groups-rings-fields

[white oxide]

ah so it does work then

my fault!

[white oxide]

why is it that two subnormal series don't have to have the same number of terms to be equivalent, but they have to have the same length? aren't they the same thing?

[dense raven]

is there a name for all groups for which the lattice normal subgroups is distributive

[crystal turtle]

Not sure but here's a relevant paper https://www.cambridge.org/core/journals/glasgow-mathematical-journal/article/groups-whose-lattices-of-normal-subgroups-are-distributive/642507CA7D388A075CDF0FD2F3D3BE30

[white oxide]

can somebody tell me what a jordan-holder decomposition is, i know the theorem but i haven't been able to find out what that means

like what does it mean for G_1 to have a jordan holder decompositoin lol, i know it means that it has a composition series

does it mean a group G that has a unique composition series or some shit?

[rotund aurora]

You can def google this or look up an algebra book

[white oxide]

well jordan holder is a theorem as far as i'm aware

google has not told me what it means for a group to have a jordan holder decomposition

[crystal turtle]

This should be in dummit and foote where they mention J-H decomp

[white oxide]

in trying to find a decomposition for finitely generated modules over a PID is it fine to just think about them as abelian groups (iso to Z/(n)

and then try to find a decomp that way

or do you run into technical problems

[topaz solar]

Yes, it is

[white oxide]

sanity check: if $R$ is a PID and $p, q$ are prime, then $R/(p^nq^m) \simeq R/(p^n) \oplus R/(q^m)$?

[cloud walrus]

okeyokay

[white oxide]

and the length of a composition series for $R/(p^nq^m)$ is $m + n$?

[cloud walrus]

okeyokay

[south patrol]

this is false for R = Z in general

But yeah ifbyou change it slightly this is just Chinese remainder

[white oxide]

Wait wat

I thought it was true if R = Z

[south patrol]

you need p,q to be coprime

Maybe that is picky but worth pointing out lol

[delicate orchid]

Death sentence

[white oxide]

wait sorry am I high

oh

so need to assume p and q are distinct primes?

[rotund aurora]

Ofc

[delicate orchid]

Cmon chat why would they be different letters if they were the same prime

[rotund aurora]

Mmh

You need coprime

Not just distinct

Because you could take q=0 🤓

[delicate orchid]

Well yeah cause CRT

[delicate orchid]

WTF!!!!!

[onyx trout]

I've been doing mathematics for far too long to still be falling prey to this and yet here we are

[crystal turtle]

🤯 🤯

[south patrol]

or 2 and -2

[rotund aurora]

how do you do the second part

Im sure its not difficult

C_P(A) is the centralizer of A

[coral spindle]

Seems like it should be an application of orbit-stabiliser but I can't quite see the right action

[rotund aurora]

mmh it might not be as easy as I thought

a bit of context: For the first part, they hint to use the following lemma. You prove it by taking C=C_P(A) and if A<C, then since both A and C are normal you can find L normal such that A<L<=C with [L:A] prime, and since A is abelian and it follows that L is also abelian and normal, contradicting maximality.

But not sure if how you prove the first part is related to how you prove the second part

the factorial makes it look so weird, because everything is a power of p

[rotund aurora]

it was just that P acts on conjugation on A \ {1} therefore you get a homomorphism P-->Sym(A \ {1})

[coral spindle]

Nice

[rotund aurora]

(i looked up a solution)

[coral spindle]

So it's an action of P on A\{1}.

[rotund aurora]

yeah like I did see we had to act on A\ {1} but idk why i didnt think of the most obvious one

[long obsidian]

When I think about a field k as a vector space over itself is it's dual vector space also itself k?

[rotund aurora]

its 1 dimensional

so yes

thinking of a field as a vector space over itself is kinda trivial tho, the fun part is when you have extensions of fields L<K and you think of K as an L-vector space

[limpid fjord]

Part 1 question 1?

Sorry if it's in frensh I can reformulate the question

Prove that A is equivalent to that matrice

[rotund aurora]

unreadable

[limpid fjord]

Forgive my goofy ahh phone

I can't afford a decent iphone

I've answer most of it because all the other parts are based on the first 3 questions but I had to skip the first one

Since it requires to study the base of the ker(A) and the Im(A)

[long obsidian]

Can someone help me with the first exercise of Harris algebraic geometry first course.

If S is a set of d points not contained in a line then S is the zero locus of a polynomial of degree d-1 and less.

I think since they aren't on a line there exists two linearly indep p,q in S

So I can extend to a basis {p,q,v3,...vn} then pick some coordinate function v=v^i has kerv contains p and q. Then for any w besides p and q in S I can extend to a basis {w,W2,..,wn} then any w^i coordinate function vanishes on w.

Then the product of w^i for each w(bad notation) and v is degree d-1 or less.

But I'm not sure how to show S contains Z(wi *v) the vanishing set so this is an equality.

Do I have to construct my bases more carefully?

[chilly ocean]

Hello

If we have a finite ring with odd number of elements

Then 2 is invertible

My solution:

[tender wharf]

by any chance is your ring an integer ring

if not, what is 2?

[chilly ocean]

A finite ring with odd number of elements. So the group (A,+) has odd order so from lagrange there is no element a ≠0 with 2a=0 so 2a=0 implies a=0. If I take the function f:A->A f(x)=2x. f(x)=f(y) implies 2(x-y)=0 so x=y so f injective and because A finite f bijective so there is c in A s.t. 2c=1 so 2 invertible

[chilly ocean]

1+1

[tender wharf]

oh like

2 cdot unity

yes that makes sense

[chilly ocean]

Is my solution ok?

Because another solution I saw it was smth like this:

Let k=charA>=2

Let k=2t+1 so 2t+1=0 so 2(t+1)=1 so 2 invertible

And k=2t+1 beacuse k divides the order of (A,+) which is odd

[hollow mica]

If the reals are the unique complete ordered field up to isomorphism, what is the analogous characterization for the rationals, if any?

[topaz solar]

Field of fractions for Z, is inside every char 0 field, something like that

[hollow mica]

so maybe like, the largest field contained inside every char 0 field?

[rocky cloak]

More like the only field contained inside every char 0 field

So it's like "the smallest" characteristic 0 field

[crystal turtle]

Analogous to Z/p being the smallest field of char p

[rocky cloak]

These smallest fields are usually called prime fields (taking their name from Z/p I guess)

[crystal turtle]

C is the unique algebraicly closed field of char 0 of it's cardinality (same with other cardinals)

[white oxide]

ok dumb question/i think this is true/i'm too lazy to prove it/just verifying, if we're trying to show that gHg^{-1} = K and we've already shown gHg^{-1} \subseteq K will g^{-1}Kg \subseteq H complete the proof

[coral spindle]

Sure but I should note that if the group is finite then you’re already done

Provided H and K are the same size :)

[white oxide]

totally not trying to show that G transitive implies all stabilizers are conjugate

[white oxide]

oh ya true

[coral spindle]

You should try showing it for all groups tho.

[white oxide]

this?

[white oxide]

or this

[coral spindle]

Yes and yes, why not

[white oxide]

damn bro ur tryna be lang, getting me to prove all results of group theory

[coral shale]

well u shud

[white oxide]

is the action of a quotient group on a set well-defined? i'm considering S as a G/N set with action given by ((gN), s) = gs. if xN = yN then y = xn for some n in N, so (xN)s = xs = yns = y(ns) = ys = (yN)s. here's the trouble tho - i'm assuming ns = s since n is in N and functinos as the identity - is this sound logic?

[white oxide]

aight time to classify all finite nonabelian groups

brb

[limpid fjord]

i have a solution anyone wanna monitor it for me see if its correct or not?

[coral shale]

uh this one isnt that hard i dont think

compared to

classify all finite groups at least...

[coral shale]

this one is actually covered in undergrad

[white oxide]

wait huh

group is either abelian or nonabelian

f.g. abelian groups were already classified right and finite abelian groups are classified

[coral shale]

yes.

