groups-rings-fields

[void cosmos]

i can see that but i dont get where do we use that char F is 0

as a contradiction hypothesis

[formal ermine]

wait I'm having a brainfart, why is this x - 1 and not x + 1

[delicate orchid]

it's just that you automatically know what (b^k)^-1a^nb^k is from b^-1ab = a^2 and (obviously) a^-ka^na^k = a^n and b^-kb^nb^k = b^n

ok maybe that's one trick conjugating powers of a generator by powers of the same generator doesn't change anything

I assume you already knew that though

[south patrol]

ok swap a for a+1 idk

you get the vibes

lol

[formal ermine]

lmao

[south patrol]

okay so if p(x) = p(x+1) like

expand that shit out

and you'll see that we get n=0 or smth

lol

like

[delicate orchid]

proof by Leviticus 19:11-12

[south patrol]

$0 = \sum_k a_k ((x+1)^k - x^k) = \sum_k a_k (k x^{k-1} + \text{other stuff})$

[cloud walrus]

potato

[south patrol]

and then pick the largest k with k < degree of p and a_k not 0 lol

[formal ermine]

it follows immediately from the fact that ( \abs{\bigoplus_{k \in S) (\mathfrak{R}^{\bF^\alpha(i)})_{i \in \mathcal{U}k}} \leq \aleph_1 ) when ( [\mathfrak{h}\mathcal{W} \cap \bF^\alpha(\bN) \neq \emptyset )

[cloud walrus]

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

[south patrol]

wait so this question i'm doing asks for me to find the simple modules over uhhhh

[formal ermine]

was tryna copy this from memory lmao

[south patrol]

$\mathbb C C_3 \times \mathbb C C_4$, but like how do you even think about that lol

[cloud walrus]

potato

[south patrol]

Like a simple module over that will be 1 dim but idk how to even think about maps CC3 x CC4 -> End(C)

[void cosmos]

yo im sorry what is this contradicting

[formal ermine]

minimality of minimal polynomial

[south patrol]

you'll see we get like

k=0 at some point

for some non-zero k

[void cosmos]

yea that must happepn

ig

[south patrol]

Like okay for simplicity let's say p(x) = x^{n+1} + a_n x^n + ... where a_n is non-zero

Then taking coefficients of x^n in what I wrote, we get 0 = a_n n

so n = 0

[delicate orchid]

am I smoking pack or is that not iso to C[C_3 x C_4] = C[C_12]

[south patrol]

Hm is it?

That's be the tensor product

[formal ermine]

what does that notation mean, like CC_3

[south patrol]

Lol I mean it was meant to match the latex i wrote

[delicate orchid]

the complex group algebra over C_3

hmmmmmmmm remains to be seen

[south patrol]

But no they can't

wrong dimensions brother

[formal ermine]

oh like a group ring innit

[south patrol]

ye

oh i should say ye like

my uni writes kG rather than k[G] for whatever reason lol

but ye

[delicate orchid]

if you can do the latter then yes that would be the quickest way

but finding out what happens when you conjugate the generators first is usually quicker

as like, b^-1ab = a^2 => ab = ba^2, which lets you swap stuff around in an awful word like (a^ib^j)^-1a^xb^y(a^ib^j)

[south patrol]

ye

[delicate orchid]

which can (will) lead to cancellation

[south patrol]

i like how uh like like

representation theory is all cute and tbales and shit

group theory is like WORDS. take them or leave them

[delicate orchid]

I'm not sure what you mean

you're forming a big chain of equalities starting from (a^ib^j)^-1a^xb^y(a^ib^j) by using the relations in that presentation - which will end up in b^?a^? eventually as the group is generated by those two dudes

[south patrol]

Anyway wew do my question instead of Joe's

[delicate orchid]

and then from there you will (hopefully) be able to see the conjugacy classes via some condition on the exponents

[south patrol]

Joking

[delicate orchid]

no because yours requires thought

[south patrol]

That like somehow roasts and compliments me

idk

I am have tired

[delicate orchid]

oh this doesn't matter, cause if you have b^?a^? you can just swap them around and get a^(2?)b^?

