groups-rings-fields

[knotty badger]

Oh I see

[south patrol]

real

[knotty badger]

I tend to perform well in exams

[elfin wraith]

(that was a joke to be clear)

[south patrol]

i just thought the Möbius transformation stuff was ufnny as most places i have seen never motivated why tf we should care

[elfin wraith]

I mean I did fuck it up, but like I dont actually have beef with them

[knotty badger]

Oh

[karmic moat]

mobius transformation kind of feels like the fundamental theorem of algebra, in the sense that it somehow appears in every class in a slightly different flavor

[elfin wraith]

I really need to start studying for exams actually but id rather do litterally anything else

[south patrol]

that was not my experience for either thing lol

[karmic moat]

[south patrol]

well okay i guess FTA does appear in a fair few things

[elfin wraith]

FTA showed up in both CAs WOW!

[south patrol]

just cause you can prove it in like 4 ways in undergrad

[karmic moat]

yea

[south patrol]

Galois theory proof was not in our course for some reason lol

Very cute

[elfin wraith]

I know of like 2 topoloical ones, and the one via Liouvilles, whats the other proofs?

Oh theres a galois one?

[karmic moat]

there's some winding number one from diff top (idk if you included that one)

i think im remembering it correctly

[elfin wraith]

Yeah thats one of the topological ones I was thinking of, its like a cororally of pi_1(Z) iirc lol

[south patrol]

surely isomorphic proof to the AT one and common CA ones in different language lol

But uh

[karmic moat]

probably

[elfin wraith]

But like yeah this is basically Liouvilles too, because CA is just basic algtop in a different package

Whats the Galois proof though that sounds interesting

[south patrol]

i am trying to remember it / work it out again as it has been like 3 years but uh

[elfin wraith]

Im supposed to be starting to study for my Galois exam today I should probably just think about it tbf

but also, eh

[south patrol]

basically you use the fact that any odd degree poly over R has a root to build constraints on the Galois group of a Galois closure of a putative finite extension of C

and some sylow theory and stuff ig

[elfin wraith]

That sounds clever and annoying

[quiet pelican]

“Mobius maps are the rational automorphisms of the Riemann sphere” or “Certain mobius maps are the isometries of hyperbolic space” are the basis for basically every single one of them

[south patrol]

Agreed lol just at the time it was like "something of the form az+b/(cz + d)" lol

[elfin wraith]

We got told the former, but for all intents and purposes just to think of the later lol

[quiet pelican]

But yeah it’s hard to motivate before CA/Algtop/AG lol

[elfin wraith]

Thinking of different proofs of the FTA though, is there any AG nonsense you can do to work it out? Ive always just assumed im doing AG over an algclosed field so idk how anything there really looks (maybe such a proof would basically just be topology again)

[elfin wraith]

Yeah I mean I did it in CA so fair lol

[south patrol]

Yeah it's like suppose L/C is a finite extension. By passing to Galois closure WLOG it's Galois. If L/C has order m.2^n for m odd then Gal(L/R) has a subgroup Gal(L/M) of order 2^{n+1} by Sylow theory and hence Gal(M/R) has odd order m which is not possible unless m = 1 (any element has a minimal polynomial of odd degree). So m = 1. If n > 0, there's (by theory of p-groups) an extension M/C of degree 2 but this is also impossible because any quadratic over C has a root

I think that's the proof

[elfin wraith]

That is clever

[quiet pelican]

Yeah

[south patrol]

I think it's cute because like the analysis is very minimal and it shows what you "need" about C

[elfin wraith]

It also shows why learning topology is good for the soul, it makes life so much more simple

[quiet pelican]

It’s both surprising and not surprising that the analytical content of FTA is just IVT

[south patrol]

yeah

[south patrol]

lol i would say this is a simpler proof idk but yeah

Just cause you can do everything very explicity and stuff which is nice

[karmic moat]

maybe i should learn galois theory this summer

[south patrol]

But yeah I like both proofs lol

well all of them gi

[elfin wraith]

Oh yeah both are nice, more proofs are always good

[south patrol]

I think it is hard to do so and I'm not sure how to use AG other than just commutative algebra but I will try to remember

I vaguely remember a cheap proof of the Nullstellensatz for C and wonder whether it uses the "n = 1" case lol, probably does

[elfin wraith]

Honestly I wouldnt reccomend. I took galois last term and going in already knowing a non-trivial ammount of algebra half of it was just either obvious or known. But I guess theres also a lot of Galois we didnt cover

But also idk how many people take multiple ring theory courses, AG and AT before Galois

[karmic moat]

fair enough

[elfin wraith]

Maybe its better if youre not as odd as me

[karmic moat]

but i feel like i see more and more Galois groups appearing as i learn more ag

[elfin wraith]

Yeah it for sure comes up in AG, its probably worth just speed running. I just personally didnt feel like I got much out of my course

[karmic moat]

like there was one example where Spec \bar Q \otimes_Q \bar Q is isomorphic to Gal(\bar Q/Q) and i realized i have no idea what Gal(\bar Q/Q) even looks like

true

[elfin wraith]

Isnt that like what number theory is

[vapid vale]

Not if you only ever work over C

[south patrol]

idk if it comes up if you only do stuff over C though

lol

amusing

[vapid vale]

lol

[elfin wraith]

I mean it depends on what were counting as Galois I guess lol

[karmic moat]

something like that yeah

[elfin wraith]

Extenstion stuff comes up but maybe thats just CA

[south patrol]

ah rip yes it does lol

[elfin wraith]

I more just mean, isnt working out what Gal(\bar{Q}/Q) is like an incredibly difficult open problem that ive seen people say "is basically the goal of number theory"

[karmic moat]

oh that i had no idea was open

[elfin wraith]

I think youre forgiven for not really knowing what that looks like

[knotty badger]

Wait how

[elfin wraith]

(I could be wrong, someone smart say something if so)

[south patrol]

You can prove "if L is of finite type over K and a field then L is finite over K" cheaply if K is uncountable but then to prove Nullstellensatz you have to use the fact alg. closed fields have no finite extenions

[south patrol]

yes lol

[elfin wraith]

https://mathoverflow.net/questions/2791/understanding-mathrmgal-bar-mathbbq-mathbbq

[south patrol]

i mean the theory of Galois representations (= representations of flavours of Galois groups like this) more broadly is a lot of number theory

[wraith cargo]

what stuff do u wanna learn

[south patrol]

this is important ye

[karmic moat]

whatever is important for AG i suppose

[south patrol]

what stuff in AG do you wanna learn

[karmic moat]

seems like galois theory gives a lot of important examples

[south patrol]

"whatever is important for NT" correct answer

[karmic moat]

i have no idea

[wraith cargo]

if u wanna do anything etale you need galois theory

[karmic moat]

eek

[south patrol]

67

[wraith cargo]

I no longer want you fr

[karmic moat]

i think NT is conceptually cool but it doesn't have an allure to me if that makes sense

[south patrol]

My partner will be happy

[karmic moat]

like i can appreciate it from a distance but it doesn't seem like something i'd want to do

[elfin wraith]

This is how I feel about PDEs

[karmic moat]

😠

[elfin wraith]

Im glad other people are doing it, because I dont want to

[karmic moat]

why is it huge all of a sudden

[south patrol]

Oh wait this isn't

[wraith cargo]

https://tenor.com/view/meme-caretaker-doctor-horse-doctor-horses-gif-16330414525219104805

[south patrol]

#1203471755449073774

[karmic moat]

this is how im starting to feel about algebraic groups

[south patrol]

lol

[elfin wraith]

Literally me

Uh yeah on topic, go to algechill etc etc

[karmic moat]

ugh i HATE MODERATORS

[south patrol]

lol

[tough raven]

Oh I think this is not quite true and everything is a bit more complicated for infinite extensions. For a Galois extension K/Q with Galois group, K (⨯)_Q K = the subalgebra {f in K^G : f is invariant under a finite-index subgroup of G}.

[karmic moat]

okay yeah maybe not isomorphic, should've said bijective on points and have the same topology

[tough raven]

Kind of interesting to muse about what it even is though, being a dominant image with the same topological space.

