds to right multiplication by a unimodular matrix in smith normal form

[glossy crag]

That's correct, I'm asking why the preimages e'_i then have to be a basis of L.

[wooden ember]

no we're starting with the e'_i

[glossy crag]

Oh shit

[wooden ember]

looking at their images

[glossy crag]

nvm

c_ie_i is in T(L), lol

[wooden ember]

yeah

[glossy crag]

I don't think so, but that's irrelevant now.

[wooden ember]

ah fair enough we need to both post and precompose by basis changes

got confused there

[warm wyvern]

Can I get a hint with this one

I feel like I'm missing something easy

[formal ermine]

$I$ and $J$ being coprime means $1 = x + y$ for some $x \in I, y \in J$. consider $z \in I \cap J$, show that it is in $IJ$

[formal ermine]

the other direction ($IJ \subseteq I \cap J$) is true in general

[warm wyvern]

I can see an argument if R is commutative hmmm

[formal ermine]

oh is R not commutative

[formal ermine]

what were you thinking of?

[warm wyvern]

xz (=zx if R is comm) is in IJ (coz it's an ideal)

And so is zy

So zx+zy=z(x+y)=z is in IJ

[formal ermine]

yeah nvm I was thinking of the same argument but in reverse lol

I cap J \ni z = z1 = zx + zy \in IJ

does need commutativity as you said though

you can find a counter example for R not commutative I think

[wooden ember]

yeah okay you definitely need R commutative here

[warm wyvern]

[agile burrow]

that's funny

[formal ermine]

Z[x,y]/(2x+2y-1), I = (x), J = (y) is a counter example I think

[agile burrow]

it's in the errata for the first printing but not in the errata for the second printing

oh maybe that's because he fixed it in the second printing 😵‍💫

[wooden ember]

that's a commutative ring

but with a free non-commutative algebra i think you can modify this slightly

[warm wyvern]

I thought i had the latest printing

[agile burrow]

lol that makes it double funny

[wooden ember]

yeah okay i think you can take what @formal ermine was saying and changing it a bit, like take something like R<x,y>/(x+y-1) with I=(x) and J=(y). You get IJ = (xy) and JI=(yx)

[formal ermine]

what does R<x,y> mean

[wooden ember]

free non-commutative algebra on x,y over R

[formal ermine]

ah

[spice whale]

there's an obvious isomorphism to T(R²)

where T(M) is the tensor algebra over a module

[white grotto]

jordan block of a nilpotent matrix

the question is why the commutants of this matrix are non cotrigonalizable

[shy turtle]

Hello. Is it true that the set of functions A → G from some set A to a group G is itself a group? Namely:

• the identity is ⊢ λ a: A. ε
• the inverse is f: A → G ⊢ λ a: A. (f a)⁻¹
• the composition is f, g: A → G ⊢ λ a: A. (f a) (g a)

For finite A, A → G is the same as a product of |A|-many copies of G. I briefly checked the laws and everything seems to be alright for any A. Is there a reference that talks about this?

[prisma ibex]

this is just the product Gx...xG

[shy turtle]

Yes, in finite case it must be isomorphous to said product.

I still have doubts because I do not remember reading about this specific fact anywhere.

[warm wyvern]

MY EYESS

[chilly radish]

It'll still be the product in the arbitrary case

[warm wyvern]

Also in the infinite case

Sniped

[chilly radish]

Specifying an element in the A-fold of G is the same information as specifying a function from A to G

In both cases, you're specifying for any a in A a corresponding element in G

This is true at the set level

But of course inherits the group structure

If you wanna be pedantic, this is how arbitrary products are usually defined

In set theory

[simple valley]

why are you using the turnstile?

that is not an appropriate use of the turnstile

[chilly ocean]

but formality

[simple valley]

no the appropriate symbol is \mapsto

[chilly radish]

The appropriate symbol is words

Use your words

[simple valley]

that too

[shy turtle]

I am not here to debate matters of taste.

