groups-rings-fields

[delicate orchid]

they haven't done the ol e^ix meme

[cloud walrus]

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

[hidden haven]

Without Zorn there can still be many you just can't prove that they exist

[chilly ocean]

its consistent that there are just 2

[delicate orchid]

there are only 2 and you cannot convince me otherwise

[hidden haven]

If they define it with standard norm |•| then it's fixed

[chilly ocean]

idk why i dont understand this theorem

i need examples

yeah its actually not easy imo to understand it early

everyone either shows the diagram or says the same shit on youtube

it eventually clicks

[delicate orchid]

first iso is fantastic once you truly understand it

[prisma thunder]

It's a pretty broken tool imo

top tier 👌

[delicate orchid]

lemme explain I wanna explain the first iso ME ME ME

[chilly ocean]

dont

[delicate orchid]

ok I won't

[chilly ocean]

let moldi

[hidden haven]

Yo first iso thread when

[chilly ocean]

[pastel cliff]

yes

[chilly ocean]

POST THE ARROWS

[hidden haven]

I need to sleep

[pastel cliff]

no

[delicate orchid]

catfan the first iso is the ONE thing I can actually explain intuitively

[chilly ocean]

hes gonna drop the diagram

LOL

[delicate orchid]

hell one might say visually

[hidden haven]

Yooo wew rant time

[chilly ocean]

ok go

wew

[pastel cliff]

make a thred

[hidden haven]

I shall come inspect your thread tomorrow

[pastel cliff]

so i can come back and read this easily :D

[chilly ocean]

first iso

[subtle ivy]

the best way ive heard it explained is that the kernel of a homomorphism is the "resolution" of the map- how accurately it preserves the structure of the domain

[hidden haven]

Wew you better not post explanation here

Post in thread

[delicate orchid]

I won't dw

[subtle ivy]

so by "dividing out" the kernel, you are obtaining a perfect preservation

no error

[chilly ocean]

Id like to think that dividing by kernel basically kills all non injective elements into just 1, anmely neutral element so since the map was surjective now its also injective so you have iso

[rigid cave]

wew lookin fresh wit da ✓

[chilly ocean]

https://media.discordapp.net/attachments/342850939306246145/939939323707527198/toki.gif

[pastel cliff]

why ✓

[subtle ivy]

i started typing before thread 😌

[hidden haven]

Tinky dinky cult

[pastel cliff]

bean cult when

[hidden haven]

Merge into tinky cult and you may not get killed before you can register

[pastel cliff]

https://tenor.com/view/no-i-dont-think-i-will-captain-america-old-capt-gif-17162888

[chilly ocean]

join 1337 gang

[hidden haven]

I'm catfan tinky cult at the same time 😌

[delicate orchid]

where is the monoids thread

[hidden haven]

Archived

[delicate orchid]

truly we live in dark times

[pastel cliff]

it's right there...

silly wew lad

[hidden haven]

wew lads ngl

[pastel cliff]

he's not lying yknow

the verbosity... it's killing me

[coral shale]

do induction

[pastel cliff]

ive said literally nothing here lol

[coral shale]

and prove it yourself

[pastel cliff]

oh ik how to prove it lol

it's not hard, i just hate having to be this verbose

[coral shale]

thats your proof?

[pastel cliff]

so far

[coral shale]

I think you can reduce words...

[pastel cliff]

i know yeah i'll send later

[hidden haven]

You can just say at the end that inverses are unique so beta is alpha inverse

Since alpha beta = 1

And beta alpha = 1

[coral shale]

couldnt u multiply the 2 things together
Then uniqueness
Thats like a 1 line proof surely

[hidden haven]

Beta must be the unique inverse of alpha

Ye

[pastel cliff]

yeah i have it all in scratch work

im just

paranoid

[wise igloo]

banana pingu 🍌 🐧

[hidden haven]

Dw bro if your prof docks your marks we'll roast him in this channel together

♥️

[pastel cliff]

https://tenor.com/view/cat-soup-nekojiru-sou-masaaki-yuasa-beating-shut-up-gif-21583840

[pastel cliff]

too verbose?

need conclusion too ofc ik

[delicate orchid]

personally I'd drop 90% of the first paragraph

looks fine though

being extra verbose when you're starting is useful to make sure you fully understand what's going on

[pastel cliff]

i mean i get this material

but like

"bad english"

[delicate orchid]

I fuckin hate this guy dunno if I've mentioned it yet

your english is fine

[pastel cliff]

i know it is

murica

[delicate orchid]

ok just because it's your first language doesn't mean you can't type utter drivel

lord knows I do

[pastel cliff]

im the one who has to sit through lectures with him

who said it was my first language

how do i even get formal about this lol

[delicate orchid]

you check the axioms duh

[pastel cliff]

inherits identity is easy

but closure

do i even need to prove closure

actually probably inherits from R too

[delicate orchid]

a line about the product of two positive numbers being positive won't go amiss

it's a subset you can't say it inherits closure

[pastel cliff]

this sign wont stop me because i cant read

R_+ is associative

proof:

im gonna assume associativity is inheritable right

[delicate orchid]

it is

thank god

[pastel cliff]

if i do well enough on everything else im slapping one of these on a homework at some point

[dreamy fiber]

Does anyone know what an Ext and Tor Functors are exactly?

My backgrounds in topology are fairly limited

Is there a good explanation of the two from the perspective of commutative algebra?

[next obsidian]

They're the derived functors of Hom and the tensor product

Explicity if you want to compute Ext^n_A(M,N) you take either a projective resolution P_i of M then look at Hom(P^i,N) and then take the n-th homology, or an injective resolution of N I^i and then look at Hom(M,I^i) and then take n-th homology

for Tor_n^A(M,N) you can take a projective resolution P^n of M, then look at the n-th homology of P^n (x)_A N, or do the same with a projective resolution of N and then look at n-th homology of M (x)_A N

Tor is named Tor because among other things, if a is a non zero divisor in A, then Tor_1^A(M, A/(a)) is the a-torsion of M

and Ext is called Ext because it classifies extensions, which are certain exact sequences with M and N on the end of length n up to some equivalence relation

[shadow skiff]

can someone plz explain what the notation (m,n) = 1 means? not looking for help with problem just unfamiliar with that expression. tyty

[next obsidian]

It’s their gcd

[delicate orchid]

It's probably shorthand for gcd yeah

bad shorthand imo

[next obsidian]

But in this case

Since it’s an integral domain

I think it just means that (m,n) = (1)

So like that ideal is the entire ring

In a general integral domain I don’t think you have a notion of gcd in general, so unless this is saying “assume they have a gcd, and it’s 1…”

[shadow skiff]

okay thanks that makes sense

[barren sierra]

I need to classify the groups of order 28

is this enough?