There are actually three objects here: the disjoint union of G copies of Spec L (with the profinite topology somehow), Spec L^G, and Spec L (x)_K L.

[sly dune]

uhh

If {A, , -, , ,B, , C} → {, -, A, , C , B,}
How about: {D, , +, | ,S , T} work?

[swift root]

folk

[swift root]

omg

youve got the new yorker reaction images too

[karmic moat]

im kind of a fraud in the sense that i just google image searched your photo and found it

i didnt have it on hand

[swift root]

💔

[tough raven]

No idea what this means or how it relates to groups, rings or fields.

[sly dune]

what

[swift root]

likewise

[sly dune]

they're groups tho

[tough raven]

No idea what this means

[sly dune]

yeahhh okay

[karmic moat]

i think they're just posting random stuff to get active role lol

[swift root]

i have no mouth and i must larp

[tough raven]

I mean if you explain we can think about it...

[sly dune]

I was asking a legit question too alongside that intention

[swift root]

yk questions usually need elaboration

[sly dune]

The variables A, , -, , ,B, , and C were rearranged

idk

anyways

oh hold on I do have a legitimate question about groups

[swift root]

LLM ascii art ahh

[karmic moat]

no it's called modern art

[rocky cloak]

Why so many commas?

[sly dune]

confusion

[rocky cloak]

Well, let's hope you get your active role soon

[swift root]

social distancing or something

[karmic moat]

once this whole covid thing blows over im gonna get a gf and go to the gym and get swole im telling you

[swift root]

does he know

[karmic moat]

:clueless:

oh

[swift root]

you can also get a gf while having a twink bod yk

[swift root]

😭

[karmic moat]

yes it's just this damn covid quarantine that's ruining everything

[rocky cloak]

You need the gf to get swole.

Otherwise whose gonna spot you when you bench?

[swift root]

good point

[swift root]

you coming outside for the first time in 2050 seeing most of the citg destroyed in a nuclear fallout:

[tough raven]

Yeah, it's "get a gf and go to the gym", not "go to the gym and get a gf".

[karmic moat]

do u think quarantine will be over by then

[swift root]

im an undertale fan, i cant read

[tough raven]

Obviously anyone can get a generating function. Very supportive partners.

[karmic moat]

but discord font is sans serif

haha get it

[swift root]

https://tenor.com/view/ands-gif-12275227633433849703

[languid trellis]

wow this is so cute and wholesome and supportive. I was expecting something along the lines of her breaking your heart so you get in the gym

[karmic moat]

yes...

[rocky cloak]

Nah, she does that after you get swole. Once you don't have that sexy twink bod

[karmic moat]

but then who is the spotter

[languid trellis]

so true

[languid trellis]

spotters are for pussies. just be stronger

[karmic moat]

i see...

[swift root]

survival of the fittest mindset

[tough raven]

most spotless*

[karmic moat]

the most spotless mindset gets eternal sunshine

[heavy swan]

Benching is for those with a weak mind

Take the CNS to failure

[chilly radish]

You should read up on infinite galois theory

And also, riemann existence theorem and the correspondence between branched covers and galois extensions of function fields

[azure cairn]

That gif has been saved

[small yacht]

I came back today, drank a monster, reread the chapter, and it’s so painfully obvious to me now

[swift root]

I'm afraid this is a pattern that will continue all throughout your math journey

lol

before sleep = "what the fuck"
after sleep = "how the fuck did I not get this"

[tardy hedge]

wow mario star cool emote good emote

ima super star mama mia

[rocky cloak]

Aprill hits hard

[knotty badger]

what's the restriction?

[rocky cloak]

Says so in the imageee

[knotty badger]

oh, no word with exactly 5 letters?

[rocky cloak]

Fun'ny joke, but I suspect it won't be fun'ny for 24 hou'rs

[elfin wraith]

I would like to at some point, it seems much more interesting. I of course didnt mean to suggest that all of galois theory is boring or trivial, I just found my course to be so (but I did probably go into it knowing a fair bit more than most)

[karmic moat]

Play smb galaxy if u havent already

[swift root]

lowkey

[karmic moat]

Nvm i thiink i misread your message

[cursive spindle]

I hit this pose this day

tfw try not to use a word of len 5

[swift root]

length

that's 6

[cursive spindle]

Cohomology

[sly crescent]

What is that?

[rocky cloak]

3d platformer by Nintendo

[tardy hedge]

[tough raven]

I can't even say "grou p" apparently 😔

Or "abou t"

[sly crescent]

There’s no b in that

[karmic moat]

😠

[rocky cloak]

Sometimes and extra b gets into our life. One way or another we're gonna have to deal with it

[restive ginkgo]

hey, looking for some direction on this: let R be a ring, and let f, g be homomorphisms from \mathbb{Q} -> R with f(n) = g(n) for all n \in \mathbb{Z}. show that f(x) = g(x) for all x \in \mathbb{Q}. we've discussed some properties of [ring] homomorphisms and some of fields, but i feel like i'm missing something

[chilly radish]

Sorry I misinterpreted R as the real numbers lol

[chilly radish]

Show that a homomorphism sends units to units, and that f(x)^-1=f(x^-1) for a unit

[restive ginkgo]

ahh I see, I was wanting to use that property but couldn't justify it. i'll give that a try 🙂

[twilit wraith]

ok am i nuts or is the location of certain ancient letter via the stated map not an element of the automorphism algebraic structure of K over F

isnt the domain of phi inverse K'

ive not seen function composition be read from left to opposite of left

[karmic moat]

should be read coleft to left

K' -{phi^-1}-> K -{si gma}-> K -{phi}-> K'

[twilit wraith]

oh yeah oops

mind stopped functioning

[karmic moat]

happens happens

[fresh gate]

What is this horrid rule

What braindead moderator put this here

Oh

Is this aapril foolls?

[chilly radish]

Yea, you can exempt yourself by choosing the role in channels & roles

[fresh gate]

Nah I won't

[chilly radish]

Are you asking what happens if you quotient by the subgroup generated by n-th powers for some fixed n?

[fresh gate]

yeah

with whom it was quotiented or something like that wouldd probably be better

[chilly radish]

So generally to be able to get a quotient from a set of relations we need to quotient by the smallest normal subgroup containing our set of relations

[fresh gate]

hmm ok, then do that in this case

[novel star]

this isnt even finitely presented i dont thinks

[fresh gate]

I mean yeah duh

Does it need to be?

If m elements is too complicated what aboutt with just a and b?

[chilly radish]

So in this case if ab is in the normal subgroup you're quotienting by, then a^-1(ab)a=ba will also be

So in the case of n=2 and m=2 you get the Klein 4 group, and in general for n=2 I think you should just be getting (Z/2)^m

[fresh gate]

The ({1,3,5,7}, x) groupp

[fresh gate]

what are aba and bab equall to

oh

sorryy

aba = baa = b

bab = abb = a

[chilly radish]

Yea

For n=2 it's nice since that automatically implies abelian

So everything simplifies

[fresh gate]

ok wait I see what you are saying

[chilly radish]

For larger n I don't know how tractable this problem but I'd imagine you can maybe get a nice presentation for n=3, m=2 at least

[fresh gate]

well theree is abab = bbaa firstt of all

Ok I have no idea what to do

[novel star]

i suspect that its actually finitely presented for all n

[fresh gate]

Bro switched up 💀

[novel star]

if i did not i wouldnt be very rational

[fresh gate]

Yeah you'd be 1/0! + 1/1! + 1/2! + 1/3!.......

[novel star]

i thinkk the size of the presentation gets big fast

[fresh gate]

Wouldd (aba)^-1 = bab be notable?

And you can find similar things by replacing a with a^-1 or b with b^-1 or both

Actually two of thesee are kind of redundant I thinkk

[novel star]

even in the case when n=2, it is infinite on more than 3 characters

so i was wrrong againn

[fresh gate]

No it really is not

The length <= 2 casess are enough

As the otherr guy explained

You can get abelianness from that

And everything else follows to give you (Z/2Z)^m

[novel star]

https://arxiv.org/pdf/0712.0139

the presentation is infinite

[chilly radish]

I'm not sure what how exactly that paper relates, but if x^2=e for all x then the group is abelian, then every element is of the form a_1^i_1 ... a_m^i_m, but you can reduce the indices mod 2, so this is just (Z/2)^m

[novel star]

hmm

i think the way i was getting relations was bloated

i was finding all squarefree words and setting their squares to 0

[fresh gate]

You can also ignore all non-squarefree wordss in the n=3 case because x^2 = x^-1

[fresh gate]

Um btw, can anyone explain this?