[warm wyvern]

Based

[chilly radish]

Based on what

[south patrol]

Random jokes go in #chill

[shy turtle]

Surely everyone here understands my question. Is it hard to say «yes, reference is this book that chapter» or «no, this is false, here is why»?

[chilly radish]

I gave you an answer

[south patrol]

Didn't shin answer it lol

[simple valley]

only like 3 people here understand your notation

[warm wyvern]

This is so unnecessarily tense lmfao

[shy turtle]

I do not see why 2 of them had to be pushing me to speak in the manner they find to their liking. If my question is not clear, I can restate it, but I have not been asked to do that.

[chilly radish]

This discussion is

1. Not relevant to this channel and
2. Not relevant to your question since it's been answered. If you have any follow up questions feel free, otherwise take this elsewhere

[simple valley]

we've had this discussion before. There are standards for mathematical communication, and if you wish to participate in it you gotta abide

most people will interpret "symbol soup" as unnecessary and pretentious

[chilly radish]

In the future, if you want help, consider that people here are helping out of their own goodwill, so it would do you well to make the discussion as accessible as possible, else people won't want to help

If you want to keep bashing this point, go to one of the discussion channels

[shy turtle]

It is not I who started «bashing this point». Indeed I expressed my unwillingness to debate this matter. I asked a question in good faith. I am however now forced to defend myself.

[rotund aurora]

[chilly radish]

This discussion is over. You got a good-faith answer, so again, I implore you to ask if you have any more questions about the explanation I gave, otherwise, this discussion is not suitable for this channel anymore

[rotund aurora]

nuke sully this and move on

[shy turtle]

I do have those questions. Is there a proof in literature that I can refer to?

[prisma ibex]

pretty much every abstract algebra textbook will talk about products of groups, even in the infinite case

[chilly radish]

What about my proof do you find insufficient? This is an elementary enough example and your question is coming at it from an awkward angle, so I highly doubt there is literature that covers exactly what you want

If you want to see how set-theoretic products and direct products of groups in the arbitrary case are defined, there are plenty of references

[rotund aurora]

These are things you should prove on your own if you are skeptical

[shy turtle]

Alright, let us look at it like so: suppose I have two models of the theory of groups in some category C, say G and G'. How can I begin talking about group structure on arrows G → G' between these models?

I want to avoid the formality of sets as far as it can be avoided.

[obsidian sleet]

[simple valley]

assuming you mean a group object wrt the cartesian monoidal product

cartesian product preserves limits

[fossil shuttle]

this works in the infinite case too.

[simple valley]

G^|A| is a limit

[prisma ibex]

Asking simple questions couched in complicated language assumes an ability to answer simple questions couched in simple language

[rotund aurora]

You are doing things wrong, you are just learning random words

[chilly ocean]

that conversation has already been had

don't try lol

[rotund aurora]

I know, but was just a remainder

[south patrol]

What is an abelian group?

[shy turtle]

I have no idea what you folks here are talking about. Complicated, random, wrong. Are you punishing me for some evil I have done, with these negative words? Please understand, I am not going to speak in your language until you talk to me without prejudice against mine. This is a location on the Internet where people seem to be talking about Abstract Algebra. I am here to talk about Abstract Algebra. Why are you all seemingly so upset?

[shy turtle]

This is more like what I expected. Clear cut answer, even if short.

[spice whale]

the main issue to me seems to be that you're using lots of formal type theoretic logical symbols that many people are not familiar with

so people are finding it hard to answer your questions

[shy turtle]

Well, it seems they could ask me what I mean or explain my notation, instead of saying how bad I am for being myself.

[chilly radish]

Please direct this discussion anywhere else

[shy turtle]

@simple valley My problem with your argument is that my category may not have infinite products.

[chilly radish]

This does not concern the topic of this channel

[spice whale]

sorry

[shy turtle]

Sure, let us redirect this to my private messages if someone wants to talk.

[shy turtle]

Alright, returning to those products:

1. Let me have a proof that a Cartesian product G₀ × G₁ of underlying sets of two groups G₀ and G₁ is the underlying set of some group itself.
2. By induction I can extend this proof to a product of any number of groups indexed by a finite set. Then, I can get an isomorphose between a function from a finite set A to G and a product of G with itself indexed by A, thus making A → G into a group.
3. However, I cannot extend this proof to a product of an infinite number of groups in the same way, because this would result in an infinitely long proof.