I feel like there's something more that I need to do, some more justification

oh I could say P is normal so PQ = G

but otherwise

idk ¯_(ツ)_/¯

[thorn delta]

you're missing an abelian group of order 28 btw

[pastel cliff]

wtf is a "nonsingular" matrix

[thorn delta]

invertible

[pastel cliff]

well that's dumb

just call it invertible

[next obsidian]

No

[pastel cliff]

speaks

[prisma shuttle]

can someone explain wut $\mathbb{R}[S_3]$ means

[cloud walrus]

JustKeepRunning

[pastel cliff]

dumb question, how can i prove that all positive reals have inverses

[next obsidian]

It’s the group ring

Look up “group ring” and you’ll see what it is

[next obsidian]

Take negative

[pastel cliff]

*under multiplication

[next obsidian]

Take 1/x

Anything more than this requires an explicit construction of R

[pastel cliff]

yeah that's my problem lol

[next obsidian]

In which case you probably have to resort to using Cauchy sequences and then using that Q is afield

[pastel cliff]

idk if should do analysis here

[next obsidian]

As long as the thing is non-zero, eventually the Cauchy sequence representing a is non-zero

So you can invert that I think or something

¯_(ツ)_/¯

[pastel cliff]

fuck this im not doing the analysis proof

but ty chmonkey

[thorn delta]

i heard somewhere that the construction of completion by cauchy sequences uses that Q is dense in R, so the definition of R as the "completion of Q" would be circular.
moral of story: break out the dedekind cuts

[next obsidian]

Lol no

You define R as being the completion

And then you get that Q is dense by construction

[uncut girder]

@fossil shuttle do you have a good motivation for group cohomology?

[fossil shuttle]

Um, yeah maybe

[uncut girder]

Id like to hear it, please

[fossil shuttle]

idk? let's give it a shot

[pastel cliff]

hi sorry dumb rq

[uncut girder]

Wait

Diligent let me open a channel

[pastel cliff]

what's a correct negation of this

[uncut girder]

Group Cohomology

[pastel cliff]

i thought "There exists a group $G$ such that $a,b \in G$ and $aba^{-1}b^{-1} \not = 1$" but that could be wrong

[cloud walrus]

nonabelian beans

[uncut girder]

#943671297311932496

[next obsidian]

It would be “there exists a,b”

It’s not a forall

[pastel cliff]

"\exists a,b in a group G such that..."

[pastel cliff]

i was thinking this cuz i was gonna use dihedral group as a specific example

[next obsidian]

The claim that’s written down is literally just that all groups are abelian

[pastel cliff]

[lavish nexus]

all the order 2 subgroups seem to be an automorphism for this F(t_1 t_2 t_3)

order 3 too

what does he mean by invariant under G_f\cap V

because it's not fixed under G_f\cap V

nvm

I see why

[chilly ocean]

for a set of givens we can just append them as "and" statements, so the negation becomes "G is not a group OR there are a,b in G where that eq is false"

ik yall settled this already just wanted to make it super clear what the negation was precisely

[desert dome]

Suppose p(x) \in F(x) and it has a root c in F, so we can write p(x) = (x-c) q(x). And q(x) is also in F(x). Is it because if it has coefficient not in F, then p(x) will not be in F(x)? Or in other words, if a \in E \ F where E is an extension of F, ab is guaranteed to be not in F if b is in F?
(Sorry if my words are confusing)

[barren sierra]

hm u right

wait but that gives 5

wait r u sure?

no other factorization of 28 satisfies classification

are these questions related in any way?

I solved 9, that wasn't bad

if I have the 4 congruence classes

[chilly ocean]

my brain isn't working right rn so I can't show it in full, but this is almost the same as claiming that the division algorithm works for polynomial rings

[barren sierra]

are the 4 homomorphisms determined by the image of 1 in C_2

[chilly ocean]

so that proof should justify what you need

[desert dome]

ohh right. I see. Thank you! I'll take a look

[chilly ocean]

me too 😂

[gritty sparrow]

There is a connection, such a homomorphism is determined by where you take the non identity element, and you can take the non identity element to any automorphism f of C_2^n that satisfies ff=id, now any automorphism of f is determined by its value at 1, say x. (because then t is forced to go to tx) and any choice of x will give an automorphism. To satisfy ff=id, it is enough that ff(1)=1, in other words x^2=1

[barren sierra]

aight bet

so use the congruence classes I found before

that's kinda slick

[gritty sparrow]

yeah, although to be honest you should prove the correspondences that i mentioned as well

[desert dome]

still want to double check this tho: if E is an extension of F, a \in E but not in F, b \in F, is it guaranteed that ab not in F?

[gritty sparrow]

as long as b is nonzero, yes. prove this by by contradiction by multiplying ab by b^-1

[desert dome]

right.. Thank you!

[lavish nexus]

First part is easy

then let g be the minimal polynomial of r, we have deg g <= p

and if K contains E the splitting field of f, since [E:F] = 3, 3|[K:F] <= p

p = 3 case is clearly false so deg g < p

[barren sierra]

yea

[lavish nexus]

and I'm stuck here

I guess E is Galois so normal and Gal(K/E) ⊲ Gal (K/F)
and Gal(K/F)/Gal(K/E) = Gal(E/F) = Z_3

but what can Gal(K/F) be

[chilly ocean]

min poly of r in F is x^p, would it be wrong to say [K:F] is exactly p?

[lavish nexus]

I'm thinking pth root of unity

in that case x^p-1 is not irreducible

and [K:F] would be p-1

[chilly ocean]

ah

tru

[lavish nexus]

so I have no clue about what the min poly actually looks like

[chilly ocean]

yeah

and whoops I meant x^p -1 even though that's still not true obv

[lavish nexus]

It would be x^p-a for some a in F and so far I showed it must split over F

[barren sierra]

hang on

the identity is 0 right?

cause the integers mod 2^n are not in general a group under multiplication

so don't I need x^2 = 0?