[delicate orchid]

cause of this

[formal ermine]

yes potato

ask simpler questions!

[south patrol]

I was doing a past paper and this was a weird question on it

😭

Oh here rep theory seems one of the most popular lol#

[delicate orchid]

yeah I've never seen a direct product of group algebras because why would you do such a thing

[south patrol]

Me neither!

the old notes are slightly idiosyncratic in that they just talk about representations of modules A as being maps like A -> End(V) for osme vector space V

[delicate orchid]

End

[south patrol]

so they say "representations of C[S_3]" and stuff

[broken stirrup]

representation theory gets better when they start talking about characters

[south patrol]

Yes characters are hot

That's what I meant is the fun bit lol

[delicate orchid]

and then much worse when they start talking about modular characters

[south patrol]

No!

[delicate orchid]

presentation theory is nothing if not long

[south patrol]

Representation theory gets so much better when representation ring comes up 🤓

[delicate orchid]

any word related problem is probably undecidable in general it's an icky field

[delicate orchid]

that's the virtual character ring u numpty

[south patrol]

Me when you can use spicy words like λ-ring lol

[south patrol]

Hm what

[delicate orchid]

the representation semiring is iso to the semiring of characters, that's like the entire point

[south patrol]

Hm don't you need to tensor w C first

Wait

Okay semiring of chracters

[delicate orchid]

semiring
nope!

[south patrol]

Not seen that lol

[delicate orchid]

[south patrol]

But fair

Yeah

[delicate orchid]

you tensor with C if you want them iso as rings yes

[south patrol]

Just like characters of actual representations

Rather than allowing for like arbitrary C-linear stuff, sure

[delicate orchid]

Z-linear only for me thanks

[south patrol]

ye fair

[delicate orchid]

ANYWAY!!! do you know any results about simple modules over a cartesian product of rings lol

[south patrol]

The stuff I have on rep theory in my essay is a bit of a pain

because everythign is unitary

which feels idiosyncratic

like unitary representatio nring

maybe it isn't too idiosyncratic though

[south patrol]

No

[delicate orchid]

wait what rep ring isn't unitary lmfao

[south patrol]

bleak

No like

[delicate orchid]

le trivial rep has arrived

[south patrol]

what lol

Unitary as in

[delicate orchid]

oh U(n)

[south patrol]

maps G -> U(H) for a hilbert space H

🤓

Yeah

[delicate orchid]

no that's not idiosyncratic every rep is similar to a unitary rep

[south patrol]

wait how do you prove that

[delicate orchid]

over finite groups anyawy

[south patrol]

finite groups

[delicate orchid]

all sets are finite

[south patrol]

Okay yeah cause you can just construct a G-equivariant inner product and shit ig

[delicate orchid]

yes exactly

[south patrol]

But no these are like infinite groups lol

compact at least

But yeah

[delicate orchid]

.... who?

[south patrol]

Compact = finite anyway

Oh yeah I asked some people about reps of compact groups and they're like

[delicate orchid]

not true as u need to write integrals instead of sums

[south patrol]

just work with finite stuff and write compact instead

jk

Uh

Well yeah but I write my sums as integrals w the counting measure anyway

[delicate orchid]

ok so what if we arty wedderburn decomp the two dudes

[cloud walrus]

Wew

[south patrol]

Wait really?

[delicate orchid]

all characters are 1 dim and Mat_1(C) = C is it not?

*reps

can u tell I'm a character theorist

[south patrol]

LOL

the 'really' wa sa joke about 3 + 4 = 7

anyway

uhhh

sure

so

[delicate orchid]

I would get this wrong

so are we not just finding simple modules over C^7 now?

[south patrol]

okay so

this is true but the thing is like we're meant to talk about them in terms of generators and stuff

[delicate orchid]

I DON'T CARE!!!!!!!!!!!!!!!!!!!