Unfortunately, I do not understand transfinite induction, so I was hoping to find a proof that A → G is a group even if A is infinite, say ℕ, that does not appeal to the construction of infinite products. What can I do?

One way is suggested by @simple valley:

1. Cartesian product preserves limits.
2. A → G is a limit. (An exponential object.)
3. So? I do not understand what product I need to look at.

Suppose we have a model of the theory of groups in a Cartesian closed category C. A group object, is this called also? Let us call the object that we build the group upon x. x ← A will denote an exponential object for some A. I think I can build a proof I want in this setting, but maybe @simple valley you can clarify what you meant first?

When thinking about this from the logical side, I understood that the theorem I am looking for does not fit into the axiomatic theory of a group because there are two different groups involved. I need to add a way to talk about functions or some kind of arrows to my theory. It seems that a little touch of Lambda Calculus is enough to put the problem forward and then the proof can be written without reference to any concrete models. Indeed I have sketched some. But I need to read up before I can explain any of it in detail.

[simple valley]

A -> G is an infinite product

you do not need any induction here

in Set, any object A is the coproduct of |A| copies of 1

by straightforward computation, A -> G is iso to a product of |A| copies of G

the proof of G^A being a group does not rely on any properties of A

I implore you to understand the Set case before heading into generalized nonsense

[shy turtle]

You can be charitable and assume I understand the Set case but look for another approach, for whatever reason.

[simple valley]

Unfortunately, I do not understand transfinite induction
does not seem to be the case

[shy turtle]

Well, I reckon few mathematicians can derive transfinite induction from the axioms of a set theory. I am trying to stay constructive, so it is even harder to find a reference.

[simple valley]

.

[shy turtle]

I may be in error, sure. I need to prove that an infinite product of groups is a group, right? But induction only lets me prove that for any n ∈ ℕ, the product indexed by n is a group.

[simple valley]

ok so how do you define an "infinite" product?

[shy turtle]

You want to say that an infinite product is a function; but I want to say that practically an infinite product is a co-inductive type and a function is an expression that I can evaluate. This is why I mentioned isomorphose and not identity. But I have no objections to having such an isomorphose.

[simple valley]

idk where you picked up that word but it's called an "isomorphism"

adj "isomorphic"

[shy turtle]

I am happy to know that you understand me. So, we found that a proof for an infinite product will be as good for us both as a proof for a function. But we are nowhere nearer that proof yet.

[simple valley]

I don't think anyone out there thinks of all infinite products coinductively

some infinite products (namely those indexed by inductive types) can be understood this way, but not all

say what's X^(N^N)

[shy turtle]

Let us restrict our attention to these well-behaved ones then.

[simple valley]

well, no, because this is adding structure that we do not need

ok so let's consider the set of functions from A to G where A is an arbitrary set and G is a group. You have already defined the identity, inverse, and multiplication on this set. What's the hold up?

[shy turtle]

Well, I was hoping someone can simply point me out to some literature that I can study, or sketch a proof that I can complete. You sketched me a proof that I cannot complete (sketch is too sparse); so also @chilly radish sketched a proof I cannot complete (seems to require transfinite induction, which I cannot prove).

[simple valley]

you just verify the group axioms, pointwise

Let $f \in G^A$, then $f^{-1} \cdot f = \lam x f^{-1}(x) \cdot f(x) = \lam x e = e \in G^A$

[cloud walrus]

mniip

[simple valley]

etc

[shy turtle]

Yep, this is about the same as what I was sketching for myself, so if we both think this works it is likely to be without fault.

Earlier today, not sure about specific time.

[shy turtle]

…But I reckon we can get a nicer proof if we allow ourselves some more tools. Suppose we have an algebra Fx → x. We can then go F (x ← A)   →   ((F x) ← A)   →   (x ← A). (Here the backwards arrow allusively denotes exponentiation.) I think we can do it with very few assumptions, but I cannot quite catch all of those assumptions. And we need to clarify how F makes x a group!