Isn'tt Z_p basically the vector spacee on Q with |2^N|| basiss vectors?

(as the groupp (Z_p, +) at leastt)

[chilly radish]

Is Z_p here Z/pZ?

Or p-adics

What's the context

[south patrol]

Lol I am confused

Maybe it means Pontryagin dual

[next obsidian]

I feel like this hat must be some sort of perfection

[south patrol]

And RHS is Qp/Zp

[next obsidian]

Either perfection or deperfection

[south patrol]

Zp already perfect tho bruv

[next obsidian]

Z/pZ

Idk

[south patrol]

Well

[next obsidian]

Actually that says p^infinity

[south patrol]

So is Z/pZ lol

[next obsidian]

Not p^1/infinity

[south patrol]

This is why I think it is this or smth

[next obsidian]

That’s true af

[south patrol]

Idk

I am a sheaf.

[next obsidian]

Don’t. Cry. I am just a sheaff

Well ok I can’t send the link

Thanks obamaa

[south patrol]

Lol

[fresh gate]

Isn'tt Z_p basically the vector spacee on Q with |2^N| basiss vectors?

[south patrol]

No

[fresh gate]

As a groupp I mean

[fresh gate]

Yes that's it

And it's p-adicss

This is what the otherr side is

Sorryy for beingg absent when you guys answered, I thinkk I was eating when you guys responded lol

[south patrol]

Yeahh that's what I thought nice

[south patrol]

Anyway, not quite. Z_p doesn't have the structure of a Q vector space – multiplication by p is not invertible (it isn't surjective)

[fresh gate]

Oh true

The question remains though

Sorryy for not providing more context

[south patrol]

1 is a topological generator of Z_p (in other words, Z is dense in Z_p). So a continuous hom f: Z_p -> C* is determined by where you send 1.

If f(1) = a then since p^n -> 0 in Z_p, we must have a^(p^n) -> 1 as n -> oo. But this is only possible in C* if a is a p^nth root of unity for some n

[rocky cloak]

The resulting group is called the Burnside group B(m, n).

It's an open question for which m, n the group is finite. You have
B(1, n) = Z/n
B(m, 2) = (Z/2)^m
it is known that B(m, 3), B(m, 4) and B(m, 6) are finite.

It is not known whether B(2, 5) is finite or infinite.

[south patrol]

Interesting

[elfin wraith]

Ooh interesting. I know free groups are weird and hard but it seems suprising that thats open

Is it expected to be finite?

[rocky cloak]

I'm assuming if it was finite, it would be reasonable small and then already been computed.

But that's just me pulling stuff out my butt

[elfin wraith]

Interesting, apparently it’s known to be finite for all m>1 and n>=8000

[quiet pelican]

Tbh this is less that free groups are weird and hard and more that presentations are weird and hard
The problem is more the thing you’re quotienting by than the free group

[elfin wraith]

Yeah sure, this is what I mean

[rocky cloak]

Groupprops tells me that the intersection of all finite index normal subgroups has index 5^34.

So I guess if it is finite it would have order 5^34, so then I don't know what one should think

[elfin wraith]

I was aware of the burnside problem but I hadn’t seen this part of it before

[elfin wraith]

That’s not so small

Probably can’t just brute force that in gap lol

[chilly radish]

OOOH

I cannot see \hat and not assume it means completion of some form

[crystal vale]

So if f: A -> B is ring homomorphism then there exists associated mapping spec(B) -> Spec (A), which maps p -> f^-1(p)

But if we take f: Q -> Q × Q given by x -> (x, 0)

[knotty badger]

yes Spec is a contravariant functor

[swift root]

this is not a ring morphism

[crystal vale]

Then taking the inverse of Q×{0} is Q

[knotty badger]

it's a rng morphism

but not a ring one

[crystal vale]

Ah my bad

[knotty badger]

ring morphisms have to preserve the multiplicative identity you see

but (1, 0) is not (1, 1)

[swift root]

the taking 1 to 1 is exactly what's stopping the preimage of a proper ideall to be the wholee ring

I HATE THIS APRILL FOOLSS THINGG

[crystal vale]

Me too

[swift root]

and yes I know I can turn it off but I do not want my name to be poo coloured

[tribal moss]

It doesn't look like the selfrole influence the coloring thing.

[swift root]

what is soundd contributor 😭

[tribal moss]

Just think of it as "very slowly rotating rainbow roles".

[crystal vale]

So I am trying to construct an counterexample for the converse of ii)

[tribal moss]

Would it work to choose A=B appropriately and let f be the identity?

(I don't know what an extended ideal is, but just on general principles ...)

[crystal vale]

For prime ideal p of B there exists ideal I of A such that < f(I) > = p

[swift root]

I have a hard time figuring out if this is aprill foolss or not lol

[tribal moss]

Ah, then my idea was probably rubbish).

[quiet pelican]

Try Z/2Z[x]_{(x)}, and squaring
It’s very hard for me to explain how I came up with this if you haven’t seen DVRs before unfortunately

[crystal vale]

What is DVRs?

[tardy hedge]

so true man so true we hate poo colour

[elfin wraith]

Discrete valuation ring, but that likely means you havent seen them

[tardy hedge]

hello there

[elfin wraith]

Theres like 11 equivalent statements of what a DVR is, and theyre variously helpful. I dont actually know what extended or contracted ideals are so I cant say which is most of use here, but the most simple definition of a DVR is that its a PID with exactly one maximal ideal

[quiet pelican]

Essentially you want a map where a prime ideal maps into a proper subideal of a prime ideal where the subideal is not contained in any other prime ideal

Consider why that is useful

[elfin wraith]

This is some AG type stuff so maybe worth noting that this is the same as a local PID which isnt a field, and this is how you get nice stuff like a curve is smooth at a point iff the local ring there is a DVR

[swift root]

contraction is preimage, extending is ideall generated by the imagee

[rich granite]

🥀

[elfin wraith]

!wordlespoilers

[flat tree]

Type ,iam 1488687375340273675 to get immunity from the wordle spoilers filter!

[swift root]

,iam 1488687375340273675

[cloud walrus]

Gave you the Not Very April selfrole.

[swift root]

bwa

[crystal vale]

I don't understand what I have to show in ii) and iii), i think i figured it out for iii) but still

[rich granite]

,iam 1488687375340273675

[cloud walrus]

Gave you the Not Very April selfrole.

[rich granite]

Yippee

[rocky cloak]

So you have a ring homomorphism
S^-1f : S^-1A -> S^-1B
then for a prime ideal in S^-1B there are two things you can do: either take the preimage with S^-1f or consider the associated prime in B, take preimage by f to get a prime in A, then take S^-1.

It wants you to show that these two give the same result

[crystal vale]

I see, thank you

[crystal vale]

So there is a bijection between prime ideals in S^-1 B and the prime ideals in B which don't intersect with f(S), right?

Maybe I am making a mistake here, do I need to consider S^-1B same as f(S)^-1 B

S^-1 B gives S^-1 A module structure but I am not sure about ring structure

[rocky cloak]

Ring structure is just
b/s * c/t = bc/st
But you can show that S^-1B = f(S)^-1B yes.