[gritty sparrow]

the identity is 0

[barren sierra]

nvm

I am dum

[chilly ocean]

NO

[barren sierra]

yes

[chilly ocean]

:(

U ARE VERY GOOD AND COOL

[barren sierra]

this class is making me feel dum

[chilly ocean]

saying the occasional silly thing doesn't mean anything really

ur feelings are valid but they are also inherently arational

just pointing it out cause I sometimes go through a similar thought pattern

[chilly ocean]

that f splits in F? wasn't it irreducible over F? so then you're done via contradiction? you're ahead of me with your understanding of several things but just was curious about the proof you were doing :)

[lavish nexus]

no that x^p-a splits in F

[lavish nexus]

basically this

[chilly ocean]

ohh the min poly of r my bad

I definitely did not follow that earlier bit but best of luck

[median pawn]

What is a linear space?

Do they just mean vector space?

[compact needle]

Vector space, unless F is just a ring, in which case they mean module

[median pawn]

F is a field, so we're good

[barren sierra]

back to this

ok so the order of C_2 x C_2^n = C_2^n+1

I can get that

but how do I show that the 4 homomorphisms give 4 unique semidirect products

[next obsidian]

I think you have to use the structure of the homomorphisms. Maybe there’s something you can argue about the orders of the elements?

I know ways to show that two homomorphisms give rise to the same semi direct product (they differ by an automorphism)

But I don’t think the converse is true

[pastel cliff]

i went w this ¯_(ツ)_/¯

[pastel cliff]

oh i never sent the picutre

[chilly ocean]

Guys I tried doing some algebra for part a but I think I might have to use the correspondence theorem

[lethal dune]

why do you need correspondence theorem for part a?

[white jackal]

is i inverse just equal to plus or minus 1?

[lethal dune]

no

[hidden haven]

use the given identities

[white jackal]

so i = jk right?

[hidden haven]

You have to construct an element which multiplies with i to give 1. ||Find this in steps by multiplying i by one thing at a time||

[white jackal]

what about -jkj?

can i use these to solve for -jkj?

[pastel cliff]

hi moldi

[hidden haven]

hello former bean

[hidden haven]

ye

[chilly ocean]

okay so i finished a

or me and onnichan did

we teamed up

[pastel cliff]

does this look aight

[hidden haven]

damn very nice

[pastel cliff]

not my handwriting

[chilly ocean]

how tf do i do part b

[hidden haven]

Nice

[pastel cliff]

lfg

,ti

[cloud walrus]

The current time for nitezba is 02:09 AM (EST) on Thu, 17/02/2022.

[pastel cliff]

fighting for my life rn

[hidden haven]

I would start by listing all elements

Bro you just need to get through this week

[pastel cliff]

just one more week bro

[chilly ocean]

so like order of 1 would be 1

[pastel cliff]

one more week is all i'll be aight

[chilly ocean]

-1 would be 2

right?

[hidden haven]

yes

[chilly ocean]

okay so

i got

order of -1 = 2

order of 1 = 1

idk how to do the rest

lol

I know i^2 = -1

But I need it to be 1

[lethal dune]

wtf is this

[chilly ocean]

Would that work for the order of I

-i

[pastel cliff]

,ti

[cloud walrus]

The current time for nitezba is 02:34 AM (EST) on Thu, 17/02/2022.

[pastel cliff]

idc

[pastel cliff]

im working w a friend we're ok w potential badness

@hidden haven do i need to show associativity?

[hidden haven]

I will see after lunch

[pastel cliff]

wait

i forgor to show the abelianness

pain

oh wait

nvm

[hidden haven]

F

[pastel cliff]

first line does it right....?

[hidden haven]

Ye

You can look at the first iso thread btw

[pastel cliff]

thats a weekend problem

[hidden haven]

I'm writing an explanation and so far it should make sense

Sure

[pastel cliff]

btw, do you mind if i ping? like in general?

i wont abuse ofc (unless i do)

asking bc i wanna study more of this on my own and you and wew lad seem to know the shit

[hidden haven]

Ye sure

I may just refuse to help though (:

[pastel cliff]

just this week bro

petition to make this a sticker

or a better version of it at least

[hidden haven]

Let me put this in suggestions and get sullied out of existence

[pastel cliff]

<3

keep me posted lol

[hidden haven]

will be unreadable. crop it a little and sharpen the edges first
Says 19eddy4

[pastel cliff]

valid

i'll remake it in the morning and post it in obsidian

[chilly ocean]

guys

when it asks give order of a coset

its just how many elements are in that coset right?

i got a quotient group with 4 distinct cosets in it

each coset has 2 elements in it so that means that the order of each coset is 2?

[paper flint]

If you're within the quotient group, your group elements are cosets so "order" might refer to the order of your coset as an element of the group.

[chilly ocean]

yeah so 2

[paper flint]

And not necessarily the number of elements in your coset.

[chilly ocean]

i honestly dont understand what that means tbh

i put 2

for this

[paper flint]

Okay, you'll agree that Z/5Z={0,1,2,3,4}, right?

[chilly ocean]

yes

[paper flint]

Here each of the elements 0,..,4 is representative of a coset

So order of the coset being represented by, say, 2, is the order of 2 in this group

(which is 5)

[chilly ocean]

so how would i approach it for my case?

my cosets are

1H

lol

[paper flint]

1H is just H, it's the identity for your quotient group and naturally has order 1

[pastel cliff]

can someone check this super quick pls

lie product defined as A#B = AB - BA

[paper flint]

For the rest, multiply coset by itself till you get the identity coset

[chilly ocean]

how would i multiply say jh

{j,-j} {j,-j} ?

how do i multiply them

is it kinda like fouil

FOIL

[paper flint]

You won't be multiplying these as sets, you choose a representative for your coset and multiply it with itself

Like we were adding stuff mod 5 in Z/5Z with our coset representatives

[chilly ocean]

so

i would do

iH * iH

?

until i get H

[paper flint]

Direct your attention to Q/H={H,iH,jH,kH} instead

[paper flint]

Precisely

[chilly ocean]

okay

so iH ^k

i know i^2 = -1

and i^-1 = -i

so ican use those right

[paper flint]

Yes

[chilly ocean]

so H would stay H

[paper flint]

Furthermore, there are only 3 possibilities for orders of these elements by Lagrange's theorem

(1, 2, or 4)

[chilly ocean]

yes

[paper flint]

Yes, that's the group identity.