Ideally this lets us state that any algebraic theory has this nice property — rings, say.

So, I was secretly hoping to gather some food for thoughts in this direction. @simple valley so can you at last explain what you meant when you mentioned limits above?

https://cdn.discordapp.com/attachments/917098403744346152/1061788629162262558/340930090231660567.png

[simple valley]

I think the more important story here is that the forgetful functor U : Grp -> Set is a right adjoint and therefore it preserves limits. Therefore the underlying set of a (category-theoretical, possibly infinite) product of groups is the cartesian product of the corresponding sets.

[fossil shuttle]

If it exists. 😉

[shy turtle]

Therefore the underlying set of a (category-theoretical, possibly infinite) product of groups is the cartesian product of the corresponding sets.

Yes, it is. …If it exists.

[fossil shuttle]

What really gets you going is that the forgetful functor is monadic, so it inherits the underlying limit

that's what really gets me going anyway

[simple valley]

this is explained in mac lane on page 111

[fossil shuttle]

[shy turtle]

How does this help us?

[fossil shuttle]

Oh, I was not interested in helping.

[shy turtle]

Ah, you helped anyway.

[simple valley]

the page I linked also deals with some considerations for when it exists

[fossil shuttle]

Oh, yes. I see that now, cool.

[shy turtle]

Is there something that helps us speak about A → G?

[simple valley]

I'll have to spell this one out won't I

[shy turtle]

Maybe one more hint will be enough.

[simple valley]

by one of the ways you can look at an adjunction F --| H, the object HX is the limit of a certain diagram involving F

[shy turtle]

I shall take some time to think about it. Good to think about either way, relevant or not.

[umbral lodge]

#precalculus

[formal ermine]

?

[tribal pasture]

questions about limits belong in precalculus /s

[toxic zephyr]

trying to understand the dicyclic group. the way it's presented is like this
$$Q_{4n}=\langle r,s\mid r^{2n}=e,r^n=s^2,s^{-1}rs=r^{-1}\rangle$$
so what would the elements be of $Q_4$? $r,s,sr,s^{-1}r$?

[cloud walrus]

nilpotent nix

[toxic zephyr]

i want to make a multiplication table but i have a hard time reading what the actual elements are from something like this...

[toxic zephyr]

for context, I'm kind of "moving up" from the dihedral group. so I'm trying to deal with it as similarly as possible

[toxic zephyr]

just realized I didn't include e, that's pretty stupid of me lol.
so i think it's {e,r,s,sr}

[coral spindle]

You seem not to have accounted for the fact that r = s^2 in Q_4

Q_4 is just the cyclic group of order 4

[toxic zephyr]

aha. now i see that

groups are so hard for me to understand lol
so much stuff is obvious when someone points it out to me, but i struggle through it without making basic observations for hours...

[wooden ember]

is there anything interesting to say about the orthogonal Lie algebra associated to the Killing form of L where L is a Lie algebra? Ie the algebra of endomorphisms alpha of L such that k(alpha(v),w)=-k(v,alpha(w)) for all v,w in L (I suspect this is more or less Inn(L) but im not quite sure)

[formal ermine]

I don't understand why [L : L^H] = |H| where L^H is the fixed field of L under H

[rustic crown]

The proof I like uses that the extension is finite, separable and so simple, then use the primitive element to bash it out

[south patrol]

Interesting way is u can show that there is there's z in L with trivial stabiliser [ig eg as a primitive element but u can also just do this more directly] and so |H| = |H.z| = |L^H(z): L^H| <= |L:L^H| where I used the fact that H acts transitively on the roots of min poly of z. the reverse bound always holds for any extension I gues

But hm im not sure if this answers the "understand" bit

Maybe the key point is just that that second equality I wrote encapsulates the fact that th galois group acts transitvely on roots of min polys here

[wooden ember]

In Cartan's criterion for solvability of a Lie algebra, one checks that the killing form satisfies $\kappa(x,y)=0 ~\forall x,y\in L$. Why can't we just check $\kappa(x,x)=0$ $\forall x\in L$? After all in this case we know that on a Cartan subalgebra $H\subset L$ we have $\kappa(x,x)=0 ~\forall x\in H$ and thus in particular we cannot have $[L,L]=L$. So $[L,L]$ is a proper ideal and by induction it is solvable, giving us that $L$ is solvable?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[chilly ocean]

how do i find the elements of order $6$ in $\mathbb{R}/\mathbb{Z}$.