[white oxide]

i know some nonabelian groups are classified but most aren't

lol "most"

idk

[coral shale]

oh i misread u i thought u wrote abelian

yh this ones the nightmare

[white oxide]

nvm this works did it in the wrong order

[white oxide]

yea lol

i wonder if it'll be completed in our lifetimes

[coral shale]

im failing to read this what.

[coral shale]

huh. its done?

all finite groups are classified

arent they

or is it only simple ones...

ah simple ones.

[white oxide]

say we define the action of $G/N$ on $S$ by $(gN, s) \mapsto gs$. if $xN = yN$, then $x^{-1}yN = N$ whence $y = xn$ for some $n \in N$. but then $(xN)s = xs$ and $(yN)s = (xnN)s = (xN)s = xs$

[cloud walrus]

okeyokay

[coral shale]

but ig if u classify the simple ones uve at least classified up to maximal quotient kinda

[white oxide]

yea i meant all finite groups ig

[coral shale]

gN . s = g . s

[south patrol]

a quotient group is just a group

so like, idk the issue

[coral shale]

then u just need to show
gs = hs
for g, h in N

[south patrol]

But yeah if you think of like

[coral shale]

or something like that

[south patrol]

an action of a group K on a set X as a hom K -> Sym(X)

then that is useful here

Like basically to define an action of G/N on S is to define a homo G/N -> Sym(S)

which is equivalent to a map G -> Sym(S) such that N is sent to 1

[coral shale]

right if ur quotienting by the kernel of an action, that always works

[white oxide]

well

i was trying to solve c

[south patrol]

i.e. an action of G on S such that N acts trivially

[white oxide]

so it naturally led me to consider S a s G/N set

yea

[white oxide]

i'm unsure what the condition N < stab(x) for some x is needed tho for..

i didn't use it in my proof, i used everything else tho

[white oxide]

yea but i wanted to check if the action made sense

or is it that group actions are always well-defined..?

[south patrol]

well like being well-defined is a vague concept right like

[white oxide]

yea but whenever i'm working with representatives, i just check just in case

[south patrol]

What you have in mind is using a G-action to define a G/N-action ig

[white oxide]

yea, basically

[south patrol]

you didn't say S was already a G-set in your original question iirc

[white oxide]

oh ya true

my b

[white oxide]

am i high or does d) follow immediately from a)

since order or orbit is equal to index of stabilizer

also

tryna show that if G/Z(G) is cyclic, G is abelian

lazy proof (?): G - Z(G) is cyclic and thus abelian, Z(G) is obviously abelian, and G - Z(G) U Z(G) = G is therefore ableian

abelian

?

does that work or nah

[south patrol]

G - Z(G)

Wdym

[white oxide]

the complement of Z(G) in G

[south patrol]

That isn't a subgroup or anything...

[rotund aurora]

try again brah

[south patrol]

Doesn't even contain the identity

[white oxide]

oh shit my fault

!

[south patrol]

And a union of abelian subgroups can be non abelian

e.g. S3 is a union of its proper subgroups

[rotund aurora]

lol true

[white oxide]

ah shit

[limpid fjord]

@white oxide if i made an endomorph function names it u for R^n associated to A
rg(u)=rg(A)=r>=1
then dim(ker(u))=n-r
after that we will consider a basis for ker(u) equale to (e1,........,er)
using incomplete basis theorem
we complete it with (er+1,.......,en)
then we get B=(e1,.....,er,er+1,......,en)
B is a basis for R^n therefore E'=vect(e1,......,er)
R^n=E'+ker(u) supplementary
we declare another fucntion u1 which is the restriction of u at E' that hold value on Im(u)
ker(u1)=ker(u) intersectioned with E'={0
u1 is injetiv
and E' and Im(u) have same dimension thus u1 is bijectiv
so u1:E'------>Im(u) is isomorph
for all 1<i<r ei in E' => u1(ei)=u(ei)
the basis (u(e1),.......,u(er)) for R^n
we complete it using trhe same incomplete basis theorem
we get C=(u(e1),.............,u(er),f(r+1),.......,f(n)) for R^n
so P is pass matrix for de R^n basis -> B
same for Q pass matrix from R^n basis -> C
then we get the result A=QA'P^(-1)
which answers the 1 question

is that correcT? i tried to translte it as accurate as possible

[coral shale]

wtf

[white oxide]

sorry bro i'm not reading allat 😭

i'm not the right person for that too cuz 1) i'm an idiot 2) my lin alg shit

[limpid fjord]

come on i tried my hardest tryna figure out that question T-T

so whom i should be asking?

[rotund aurora]

its like you just made up completely new notation

[limpid fjord]

im a frensh student

[rotund aurora]

you should get better at math communication

[limpid fjord]

not english

[rotund aurora]

Im not english either

[limpid fjord]

i study maths in frensh not english

ive never stduied it in english

[rotund aurora]

bruh Im pretty sure its "french" not "frensh"

[limpid fjord]

just try and get along please bear with me

sorry im dyslexuc

dyslexic*

[void cosmos]

motif

[rotund aurora]

funny

the message I mean

[limpid fjord]

did i told u ajoke?

i see no reason u should be laughing

[rotund aurora]

ok

[void cosmos]

true

[limpid fjord]

so is there anyone who can help me

[void cosmos]

yea king there are tons of people who can here

but

problem is

we cant understand u much

do u have a specific like

problem

[limpid fjord]

ill try and explain

[void cosmos]

in a textbook or pdf

u can screenshot

[limpid fjord]

i already sent it

but its a low qulity gofy aah post

[rotund aurora]

how do you know the word gofy

[limpid fjord]

goofy?

[rotund aurora]

yeah

[limpid fjord]

wdym?

[rotund aurora]

idk the french thing sounds like an excuse

put some work in being clear, that is all

dont send shity pictures no one (not even you) can read

[limpid fjord]

bro me knowing good english doesnt have anything to do with me sucking at terminologies

well not everyone has iphones you know T-T

[rotund aurora]

yeah again

[limpid fjord]

ill try and resend it form another better phone

[rotund aurora]

phones below 100 euros have good enough camera to take a photograph of a text

[hollow mica]

only nontrivial right

[limpid fjord]

my currenecy is way weaker then the euro or a dollar

[rotund aurora]

didn't you live in France?

[limpid fjord]

im moroccan bro our economy is shit T-T

well no i didnt said so

[rotund aurora]

ok but you study in french, right

[limpid fjord]

Hopefully this one is clearer

[coral shale]

France is not the only french speaking country

[limpid fjord]

no

i study in a french uni

[rotund aurora]

so you live in france?

[limpid fjord]

in morocco for engeneering called CPGE

[limpid fjord]

i've never said that

[rotund aurora]

I just asked

[limpid fjord]

how was that last one?

[rotund aurora]

your phone can't rotate pictures because is not an iphone?

[limpid fjord]

this one is way better no?

no i was in a hurry

sorry

cant you rotate it on ur phone?

like save the pic and edit it rotated

[rotund aurora]

mmh I think you can rotate with a discord command

,rotate

[cloud walrus]

Couldn't find an attached image in the last 10 messages.

Sep 23, 2023, 10:05 PM

[rotund aurora]

too late

[limpid fjord]

sorry im a such a burden to deal with all i want is my asuestions answered

[rotund aurora]

,rotate

[cloud walrus]

Couldn't find an attached image in the last 10 messages.

Sep 23, 2023, 10:05 PM

[rotund aurora]

fuk

[limpid fjord]

ill resend it and rotate it

.rotate

Huh.??