[chilly ocean]

ohh

so i can lowkey just check for order 2

and it doesnt finish the job

so it must be 4

right?

@paper flint

[paper flint]

Exactly!

[chilly ocean]

awesome man

its very clear

[paper flint]

[chilly ocean]

hey @paper flint

can i ask u one question for part e?

cant i just say that the order of Q and Q/H are not equal

so not isomorphic

since |Q| = 8

[paper flint]

Yeah absolutely

[chilly ocean]

and Q/H has 4

thats valid right?

[paper flint]

An isomorphism is meant to be a bijection, you can't have a bijection between sets with different cardinalities.

So your argument is completely valid.

[chilly ocean]

cool

thanks!

[paper flint]

No worries, goodluck.

[hidden haven]

Yes

lol

[coral shale]

https://i.imgur.com/s4dF6Az.png

If L : K is an algebraic field extension

It doesn't follow K[x] = K(x) yes?

^this would just be nonsense

[hidden haven]

Yes

[next obsidian]

It totally does

[coral shale]

But (forall a in L) K[a] = K(a)

[next obsidian]

If a is algebraic over K

Then that equality holds

[hidden haven]

For all A in L

[coral shale]

yah, but im writing the polynomial ring/polynomial fraction field

Just clearing up this confusion in my head between the 2

[next obsidian]

Well in that case L is completely irrelevant

[coral shale]

Like there's no way K[x] = K(x)

[hidden haven]

Ye if x is a variable then those aren't equal

[coral shale]

yh yh

argh.

[hidden haven]

Because x transcends

[coral shale]

oooooohhhhhhhhhhhh

Ok ok heh heh

Still wrapping my head around what exactly this is

[lethal dune]

[hidden haven]

Field of fractions of F_p[x]

[coral shale]

im getting there yh... thanks 😅

[frail zealot]

don't most people take F_p to mean integers modulo p, p is prime

[coral shale]

yes

idk, I'm just mixing up ideas in my head.

I am thinking of stuff like K[x] / (f) at the same time

[hidden haven]

Remember to take t!cat breaks in the middle of study 😌

[coral shale]

K[x] / (f) is K[x] mod f

F_p[x] is Z[x] mod p

ones a number, ones not 😄

[frail zealot]

ah

[coral shale]

Thanks 😄

[hidden haven]

What is F[x]

[frail zealot]

i have two modes: study way too much, take a break that lasts weeks

[hidden haven]

You mean Z[x]

[coral shale]

polynomials in Z, but u mod the coefficients

yh ok

[hidden haven]

And it is not just a kinda btw

Z[x]/(p) = (Z/(p))[x]

[coral shale]

OH

AHHHHHHH

ahahahahaahaha

godamn

i was looking at a specific case all the time smh

[hidden haven]

[coral shale]

so before u said that

i was thinking to myself what if we mod p and mod f at the same time

(Z[x]/(p)) / (f) or something 🤔

[hidden haven]

That is isomorphic to Z[x]/((p)+(f))

Third isomorphism theorem

R/(I+J) = (R/I)/J, where the J on the right side really means J/I

[coral shale]

right, cus that (f) is mod p

[hidden haven]

ye

write → rihgt → right

[coral shale]

english language is a multifunction or something

Ok cool stuff, thanks lel

[coral shale]

Let $B$ be a basis of vector space $V$ over $F$. Then

$$V = \left{\sum_{k\in I} a_kb_k : I \subset\bN, |I| < \infty, (\forall k\in I)(a_k\in F, b_k\in B)\right}$$

[cloud walrus]

Shuri2060

[coral shale]

Uff thats ugly

But checking --- we only take finite linear combinations for an infinite basis right (me attempting to express that with symbols ^ )

[lethal dune]

$\sum_{\text{finite}} a_kv_k$

[lethal dune]

is what I use

[coral shale]

Words to the rescue 👀

but ok, yh, ty

[hidden haven]

Why I ⊂ ℕ

[next obsidian]

SO TRUE MOLDI

[coral shale]

?

[next obsidian]

You don’t have to have a countable basis

[hidden haven]

Where ℕ come in

[coral shale]

But you can only have a finite combination

[hidden haven]

Even if countable basis

[coral shale]

(was my original Q)

Is it not by definition, we can only take finite linear combinations ?

[hidden haven]

Oh ok I see

So you're indexing then yourself

[next obsidian]

Yeah Moldi that was what I realized too

[hidden haven]

In that case you can also sum from 1 to n and say n ∈ ℕ

[next obsidian]

Oh so true

For arbitrary n

[coral shale]

that sounds less symbols so yes !

$$V = \left{\sum^n_{k=1} a_kb_k : n\in\bN, (\forall k)(a_k\in F, b_k\in B)\right}$$

[cloud walrus]

Shuri2060

[coral shale]

or just this 🤔

[frail zealot]

isn't it a given that the n on top of the sum is in N

[coral shale]

If I want to be accurate, the n can be different for each combination, so I want it after the colon regardless to be clear about its scope

[frail zealot]

in the very beginning of the paper say n\in N = {1,2,3,4,...} and then forget about it

[coral shale]

I'm not writing a paper, but that aside, if I omit n after the colon, it may be assumed to be a constant

[frail zealot]

then use words or just use the finite one

[coral shale]

Let $n \in \bN$.\

$$V = \left{\sum^n_{k=1} a_kb_k : (\forall k)(a_k\in F, b_k\in B)\right}$$

[cloud walrus]

Shuri2060

[coral shale]

This is very different

so yh - either words/finite, or what was before to be accurate about the meaning

[frail zealot]

as sad as it is words are some times clearer than symbols

to humans

[coral shale]

$$F_3[x] / (2x + 1)$$

[cloud walrus]

Shuri2060

[plush wasp]

Hmm

[coral shale]

Claim from me : (2x + 1) = (1)

I'm just thinking how to show this immediately uff

2x + 1
4x + 2 = x + 2

@plush wasp In practice, euclidean algorithm with 2x+1 and 3 should be in the ideal...

[plush wasp]

*4x + 4 = x + 4

[coral shale]

why

2(2x+1) is in the ideal

[plush wasp]

yes, oh right, sry 🤦‍♂️

[coral shale]

If I didn't interpret this wrongly though, it should be true

[plush wasp]

Imma just shut and listen

[coral shale]

Ansh how do you do gcd on this (2x+1, 3) (ive forgotten)

[plush wasp]

no clue lmao

[coral shale]

eh???