[toxic girder]

ringses

[elfin wraith]

POV youre gollum:

[slate oriole]

Can anyone draw me D8 ) the diedergroup

like if i ask really nicely will someone draw me the diedergroup D8

with 8 = 2*4

[elfin wraith]

What do you mean exactly? Its usually thought of the symmetries of the octagon (or the square depending on your convention)

[slate oriole]

i have a exam tomorrow on (in germany called Algebra) and i cant find a picture on the structure of all the subgroups of D8

[elfin wraith]

Ah, I mean this is something you should be able to compute, but the term for the picture youre probably looking for is the subgroup lattice

[slate oriole]

yeah i think its called like that in english, i dont want to calculate all the (IDK HOW ITS CALLED n in N so that g^n is 1) for each element

[rocky cloak]

https://groupprops.subwiki.org/wiki/Subgroup_structure_of_dihedral_group:D8

[elfin wraith]

https://groupprops.subwiki.org/wiki/Subgroup_structure_of_dihedral_group:D8

Ah

[slate oriole]

thanks

[elfin wraith]

Jagr snub

[rocky cloak]

The first result when you Google "subgroups D8" for me

[slate oriole]

yeah i googled in german

didnt find any results...

[deft light]

Hey guys can you recommend me a book on the algebraic structures

[elfin wraith]

Dummit and Foote is the classic introduction, Artin is another option. Here :#book-recommendations message is a more full list with some comments about them

[deft light]

Yeee thanks dude

[restive sparrow]

hey everyone i'm reading d&f chapter six (adv topics in grp theory)

specifically i'm trying to show that the subgroups and quotient groups of a nilpotentgroup are nilpotent

[rocky cloak]

What do you have so far?

[restive sparrow]

starting with subgroup -- if H is a subgroup of G and we use the uppercentral series defined by Z0 = 1, Z1 = Z(G) and Zn+1/Zn = Z(G/Zn) presumably you take the intersection of H with each of the elts in the series

god this five letter cap is really twisting my mind

if G is "c-nilpotent" (so that Zc = G) then obviously H is at most c-nilpotent, so it remains to show that the intersections provide another uppercentral series

[elfin wraith]

!wordlespoilers

[flat tree]

Type ,iam 1488687375340273675 to get immunity from the wordle spoilers filter!

[restive sparrow]

,iam 1488687375340273675

[cloud walrus]

Gave you the Not Very April selfrole.

[restive sparrow]

tysm king <3

[rocky cloak]

Yeah, so all you need is to show that Zn(H) contains Zn(G) intersect H, right

[restive sparrow]

can you talk me through how you came to that conclusion?

so we are to show that Zn+1(H)/Zn(H) = Z(H/Zn(H)) correct

oh wait no i'm stupid

@rocky cloak coz the reverse containment is obvious, so we are left to show the forward inclusion?

sorry i'm new to this whole process

[rocky cloak]

Well, the reverse containment isn't necessarily true, but. If
Zc(G) = G, then Zc(H) containing Zc(G) intersect H means it contains H. So Zc(H) = H

Which is saying H is nilpotent with nilpotency class <= c

[sly rock]

,iam 1488687375340273675

[cloud walrus]

Gave you the Not Very April selfrole.

[limpid ferry]

.

can you have a look at why addition is preserved? I cannot see why

[rocky cloak]

It's not guaranteed that it will really. There are more isomorphisms F^x -> G^x than F -> G

So it depends which generator you send it to

G^x should have 16 possible generators, but there are only 2 isomorphisms F -> G.

So only two of the possible 16 choices will give you something additive

[limpid ferry]

do thosse two satisfy certain conditions or they are just generators just like the otther 14?

they are blocking 5 letter word I am so pissed

[rocky cloak]

Well they will satisfy the same minimal polynomial as the generator you choose for F^x

Otherwise there's nothing special seperating any generator from another

[rocky cloak]

!wordlespoilers

[flat tree]

Type ,iam 1488687375340273675 to get immunity from the wordle spoilers filter!

[limpid ferry]

,iam 1488687375340273675

[cloud walrus]

Gave you the Not Very April selfrole.

[restive sparrow]

wahoooo

is there a typst bot

[tough raven]

Let V and A be vector spaces and a: V → A (⨯) V a linear map. For a linear subspace U of V, how can I find the smallest linear subspace U' of V such that a(U) ⊆ A (⨯) U', if that exists (and it should if A is a Hopf algebra and a is a comodule structure on V)?

[rocky cloak]

I guess pick a basis {f} for A^* then take the subspace generated by f(a(U)) < V

[limpid ferry]

is there a simpler way to explain without talking about that minimal polynomial?

[rocky cloak]

I don't really think so.

What do you know about fields / finite fields?

[limpid ferry]

it has prime characteristic, that multiplicative group is cyclic.

[rocky cloak]

Have you proven that there is a field with 49 elements at all?

[limpid ferry]

Yes, Z_7[x]/<x^2+1>. I basically want to prove every finite field with order 49 is isomorphic because my homework asked me to prove that every field with 49 elements is isomorphic to the one I defined

[rocky cloak]

Then a good approach would be showing that any field of order 49 has a square root of -1.

This would then give you a map Z7[x]/(x^2 + 1) -> F, which is an isomorphism just by cardinality reasons

And you can prove -1 has a square root using that the multiplicative group is cyclic

[restive sparrow]

@rocky cloak ok i've wrapped up subgroup, can you help me with quotient xx

[rocky cloak]

Sure, should be basically the same.

Compare Zn(G/N) with Zn(G)N/N

[restive sparrow]

bro where do you get that from

[rocky cloak]

Like I'm taking G/N to be a quotient group.

Then taking the central series for G/N

That's it

Comparing it with the image of the central series of G

[restive sparrow]

oh so the image of Zn(G) in G/N?

oh so it's a projection instead of a injection

whoa okay

[tough raven]

Of course - just use a base for A rather than V. Although I feel it to be cleaner to (following the idea in the previous sentence) take a dual base of a base of A rather than a base of A^*.

Can this be done without a base of A? Say if A is flat over the base commutative ring.

[rocky cloak]

So if A is fg projective, then
A (x) V = A** (x) V = Hom(A*, V). Then adjunction gives you
V (x) A* ->V
and then you just take the image of
U (x) A*

If A is flat I guess it's a filtered colimit of fg projectives, so you can use that

Well, idk how exactly

[tough raven]

Even otherwise you can map A to A^** and use that. Thus map a(U) to a subspace of Hom(A*, V). Thus map a(U) (⨯) A^* by evaluation to get U' ⊆ V. This is equivalent to taking U' = ∑_{f in A^*} (f (⨯) 1)(a(U)), i.e., the idea is "it appears like it's in U' whenever we replace the coefficients in A by scalars".

This should work whenever A → A^** is injective.

Or simpler: use the map A^* (⨯) A (⨯) V → V by evaluation and take the im of A^* (⨯) a(U).

[rocky cloak]

And I guess you can get situations where there is no minimal.

Like working over Z, take V = Z, A = Q and a(1) = 1(x)1.

Then a(V) < A (x) nZ
for all n. So no minimal

[tough raven]

Hmm that actually make s more sens e

I have an extr a assumption that was not getting used

Well but it's complicated

[tough raven]

Basically this Hopf algebra hypothesis

Eh perhaps I'll just remain over fields this time

[rapid cave]

Let A,B two integral domains with injections A->B and B->A. Is A isomorphic to B?

Maybe at least in the finitely generated case

[velvet hull]

I misremembered the statement, I don't think what I said was right

[rocky cloak]

Certainly not in the infinitely generated case, like C -> C(x) -> algclosure C(x) = C

[wraith cargo]

k[t]->k[t^2,t^3]->k[t]

[clear ingot]

Do you mean over a ring R, finitely generated R-algebras?

then k[t] -> k[t^2, t^3], t -> t^2 gives an injection, and k[t^2, t^3] embeds into k[t] as k algebras is a good counterexample

[rapid cave]

Thanks

I meant as k-algebra

[crystal vale]

so let take an prime ideal in f(S)^-1 B say q = f(S)^-1 p, where p is a prime ideal in B not intersecting f(S)

now i have to show the (S^-1f)^-1 (f(S)^-1 p ) = S^-1 f^-1 p

one inclusion is easy, now if a/s in (S^-1f )^-1 q then f(a)/f(s) in f(S)^-1 p, from this i don't get how do i show a/s in S^-1 f^-1 p

[crystal vale]

So if q is one of the prime ideals of the form r(a:x) then q = p_i for some i ?