[rotund aurora]

I think you are trolling

if you are not I apologize

but Im stepping out of this

[limpid fjord]

bro i swear to whatever u beleive in im not trolling u

,rotate

[limpid fjord]

there we are

is it better now?

just help me out i would apreciate it very veery much

[limpid fjord]

here is my answer

bear with me @rotund aurora please

or not you do you

[white oxide]

so let $K$ be a p-subgroup of $G$. idk if this works, but if we could show that $[G: N_G(K)] = 1$ then we'll be done since that'll imply that $K$ is normal in $G$. so i'm considering $[N_G\bigl(N_G(K)\bigl): N_G(K)] \equiv [G: N_G(K)] \text{mod } p$, which yields $[N_G(K): N_G(K)] = 1 \equiv [G: N_G(K)] \text{mod } p$. is this the right line of attack? it feels wrong for some reason

[cloud walrus]

okeyokay

[white oxide]

the second Sylow theorem doesn't really tell me anything since the Sylow $p$-subgroups of $G$ are well $G$ itself right

[cloud walrus]

okeyokay

[white oxide]

here I just took $H = N_G(K)$

[cloud walrus]

okeyokay

[rotund aurora]

My solution: || it is enough to show it has a normal subgroup of order p, but p-groups have non-trivial center (use the class equation)||

Hopefully I didnt clown

[white oxide]

also i used this theorem

[white oxide]

ooh ok i'll take a look if someone says that my solution fails per usual

Ok, I think my solution works: Let $K$ be a $p$-subgroup of $G$. Take $H = N_G(K)$ as in Lemma 5.5; then $[N_G\bigl(N_G(K)\bigl):N_G(K)] \equiv [G:N_G(K)] \text{ mod }p$. But $[N_G\bigl(N_G(K)\bigl):N_G(K)] = [N_G(K):N_G(K)] = 1 \equiv [G: N_G(K)] \text{ mod } p$ . Since $[G:N_G(K)] \geq 1$ and divides $|G| = p^n$, we cannot have $[G: N_G(K)] = kp + 1$ for $k \geq 1$. Hence $[G:N_G(K)] = 1$ and $K$ is normal in $G$.

[rotund aurora]

Why do you have N(N(K)) : N(K) = G: N(K) mod p ?

[cloud walrus]

okeyokay

[white oxide]

wdym

i put an \equiv there

didn't i or am i high

[rotund aurora]

Yeah, how do you derive G : N(K) =1 mod p?

[white oxide]

you mean congruent to 1 mod p?

[rotund aurora]

Wait its the lemma you shared

[white oxide]

Ye

[rotund aurora]

But K is not a Sylow subgroup

[white oxide]

Ah shit

Ofc….

[rotund aurora]

So you dont know N(N(K))=N(K)

[white oxide]

Oh well. Guess I’ll return to this problem tmrw then ig

[rotund aurora]

<@&268886789983436800>

[chilly ocean]

@rotund aurora
?

[topaz solar]

Okay bro, can you find a normal subgroup of G for me?

[rotund aurora]

I'm not sure what your strategy even is tho, you want to show normal p-subgroups exist for all p^b less than the order of the group, but you are starting with a given p-subgroup K of undefined order. You wanted to show that every p-subgroup K is normal, which doesn't smell true, although idk

[topaz solar]

It’s order is p^n for some n yeah? So, didn’t I help you out with a similar sounding exercise before?

Namely, if its index is the smallest prime (though this is just a first guess, idk if you can easily find one oop)

[rotund aurora]

What is an example of a nonabelian group where all its subgroups are normal

If it exists at all

[obsidian sleet]

Q8

fck

QUaternion group

8 elemnts

[rotund aurora]

Mmh interesting

[topaz solar]

That, or finding an order p = p^1 subgroup, which goes from the bottom since you need one of those anyway. And you can find one since that one thing about finding a cycle for prime divisors

[rotund aurora]

Yeah Pauli matrices

[rotund aurora]

Yeah you can do that

[crystal turtle]

Yeah Q8 is a fun example of that, I asked that question in my group theory class and the prof added it as an assignment lol

[south patrol]

Hm let $R$ be a commutative ring, $n \ge 1$ an integer and $E_{ij} \in M_n(R)$ be the matrix with $(i,j)$th entry $=1$ and all other entries $0$. I'm trying to show that if $Y$ is a left $M_n(R)$-module, then $Y \cong X^n$ where $X = E_{11} \cdot Y$ and $M_n(R)$ acts on $X^n$ by matrix multiplication (on the left). I can show that $Y = \bigoplus_i E_{ii} \cdot Y$ and that there are isomorphisms $E_{II} \cdot Y \cong E_{11}\cdot Y$ of $R$-modules, but I'm struggling to show that this assembles to an actual $M_n(R)$-module isomorphism - any ideas?

[cloud walrus]

potato

[south patrol]

I mean the isomorphisms $E_{11}\cdot Y \to E_{ii} \cdot Y$ are just given by multiplication by the matrix $E_{i,1}$, so we have a map $X^n \to Y$ given by adding together the maps $E_{i,1}$, but idk how to show that is an $M_n(R)$-module hom

[cloud walrus]

potato

[rocky cloak]

Are you able to show that
M_n(R) is the direct sum of M_n(R)E_ii as a left module?

If so you have that Y = Hom(Sum M_n(R)E_ii, Y) = Sum E_ii Y

[south patrol]

Hm I already have that

[rocky cloak]

Well then your done

You could also just do it explicitly, like write out what the matrix E_i,j does to both sides, and see that this is a homomorphism

[south patrol]

Hm how am I done

Okay sorry I now realise I wrote it unclearly (probs cause i am too tired)

I meant I'm struggling to show that the maps all assemble into an isomorphism X^n -> Y

where i mean like you send (E_{11} . Y )^n -> sum E_ii Y = Y via the various isos

However, I've now done it - I just used the fact that that map is R-linar, so it suffices to check it on the various E_{ij}

and that removes much of the clutter :)

[terse crystal]

Any subgroup K of order p^k, the orbit X containing K (under conjugate operation by G) , you have either |X|=1 or p| |X|.

G conjugate acts on subgroups of order p^k I meant

[white oxide]

I’m afraid to read this since your hints are usually solutions

[crystal turtle]

[terse crystal]

But you don’t need it to be a Mn(R) module isomorphism. X is not even a Mn(R) module. It’s X^n that is a Mn(R) module by matrix multiplication. R modules isomorphism E11 Y -> Eii Y is sufficient for us to have X^n isomorphic to sum Eii Y as Mn(R) modules since Mn(R) acts as matrix multiplication on both sides…
It’s like R module isomorphisms Xi -> Yi gives us Mn(R) module isomorphism sum Xi -> sum Yi I think?
I am not sure about your multiplication by Ei1 part… I multiply by P=I+E1i+Ei1-E11-Eii

[terse crystal]

I am not sure. I only revealed orbits either have 1 or have tp many elements. Now recall what normal subgroup means

[delicate orchid]

11 hours late but I don’t care

Q_8 is basically the ONLY example of a Hamiltonian group

I say basically because you can direct product with an elementary abelian 2-group and a finite abelian 2’-group to get more but that’s cheap

[topaz solar]

I know the pi \pred \rho shenanigans can be applied elsewhere and all, but what the heck would the \subset one look like if we don’t have 1 on the left?

I feel like this is the wrong direction to not know what’s going on but I’m kinda working backwards so

[void cosmos]

yo yo

is a not necessairly direct sum

of projective R-modules projective?

no right?

cuz i dont have that universal property

[south patrol]

i mean showing you don't have the UP is equivalent to showing it's not projective

Trying to think of an explicit counterexample

[void cosmos]

what

what up

universal propertty?

okay

yea thats what i thought cuz the proof is literalyl just the "up"

over Z_6 , Z_2 and Z_3 are both projective R-modules

what is Z_2 + Z_3 haha

[south patrol]

it is still projective

assuming you mean like

3Z/6 and 2Z/6 as subgroups of Z/6

[void cosmos]

yea

its on an exam paper sadly

so its probably a typo

R is a ring with 1 does that do anything?

[south patrol]

to most people rings always have 1 so doesn't really change it lol

[void cosmos]

yeah ig

for me its literally a heaven moment

[delicate orchid]

Trying to slam the diagrams together in my head

[south patrol]

https://math.stackexchange.com/questions/4257596/is-the-internal-sum-of-projective-submodules-projective?rq=1

[void cosmos]

cool cool

tysm everyone

except ttpa

[summer path]

How hard do you need to slam your head to slam the diagrams together in your head

[south patrol]

only thing is some of these have infinite sums

but the 2nd answer is nice

Ax + Ay as an A-module where A = k[x,y]

[void cosmos]

how is that not projective

wait

let me rea it

read*

[south patrol]

not free

and projective <=> free over k[x,y]

[delicate orchid]

40kN

[void cosmos]

trueeeeeeeeeeeeeee

yea that was nice

ty

[delicate orchid]

Now the question becomes “when is it true?”