[south patrol]

of C[C_3] x C[C_4]

and i have no idea how you think about that rly

like

[delicate orchid]

oh ok

[south patrol]

obviously you wanna say like

they're determined by like (g,1) and (1,h) with g,h generators

but then because we aren't working w the tensor product it's not gonna work out that nicely

i guess

so uh

[delicate orchid]

not quite...?

unless I'm misinterpriting what you're saying

[south patrol]

or (g,0) and (0,h)

etc

but no i mean i'm saying that's wrong lol

But if we had a tensor product it'd be way nicer

[delicate orchid]

right ok

[south patrol]

So uh ye i'm confusion

Maybe this is just smth from the old lecture notes they used to cover lol

[delicate orchid]

C_3 = <g>, C_4 = <h>
C[C_3] has a basis {e, g, g^2}
C[C_4] has a basis {e, h, h^2, h^3}
as these are abelian groups their direct product is their direct sum (I'm doing this because it's easier for me to think about them like this)
is the basis of C[C_3] x C[C_4] not just the union of the two basis sets lol?

[delicate orchid]

before they realised it was just TOO silly

[south patrol]

lol

I'm gonna be really silly

[delicate orchid]

we could also try factoring maps from C[C_3] x C[C_4] -> End(V) through the tensor product using the universal property

[south patrol]

this is true i mean also just forget the ring structure + then clear by taking bases of each lel

yah

[south patrol]

hm but that'd be bilinear maps rather than linear right

but ye

[south patrol]

i hate it here

the funny thing is the rest of this question seems like way easier

[delicate orchid]

however potato if u just do like

[south patrol]

like they asked us to do C[C3 x C4] and C[S_3]

both of which are like super standard

and this came before both of those lol

BLEAK

[delicate orchid]

nah nvm

[delicate orchid]

maybe they do mean the tensor product lol

[south patrol]

lel

[delicate orchid]

their direct product is their direct sum
literally add the characters...?

irreducible characters are of the form chi_i + phi_j, chi_i \in Irr(C_3) phi_j \in Irr(C_4)

might be chatting shite

[south patrol]

STOP DOING CHARACTERS

TRACES WERE NEVER MEANT TO BE COMPUTED

[delicate orchid]

I'm having to translate literally everything we're talking about into character theory and then back into module fuckery

[south patrol]

this problem was so screwed up honestly like

there wasn't a single question in this paper on characters

introduction to rep theory

[delicate orchid]

I am chatting shite, if we do the arty wedderburn decomp then these corrispond to 2 dim subspaces of C^7 which are obviously not simple

what a bunch of BALONEY

[hot lake]

what's happening in here

[delicate orchid]

simple modules over $\bC[C_3} \times \bC[C_4]$

[cloud walrus]

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

[south patrol]

Latex error in chat guys

[delicate orchid]

good no I'm glad I missed that bracket yup mm hmm

[hot lake]

ah ring morphisms from R1 x R2 into End(V) ?

[delicate orchid]

yus

[south patrol]

I guess more generally that'd be good to know about lol

[formal ermine]

what does this have to do with tensor products

[delicate orchid]

there will be some easy theorem I simply don't know

[south patrol]

i guess right

well no ignore me

Hm

[hot lake]

I think they just correspond to pairs of commuting representations

[south patrol]

Makes sense

[delicate orchid]

these aren't that bad

[south patrol]

Hm

[hot lake]

(a,b) = (a,1) * (1,b) = (1,b) * (a,1)

[south patrol]

I mean illum that chapter isn't all about tensor products right

[formal ermine]

exact sequences + tensor products

[south patrol]

Anyway this stuff is actually like lol more key than I realised

[delicate orchid]

yeah I was thinking about taking maps (f, id) but I didn't think about it hard enough apparently

I think I missed the fact they commuted

[hot lake]

but the actual ring itself looks kinda weird uuh

[south patrol]

Like the p[x] stuff arises naturally when you study dimension theory

Which is coolio

[hot lake]

uh

[south patrol]

Yeah I might look at uh examiner's report to see if they say anything about this

[hot lake]

uuuh

[delicate orchid]

inb4 they actually meant the tensor product

[south patrol]

Seems randomly tricky compared to like, the rest of the paper

Lol

Or at least unfamiliar

[south patrol]

Yeah this seems most likely lol

[void cosmos]

Let p be a prime number. Let Fp denote the finite field with p elements. Let f(X) denote the polynomial X^p^4-X
Prove that all irreducible factors of f are distinct. How many irreducible factors of each degree does
f have?