[small yacht]

I'm a bit lost in the sauce, how do 7.4 and 7.6 show that for every subgroup?
7.4 specifically talks about the equivalence class of e_g, which isn't necessarily the same as H

[next obsidian]

It is the same

H corresponds to the relation x ~ y if xH = yH

Given a relation ~ if you let H be the equivalence class of e, then ~ is the same as the relation induced by H

And vice versa if you start with H and get the relation ~ from that, then the equivalence class of e is just H

So the two processes are inverse

[knotty badger]

what's $\dagger$ here

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[next obsidian]

Some list of properties that make this tru

[small yacht]

a ~ b => ga ~ gb for all g in G

[knotty badger]

wait

i thought it'd be $ga \sim gb$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[small yacht]

Oh wait it’s on the left

[small yacht]

You’re right

[knotty badger]

right right, so the relation is "left-invariant" (or perhaps "left-equivariant" you could say)

[next obsidian]

Anyway idk if that clears it up

[small yacht]

I’m stupid,,,

[next obsidian]

I just basically restated what’s said there but idk where you’re stuck

[small yacht]

So not really

[next obsidian]

What do u disagree with

[small yacht]

How does a ~ b iff a^-1b \in H imply that the [e_g]~L= H

[next obsidian]

I mean you’re just asking when is eH = aH

This is just exactly when a is in H

[small yacht]

Ohhhhhhh

[small yacht]

æ æ æ æ æ æ

Anyhow thx i got it

[next obsidian]

Ok

Swag

[small yacht]

WOWIES
It’s called Z/nZ because it’s Z modulo nZ

[narrow monolith]

Is this a typo? I think they should have S instead of T

[next obsidian]

No

Ok yes

[narrow monolith]

Yes?

[next obsidian]

Yes

[next obsidian]

Why use many word when few do trick?

[crystal vale]

I don't see the connection between them

Any hint?

Should I ask this question in advanced algebra ?

[chilly radish]

So first try to write Spec(A/a) as a finite union of closed sets. If you have an irred. Component it can only be contained in one of these, so show that each has finitely many irred components

[chilly radish]

Yea probably

Also remember what irreducible components of an affine variety look like

[crystal vale]

Sorry i don't know about it

[crystal vale]

So if a = \cap q_i then Spec (A/a) = \cup V'(q_i) , where V'(q_i) = { p/ a | p in V(q_i) }

[chilly radish]

not /a but rather mod a, but yea

[rocky cloak]

You might want to show that any irreducible closed set looks like V(p) for a prime ideal p

[chilly radish]

And then, remember that an irreducible component is a maximal irreducible closed set

[crystal vale]

So are these V(q_i + a) irreducible?

[crystal vale]

Why can components only be contained in one of these?

[chilly radish]

If it were contained in more than one you could write it as a nontrivial union of closed sets

[crystal vale]

I see what you mean

[swift root]

the radical of a primary ideal is prime

[crystal vale]

Yes they are

[swift root]

and the radical of an intersection of ideals is the intersection of radical

[crystal vale]

Yes

Finite or arbitrary?

[swift root]

furthermore, minimal primes correspond to irreducible components

[swift root]

finite, but a primary decomposition is always finite

[crystal vale]

Yes

So I have Spec(A/a) = \cup V(q_i + a)

[crystal vale]

I have to show it

I know number of minimal primes are finite

[swift root]

okay, then do you know the correspondence theorem?

[crystal vale]

Yes

[swift root]

just q_i is fine, a is always contained in q_i

oh right nevermind

youre taking the quotient of q_i by a

[chilly radish]

Once you show that irreducible is equivalent to prime, maximally irreducible is equivalent to minimally prime

[crystal vale]

Yes

Let me show V(q_i + a) contains finite maximal irreducible components

[chilly radish]

That would show (correctly) that they are themselves irreducible components (component means maximally irreducible)

[chilly radish]

Btw, if you go to the algebra side, what's happening here is that any prime contains 0, so it contains the intersection of the q_i+a, this it must contain one of them

[crystal vale]

Sorry i don't get this

Equivalent in which sense?

[chilly radish]

Ok so, if V(p) is a maximal irreducible component, that means that it isn't properly contained in any other irreducible closed set.

If p is a minimal prime, that means it doesn't properly contain any prime ideals.

We can go back and forth using

V(I(X))=\bar X, and I(V(J))= sqrt(J), which reverses containment, so V(I)<=V(J) iff sqrt(J)<=sqrt(I).
For prime ideals we have I(V(p))=p

So once you know irreducible closed is of the form V(q) for p prime, you get

V(p)<=V(q) iff q<=p

[crystal vale]

What is I(X)?

[swift root]

the intersection of ideals in X

equivalently it is the radical ideal corresponding to the closure of X

[chilly radish]

If you don't know this stuff you shouldn't really be doing this problem

This is entirely a problem of translating algebra to geometry

[crystal vale]

I am doing Atiyah, and I didn't skip any theory yet so let me try

[chilly radish]

I didn't assume you did, I just find it very odd it would give you this exercise without giving the proper language first

[crystal vale]

I am just not aware of notations and terminology

[swift root]

doesn´t atiyah use that notation=

?

[crystal vale]

I don't think so, because I am seeing the first time V(I(X) ) = \bar X, no idea what is bar X ?

[chilly radish]

Closure

In the Zariski Topology

[languid trellis]

I'm pretty sure all of this is defined in the exercises to chapter 1

If you're looking for a reference

[crystal vale]

So X is Spec (A)

I don't see he defined I(X) in Atiyah' s first chapter

[languid trellis]

It's exercise 27

[crystal vale]

But isn't it different from the intersection of all ideals in X ?

[swift root]

X is a subset of Spec(A)

[swift root]

its defined as the intersection of all x in X no?

[stone sky]

Has anyone seen a proof of the invariant factor form of the fundamental theorem of abelian groups?

[crystal vale]

I think I lost here

[rocky cloak]

Sure.

Do you already have another form, because it's not so hard to translate between them

Anyway, the usual proof is take some presentation of your group
Z^n -> Z^m -> A -> 0

Change to smith normal form and you're done.

The proof for how you can change to smith normal form is in the algorithm section on Wikipedia
https://en.wikipedia.org/wiki/Smith_normal_form

[crystal vale]

So I have to show any irreducible closed subspace of A/a is the form of V(p), where p is prime ideal in A/a.

And any maximal irreducible component of A/a is form of V(p+ a) , where p is minimal prime ideal in A containing a

[chilly radish]

Yes

[stone sky]

I did not use this. I constructed something else.

Took me a long time.

[crystal vale]

Okay, let V(I) be the closed irreducible subspace of A/a.

We have to show I is a prime ideal.

Let xy in I but x and y not in I.

So V(I) = V(I) \cap V(x) \cup (V(I) \cap V(y) )

We have to show both V(I) \cap V(x) is proper set

[chilly radish]

I is not necessarily a prime ideal by itself

[crystal vale]

Okay what if I is radical ideal

[chilly radish]

But V(I) can be written in the form V(p) for a prime p

Then yea

[crystal vale]

Okay V(I) = V(r(I) )

So take I as radical ideal

I = √J

[chilly radish]

Ye

[crystal vale]

Then how do I conclude I is prime ideal?

[chilly radish]

So V(x)\cap V(I) would be the set of all prime ideals containing x and I, i.e. x+I right? Same with V(y)\cap V(I).

You are assuming that V(I) is irreducible, so one of these is trivial and the other is equal to V(I), i.e. V(I)=V(I+x) wlog.

Can you conclude from here?

Also you don't need contradiction, just start with xy in I and show x in I or y in I

[crystal vale]

I am dumb, i realised i already done this exercise before

So actually V(I) is an irreducible says Spec(A/I) is irreducible

So nilradical of A/I is prime ideal, say p', then take inverse image of p' under canonical mapping gives prime ideal p

And then V(I) = V(p)

[crystal vale]

@chilly radish thank you, finally I got it

It's just that I am dumb, i already did this exercise before. If A is a ring and X = spec A then the irreducible components of X are the closed sets V(p) where p is a minimal prime ideal of p.

I don't know how to remember all such exercises

[chilly radish]

Eh, It's always good to resolve

[crystal vale]

Do I need to consider a is decomposable ideal?