[cloud walrus]

isomorphism

[chilly ocean]

i got 1/6 + Z, 2/3 + Z

[formal ermine]

the second element has order 3

a/b + Z with a in Z, b in N and gcd(a, b) = 1 has order b

[chilly ocean]

so what are the elements then

1/6 + Z, . . .

[formal ermine]

you want (a + Z) + ... = 0 i.e. 6a in Z

[chilly ocean]

yeah

[formal ermine]

,, 0 \leq a < 1

[chilly ocean]

yes

[delicate orchid]

but, critically, you also don't want 2a or 3a to be in Z

[chilly ocean]

so 1/6 + Z is one such element because it is composed with itself 6 times to yield 1 + Z = 0 + Z

but what other elements are there

[delicate orchid]

the group generated by 1/6+Z is iso to C_6, there are two different generators for C_6

see if you can think about what the other one should be

[formal ermine]

||this gives 6a in {0,1,2,3,4,5} but what wew noted reduces it to {0,1,5} meaning that the elements could be 0/6, 1/6 and 5/6. it's obviously not 0, checking gives us that the result is 5/6 and 1/6||

don't unspoiler until you've tried what wew suggested

[delicate orchid]

||alternatively, if g is a generator for a cyclic group then so is g^-1||

[formal ermine]

this is big brain

[chilly ocean]

just solve 6x = n where n in Z for 0 <= x < 1 and see which elements have order 6

since n + Z is ultimately Z

[formal ermine]

that's essentially what I said

[chilly ocean]

ty @delicate orchid @formal ermine @chilly ocean

[rustic crown]

find primitive 6th roots of unity in S^1 = R/Z

[chilly ocean]

i also have this question

[formal ermine]

which one

[rustic crown]

i hope this is just a practice exam >.<

i get scared when i see [6] on the right

[formal ermine]

this would be for the multiplicative group though, no?

[chilly ocean]

$a \in R$ is a unit if it has a multiplicative inverse. Similarly, $R^\star$ is the unit group of the ring i.e. $$R^\star = {a \in R : \text{a is a unit}}$$

[delicate orchid]

that's just the citation det dw

[cloud walrus]

45

[chilly ocean]

its not a practice exam

its a past paper

[karmic moat]

not a practice exam, just an exam

[chilly ocean]

I am more confused with part ii)

show the group axioms hold

[rustic crown]

yea, but my eyes just see the isomorphism as an equality >.<

[delicate orchid]

isomorphism is an equality. Cope more

[chilly ocean]

can you help me?

yes

[formal ermine]

not for the galois correspondance

[chilly ocean]

show closure, associativity, inverses, nonemptiness (or alternatively, existence of an identity)

[delicate orchid]

galois
mans not bovvered

[formal ermine]

wew

I might do rep theory next semester

[delicate orchid]

non-emptiness is not equivalent to the existence of an identity, one is a much stronger assertion

[delicate orchid]

a grave error

[formal ermine]

idk yet tho

I'm gonna take algebra 2

my prof said he's either gonna do rep theory or com alg/alg geo

[chilly ocean]

you need com alg for alg geo

[delicate orchid]

you need com alg for rep theory

[chilly ocean]

or at least a lot of com alg motivates alg geo

[formal ermine]

yes that's why it's a combined course

[formal ermine]

wait

really

I thought it just needs like

linear algebra

rings

modules

[delicate orchid]

rep theory is the study of modules over the group ring of course you need com alg