[robust pollen]

If $A$ is an algebra over a comm ring $k$, and there is a ring hom $\epsilon \colon A \to k$, then it makes sense to look at the tensor products $\overline{M} = M \otimes_A k$ for any right $A$-module $M$, where $A$ acts on $k$ from the left via $\epsilon$. Let $A^+ = \ker\epsilon$.
Why do we have $\overline{M} = M/MA^+$?

I see the obvious map $\pi \colon M \to \overline{M}$, the only thing I'm missing is the inclusion $\ker \pi \subset MA^+$.

[cloud walrus]

expectTheUnexpected

[coral shale]

ansh

[wind steeple]

If you have an exact sequence 0->A+->A->k->0 (it is surjective bc A is a k-algebra) by tensoring by M you get the exact sequence M(x)A+->M->Mbar->0, the image of the first morphism is by definition MA+

[robust pollen]

Ah, very good. Thanks 🙂

[prisma shuttle]

can someone explain why $\mathbb{N}_0$ satisfies the ACCP condition

[cloud walrus]

JustKeepRunning

[chilly ocean]

whats N_0?

[full panther]

I can't seem to verify this result

i get different answers for sigma(x), ...

and this is an example of automorphism group of a lie algebra

unless im mistaken because those results arent even in the group SL(2,F)

i got sigma(x) = exp(h), sigma(y) = (exp(h))^2, and some completely different matrix for sigma(h)

[hot lake]

how did you get an exp of a lie algebra element ?

exp (ad x) = 1 + (ad x) + (ad x)²/2 + ....

exp (ad x) (y) = y + [x,y] + [x,[x,y]]/2 + ...

for this special case, all those "infinite sums" are finite

so it should be easy to get the matrices of exp (ad x) and exp (ad (-y))

if I call fx = exp (ad x)

I get fx(x) = x, fx(h) = h-2x, fx(y) = y+h-x for example

no exp anywhere

[rustic crown]

btw i had a question, so i was wondering when we could say the following,
R --> S is a map between rings then SL_n(R) --> SL_n(S) is surjective

I could prove it for Z --> Z/NZ and SL_2 but don't see anything nice in general

in general we can't expect GL_n(R) --> GL_n(S) to be surjective, because if it were, then so would be R* --> S*, which doesn't hold even in nice conditions like Z --> Z/5Z

[hidden haven]

Is this even true for any field homomorphisms?

Doesn't seem like it

[rustic crown]

like one necessary condition i would expect is R --> S surjective which won't happen for fields

[hidden haven]

Fair

[rustic crown]

the proof for Z/nZ was also pretty bad

say a matrix [a b \ c d] which is sent to something with det 1 in Z/NZ, we need to be able to tweak a, b, c, d by Ns so that the new matrix [a' b' \ c' d'] has det 1

the main idea was the following if gcd(c, d, N) = 1 then we can find c' and d' such that c = c' (mod N) and d = d' (mod N) such that gcd(c', d') = 1

now we can tweak a and b, and since c' and d' are coprime, we could be able to prove what we needed.

don't see how this idea would generalize, showing coprime cofactors would be annoying?

[full panther]

Oh i forgot the y in the start of the sum in this case lol

[full panther]

i think i was just doing exp(ad x(y)) lol

[coral shale]

Claim: if f in R[x] is irreducible, (f) is maximal

This is true if R is a field. Is it true if R is a ring?

[lethal dune]

in Z[x], x is irred but (x) is not maximal as (2, x) is a proper ideal containing (x)

[delicate orchid]

it's true in PIDs

[prisma shuttle]

its true in PIDS in general

can someone explain wut this diagram is syaing

like why are there two V--> W arrows idk wut is the point of that

[coral shale]

thanks

[delicate orchid]

it's saying T(av) = aT(v)

[prisma shuttle]

and i don't see how that is equivalent to the condition T(a * v ) = a * T(v)

[prisma shuttle]

wait how

[delicate orchid]

follow the arrows

[prisma shuttle]

also wut do $\rho_1$ and $\rho_2$ represent

[cloud walrus]

JustKeepRunning

[lethal dune]

row(a) means scale by a

[delicate orchid]

top line is multiplication by a first, and then T

i.e. T(av)

bottom is T first, then multiplication by a

which is aT(v)

the fact that the diagram commutes is equivalent to saying these two compositions are always equal

[lethal dune]

this sticker is like the "QED" of any proof contacting the word "diagram"

[delicate orchid]

[prisma shuttle]

ok thx for explaining uguys

[delicate orchid]

why you'd ever phrase that as a commutitive diagram I will never know

[lethal dune]

true

obsessed category theorist

[rigid cave]

Associative diagrams when

[hidden haven]

Just you wait till I finish this fire pasta

[hidden haven]

I mostly said that as a joke but there's also proper reasons

[delicate orchid]

when your supervisor writes this on your coursework

when your supervisor writes this on your coursework

[hidden haven]

I just wrote so true on an assignment I was grading

[hidden haven]

Because the guy wrote some deep stuff

Can't wait to embarrass him

[delicate orchid]

my proofs using diagrams are still very sloppy apparently 🤨

[hidden haven]

Lol diagrams aren't a get out of sloppy jail card

[delicate orchid]

yeah I know, I thought I was being rigorous

it's fine it's just like, I didn't explicitly explain how something was defined - which I'll make sure to do in the future

the actual proofs were correct

mostly

I might run a few of my proofs for the next exercise sheet through here if no one minds (once I TeX them)

[hidden haven]

So the reason here is that if you look at the theorem statement it goes like
For every f: G → H there is a unique injective map g: G/ker f → H such that gp = f, where p is the quotient map. How do you understand this statement, in particular the equation? You draw this diagram in your head lmao unless you're a witch

This is like the most basic reason

[delicate orchid]

no moldi NO

not THAT ONE

[delicate orchid]

THIS one

[hidden haven]

Oh bruh

Oh bro

[delicate orchid]

monad moment I think

[hidden haven]

That one's even better

Absolutely monad moment

My phone doesn't have enough battery for what's about to happen

[delicate orchid]

moldi you're infecting me stop it

[hidden haven]

(I'm kidding I don't plan on typing that much)

[delicate orchid]

"my proof is marvellous but my phone battery is too short to write it"

[hidden haven]

But am I infecting you in a good way though

[delicate orchid]

are you sure about that

[hidden haven]

As they always say

Moldilocks knows best 😌

Just close your eyes and give me the steering wheel 😌

Preferably in the opposite order

[delicate orchid]

no promises

[chilly ocean]

the order of a cycle decomposition is the # of elements in it rite

or is't

smallest common multiple of permutations

[lavish nexus]

yeah

lcm

[paper flint]

If you write your cycle as a product of disjoint cycles, then yes, its order is the lowest common multiple of all the cycle lengths.