[chilly radish]

I mean this is true. You can also do it directly like you were trying to. V(I+x)=V(I) so every prime ideal containing I contains x, i.e. x is in rad(I)=I

[crystal vale]

I see

Yeah got it

[crystal vale]

I think yes otherwise, if I take A = K[x1, x2,....], and a = (x1) then there are infinitely many prime ideals containing a but not minimal prime ideal

[narrow monolith]

In this definition of the completion of a commutative monoid the completion is defined as a terminal object, is this correct or should it be initial instead? Gpt says it should be initial

[proud vigil]

here's a diagram to visualize

if you ask gpt "is x true" itll often just agree with you tbh

[primal junco]

what?

[tough raven]

It should definitely be initial.

[primal junco]

group completions are initial objects

if $G(M)$ were terminal, it would have to be the trivial group. if you take some massive group $G$, there's no reason there should be a unique map $q: G \to G(M)$ that makes the diagram commute unless $G(M)$ is trivial

[cloud walrus]

okuu

[proud vigil]

should q be reversed then?

[primal junco]

yea

[narrow monolith]

Thanks

[primal junco]

the diagram should be $F: M \to G(M), f: M \to G, q: G(M) \to G$

[cloud walrus]

okuu

[primal junco]

what textbook is this lmao

[narrow monolith]

It's commutative algebra by Peter Clark

https://share.google/TIqyll7ccO5LiQNv7

[next obsidian]

I knew this exists

I am surprised someone is using it

It’s not really easy to find lol

[limpid ferry]

I just learnt about the definition of minimal polynomial, am I far from proving two finite fields of same order is isomorphic? I just dont know what to do next

[rocky cloak]

Yes, so I can outline the proof for you:

Say F has order q = p^n.

||Then every element of F is a root of the polynomial X^q - X (can you see why?).||

||A polynomial of degree q has at most q roots, so F consists of all the roots of X^q - X.||

||An element is a root of a polynomial if and only if it's minimal polynomial divides it. ||

||Now if you pick a generator z for F^x then F = Fp(z) = Fp[x]/(f(x)) where f is the minimal polynomial of z.||

||Then to get a map F -> G you just need G to have a root of f.||

||Since every element of F is a root of X^q - X you have that f divides this. ||

||Since G contains all the roots of X^q - X it also contains all the roots to any polynomial dividing it. So it contains a root of f.||

That's it, that's the proof

[crystal vale]

|| i think one can also prove this by showing the splitting field are isomorphic to each other, right? ||

[rocky cloak]

Sure, but then you need to know what splitting fields are

[crystal vale]

okay

[crystal vale]

i am trying to show in C(X), where X is compact Hausdorff space, then any prime ideal contained in unique maximal ideal

say p is prime ideal contained in m_a and m_b, how do i show a = b

a and b are points of X

[chilly radish]

X also needs to be infinite I'm pretty sure

[crystal vale]

why?

[chilly radish]

That's the only context in which I've seen the result. I'm trying to see if you I can find a counterexample for finite spaces

It's possible it's unnecessary

In the finite discrete case every prime is maximal, so this holds

Oh but finite hausdorff (even T_1) is discrete

So this is just trivial for finite spaces

[frail shoal]

i think until then you only needed compactness to get that they're all of the form m_a for some a ∈ X, but we need hausdorffness to show that m_a ∩ m_b never contains a prime ideal for a ≠ b

[crystal vale]

yes

[frail shoal]

i wont lie my topology knowledge isn't the greatest so idk the details but i think the gist is we use separation to get witnesses to not being prime

so hopefully this serves as a good hint

[chilly radish]

The problem is you know f(a)=f(b)=0 for all elements in P, but if you make a pair of functions whose product satisfies that, you don't know that it'll be in P

[frail shoal]

oh oops

sorry man i think i gave bad advice

[chilly radish]

This would prove that m_a n m_b itself isn't prime

[frail shoal]

so yea @crystal vale actually ignore me cuz idk what im talking about

nvm i think im thinking of a way to salvage it

||can you like, produce a pair of functions with disjoint support such that one is nonzero on a and another is nonzero on b||

[frail shoal]

i will admit half of the reason i responded was cuz i was like "oh that's a cool exercise imma work it out rn"

[chilly radish]

||I don't think so generally, no. Urysohn's lemma which is what you'd invoke can only guarantee that a function is 0 on some closed set and 1 at a point (or vice versa). I don't think there's a way, even with hausdorff, to get disjoint supports, since you don't know what happens away from these||

[frail shoal]

figures 😔

yeah im stumped

[chilly radish]

I was looking at the proof for the case C[0,1], but it quite explicitly uses order properties and the fact that [0,1] is a subspace of R

But I wonder if you can somehow try to reconstruct this argument

[tribal moss]

Choose disjoint open neighborhoods A and B of a and b; select f to be 0 on X\A and 1 on {a} and g to be 0 on X\B and 1 on {b}?

[proud bear]

||for each f in p, let Z_f=f^-1({0}). these are closed subsets of X. for any finite number of f_1,...,f_n in p, (f_1)^2+...+(f_n)^2 is in p, so it has to vanish somewhere to not be a unit. so Z_f_1\cap...\cap Z_f_n is non-empty. then the set {Z_f : f in p} has the finite intersection property, so there is a point in the intersection||

[chilly radish]

||FUCKING BAER CATEGORY THEOREM, OF COURSE||

[chilly radish]

You cannot guarantee that fg will be in P since you can't control the values outside of those sets though

Wait

[tribal moss]

Sorry, I didn't read far enough back and reacted just to Keith's specification of the two functions he wanted.

[chilly radish]

Ah

[tribal moss]

Didn't consider whether he would be able to use them for something useful for the original problem.

[chilly radish]

Actually, I think fg=0, since if you're in A, then g=0, if you're in X\A then f is 0, but neither are in m_a n m_b

So this should work

[tribal moss]

Wait a moment, fg will be in P because it is 0 when the supports are disjoint.

[chilly radish]

I jumped the gun it seems, apologies

[chilly radish]

You're right

@frail shoal I was wrong, you can have disjoint supports as Tropo demonstrated. I had smth different (that doesn't work) in mind

Ok so this finishes the argument

I was trying to use regularity/normality (given by compactness) when you should just be using hausdorff

[tribal moss]

Well, that does come in when we appeal to Urysohn's lemma.

[chilly radish]

Yea sure, I meant to construct the closed sets I want to separate, specifically

[tribal moss]

👍

[frail shoal]

why can we take disjoint open neighborhoods?

that's what i got stuck on

wait

i feel so dumb

[tribal moss]

Felix Hausdorff says so.

[frail shoal]

i conflated T2 with T1

im gonna explode myself

ok yea then in conclusion ||the definition of hausdorff + urylson lemma literally does give functions with disjoint support||

[frail shoal]

for this reason

[tribal moss]

Yeah, I agree that solves Notknow's problem.

[frail shoal]

glad to know i gave them a good hint after all

[crystal vale]

Yes

But now I am stuck on new problem

[tribal moss]

Math is such an ingrate.

[coral whale]

I was looking at the definition of the vector additive and scalar multiplicative identity in a vector space and they tend to take the form
I + v = v
J*v = v

Where I is the additive and J the multiplicative identity

The group identity axiom i have here, on the other hand, is
eg = e = ge

the inverse axiom says
gg^-1 = e = g^-1g

this caught my attention because this would seem to imply the identity element wrt the group operation functions differently in a group as opposed to the identity wrt scalar multiplication and vector addition in a vector space.

It seems according to the definition given here that the “identity element” (I don’t know if this is the correct employment of the term) of any group G is an element that always yields itself when “operated with” any other element of G.

The definition of the identity element of a vector space V wrt any of the two “seminal” operations, however, seems to imply that the identity element is that element of V which always yields the “other” when “operated with” any other element of V.

Am I missing something? Are my resources flawed? If not, are these respective “approaches” equivalent or analogous in some way?

[rocky cloak]

There's a typo in your notes. Should say
eg = g = ge

[coral whale]

ah

[rocky cloak]

That's generally what identity element means. When you combine it with something you just get that thing back

[coral whale]

Got it

[coral steeple]

Is this not a direct application of 9.4.1 or am I missing something??