PID is fairly obvious I think

[south patrol]

internal sums are horrible

[delicate orchid]

There’s a difference?

[south patrol]

sum, not direct sum

[delicate orchid]

Oh well nobody cares about internal sums

[south patrol]

direct sum would be easy

[delicate orchid]

Ok yeah this question sucks

[south patrol]

ye

lol

[delicate orchid]

I wondered why it wasn’t just “by the universal property of the coproduct ur chatting shite”

[south patrol]

But this has reminded me of the Quillen-Suslin theorem

idk the proof

[delicate orchid]

idk the theorem

Got the word sus in it that’s kinda funny

[south patrol]

projective = free over poly ring

[delicate orchid]

Ah ok

[south patrol]

apparently there is now a nice proof but originally v hard it seems

well like part of quillen's fields lol

[void cosmos]

yea i just did it but i juist realized

that i used the UP of the direct sum

and that like h o i = g o f o i --> h= g o f something like that

which is the UP right?

[south patrol]

Not really

Depending on how you phrase it lol

[delicate orchid]

No not really

[low wyvern]

👍

[delicate orchid]

If you know what a set and a function are you’re good to go imo

[low wyvern]

good stuff, I think I know what those are 🙂

[void cosmos]

can i do that tho?

[south patrol]

Idk what you are saying tbh lol

What is i for starters

[void cosmos]

inclusion

[south patrol]

Which one

[void cosmos]

😄 what

the one that comes with the direct sum

[south patrol]

[delicate orchid]

any includers ?

[south patrol]

Then your UP is just saying i is a monomorphism which isn't particularly enlightening

But there is one inclusion for each summand

[void cosmos]

yea truee

[delicate orchid]

oh Universal Property

[south patrol]

so that's why it is odd to say "the inclusion"

[void cosmos]

oh got it

what i said was

since for every P_i u can find a map to A ( projective )

u have a map from the direct sum to A

such that ....

then i got to g o f o i = h o i

( i want to show g of = h )

[south patrol]

Yes you can lift individually and then glue together

[void cosmos]

🏋️‍♂️

[coral shale]

equivalence relations, quotient set.

[low wyvern]

Ok thanks

[rocky cloak]

I think if R is a Dedekind domain than any (finite) sum of projective modules is projective. At least assuming the projectives are finitely generated

Nakayama algebras also has this property, but for stupid reasons.

[mighty kiln]

Elementary number theory

[low wyvern]

uh well I know none of that 😦

[rotund aurora]

good observation, specially the arithmetic in Z/nZ

[mighty kiln]

(actually-elementary NT, not the field "elementary NT")

Learn modulo arithmetic (up to inverses)

[low wyvern]

ok

[mighty kiln]

You may also encounter stuff like Euler totient function, Fermat's little theorem, Chinese remainder theorem etc

[low wyvern]

in pure maths anyway

[mighty kiln]

Tho treatment of those in groups/rings may look different from treatment in NT or oly math

[low wyvern]

ok

I guess I have a week and a half to pick up some number stuff 🙂

[south patrol]

hm you probably won't need much elementary nt right

[delicate orchid]

you don't really

[south patrol]

like most of those things can be easily proven using groups

i didn't use any in my first course for example

[delicate orchid]

the only thing that comes to mind is bezout's

[south patrol]

oh we just proved bezout as part of the course though lol

[mighty kiln]

Dis prob not prerequisite ye

[delicate orchid]

maybe knowing about modular arithematic for examples would be good

[delicate orchid]

yeah, same for mine

[south patrol]

but yes that is true ^

[mighty kiln]

But may encounter

[low wyvern]

ok

[south patrol]

i think understanding modular arithmetic is good

[low wyvern]

I can just get this from a disctrete maths book right?

[mighty kiln]

Ye cuz many example come from mod

[livid willow]

D&F chapter 0 covers it iirc

[white oxide]

does this problem require any linear algebra?

[rocky cloak]

Not really no.

Just requires you to know about modules over a pid

[white oxide]

okay that's good to know, i'm certain the next problem does

yea i have to review that, that concept is very weak for me

i have a vague idea (?) of how to approach this - would it be similar to classifying a finitely generated abelian group in that you look at the order of the elements?

[delicate orchid]

the finitely generated abelian group case is the PID-module structure theorem but just specifically over Z

[white oxide]

lol ya i guess that makes sense

all abelian groups are Z-modules right

[delicate orchid]

yus

[south patrol]

lol "thinking of x as a prime element"

[white oxide]

sorry but what does that even mean

[delicate orchid]

it just straight up is a prime element lol

[white oxide]

oh wait that's fax

wait dumb question, by $\frac{k[x]}{(x - 1)^2 x^3)}$ he means $k[x]/((x - 1)^2 x^3)$ right

[cloud walrus]

okeyokay

[crystal turtle]

of course?

[delicate orchid]

have some confidence in yourself dude

[white oxide]

i don't have any confidence in myself when it comes to math

[open sluice]

you should probably fix that

[white oxide]

hey at least i can list the axioms for a group

[coral shale]

lets hear it.

time to shake you up on this one

[white oxide]

Certainly! The axioms for a group are the fundamental properties that a set G with an operation • must satisfy to be considered a group. There are four axioms for a group:

Closure Axiom (Closure Under •):
For all elements a and b in G, the result of the operation a • b must also be an element of G:
∀a, b ∈ G, a • b ∈ G.

Associativity Axiom (Associative Property):
The operation • on G must be associative, meaning the grouping of elements does not affect the result of the operation:
∀a, b, c ∈ G, (a • b) • c = a • (b • c).

Existence of an Identity Element:
There exists a unique element e in G, often denoted as the "identity element" or "neutral element," such that for any element a in G, the following equalities hold:
∀a ∈ G, e • a = a • e = a.

Existence of Inverse Elements:
For every element a in G, there exists a unique element a⁻¹ in G, often called the "inverse" of a, such that the following equalities hold:
∀a ∈ G, a • a⁻¹ = a⁻¹ • a = e.

[open sluice]

meaning the grouping of elements

circular definition, try again

[chilly ocean]

the "Certainly!" makes this read like chatgpt

[coral shale]

the closure axiom. cringe

[white oxide]

no? I typed this out myself!

[coral shale]

There exists a unique element e in G
Interesting choice, but this is not standard, let's say.

[chilly ocean]

i said it "read like chatgpt" not was chatgpt

[coral shale]

but I think I prefer this possibly

cant remember why for now

For every element a in G, there exists a unique element a⁻¹ in G
I dont agree with (as in like) this, however 🚫

too strong

[white oxide]

why can't i react to this

[coral shale]

someones blocked

[white oxide]

😹

[coral shale]

Let G a set, * : G x G -> G, an associative binary operation.
(G, *) is a group iff the identity and inverse axioms hold.

The weaker form I'm used to seeing is:

• (exists e)(forall g in G) eg = ge = g
• (exists an identity e)(forall g in G)(exists h in G) gh = hg = e

Then your 1st 2 props will be showing identity unique, inverse unique

[white oxide]

also isn't it enough to say left identity and left inverse exist and then you can prove existence of their right counterparts

[coral shale]

I asked about this a good while back tbh #groups-rings-fields message

[coral shale]

possibly, not so sure.

gut instinct... that doesnt quite sound right to me ngl

in a monoid, you can have things with left inverses, no right

[uncut rune]

Any tips on blackboxing what an inverse limit is? for someone who doesn't care about algebra

[white oxide]

this is p cool

[summer path]

what are you looking for when you are asking for "tips" to blackbox

[uncut rune]

just a working definition that I can use for probability

[rotund aurora]

the definition is what is in the middle no?