[desert dome]

Hi, I don't quite understand the (2) -> (1) direction, as I don't know why sigma permutes the roots? I know sigma has to map roots to roots, but why is it injective? Thank you
(F^alpha is just the algebraic closure)

[terse crystal]

Homomorphisms from field are injective…

Fields don’t have non trivial ideals

[desert dome]

right.... I forgot again. Thank you!

[coral shale]

For a group, G, all subgroups of the center Z(G), are normal. But Z(G) isn't necessarily the largest normal subgroup... or is it?

Also, does G/[G, G] have anything to do with Z(G)

[G, G] being the commutator subgroup

[chilly ocean]

Z(G) is not necessarily the largest

And well, G/[G,G] is a commutative group, it is isomorphic to Z(G) when G is abelian, not sure what else youre looking for

[hidden haven]

When G is abelian

[terse crystal]

Center of S_n is {1} when n>=3…

[delicate orchid]

it's isomorphic to Z(G) because Z(G) = G for G abelian

top ten wacky results

[hidden haven]

catfacts1337

[delicate orchid]

anyway yeah uhh

something something all normal subgroups are unions of conjugacy classes so just pick a conjugacy class consisting of elements not in the centre and you'll get a bigger one something something

[coral shale]

Thanks. No idk if they were related (just was wondering)

[delicate orchid]

just realised this is why all normal subgroups are intersections of the kernels of some characters

time to add another lemma to my dissertation!

[coral shale]

Perhaps if [H, K] is normal, you can say something meaningful about Z(G/[H,K]) ? 🤔

[chilly ocean]

what's [H,K]?

like commutator but elements from different subgroups?

[coral shale]

yes

$[H, K] := {h^{-1}k^{-1}hk:h\in H,k\in K}$

[cloud walrus]

Shuri2060

[coral shale]

from my old notes, for H, K subgroup G

[chilly ocean]

Schuri's Lemma

[coral shale]

uhh how do I put it --- the intuition I had from the course like a few yrs back was as [H, K] gets 'bigger' Z(G/[H, K]) also gets bigger (changing H or K while keeping [H, K] normal)

[chilly ocean]

yeah intuitively it makes sense

[coral shale]

Well it's completely unrelated to what I'm doing rn, but something someone said a few days ago brought this back to me

[woven obsidian]

In a commutative ring, if $P$ is a prime ideal, and $x\in P\setminus P^2$, then how do I show $x^n \notin P^{n+1}$?

[cloud walrus]

AoiKunie

[woven obsidian]

If this is not true in a general ring, how can one show it in a Dedekind domain?

[next obsidian]

You want to localize at P, and then A_P is a DVR. Now it follows because the valuation of x is 1, and then the valuation of x^n will be n so that it isn’t in P^n+1A_P

It then follows that x^n is not in P^n+1

For a general ring I’m not sure it’s true, for example take k[x]/(x^2) and take P = (x) and the element x

Then x is in P\P^2, but x^n is in (x)^n+1 for all n > 1 because x^n = 0

[woven obsidian]

Oh I see

I guess one can also look at $A/P$ as it is also local

[cloud walrus]

AoiKunie

[woven obsidian]

Ah, but the $A_p$ thing works in all integral domains?

[cloud walrus]

AoiKunie

[next obsidian]

No

This specifically uses that A is a Dedekind domain

You need the localization to be a DVR

I can’t think of a counterexample when A is an integral domain off of the top of my head, but my proof at least required A_P to be a DVR

[woven obsidian]

I see, I will have to work out the details

[next obsidian]

Okay here’s a counterexample with an integral domain

Take k[x,y]/(x^2-y^3), this is isomorphic to k[t^2,t^3]< k[t] so is an integral domain

Let P be the image of (x,y), then x is in P\P^2

However note that x^2 = y^3, and so x^2 is in P^3

[woven obsidian]

Ah thanks

[next obsidian]

[tender mist]

Why does this follow?

Sorry for the question, this topic is interesting

[next obsidian]

Because the valuation of x^n is n

In a DVR with maximal ideal m, you have a valuation v(x) which is the unique n so that x is in m^n\m^n+1

It’s basically just the largest power of the maximal ideal you’re in

Part of the axioms of a valuation is that v(xy) = v(x) + v(y)

From knowing that x was in P\P^2 you know that v(x) = 1 from which it follows that v(x^n) = n

Thus x^n is in P^n\P^n+1

Buried in here is some of the theory, I make no attempt to justify why this function v exists

Or why v(xy) = v(x) + v(y) because it’s kind of annoying

But it is true

Anyway, in our specific case the DVR is A_P and the maximal ideal m is the ideal PA_P

[woven obsidian]

Ok I think I found a theorem that gives me the valuation

If I can show $$A_p$$ is Noetherian, Normal, domain with only two prime ideals

[cloud walrus]

AoiKunie

[woven obsidian]

Only the normality is not immediately clear I guess

[next obsidian]

It follows almost definitionally that A_P is a DVR when A is a dedekind domain

In fact one definition is a dimension 1 Noetherian integral domain all of whose localizations are DVRs or a field

What’s your definition of a Dedekind domain?

[woven obsidian]

Noetherian domain with all primes maximal

[next obsidian]

This isn’t enough

[woven obsidian]

Oh wait

Normal also

And I seem to recall localising something normal gives normal

[next obsidian]

Then A_P is normal because the localization

Is also normal

Indeed

[woven obsidian]

Been a while since I did commutative algebra

[next obsidian]

Also this is non-zero :)

Non-zero primes, I mean

[woven obsidian]

Taking a course in algebraic number theory now

Ah yeah

[next obsidian]

Yup

[woven obsidian]

Otherwise you got a very nice ring 🙂

[next obsidian]

Yeah lol

I mean your problem becomes vacuous in that situation, lol

[woven obsidian]

yeah

[tender mist]

Thank you Chmonkey, it was very useful

[wraith obsidian]

completely unrelated – I know of Nielsen equivalence of generators of a group. Is there some notion of „nielsen distance“ generalizing that, e.g. S₁ nielsen equiv. S₂ iff d(S₁, S₂)=0? like in, how many „detours“ through non-generating sets do you have to take?