More context

[swift root]

as far as i can tell lol yeah

thats it

[coral steeple]

My understanding of Galois theory is profoundly deepened by this exercise...

[elfin wraith]

I guess it’s a question of “can you not get scared by all the symbols and just apply the result” lol

[sudden condor]

such questions are typically an "are you awake" check

[small yacht]

What does he mean by cosets of the matrices?

[azure cairn]

$PSL_2(\bZ) = SL_2(\bZ)/\sim$, meaning that for every $A, B \in SL_2(\bZ)$, if $A \sim B$ then they belong to the same coset. so a coset is just the sets of matrices that are equivalent under $\sim$.

[azure cairn]

it's basically saying to show $PSL_2(\bZ)$ is generated by the coset of $S$ (${S, -S}$) and the coset of $U$ (${U, -U}$)

[small yacht]

Like every gS and gU for every g...?

Or is it that PSL_2(Z)/H for any subgroup H, it equals to SH and UH

sigh confusion

[azure cairn]

are you familiar with what cosets are?

[small yacht]

I think so yeah
for some subgroup H of G, a left coset of H is aH for some a \in G

where aH = {g in G | g = ah h \in H}

[karmic moat]

in this context, it's likely equivalence classes

show that PSL(2, Z) is generated by the (equivalence classes) of those matrices

anyway the center of SL(2, Z) is the normal subgroup {id, -id}, and PSL(2, Z) = SL(2,Z) / {id, -id}

so if you'd rather think of cosets you can just do it in that way

[small yacht]

chmonkey im back to being confused with this...
how are the equivalence classes of every element of the form aH, then
like if a \notin H, how is it equivalence class a left coset of H

wait

a ~ b iff a^-1b \in H

a ~ ah since a^-1ah = h \in H

thanks chmonkey!!!

[next obsidian]

You’re welcome

I did this

[small yacht]

thanks for telepathically helping me

[next obsidian]

This why H gotta be subgroup

So h = e

Or something

Idk

[maiden crater]

want to be sure I'm getting this right

$(2)= {2 p(x)}$ where $p(x) \in \Z[x]$

[cloud walrus]

Wai

[maiden crater]

Cool, so $(a) is basically just (2) and the set of polynomials with odd coefficients quotiented with (2)

[velvet hull]

There is a better way of describing this ring

[maiden crater]

hmm?

Isomorphic to something?

[velvet hull]

Yes

[maiden crater]

well, its char 2 so Z/2Z seems like one option

ooops

(Z/2Z)(x)

[maiden crater]

I will want to find something this is iso to as finding what (2,x^3+1) is would probably be futile

it's 2(p(x)+ (x^3+1)q(x) yes, but that doens't make finding the ideals any easier

[vocal pebble]

Z[x]/(2,x^3+1)=(Z/2)[x]/(x^3+1)

[maiden crater]

how

[vocal pebble]

Z/2 is a field, so (Z/2)[x] is a PID, and you want to find ideals containing (x^3+1) in this PID (think about factoring x^3+1 in (Z/2)[x])

[maiden crater]

yea, but how are they isomorphic

[vocal pebble]

I mean R/(a,b)=(R/(a))/(b') where b' is the image of b in R/(a)

[maiden crater]

okay, thanks

[maiden crater]

How does it follow that p must be associate to one of the irreducibles

[rapid cave]

We have ab = pc.
p is in the decomposition into irredcucibles of the RHS, so by unique factorization it should appear in the decomposition of the LHS, up to asssociates.

[maiden crater]

I don't get this

p is in the decomposition into irredcucibles of the RHS

oh

[maiden crater]

Well, what if c is reducible

I suppose that doens't matter here though

okay,got it

thanks

[faint knoll]

@pearl birch

[chilly radish]

Why did you ping a random pre-uni person who can't even see this channel?

[swift root]

first message this dude has ever sent 💔

[tribal moss]

I think everybody can see the advanced topic channels; it's just posting that requires jumping through hoops.

[swift root]

the hoops being clicking one button in the roles tab?

[rocky cloak]

I think it might be two buttons actually

[faint knoll]

i was just testing my keyboard

[chilly radish]

Don't do that in a channel next time

[south patrol]

(Esp. whilst pinging someone)

[small yacht]

i'm deleting that to avoid the shameful shaming

you guys saw nothing

[azure cairn]

real af

[small yacht]

how do i prove that [G, G] is normal?
i think the fact that gamma_g is an automorphism might be relevant, but i don't know how to go from there

[warped pilot]

Honestly the hint is the slick way to see things but it seems to be confusing so maybe worth saying that you can give a very concrete commutator that works here.

[chilly radish]

Ok so if you use that fact, and write

$\gamma_g(aba^{-1}b^{-1})=\gamma_g(a)\gamma_g(b)\gamma_g(a)^{-1}\gamma_g(b)^{-1}$

What is this?

[cloud walrus]

ShiN

[small yacht]

an element of the form aba^-1b^-1!

[chilly radish]

Exactly

So it's in [G,G]

[small yacht]

Thank you!

[chilly radish]

The same is gonna be true for any product of commutators

Yw

[azure cairn]

what book is this?

[small yacht]

Aluffi's algebra chapter 0

[azure cairn]

oh interesting

[next obsidian]

@small yacht okay there’s another way which isn’t necessarily rigorous but nonetheless I will explain

[small yacht]

explain waht

[next obsidian]

This subgroup is canonical which means more than being normal (which you can say means it’s fixed by all interior automorphisms which by definition is just conjugation), but it’s in fact characteristic meaning it’s fixed by all automorphisms

Now what I mean by canonical is that it’s defined by some properties which make no specific reference to specific elements

Let me list a few,
The center Z(G) references no specific elements
[G,G] references no specific elements
The unique subgroup such that…
The largest subgroup such that…

The idea is that an automorphism preserves all group theoretic properties of a group, so it needs to send something defined purely group theoretically to itself

Think of it this way, for the center Z(G)

Membership in the center is dictated by satisfying a bunch of equations, for every element g in the group you need to satisfy gxg^-1x^-1 = e (if you clear the inverses you get gx = xg)

[small yacht]

hheh i have no idea what that means

[next obsidian]

A group automorphism preserves that equation, so if x satisfies that equation over all g, then it satisfies the equation f(g)f(x)f(g)^-1f(x)^-1 = e

When you let g vary over the entire group G, then f(g) also runs over the entire group cuz f is bijective

[frail shoal]

oh hey i learned out of that one

[next obsidian]

So you see that gf(x)g^-1f(x)^-1 = e for all g, and so f(x) is in Z(G) because membership in Z(g) is based upon satisfying that equation enumerated over all of G

This process is also reversible (cuz f is an automorphism) which tells you if f(x) ended up in Z(G), then x was in Z(G) to begin with so Z(G) is sent to itself

What I wrote above is literally just a rigorous way to write down “Z(G) is the elements that commute with everything, and an automorphism needs to preserve that property)

[frail shoal]

this reminds me of a stupid example of a canonical subgroup, i call it stupid cuz it almost never comes up, but i think it's funny. you can consider the subgroup generated by all elements of the form x^2

it's also irrelevant here

[next obsidian]

Similarly [G,G] is generated by a bunch the things of the form [g,h] over all g,h in G

And when you apply an automorphism that gets sent to [f(g),f(h)]

[next obsidian]

Again cuz f is bijective these elements run over all pairs so you get the same thing

It’s okay if you don’t understand now

One day you will.

And you’ll go.

Wow, Chmonkey was right all along. He’s so smart.