[coral shale]

thats interesting

right, i believe the all elements having a left inverse forces it

or at least, thats the only difference i can discern from the monoid case

[uncut rune]

Sorry, idk I'm just trying to know how to think about them, definition on wikipedia is not helping

[rotund aurora]

I think you can draw the diagram

[delicate orchid]

if you keep throwing out structure you can have left/right identities too

[uncut rune]

nvm they have an appendix on this stuff that I need to read through

[open sluice]

i learned that this kind of result existed from https://www.youtube.com/watch?v=KufsL2VgELo but i didn't know the precise statement

[white oxide]

ok so brief overview/attempting to remember the proof for structure theorem for f.g. modules over a PID: 1. show that every f.g. module over a PID is equal to a direct sum of a torsion module and a free module 2. show that this torsion module is equal to a direct sum of A(p)s where A(p) is the subgroup of all a in A such that some power of p kills a 3. show that each of these A(p)s is a direct sum of cyclic R-modules (this follows from the assumption that every a in A(p) is of order a power a prime) 4. combine everything and enjoy

[white oxide]

oh woah

that's sick

[rocky cloak]

In general, if an element has a right inverse and a left inverse, then the two inverses are equal.

(and are two-sided inverses)

[coral shale]

am aware

[south patrol]

Ye

[white oxide]

now to actually see if i can apply the theorem to classifying such modules..

this proof has so much machinery involved lol

[white oxide]

why is it that C contains an element of order n2/why do we need that condition? why is it not C contains an order of p^n_2?

[south patrol]

seems a typo to me

[white oxide]

thank god i'm sane

[white oxide]

wait i'm confused, how does (i) hold? it seems that all of the proofs have led up to condition (ii) - does it have to do anything with condition 6.11?

[pale jasper]

Is my proof okay?

[chilly ocean]

why does x being non-zero in R imply that it has an inverse?

[pale jasper]

R is a ring

[night onyx]

it doesn't in general, but if x isn't in I then consider the ideal (I, x) generated by I and x, and consider that I \subset (I, x) but I is maximal

[open sluice]

consider R = Z

[pale jasper]

oh fuck

I'm rusty

I got field and ring mixed up

[chilly ocean]

Hello

If we have a ring A with x^3 is in {0,1} for every x in A, find A

We dont know if A is finite or not

We can see Z_2 and F_4 fields work

I picked x which is not 0 and not 1

Oh another thing 1+1=0 so char A =2

And I showed x^2=x+1, (x+1)^2=x and x^3=1

[rocky cloak]

Yeah, so the module is a direct sum of modules of the form R/p^n , using lemma 6.11 you can group all those with the largest exponent together into one cyclic module, which will be R/rt

[chilly ocean]

Now I want to show that A doesnt have order >4

[white oxide]

ohhh, right! thank you

[chilly ocean]

I picked y in A different by 0,1,x,1+x

Again y^3=1, y^2=y+1 and (y+1)^2=y

How I can show A commutative

Beacuse if A commutative I can get a contradiction

We dont know if it is a boole ring right?

If x=x^2 then x=x+1 and false

So it is not a boole.. if there is x≠0,1

[rocky cloak]

Are you missing some assumptions?

Because F2[x,y]/(x, y)^2 has order 8.

[chilly ocean]

Yes but doesnt have the proprety

That every x in the ring

[rocky cloak]

Wait, yeah I'm being silly

[chilly ocean]

x^3 is 0 or 1

[cloud walrus]

okeyokay

[primal tusk]

not to distract from the thread above but are homomorphism between rings unique?

[rocky cloak]

Not usually, no

[primal tusk]

i cant think of a counter example

[rocky cloak]

R[x] -> R[x]

Given by g(x) |-> g(f(x))

Is a ring homomorphism for any choice of f

For example

[open sluice]

complex conjugate

[night onyx]

Or just x -> nx in Z

[rocky cloak]

For any ring there is a unique ring homomorphism Z -> R though

[primal tusk]

wait so is it only between the same ring that this holds?

[rocky cloak]

No that was just a random example.

For example

Z[x] -> R
g(x) |-> g(r)

Is a ring homomorphism for any ring R and element r in R

[primal tusk]

ohh true thank you

those are subrings

so what about rings that arnt subrings?

[rocky cloak]

?

[primal tusk]

wait

[rocky cloak]

Which rings are subrings?

[open sluice]

when you say homomorphisms between rings you mean two homomorphisms f, g between the same two rings R and S?

[primal tusk]

yea

[primal tusk]

i confused R with Z (the integers)

i guess i just see how R could be Z and so Z is a subring. Also since theres always a unique homomorphism between Z and R im trying to think about what the limitations are of this

i barely understand any of this i feel like i never really got a deep understanding of homomorphisms. Like what is the connection between distinct homomorphisms between two rings?

or am i just thinking this is some possibly deep thing and its not important

[rocky cloak]

I'm not sure I understand your question, like you're asking how two homomorphisms can be related?

[primal tusk]

yea like if i look at the set of all homomorphisms between two rings is there some interesting connection between these homomorphisms

[topaz solar]

Consider Z[x] -> R

Pretty sure each of those are determined by where it sends x (as long as you’re sensible with unital rings)

[echo path]

Can someone help

Composite function

[primal tusk]

yea that makes sense

[open sluice]

there are exactly two field automorphisms on C (isomorphism from C to itself) that preserve the real line
namely the identity function and the complex conjugate

[topaz solar]

Complex automorphisms which don’t preserve the real line

[rocky cloak]

I mean there are ways that homomorphism can be related, like they can be conjugate. There are also rings for which the set of homomorphism from that ring has some nice interpretation. Like

Hom(Z[x], R) can be identified with R and
Hom[Z[x, x^-], R) can be identified with the unit group of R

[rocky cloak]

And there is exactly one field automorphism on the real line.

[primal tusk]

what is Z[x, x^-] ?

but that does make sense

[open sluice]

do they exist
idk if they do this is the limit of my knowledge

[primal tusk]

to me learning that isomorphic rings are the same ring was super awesome and im really trying to get the same understanding with homomorphisms. The best analogy i can make is that they are somewhat similar to going from an abstract vector space to R^n but thats an isomorphism so...

[rocky cloak]

Polynomials where you allow negative powers of x

[topaz solar]

Something something transcendence basis & choice

[open sluice]

pain

[topaz solar]

But they’re discontinuous

[white oxide]

sorry, i forgot to post the question

anyone?

[topaz solar]

Think of it more like a linear map

And A -> B homomorphism can sometimes be thought of as an interpretation of one in the other

Think first iso, a surjective A -> B gives you an iso A/ker <-> B

So you’re interpreting B in terms of A, in that sense

Or, if you know B well, each map A->B can give you a bit of information on A, since A/ker f <-> image of f

So this is interpreting A in terms of B in a sense, but that’s a thing from looking at all the maps (and will not always tell you everything about A)

[rocky cloak]

Maybe you want to start by splitting M into indecomposable cyclic modules

[topaz solar]

also I still have no idea how the subset containment thing should look even after napping L

[white oxide]

do you mean M or M(x)?

[primal tusk]

this might be the dumbest thing ill ever say in this server, but do you think that this information is related somehow to information theory?

[topaz solar]

No

You could probably force it to be in some cases

But no

[rocky cloak]

Presumably

f < g if

||f(x)ksi - g(x)ksi|| < epsilon ksi

[topaz solar]

Right but that’s the curvy \pred

[rocky cloak]

M first, then use that to find M(x)

[white oxide]

hm ok

[topaz solar]

Should it be like, fk - gk = 0 or smth

[primal tusk]

oh yea since all maps dont tell you everything about A

[topaz solar]

For some \xi or wtv

[primal tusk]

what if there existed a ring R not isomorphic to A such that if you knew all maps from A to R then you would understand A completely ?

[topaz solar]

I know some characterization of \pred in terms of ideals but idk about \subset

[rocky cloak]

Seems weirdly symmetric though...