Ah nevermind it kinda leaves the number of generators constant, which is not what I'm looking for

[lavish nexus]

why is the highlighted part true

[prisma shuttle]

$\mathbb{N}_0$ is the nonnegative naturals

[cloud walrus]

JustKeepRunning

[chilly ocean]

but thats not a ring so how can we talk about ideals?

[prisma shuttle]

no its a numerical monoid

idk maybe there's a different definitin for ACCP on monoids?

[chilly ocean]

I dont know how you define ideals on monoids

[prisma shuttle]

https://artofproblemsolving.com/community/c2614813h2779312p24398022

idk in that link that's what they are saying

[chilly ocean]

Can someone check if my proof holds

Lmk if you need the other parts

[lavish nexus]

your definition isn't well defined

[lavish nexus]

(there's no way to have it well defined but I don't get your argument that it is not surjective)

what subgroups of order 4 are in Q8?

Q8 has 3 of them and they're all isomorphic to your group A

but A has an element of order 4

does Q/H have any element of order 4?

[chilly ocean]

hmm

yes Q/H does

|iH| = |jH| = |kH| = 4

@lavish nexus

[lavish nexus]

iH*iH=?

[chilly ocean]

-1H

[lavish nexus]

is -1 in H?

[chilly ocean]

no

which is why the order of iH is 4

[lavish nexus]

read c) again

[chilly ocean]

yeah it normal

[lavish nexus]

H=?

[chilly ocean]

1 , -1

[lavish nexus]

ok what is iH*iH

[chilly ocean]

i^2H

[lavish nexus]

which is?

[chilly ocean]

-1h

-1h

[lavish nexus]

but -1 is in H

[chilly ocean]

-1H

so

wym

[lavish nexus]

-1H=H

[chilly ocean]

OHHHHH

so order of iH is 2?

[lavish nexus]

yeah

[chilly ocean]

rest would be 2 as well right?

[lavish nexus]

yes

G/H is isomorphic to V4

[chilly ocean]

i get it

so i had part d wrong

i put all of them as 4

its 2

oka

so how do i do 2e?

im gonna eat lunchbrb

[lavish nexus]

<i>=<j>=<k>=Z4
can't be isomorphic to V4

[chilly ocean]

be right back

[lethal cipher]

Okay, so I am having a little trouble with this. Although I think I can figure it out on my own no problem, I am not too sure how to represent a polynomial in F[a_1,...,a_n].

[chilly ocean]

How is this statement true

Whats V4

[lavish nexus]

klein 4 group

doesn't matter

[next obsidian]

Isn’t this definitional?

Like F[stuff] should be the smallest subring containing F and containing stuff

Like definitionally

[lavish nexus]

|G/H|=4

[next obsidian]

And likewise for F(stuff) but as a field

[lavish nexus]

and all groups of order 4 in Q8 are <i> <j> <k>

[next obsidian]

But anyway, if you’re taking it as polynomials in those elements

You can take it as the image of the following map, F[x1,…,xn] -> L defined by sending the polynomial f(x1,…,xn) to f(a1,…,an)

[lavish nexus]

but there are order 4 elements in <i> <j> <k>
and none in G/H
so they cannot be isomorphic

[next obsidian]

And likewise for F(a1,…,an) except using rational functions instead of polynomials

[lethal cipher]

So how do we express a polynomial in f(x1,...,xn)?

[lavish nexus]

like a normal polynomial

f(x1,...xn) is the field of rational functions in x_1....x_n

polynomials would still be polynomials

[lethal cipher]

I think there is a misunderstanding of what I am asking. How do I write an expression for a polynomial with n variables? I know examples of them, but I am not so sure of a general form for them

[lavish nexus]

honestly p(x_1,...x_n) is enough for the purpose of this question

[lethal cipher]

Really? Why is that Iteribus?

It seems like I need the more general form

[chilly ocean]

@lavish nexus so we are comparing the order of elements in Q vs the order of elements in Q/H ?

[lavish nexus]

I mean you're not trying to prove F[x_1,..,x_n] is a ring

but if you are you can always consider it as F[x_1,...x_n-1][x_n]

and write it like a single variable polynomial with coefficients in F[x_1,...x_n-1]

[lavish nexus]

yes this is a common strat to prove two groups are not isomorphic

[chilly ocean]

what do you mean when you said there are 4 elements in <i> <j> <k> but none in G /H

[lethal cipher]

Oh I misunderstood what you were saying Todd.

[lavish nexus]

I said there exist elements of order 4 in them

like i, j, k, -i, -j, -k

but in G/H all nonidentity elements have order 2

[lethal cipher]

Nice. Latex is not working rn

[lavish nexus]

$

[cloud walrus]

dackid

[lethal cipher]

There we go.

Where m is just the total number of terms

Maybe more variables then needed, I don't really know

[barren sierra]

I have absolutely no idea how to do 3 can anyone help?

[lavish nexus]

take a \phi : F[x_1,...x_n] to L by sending x_i to \alpha_i
this is obviously (handwaving is enough) a homo and the image is F[\alpha_i,...\alpha_n] contained in L
then because F[x_1,...x_n] is a ring
the image has to be a subring

[prisma shuttle]

can someone explain why the red part is true

why does schur's lemma imply constant

[chilly ocean]

how do i use the correspondence theorem to find subgroups of G/H ?

im assuming i use the correspondence theorem

[lavish nexus]

find the subgroups of G that contain H

say A

then A/H is a subgroup of G/H

[chilly ocean]

so if Q/H is 4 then the possible orders for subgroups would be 1,2,4?

[lavish nexus]

yes

[coral shale]

wut is 4 🤔

[delicate orchid]

the representations are irreducible right? So schur's lemma lets us know that any homomorphism between them is either an isomorphism or the 0 map
I'm guessing n isn't equal to m?