[hot lake]

actually I'm not sure of anything

[delicate orchid]

it's semisimple at least

[void cosmos]

yo is this just the finite field of p^4 elements cuz the factors would be the roots?

[hot lake]

acting by (x,0) and (0,y) give representations of the C[C3] and C[C4] right ?

[south patrol]

wdym "Is this"

[void cosmos]

the irreducible factors

[delicate orchid]

yes

[south patrol]

Its splitting field is that sure

[void cosmos]

yea thats what i meant

[south patrol]

So yeah zef thing is like uh

[void cosmos]

How many irreducible factors of each degree does f have? this can be done by just that theorem from cyclic groups?

[hot lake]

but they annihiliate each other cuz (x,0).(0,y).v = (0,0).v = 0 ?

[south patrol]

Hm so like it's clear we will have reps but then uh

[formal ermine]

isn't this just like (A/a)[x] = A[x]/a[x]

[south patrol]

Lol

[south patrol]

Yup

It's very nice

[formal ermine]

ic

[south patrol]

And so in fact you can say smth stronger for the bit about m[x]

[hot lake]

(1,0) and (0,1) are projections

[formal ermine]

huh?

[south patrol]

Like

Is m[x] maximal?

[formal ermine]

no

[south patrol]

Yeah I meant the interesting thing is uh

How it's never maximal

Rather than only sometimes or smth lol

[formal ermine]

oh that's what you meant lmao

that seemed obvious to me

[south patrol]

Yeah I mean it is

[delicate orchid]

it is

[south patrol]

But cool

I guess this also shows uhhh

[formal ermine]

isn't a polynomial ring over a field local or smt

[south patrol]

so if you have a chain of n prime ideals in A then you can get a chain of n+1 prime ideals in A[x]

Which kinda anticipates stuff from uh the final chapters

[hot lake]

so an A x B - module rather is a direct sum of an A-module and a B-module

[delicate orchid]

potato what did the examiner's report say

[south patrol]

[hot lake]

yeah

[delicate orchid]

oh was I right then lol

[hot lake]

if M is an AxB - module, (1,0).M is basically an A-module, (0,1).M a B-module, and M is their direct sum

[south patrol]

hm sure, so how can we really describe A x B modules in terms of generators of R, S hm

Like in this case

Feels a lil weird

[void cosmos]

anyone boys and girls

algebra monsters

[hot lake]

generators ?

like when you say a C[C3]-module is a vector space with an endomorphism f with f³ = 1 ?

well here you need to include projections

p1+p2 = 1

p1p2 = p2p1 = 0

f1³ = 1, f1 p1 = p1 f1

f2⁴ = 1, f2 p2 = p2 f2

uh

[south patrol]

Oh tbh this is weird like they said to describe in terms of "algebra generators"

[hot lake]

maybe f1 p2 = 0

[delicate orchid]

do they want a full on list of relators like zef's given or can you just say "it's generated by blah blah from the duality blah blah morphism blah"

[hot lake]

lol I'm getting confused again

[delicate orchid]

this problem is very weird dw about it

[hot lake]

f1 f2 = 0 I think yeah

anything1 anything2 = 0

[delicate orchid]

yur, (f, 0) \circ (0, g) = 0

[delicate orchid]

wait a minute why do I recognise this

are these (p_1, p_2) block idempotents

[hot lake]

p1 = the action of (1,0)

and generators x of A get pulled to (x,0) of AxB

generators y of B to (0,y)

so (x,0) * (0,y) = 0

so you add 1 to your generators of each factor ring, then put everyone together and say that product of generators from different factors is always 0, and also that 1 = sum of the 1 from each factor ring

and then copy paste your relations

[hot lake]

isn't that like counting how many conjugates people in Fp⁴ can have

[void cosmos]

idk tbh

i just thought about orders of the elements inside the cyclic gruop

[hot lake]

the roots of X^p^4 - X are all the elements in Fp⁴

[void cosmos]

yea

[hot lake]

each with multiplicity 1

and how do you group the roots to make irreducible factors ?