[formal ermine]

and not noetherian level stuff

[delicate orchid]

tesnor products

[elder wave]

tesnor products

[delicate orchid]

sorry, *plopducts

[rustic crown]

i would say a lot of stuff in alg geo reduces to com alg and not really a statement about motivation

[chilly ocean]

in my uni you don't need any of that for rep theory

only lin alg and some groups/rings/fields

[formal ermine]

my prof said that he doesn't wanna do com alg because it's very dry

[rustic crown]

nice

just do ag directly >.<

[chilly ocean]

i have some practice problems i can send you for com alg later if you wanna see what it's about

[elder wave]

the prof in my uni who does rep theory states only lin alg as prereq and does all the necessary comm alg in lin alg

[chilly ocean]

from my uni

as long as you don't share them

[delicate orchid]

there are many different approaches to rep theory tbh it's hard to give a universal pre-req other than lin alg + groups

[formal ermine]

yeah that's what I'm planning on doing anyway maybe this autumn

[chilly ocean]

closure meaning if a, b in R* then ab in R*?

also does it suffice to say multiplication is associative because a(bc) = (ab)c

[formal ermine]

that's associativity not commutativity

[chilly ocean]

right

fixed

[chilly ocean]

@delicate orchid could you help me go on from here?

im not sure how to show inverses/identity

the identity is in $R^\star$ because $1 \cdot 1 = 1$ i.e. $1$ is a unit and thus in the unit group of the ring

[cloud walrus]

45

[barren sierra]

And then inverses should be from definition of R*

[chilly ocean]

you are correct regarding associativity because multiplication is associative

[chilly ocean]

yes

[chilly ocean]

^

definition of R* implies existence of inverses because by definition, R* is the group in which every element of the ring has an inverse

now closure is a bit tricky

is $xy \in R^\star$? think about it, do we have a formula for the inverse of this "product"?

[cloud walrus]

45

[chilly ocean]

$y^{-1} x^{-1}$ is the inverse of $xy$ so $xy \in R^\star$?

[cloud walrus]

isomorphism

[chilly ocean]

yes

thank you

my last question is for b) i) and then I'll try b ii) and iii)

[barren sierra]

What are the ring axioms 👀

[chilly ocean]

ok so

no

showing this is a ring is a very tedious process

much easier to show it's a subring

of the set of 2x2 matrices with coefficients in Z

[barren sierra]

Oh sure

Then do you know the conditions for a subring

[chilly ocean]

i don't remember them

first off you establish $R \subset M_{2 \times 2}(\mbb{Z})$

[cloud walrus]

45

[formal ermine]

for something to be a subring it has to be a ring itself

[chilly ocean]

M_2(Z) is a ring

[chilly ocean]

also that's not the point

showing something is a subring is a much easier way of showing that it is a ring in its own respect

because you have less conditions to fulfill

Let $R$ be a ring. A subset $S \subset R$ is a subring iff it satisfies the existence of the additive and multiplicative identity, additive closure, multiplicative closure, and additive inverses.

[cloud walrus]

45

[chilly ocean]

otherwise you'd have to show stuff like distributive property over addition which is a tedious process

which is otherwise inherited from the main ring

associativity, too

which would be a pain in the ass given that we're talking about matrices here

[formal ermine]

by definition of matrix multiplication and addition...

[chilly ocean]

why take the harder route

[formal ermine]

I feel like this is not the route their teacher expected them to take

[chilly ocean]

you can replace all of that simply by stating $R \subset M_{2 \times 2}(\mathbb{Z})$

[cloud walrus]

45

[chilly ocean]

if you want to verify all ten axioms go ahead

i won't stop you

[chilly ocean]

im gonna be sipping tea after im done showing these

[formal ermine]

that's probably what their teacher expected them to do considering they've just been introduced to rings

[chilly ocean]

usually subrings are covered second to its definition and examples

but that's a question of pedagogy

[delicate orchid]

showing matrix multiplication is associative
I'd rather die

[chilly ocean]

exactly

why waste time

im guessing it's a 2 hour exam

if you wanna waste a precious 10 minutes on six marks, that's not how i'd do it at least