Such confusing notation

[rocky cloak]

Welcome to representation theory my friend

[topaz solar]

That’s why I’m confused

[rocky cloak]

Yeah me too

[primal tusk]

no idea what that is other than people saying its hard lol

[topaz solar]

F.1.1 on the latter & F.4.4 on the former is kinda what I’d like to see something akin to

[ivory trail]

the idea of maps losing information can be used to think about "observing" a hidden state via a map A -> B where you can directly see B

[white oxide]

i'm sorry, i have no idea how to do this/start this; i know that in the case of when R = Z, and we're working with a finite abelian group, you just find the prime decomposition of the order and check the order of the elements. here I have no idea how to start, since I don't even think M is finite

[ivory trail]

which is kinda used for hidden Markov models for example

the vague idea is related

[rocky cloak]

Hmm, so 1_G will just be like the trivial representation. And this condition is that pi has the trivial representation as a submodule. But the dimensions don't quite match, so maybe some multiplicity stuff...

[topaz solar]

Ye 1 is the trivial rep

They do some “functions of positive type” stuff but I’m not familiar with this subject so

This is mostly stuff in terms of the representations rather than some modules

[rocky cloak]

So you do exactly the same thing, but instead of considering the primes in Z you look at the primes in k[x]

[topaz solar]

I know that the \pred one doesn’t depend on multiplicities though

[white oxide]

hell i'm having trouble seeing that M is finitely generated 💀 bc (x, x) is not a generator and same with (1, 1) since it's over R

[rocky cloak]

Finitely generated does not mean generated by a single element. It means generated by finitely many. For example 2 in this case

(1,0) and (0,1)

[white oxide]

yea but isn't it over R? so if we're multiplying by elements of R how do we get say (x^7, 2x^5)

are we viewing it as a k[x] module?

[primal tusk]

thats really interesting im gonna look more into that

[rocky cloak]

If you multiply (1, 0) by x^7 you get (x^7, 0)

[white oxide]

oh so the scalars do come from k[x]

[rocky cloak]

Yes

[white oxide]

oh i thought it was just some arbitrary PID R

ok hopefully that makes life easier

thanks

so how are we supposed to characterize M if we don't know it's order?

oh would it just be the decomposition of (x-1)^2x^3 into irreducible polynomials so like R/(x - 1)^2 and R/(x^3) and then direct sum with the other decomposition?

so I could just think of it like

decompose $\mathbb{Z}/(p_1^{n_1}p_2^{n_2}) \oplus \mathbb{Z}/(p_3^{n_3}p_4^{n_4})$ into a direct sum of cyclic modules?

[cloud walrus]

okeyokay

[white oxide]

well here p3 and p4 would be equal since they're both x

ok, so $k[x]/((x - 2)x^2))$ is either isomorphic to $k[x]/(x - 2) \oplus k[x]/(x^2)$ or $k[x]/(x - 2) \oplus k[x]/(x) \oplus k[x]/(x)$. but $k[x]/((x - 2)x^2))$ contains an element of order $x^2$ (namely $(x - 2)$) so we must have $k[x]/((x - 2)x^2)) \simeq k[x]/(x - 2) \oplus k[x]/(x^2)$

[cloud walrus]

okeyokay

[white oxide]

right?

so $k[x]/((x - 1)^2x^3) \oplus k[x]/((x - 2)x^2) \simeq k[x]/(x - 1)^2 \oplus k[x]/(x^2) \oplus k[x]/(x - 2) \oplus k[x]/(x^2)$?

[cloud walrus]

okeyokay

[crystal turtle]

[white oxide]

pls 😭😭😭 if someone could verify I would be very grateful

[rocky cloak]

You swapped an x^3 for an x^2 there, but otherwise correct

[white oxide]

Thank god

Thank you so much

[warm saffron]

Had a question I was hoping someone else can confirm for me. In the following screenshot, isn't the "g in G" before the sentence beginning "show that", actually redundant.

I'm saying that "Let G be a group and g in G" should really just be "Let G be a group."

[crystal turtle]

nevermind

[open sluice]

i think the g in the definition and the g in the problem statement aren't the same g, this is an abuse of notation

[warm saffron]

yea it's redundant because it's defined in Z(G)

[crystal turtle]

I didn't read that it was center. In fact yeah this is bad to have the g in G before

[open sluice]

but that first g doesn't make sense i think...

[warm saffron]

okay thanks for the cinfirmation

believe it or not this came from a textbook

lol

[crystal turtle]

was it D&F

[warm saffron]

nah it's called Abstract Algebra Theory and Applications by Thomas Judson

and then I had another question too I was hoping you guys could confirm for me as well

[open sluice]

o judson

i didn't use that but i've seen people recommend it

[warm saffron]

I mean it's a good text but yea these kinda mishaps aren't helpful ahaha

anyway, my other question is:

I'm asked to prove

If $(xy)^2=xy$ for all $x,y\in G$, show that $G$ is abelian.

[cloud walrus]

logician

[warm saffron]

that's literally the problem statement

I can assume G is a group, right?

it doesn't make sense otherwise

[crystal turtle]

Unless your text is considering more general stuff like semigroups, then yeah and then it would be bad to use G for semigroup

But in a group theory section, G = group

[warm saffron]

yea that's the notation they've stuck to very on and off

[warm saffron]

I think I can assume G is a group here because I mean it doesn't really make any sense otherwise. I won't be able to prove associativity or anything else since the operation isn't clear

[crystal turtle]

yes G is a group

[warm saffron]

thanks, god I love this textbook

https://tenor.com/view/dog-meme-gif-25326537

lol

[crystal turtle]

I mean eventually this gets absorbed into your assumptions lol. Once you mode past basic defs of binary operations, you'll always be working with some specific structure. And it's probably in these sections gonna be a group structure; if it wasn't, they'd make it clear

[warm saffron]

that's what I was thinking. We're in the subgroup section rn so we should already know group definitions

thanks for the confirmation and help

[rocky cloak]

This is kind of a weird problem, since the trivial group is the only one that satisfies this

[crystal turtle]

yeah

[warm saffron]

I mean the trivial subgroup group {e}=G satisfies it for sure, but wait how do we know it's the only group that does?

[rocky cloak]

Let y equal e, then x^2 = x, cancel one x gives x = e

[warm saffron]

hmm

that was actually a question on one of my quizzes lol

it said show that given a group G, x^2=x iff x=e. super easy

but yea it's a special case of the other problem

[rocky cloak]

(xy)^2 = e would have been at least a little more interesting

[warm saffron]

true. I'm gonna ask my prof tmrw if any other groups G satisfy (xy)^2=xy for all x,y in G.

It'll be interesting to see if he comes up with anything else other than {e}

[crystal turtle]

It's just {e}

[crystal turtle]

for this reason

[delicate orchid]

<x,y | xyxyy^-1x^-1> there

[crystal turtle]

since this shows any arbitrary element x must in fact be equal to e

[warm saffron]

hmmm

[open sluice]

that problem statement seems a bit stronger than intended

[warm saffron]

yea

[crystal turtle]

I mean

[delicate orchid]

just incase you can't see this, pick whatever element "z" you want and then set y = x^-1z

[warm saffron]

I'm gonna let this stew in my mind for a bit. x=e iff x^2=x makes sense to be rn, but not much else since it seems like that's only when y=e

[rocky cloak]

Yeah I feel like (xy)^2 = e would have been a much more natural exercise

[crystal turtle]

what's the point of having (xy)^2 = xy for all x,y anyways? Why not just show x^2 = x for all x implies trivial?

All around strange

[warm saffron]

I love this textbook

[delicate orchid]

noooo my wholesome idempotents noooo

my block decompositionrinos noooo

[warm saffron]

wait now I get why it's only {e}

because if y wasn't equal to x=e, then y=xy=(xy)^2=y^2 which is the same as x^2=x since x=e.

meaning every element equals the square of itself

[rocky cloak]

The point is that if something is true for all x and y, then it is in particular true when y=e

[warm saffron]

right

[warm saffron]

I was using the fact here that if G was a group containing e and some other element y, then since G is closed under the operation y=ey=(ey)^2=y^2. Which only happens (based on the problem I proved on my quiz) exactly when y=e.

since y=e is both a necessary and sufficient condition for y=y^2

aight yea now I see it

[crystal turtle]

ring multiplication allowing for nontrivial elements to satisfy x^2 = x

[warm saffron]

don't scare me away from rings lol...I'm still in group theory

obviously

LMAO

[rocky cloak]

Scary things like 0

[open sluice]

oh judson does groups before rings

[crystal turtle]

non-0,1 idempotents

[warm saffron]

good old judson

https://tenor.com/view/byuntear-meme-react-funny-dog-gif-25019867

[rocky cloak]

Still not scarier than 0

[crystal turtle]

but 0 is the most understood element!