[void cosmos]

yea and they are distinct cuz the multiplicative group is cyclic

idk lmao yea

i just thought that

or nvm

any hints for that?

[south patrol]

"algebra generators" lol

etc

Kinda sus

[delicate orchid]

I would read that as needing the full list of relators

[formal ermine]

what are examples of absolutely flat rings

[hot lake]

I would give explicit actions of enough elements and then write "those generate A believe me bro"

[formal ermine]

ok so apparently I get a bunch of results when searching for von neumann regular ring lmao

[delicate orchid]

Q

hope this helps 👍

[formal ermine]

how do we show that local rings are domains

[delicate orchid]

are they?

[lethal dune]

they arent?

[formal ermine]

yes

[delicate orchid]

Z/4Z lmfao

[formal ermine]

wait fuck

I misread this one statement

[lethal dune]

it says regular local rings

[formal ermine]

yes I misread something else

[delicate orchid]

ah ok

[formal ermine]

lmao

an absolutely flat local ring is a domain

is that right

wait that's stupid lmao

an absolutely flat local ring is a field

🤦

[delicate orchid]

are fields domains, chat?

[formal ermine]

domain is absolutely flat <=> field
local ring is absolutely flat => field

[lethal dune]

no 0×0=0

[formal ermine]

I thought at first like

it wasn't absolutely flat or something

the local ring

[formal ermine]

ringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzeroringsdontcontainzero

[south patrol]

My favourite is finite local domins

[formal ermine]

since when did finite rings exist

I thought finite rings = {}

[delicate orchid]

all rings are finite actually

[void cosmos]

yoo

algebra qualis are so much fucking easier than the analysis ones

wtf

[south patrol]

Lol nic

e

Are these like grad quals

[void cosmos]

yea

i just write phd algebra quals for university [x]

the hardest ones are the real analysis ones

yo

Let R be a commutative ring with identity and I ⊂ R an ideal such that R/I ≃ Z. Show that
there are infinitely many maximal ideals in R that contain I.

[south patrol]

yo

[void cosmos]

this is literally just iso theorem

right?

[south patrol]

Correspondence theorem

[void cosmos]

the correspondonce one

yea

[south patrol]

Yes

[void cosmos]

its like the third

or second

dk

but yea this is was on a quali exam

[south patrol]

I'd not say it's an isomorphism theorem really

[void cosmos]

it is tho

its the second done

second one*

or third

[south patrol]

Hm those just give certain isomorphisms though

[void cosmos]

wouldnt A --> A/M work?

[south patrol]

the correspondence theorem gives more structure

Wdym

[void cosmos]

nvm nvm

anyways do i just cite it

[south patrol]

Yes

[void cosmos]

and then given the maximal ideal in R

u would find many in Z

[south patrol]

By the correspondence theorem, maximal ideals of R containing I correspond to maximal ideals of R/I

[void cosmos]

that contains?

yea

and there are infinitely many in Z

lmfao

[south patrol]

which correspond to maximal ideals of Z

yes

I thought this was gonna be smth insane

Well

I have done some interesting questions about infinitely maany primes in comm alg

[void cosmos]

the char(F) not equal 0

took me half an hour and still couldnt do it

sad

was also from a quali

[south patrol]

Do you understand that one now though?

[void cosmos]

same university quali

[south patrol]

Sure

[void cosmos]

yea i think

but yoo

the top 5 unis have like

IMPOOSSIBLE qualis

[south patrol]

Oop

[void cosmos]

like standford algebra quali was like

lmfaaaaaaao

i couldnt do a single shit problem

its like there are no free problems

no standard problems

no nothing

like for example this

Assume F is a field and F(α) is a finite field extension of F of odd degree. Prove that
F(α
2
) = F(α)

this is from dummit and foote

and its in the other uni quali

i know how to do it

nothing like this in the other uni qualis like no free bono

[formal ermine]

the first thing I see is an exercise from atiyah macdonald that I just did lmao

[void cosmos]

:ANGRY:

are you challenging me

they hard as shit tho for me tbh

[formal ermine]

bro the last exercise is algebraic geometry

[void cosmos]

the exams are like split into morning and afternoon

[formal ermine]

oh?