[small yacht]

i agree

right now i'm more concerned wit

h

how to prove that G/[G, G] is commutative

[frail shoal]

hint: you want to prove that ab = ba, modulo [G, G]

[small yacht]

Yes that is what commutativity is

and what i'm trying to figure out

[frail shoal]

if you're stuck, try ||moving all the variables to one side||

[small yacht]

Well, i could start with ab[G, G] = ba[G, G]
multiply both sides by a^-1b^-1 to get
aba^-1b^-1[G, G] = [G, G]
So it is true for all a, b. Though i feel like I could find a more elegant way

oh

well

could just do it in reverse

[frail shoal]

this is essentially what all proofs have to look like

[small yacht]

starting with an equality is bad i thought
because you could do something like multiply by 0 on both sides, and if you do not catch it, it will seem liek you proved it

[frail shoal]

ok that's true. a philosophical opinion i have is that the reason why groups are so popular is that it's actually impossible to make this mistake in some sense

[small yacht]

start with [G, G] = aba^-1b^-1 [G, G]
multiply by ba on both the right
[G,G]ba = aba^-1b^-1[G, G]ba
because [G, G] is normal, [G, G]ba = ba[G, G]
therefore
ba[G, G] = ab[G,G]
qed

[small yacht]

really?
oh, i see how

[frail shoal]

yea

cuz you can always go in reverse

like you just did

[small yacht]

thx that's actually very good to know

[frail shoal]

np

[frail shoal]

i will say you can avoid using normality by doing it a little more carefully

[small yacht]

hmm

[frail shoal]

well ok i guess. normality is needed to even be able to mod out by it

otherwise it's not well defined

[next obsidian]

Maya

Make an equation that’s like this

[small yacht]

easy just divide both sides by [G, G] \j

[next obsidian]

Something involving a and b = e

Where this is true iff a and b commute

Then u can just ask yourself if this is 0 in the quotient

[frail shoal]

side note: there is something called the normal closure of a subgroup, which is the smallest normal subgroup containing a subgroup. if you try to fix the obvious issue you end up getting this concept i think

this is mostly irrelevant to the topic at hand tho

i just love tangents

[small yacht]

nooo 0 doesn't have an inverse with multiplication

[next obsidian]

If it’s e

Lol

[small yacht]

hwhwat

well if it's e then

it's it's own invers

[next obsidian]

No I mean

The LHS is something involving a and b

[small yacht]

i dont knwo waht you mean chop chop monkey

[next obsidian]

Like okay x = y

[small yacht]

indeed

[next obsidian]

Iff xy^-1 = e

[small yacht]

yes

[next obsidian]

So do something like that

[frail shoal]

yeah chairmonk is spitting

tbh

[next obsidian]

Where you say “something = e”

Iff a and b commute

[small yacht]

ab = ba
iff
aba^-1b^-1=e
or
a^-1b^-1ab=e

[next obsidian]

Okay

Now is that equation true in the quotient

[frail shoal]

it's cool how every equation in the language of groups can be rephrased as "blah = e"

by language of groups i mean like. multiplication, inverses, identity

you can also sorta cycle stuff around

[next obsidian]

Me when I invent model theory

[small yacht]

i think so
cuz the left side is in [G, G] so in the qutioent this amounts to saying e = e

[next obsidian]

Yup

Wowza

[frail shoal]

i hope to do a model theory phd over the next 5 years orz

[next obsidian]

So a and b commute in da quotient

This holds for all a and b

[small yacht]

I'm so sleepy should i be doing math this sleepy

[next obsidian]

Wowza^2

[frail shoal]

eepy sleepy

relatable

[small yacht]

yowza

[broken wolf]

i have a group theory module next year im looking forward to it but need a book to get started over the summer :p

or a libretexts course they are fire

[next obsidian]

I’d probably just use whatever text your course will use

Ask friends or something if they know what the course usually uses

[broken wolf]

ngl this looks complicated

[next obsidian]

Lol

[broken wolf]

i prefer applied maths as im doing a joint honours but this looks like the more formal side

[next obsidian]

Yeah

I mean it’s probably a proof based algebra course

You’re just gonna have to get used to it by doing it

Algebra is pretty abstract as a subject, maybe moreso than analysis given how it’s usually taught

[knotty badger]

It is literally called “abstract algebra”, no?

[next obsidian]

Idk some ppl call it that

I call it algebra

Once you grow up and become a big kid everyone knows you aren’t talking about y = mx + b

[copper kestrel]

i love proving polynomials of degree 4 are irreducible it brings me so much joy

(sarcasm)

[next obsidian]

Which ones

Over what ring

[copper kestrel]

trying to show 1+i is either algebraic or transcendental over R

[next obsidian]

Wut

All of C is algebraic over R

[copper kestrel]

right

[next obsidian]

It’s finite dimensional

[copper kestrel]

i have to prove it tho :')

i'm just trying to show a^4 + 4 is irreducible

[next obsidian]

I mean you don’t need to show the polynomial is irreducible lol

If it’s a zero of that polynomial

[copper kestrel]

right but if it is algebraic i need to find deg(1+i, R)

i'm guessing its 4

[next obsidian]

Why

It definitely isn’t

C is deg 2 over R

[copper kestrel]

question 10

[next obsidian]

C is 2 dimensional over R

Because it’s generated by i so it’s R[x]/(x^2 + 1)

[copper kestrel]

sigh

[next obsidian]

So if you let a be anything in C\R you have the following inclusions

R < R(a) < C

The first < is proper and since C/R is degree 2 this forces R(a)/R to be degree 2

So R(a) = C and deg(a,R) = 2

[restive quail]

Get real

[copper kestrel]

hmm interesting

[next obsidian]

There just isn’t enough room

[copper kestrel]

ah i see, so since C is a 2 dim vector space over R the extension is finite, so it must be an algebraic extension thus 1+i is algeraic

[next obsidian]

Yeah

[copper kestrel]

and then somehow show that dim(a, F) is 2

[next obsidian]

I mean you’re forced to have R(a) = C

[copper kestrel]

like you said

[copper kestrel]

true

[next obsidian]

And then the degree is 2

[restive quail]

R[x]/(x^2+2)

1+i solves

🐱

[next obsidian]

I don’t think so…

[restive quail]

Isomorphic to C

Wha

Im isnane i need to go back to bed

[next obsidian]

It’s some dumb shit like

[copper kestrel]

have a good sleep :]

[next obsidian]

x^2 - 2x +2 or something

[restive quail]

Aw

Ohhh yeah

[next obsidian]

It’s definitely x^2 -2x + c

[restive quail]

Cuz youd be squaring the 2

[next obsidian]

And I’m too lazy to compute the c lol

[copper kestrel]

real

[next obsidian]

I mean it’s just that you end up with 2i when you square 1 + i

[copper kestrel]

i need to remember what F(a) as a field is

[copper kestrel]

yeye

[restive quail]

Yea

[next obsidian]

And the linear term is the only way it can die

[next obsidian]

Fractions with a

But if a is algebraic the it’s the same as F[a]

Where you allow polynomials in the a

[copper kestrel]

hm

[next obsidian]

Sorry I guess like rational functions in a

But a isn’t formal here it’s the actual element

That’s why you need a to be inside a field extension L

Cuz the only way you know how to make sense of those expressions is by using the operation in L

[copper kestrel]

yeye

fields are hard bleh

[restive quail]

Field impawsible

[copper kestrel]

literally :(

[restive quail]

I was chatting with my professor about structure of padics and half of the stuff he explained i forgot within the next minute or so

There so many objects floating around

[copper kestrel]

so real omg

[restive quail]

Yk what else is so real

Pi

[next obsidian]

Haha padic valuation tells you the pness of an element

[copper kestrel]

true!!

[restive quail]

So many of the properties for padics have to change for p=2 and its so unintuitive to me

[next obsidian]

Even

Lowkenuinely it’s just that 2 shows up too often

And u gotta divide

[restive quail]

Like how filtration of multiplicative group of the integers starts at 1+4Z2

[next obsidian]

Even quadratic formula be like bleh in char 2

[restive quail]

Ohhh

[copper kestrel]

lowkenuinely 😭😭😭

[next obsidian]

I’m one of the greatest to ever do it

Liluzivert

[restive quail]

What the one uzi vert song

Uhhh

Push me to the edge

All my friends are dead

[next obsidian]

Allmyfriendsrded

Xotourlife

[restive quail]

Xo tour life

Yes

[next obsidian]

I listened to that a lot in 2017 post breakup

[restive quail]

Bwahaha that sounds so real

2017 i was but a kiddo

[next obsidian]

In reality I’m 5’4, stand on my money now I’m 6’6

I was 18

[copper kestrel]

truth!!

[next obsidian]

A kiddo to me now