[oblique river]

matrix multiplication corresponds to composition of linear maps which is clearly associative

[chilly ocean]

sorry I'm back

[chilly ocean]

it's one hour long

LMAO

[delicate orchid]

proving the correspondence is only slightly better than just doing (AB)C = A(BC)

[chilly ocean]

all you have to remember is identity, closure, inverse

Both for subgroups and for subrings

[delicate orchid]

inverse
☣️

[rustic crown]

first implies the second, which is sad the other way

[delicate orchid]

yeah I know det I meant better from a "this is really annoying and sucks" perspective rather than "which one is stronger"

[chilly ocean]

I understand parts i) and ii) thanks to @chilly ocean but I'm struggling with showing that the unit group does not have elements of order $3$

[cloud walrus]

isomorphism

[rustic crown]

also isn't proving correspondence actually more work?

[rustic crown]

my perspective was, if you're doing some annoying work, then might as well do the stronger annoying work :p

[delicate orchid]

true

[oblique river]

is it really that annoying? you only need two facts, 1) the jth column of the matrix associated to a linear transformation T is just T(e_i), and 2) the jth column of a product AB is just A*(jth column of B)

and both of those are just from the standard definition of how matrix multiplication works

[rustic crown]

i don't consider these annoying after the checks i do for ag

[oblique river]

then if S is represented by A and T is represented by B, then the jth column of AB = A*(jth column of B) = A*(T(ej)) = S(T(ej)) = (ST)(ej)

and that shows that AB represents ST

[chilly ocean]

can anyone help

[oblique river]

you can use the Nike method for that part

[chilly ocean]

what is that

[oblique river]

Just do it

[chilly ocean]

uh

[elder wave]

lmao

[rustic crown]

ig you can use cayley hamilton to save some work

[chilly ocean]

what is that

[rustic crown]

or just use some eigenvalues

for an upper triangular matrix the diagonals are eigenvalues

[oblique river]

you can just pick a matrix in R and raise it to the 3rd power and then try to solve the equation "that = Identity"

and you'll find that you cant

just prove it directly

[rustic crown]

and if it's a matrix of order 3, then the eigenvalues are roots of x^3 = 1

[oblique river]

you can also use more theory

like what det is suggesting

which would save some computation

[chilly ocean]

but its asking for R*

[rustic crown]

since diagonals are integers, they're integral roots of x^3 = 1, so x = 1

[oblique river]

yes.....

elements of R* are still elements of R

[oblique river]

so if R has no elements A with A^3 = I

then R* certainly has no such elements either

(even beyond that, if A is in R and A^3 = I, then A is automatically in R* anyway)

[chilly ocean]

aren't elements of R* matrices where ac != 0

because they have to have inverses

[oblique river]

yes

but that has nothing to do with what I'm suggesting you do

[rustic crown]

and so the matrix looks like

[1 *]
[0 1]

only way this cubes to I is if 3* = 0 so * = 0 but this has order 1 and not 3

[oblique river]

take an element of R and raise it to the 3rd power

[chilly ocean]

I understand

[oblique river]

and try to set that matrix equal to the identity matrix

[chilly ocean]

and for iii) I do the same thing but show its not commutative

[oblique river]

and see if you can find solutions or not

[chilly ocean]

so I show for some A, B in R AB != BA

[oblique river]

yes

althoguh for that one you want to make sure that A and B are actually in R*

[chilly ocean]

wdym

[oblique river]

it asks to show that R* sin't commutative

[chilly ocean]

yes

[oblique river]

so you need to find elements of R* which don't commute

[celest furnace]

Just a question: for part ii) would this be formal: det(M) would have to be 1/3 which is not attainable by multiplying / subtracting integers, so M can’t exist

[oblique river]

if your A and B aren't in R* then you haven't done what the probelm asks for

[chilly ocean]

how do i make sure they're actually in R*

[oblique river]

no

[chilly ocean]

I say "ac neq 0"

that's all

right

[oblique river]

det(A^3) = det(A)^3, not 3*det(A)

[celest furnace]

Ah