[chilly ocean]

so im basically looking for subgoups of G that are order 1,2,4 that contain H

[lavish nexus]

no

any subgroups of G that contain H

[chilly ocean]

and by contain H you mean the whole group right

[lavish nexus]

only after modding by H do you get order 1 2 4

yes

[chilly ocean]

what would the operation be

mult?

[lavish nexus]

yeah

[chilly ocean]

okay so I have Q8

and H = {1,-1}

so obviously the trivial subgroup works

but say i do

{1,-1,k,-k}

that wouldnt worjk

[lavish nexus]

that works

[chilly ocean]

how

[lavish nexus]

it's <k>

[chilly ocean]

k *-k

[tribal moss]

Formally you'd say {{1,-1},{k,-k}} -- a set of two cosets of H.

[lavish nexus]

that's <k>/H

[chilly ocean]

H only has {1,-1}

[tribal moss]

Sorry, I seem to have missed the direction of the original question.

[chilly ocean]

i still dont see how the bottom 3 work

wait

i already have Q/H

[barren sierra]

I am dum dum is the operation on $\phi$ here composition?

[cloud walrus]

Spamakin🎷

[tribal moss]

So Q/H is just V4 and we know what its subgroups are.

[lavish nexus]

<i>/H = {H, iH}

[chilly ocean]

so

[lavish nexus]

<j>/H = ?

[chilly ocean]

Q/H = {H , iH, jH, kH}

[tribal moss]

Yes, since it is introduced as an element of the automorphism group.

[chilly ocean]

im so lost man

[lavish nexus]

the subgroups of G are

G, <i> <j> <k> H {1}

the first five contain H

so by correspondence the subgroups of G/H are

[barren sierra]

ok figured but idk why they wrote it without the composition symbol lol

[lavish nexus]

G/H, <i>/H, <j>/H, <k>/H, H/H

[chilly ocean]

how is <i> a subgroup of G

[tribal moss]

That's just a matter of writing the group operation (in this case of the automorphism group) multiplicatively. Happens all the time in higher algebra.

[lavish nexus]

it's the cyclic group generated by i

it must be a subgroup of G

i, i^2=-1, i^3=-i, i^4=1
repeat

[chilly ocean]

oh so it creates the whole group

since i^2 = j^2 etc

[tribal moss]

By definition <i> means the smallest subgroup of G that contains i.

[chilly ocean]

okay

[chilly ocean]

how did u calculate this

left cosets?

[lavish nexus]

yes

[chilly ocean]

how did u get the first H

[lavish nexus]

<i> contains H

[chilly ocean]

{H,iH}

ohhh

how would H/H look like

[lavish nexus]

{H}

[chilly ocean]

H

[chilly ocean]

hhhhhhh

[kind temple]

hurb

[delicate orchid]

c squared you've done it again 🥳

[kind temple]

lets goooo

when did i do it the first time lol

[delicate orchid]

I don't know

[pastel cliff]

https://tenor.com/view/hng-gif-4501621

[delicate orchid]

what's wrong?!?!

nitezba you wouldn't to have a group theory question for me would you

[pastel cliff]

HAH

you wish

[delicate orchid]

fuck guess I gotta stop procrastinating

[pastel cliff]

reading a paper on geodesics rn

i love big looking words that arent big

dw tho i'll be here to get degraded over the weekend

https://tenor.com/view/dies-cringe-math-dies-from-math-death-gif-22822700

[kind temple]

me too

[proud bear]

are you british wew?

[kind temple]

think so. hes like six hours behind me lmao

[proud bear]

i see

[kind temple]

this was actually hilbert's 25th question

[delicate orchid]

this is why I was procrastinating

time to do the even length case

[barren sierra]

Extensions are still a mystery to me

so for (3) I've shown G is a group, that's fine

[pastel cliff]

the fuck

[barren sierra]

so if it's an extensions of K by H then that means H is normal in G and K = G / H
or
H injects into G, G is surjective into K, and the image of the injection is the kernel of the surjection

idk how to do either of these

with the given info

[upbeat swift]

I had a quick question about a problem. I'm not sure how to go about this problem

And this is F, just for more info

[lavish nexus]

umm

2 and 5 are primes dividing 10

so you have to have an element of order 2 and an element of order 5

[upbeat swift]

I understand that, but I'm not sure if it's asking for some sort of explanation

[prisma shuttle]

why is it true that "homomorphism is either an isomorphism or the zero map." The statement in the book is that its multiplication by a scalar

[delicate orchid]

lemme check my notes

[delicate orchid]

Indeed you're right, my apologies

wait wait wait

do we know that these are isomorphisms or just homomorphisms

cause my notes say that if it's a homomorphism of representations it is indeed either 0 or an isomorphism

but if it's an isomorphism then your statement is correct

[prisma shuttle]

its a homomorphism of representations

[delicate orchid]

ah ok, then the map is either 0 or a scalar multiple of the identity (the latter iff it's an isomorphism)

lemme take a look at the original question again

the commutitive diagram is confusing me (what a surprise 🙄) - what does it mean for a map to be constant? That the composition always results in the same overall map regardless of the choice of T?

[prisma shuttle]

this is my quesiton lol

idk wut it means

[delicate orchid]

damn

[prisma shuttle]

yea its confusing 😦

[open pilot]

Are these field extensions?

[barren sierra]

group extensions

yea

[delicate orchid]

even length case done

no way did that take an HOUR

[barren sierra]

damn u be crossing shit out

I omit that and leave that as an exercise for the grader

"lol figure out how I got from this line to the next it works"

[delicate orchid]

the poor bastard has to sit though my proofs that aren't just straight computation, I feel like I'll go at least a little easy on them where I can

[barren sierra]

fair enough

speaking of not just straight computation

uh

the question I sent above

[delicate orchid]

like I'm throwing around all sorts of fraudulent definitions

[barren sierra]

oof

[delicate orchid]

which one

[barren sierra]

number (3)

[barren sierra]

this

#2 defines what a characteristic map is

but then like I've shown G is a group

idk how to show it's an extension

[delicate orchid]

I'll have a quick look but it is nearly 2am

[barren sierra]

if u can't it's ok

[delicate orchid]

Outer automorphisms
🚪 🚶‍♂️

[barren sierra]

appreciate it tho

yea

I have like 4 problems in this topic in this HW

I did the other 8 problems tho so we good-ish

[delicate orchid]

knowing the definition of a group extension might help me here, one mo

[barren sierra]

I sent

[barren sierra]

group extension

so I have to show that H injects into G