[south patrol]

Not just redundant, confusing

[warm saffron]

indeed

I love the textbook we're using

[white oxide]

here, would $M(x) = k[x]/(x - 1) \oplus k[x]/(x - 2)$?

[cloud walrus]

okeyokay

[white oxide]

and the generators would be $(x - 1)$, $(x - 2)$?

[cloud walrus]

okeyokay

[white oxide]

wait a minute.. that wouldn't be possible since (x - 1) isn't even in M(x)

wait is it just that $M(x) = k[x]/(x - 1)^2 \oplus k[x]/(x - 2)$, with generators $(x - 1)^2, (x - 2)$?

[cloud walrus]

okeyokay

[white oxide]

since $f(x)x^i \in \bigl((x - 1)^2x^3\bigl)$ implies that $f(x)$ is a multiple of $(x - 1)^2$

[cloud walrus]

okeyokay

[white oxide]

actually wait no

okay, here's my thought process: so $M(x) = \bigl((x - 1)^2\bigl) \oplus \bigl((x - 2)\bigl)$. now Ann($(x - 1)^2$) = $(x^3)$ and Ann$(x - 2)$ = $(x^2)$. thus we have $M(x) \simeq k[x]/(x^3) \oplus k[x]/(x^2)$, right?

[cloud walrus]

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

[white oxide]

<@&286206848099549185>

😭

sorry i don't want to be annoying anyone else can use this channel ofc

just don't want to die on my midterm soon

[noble saddle]

I "know things about modules" and can help you, but I don't recognize the notation M(x)

what do you mean by that?

[white oxide]

thank you!

yea it's kinda weird notation

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

[cloud walrus]

okeyokay

[noble saddle]

Okay so you need conditions for when a polynomial is killed by some power of f(x)

you're on the right track, but you were checking for things that are killed in either the left or right factor of the direct sum. It needs to be killed in both

So (x-1)^2 is not annihilated by a power of x

[white oxide]

ohhh i see...

actually yea that makes sense

ok! i'll try working on that now then

[white oxide]

oh so if I had my generators for M(x) as (x - 1)^2 and (x - 2) separately, then there would be some f(x)(x - 1)^2 that gets killed by some x^i, but this wouldn't necessarily get killed in the right factor, if i understand correctly?

[noble saddle]

Yes, and you could even take f(x)=1. x^n (x-1)^2 will never be 0 in k[x]/((x-2)x^2)

[white oxide]

oh sweet yea i just literally wrote down that counterexample

ok thanks, that makes a lot more sense then

so our generator would just be (x - 1)^2(x - 2) then

[rapid junco]

how does one prove g^a^b = g^b^a for groups?

[noble saddle]

wait sorry I'm being an idiot. ((x-1)^2, 0) is an element of M(x)

you were originally correct

M(x) is generated by ((x-1)^2,0) and (0,x-2)

[white oxide]

ohh ok no worries

oh yea i forgot about components and everything lol

so $M(x) \simeq k[x]/(x^3) \oplus k[x]/(x^2)$ then?

[cloud walrus]

okeyokay

[open sluice]

(put parentheses because it looks like g^(a^b) = g^(b^a) rather than (g^a)^b = (g^b)^a)
you could probably formally approach it by induction
but it's easy enough to just expand it: b copies of g^a vs a copies of g^b

[rapid junco]

yeah

induction is just pain though with this

so boring

[noble saddle]

you can prove this with the 1st isom theorem, yes.

[white oxide]

Since every cyclic module C over a PID R is isomorphic to R/(r) where (r) = Ann(a), and (a) = C

oh hokay

yeah this result ^ is pretty easy to see with firs tiso

first iso

[white oxide]

yo I'm a fucking idiot, I'm assuming that every short exact sequence of unitary $R$-modules is split exact, and trying to show that every unitary $R$-module is hence injective. So I've let [\begin{tikzcd}
& I \
B & A & 0
\arrow[from=2-3, to=2-2]
\arrow["g", from=2-2, to=2-1]
\arrow["f", from=2-2, to=1-2]
\end{tikzcd}]
be any diagram with bottom row exact, the issue is that I can find things mapping from $I$ but not things mapping to $I$ if that makes any sense. would appreciate any hints

[cloud walrus]

okeyokay

[white oxide]

it's also annoying since the bottom row has to be any arbitrary exact sequence, so I can't put a factor of I into there or do any tricks

pls tell me this is nontrivial this problem is making me feel like a moron

[sonic coral]

I need help showing that if G is a finite group then the other or a is odd if and only if a = (b_k)^2^k for some b_k in G. I have done the direction where i’ve supposed that the other of a is odd

[rocky cloak]

So you assume that every exact sequence splits, is there perhaps an exact sequence in your diagram?

[rocky cloak]

Pick k such that 2^k doesn't divide the order of G. What can the order of b_k be? What will then the order of (b_k)^2^k be?

[summer path]

Why does Fulton and Harris use $\mathfrak{S}_n$ for symmetric group :c

[cloud walrus]

:3c

[delicate orchid]

that's fairly common tbh

[summer path]

I keep reading it as a G

What is G_n

[crystal turtle]

Common ≠ good

[night onyx]

to be fancy of course

[delicate orchid]

well for n = 36 it's this ;3

[open sluice]

how the fuck is that an S

[mighty kiln]

Fraktur be like

$\mathfrak A$

[cloud walrus]

Arki

[mighty kiln]

21

[sonic coral]

i was told to use the contrapositive and show a contradiction where the order of the element is larger than the order of the group

[topaz solar]

Fraktur y is an abomination

[mighty kiln]

Let's play: guess the letter

[ashen heron]

V?

[rocky cloak]

B?

[rocky cloak]

That works too I guess

[summer path]

$\mathfrak{Y}$

[cloud walrus]

:3c

[summer path]

What

[delicate orchid]

D

[open sluice]

why even use mathfrak

[delicate orchid]

medieval monk aesthetic

[mighty kiln]

How

[old spire]

Why are the exceptional root systems labelled using three different letters, instead of a single letter?
G2, F4, E6, E7, E8
Is there any reason for using three different letters E, F, G
Why not E2, E4, E6, E7, E8?
GPT-4 says there's no reason at all, and Google only gives me clickbait
(Google is useless)

[delicate orchid]

GPT-4
hahahahaha

[old spire]

why?

[delicate orchid]

just look at the coexter graphs

[crystal turtle]

Don't ask gpt math lmao

it is never good

[delicate orchid]

it's pretty obvious why they're labelled in 3 different families

[old spire]

It sometimes helps, actually

[old spire]

I'm responding to this

[old spire]

Well, not really, because you can't generalise from finitely many examples

"Problem of induction" it's called.

[delicate orchid]

what

[broken pollen]

huh?

[old spire]

I meant finitely mean

Typoed

[delicate orchid]

the top 4 are literally infinite families

I have absolutely no idea what you're talking about

[south patrol]

Are we going Hume

[old spire]

That's not what I said

[south patrol]

That's not really what the problem of induction is

[delicate orchid]

there is no "E9"

[old spire]

Basic critical thinking and rigour: You can't generalise an infinite rule from finitely many examples.

[delicate orchid]

there is no "F5"

[old spire]

You're trying to do that by looking at the Dynkin diagrams, which is nonsense

You can't generalise other Es, Fs and Gs, and that way justify the naming.

[delicate orchid]

you're making literally zero sense lmfaoooo these ARE the entire families

there is only one F, one G, and three Es

these were classifed about 100 years ago

[old spire]

I said that there's no reason to use three different letters instead of just one

[delicate orchid]

they have completely different structures

[old spire]

What are they?

[delicate orchid]

the three familes in question

[old spire]

What are they? Why use three letters?

You're just being rude and not helping

And you're agitating me

[delicate orchid]

all of the Es are simply laced for example

[delicate orchid]

you're the one who called me an idiot

[old spire]

I blew my gasket. Sorry

[delicate orchid]

apology not accepted

reflect on your behaviour

[old spire]

🤷‍♂️

Anyway, there might be some logic to a naming scheme that uses three different letters, but looking at Dynkin diagrams isn't enough to find it.

I mean, it's suggestive.

But like, Hume's problem of induction, and all that.

[coral shale]

hell is hume doing in aa

get him out, get him outa here

[rocky cloak]

At least in the simply laced case, every graph corresponds to a root system, so the Es do fit into an infinite family. But E9 and above are not finite root systems.

I think you can define root systems for generalizations of F and G as well, but I'm not quite sure how that works

[south patrol]

I don't see what Hume has to do with any of this

[delicate orchid]

Hume's theorm of induction is a critism of the scientific method

[south patrol]

Nowhere are we justifying anything using finitely many examples anyway

[delicate orchid]

not mathematicians noticing structual patterns and chosing names to reflect them

the names are descriptive not prescriptive

[coral shale]

if we did what hume laments against, we would have already proven collatz and riemann. They would be uh laws, just like physical laws

open to be disproved but taken to be true

[old spire]

But are the names derived from some general (infinitary) principle?

It seems someone suggested there are contexts in which something like E_n exist for general n, but those wouldn't be root systems

Because using 3 different letters suggests there's an infinitary generalisation for the Es, Fs and Gs.

[rocky cloak]

You even get a corresponding lie algebra

https://en.m.wikipedia.org/wiki/Kac–Moody_algebra

Though it will be infinite dimensional if it's not a Dynkin diagram

[old spire]

Otherwise I don't know why things are that way.

[coral shale]

i dont even know what this is, but its clear to me the subscripts refer to the graph degree

[delicate orchid]

also rip my boys H_3 and H_4

[old spire]

The letters, not the subscripts, damnit

[coral shale]

then ur gonna find yourself with some clashes

if u insist theyre all A_n

[old spire]

I'm not saying that

[coral shale]

because there are multiple possibilities for a 4-graph

[old spire]

Like, why are you here and trying to argue with me?

There's a good reason for A_n, B_n, C_n and D_n. No one's arguing about that.

That's obvious.

[coral shale]

did u just want E678 to have different letters

[delicate orchid]

@rocky cloak do the extended familes have coexter-esque systems associated with them? reading the wikipedia page their presentations are very coextery

and if they do are their coexter graphs the obvious extenstions of the finite cases

[coral shale]

lol and its worth noting ABCDn dont even make sense in the way you want it because they have different restrictions on n

[delicate orchid]

wdym

oh right yeah the n > garbage

[coral shale]

theyre probably complaining about the subscripts implying the objects are defined for all natural N

but thats not the case

defined for some subsets

[crystal turtle]

A starts at 1, B starts at 2, C starts at 3, and D starts at 4. Makes sense.

[old spire]

Is there a classification of these?

[open sluice]

where's 0

[crystal turtle]

0 is not real. The mathematicians who came up with 0 are delusional

[rocky cloak]

I believe so, though I don't know so much about that side.

What I know is for a simply laced diagram, if you consider the representation of a quiver with that underlying graph then the dimension vectors of indecomposable representations are exactly roots in the root system

[broken pollen]

0 is imaginary 😔

[rocky cloak]

There is also some generalizations of this, which use representations of valued graphs instead of quivers, which then also covers multiply laced graphs

[old spire]

But that wouldn't give you other Es other than the 3 possibilities

So the use of two more letters is not really explained.

[rocky cloak]

And then there's this
https://arxiv.org/abs/2302.01866

[old spire]

Gabriel's theorem uses a subset of the Dynkin diagrams that classify root systems.

So there's no general principle for deriving E, F and G letters.

[delicate orchid]

right so the quiver side works the same

[rocky cloak]

finite root systems

Kac generalizes this to infinite root systems for arbitrary graphs

[old spire]

Are there more Es then?

Or Fs and Gs?

[rocky cloak]

The pattern for how to define E_n is pretty clear

[old spire]

Does it extend to infinitely many Es? Or does it use the octonions or other non-associative algebra in some way analogous to the A, B, C, D series?

[delicate orchid]

"simply laced" or as I like to call it, an alkane

[old spire]

I'm not seeing it, btw. I look at the Freudenthal magic square construction, and F4 should really be called E4.

https://en.wikipedia.org/wiki/Freudenthal_magic_square - Look at the bottom row of that table.

Really, F4 should be called E4.

[delicate orchid]

lets have a peep

that is a good argument

[sturdy mirage]

Question: Ker(phi) is a subset of M which is finitely generated... why isn't this obvious? I must be missing something here

[rocky cloak]

I guess ultimately it's arbitrary how they're divided, but to me it's very useful that the simply laced an non-simply laced are in different families. So you can say stuff like "ADE diagrams"

[chilly ocean]

submodules of finitely-generated modules need not be finitely-generated in general

these are not vector spaces

i think it's always true when your module is over a noetherian ring

[ashen heron]

#bots message

[sturdy mirage]

oh really huh. thanks. I don't know what a noetharian ring is yet

[chilly ocean]

it's a ring whose finitely-generated modules have finitely-generated submodules

[delicate orchid]

fun exercise

there's a really slick ||SES argument|| you can do

[old spire]

There is actually an infinitary generalisation of the E series beyond just E6,E7,E8: https://en.wikipedia.org/wiki/En_(Lie_algebra)

And in fact, E2 and E4 used to be used as names for G2 and F4 respectively.

[sly crescent]

Even though E4 is actually the same as A4

[old spire]

That's why the names E2 and E4 stopped being used for the other exceptional root systems.

[sly crescent]

To add further confusion, the classification of finite Coxeter groups is the same thing except that Bn and Cn are the same Coxeter groups and there are two new exceptional Coxeter groups (H3 and H4) and there’s a new infinite family (I2(n)) which includes G2 as a special case.

[old spire]

In the finite Coxeter group case, the corresponding Dynkin diagrams look undirected

If you take the Dynkin diagrams for root systems, and remove the directionality, then B_n and C_n become the same.

Also, the edges in the Coxeter case are weighted.

[errant shadow]

Can you even express Q as a finite union of proper subspaces? Can I argue with that?

[delicate orchid]

the only proper subspace of Q is the zero space

this is not true for Q^2

Q^2 has an infinite number of proper subspaces

[errant shadow]

okay, then I can't ig

[errant shadow]

wait, how do I show that?

[delicate orchid]

it's pretty trivial to show

[rocky cloak]

Just list all the subspaces

[delicate orchid]

all the subspaces are 1 or 0 dimensional

we'll ignore the zero dimensional one

[errant shadow]

oh wait is it like

Q[\sqrt{2})]

[errant shadow]

ah, okay that makes sense

[delicate orchid]

sure, if you can show that there are infinite distinct field extenstions - but that's straying a bit far from the vector space perspective imo

[delicate orchid]

what does it mean for a space to be one dimensional?

[errant shadow]

Isn't that just, Q?

[delicate orchid]

They’re isomorphic to Q

The answer I was looking for was “spanned by a single vector”

[errant shadow]

ah, right

[delicate orchid]

So if you take a finite number of sub spaces which is just unioning spans of individual vectors (because obviously if you take the span of two at once you’ll probably get all of Q^2), try and construct a new vector that doesn’t lie in any of your spans

There we go wow I could not word that right to save my life

[errant shadow]

gotcha thanks

[delicate orchid]

Geometrically this should be quite clear - you can’t colour in an entire plane with a finite number of lines!

[errant shadow]

yea, that was the first thing that popped up in my head

I just need to write this down ig

[night onyx]

similar result, you can't write R^2 as a countable union of proper subspaces!

[errant shadow]

oh right, that's neat