[void cosmos]

i couldnt solve shit in the afternoon

next year they add night mode

and u just get fucked

lmfao

[formal ermine]

oh I see

[formal ermine]

night mode?

[void cosmos]

yea just a joke

yea ig being good at courses corelates with research

the corelation is higher than i thought

[formal ermine]

I could probably only do like one or two of these lmao

the galois one is free tho

[void cosmos]

which exam

are u looking at

2022?

[formal ermine]

2022

yes

[void cosmos]

which one

spring or auttmn

https://drive.google.com/drive/folders/1V7Z3Bk4_WxhUKQb1XUMKcRTOZTdIA955

[formal ermine]

spring

[void cosmos]

oh yea you mean the first problem?

[formal ermine]

ye

[void cosmos]

yea ig its free

or like standard

by studying you should be bale to do it easy

able*(

[formal ermine]

it's like a standard exercise

[void cosmos]

also M3

M3 is also standard

[formal ermine]

yeah m3 was the am exercise I just did

[void cosmos]

oh wow

[formal ermine]

or at least (i)

[void cosmos]

yo A4 is easy too lmfao

in the afternoon

its p standard

even for like an undergrad ig

[formal ermine]

yeah

[void cosmos]

wtf is A5 tho

arent varities a geometry thing

[formal ermine]

yeah

[void cosmos]

oh lmfao yea

so uptill now

i can do like

3 problems

or 4

at moax

i still get that juicy F

only thing that WOULDmake this fair is passing is from 10 points

lmfao

[formal ermine]

lmao

I'm ingame right now will take a closer look at the problems in a sec

[void cosmos]

xd the analysis ones are literally impossible

[void cosmos]

gl

[formal ermine]

easy win

[formal ermine]

what the actual fuck

[void cosmos]

yea

fucking curse

part 2 is just ??

u sholud get ur PHD if u pass those

not the other way around

[formal ermine]

I can do like 4

maybe 5

[formal ermine]

bro

"Do all five problems. Write your solution for each problem in a separate blue book."

[void cosmos]

i think should be enough to pass

or it dpeends on which ones u can do cuz points are not all the same

[formal ermine]

4 looks doable

[summer path]

Don't you generally want to do well on quals and not just pass, in order to get a better advisor?

[void cosmos]

i have no idea

how these things work

but ik they are PASS/FAIL

[formal ermine]

does every uni require you to do one of these qual exams for a phd

[void cosmos]

yea

99%

and most of the time its a written

some are oral

[formal ermine]

oke

[void cosmos]

but there are easier ones

[formal ermine]

this will be a problem for future me then

[void cosmos]

not really

like

its not the type of things where like

u study hard but still do not pass

like i think if one studies , memorizes the proofs of important theorems

u can pass

[summer path]

Some require you to just pass a set of classes

[void cosmos]

oh yeah that or that

you can choose this ro that

[formal ermine]

what if I didn't do my bsc in the us

[void cosmos]

there are qualis in canadian unis

thats all ik

Let R be a unique factorization domain in which there exists a unique irreducible element up
to associates.
(a) Prove that R is a principal ideal domain.

is this saying that every element is expressed as r1 * u1r1 * u2r1 * ....

where u_is are units?

[summer path]

i was looking at the quals i'd have to take next year and the algebra ones are like the prereqs of whatever that stanford qual required LOL

[formal ermine]

lmaoo

can you send a screenshot perhaps?

[void cosmos]

send

[summer path]

also the algebra qual just looked super free

[void cosmos]

which one

yours or the standford one?

[summer path]

the one i have to take next year

not the stanford one obviously lmfao

[void cosmos]

ig thats great news for you haha

[void cosmos]

yeaa

u have to take only one?

or ig two?

[summer path]

i need to pass 2/4

[formal ermine]

@void cosmos btw I thought you're not a math major

[void cosmos]

i am not

what made you not not think i am not a maht major

haha

[formal ermine]

you told me once

[void cosmos]

i told you i was a math major?

or not

[void cosmos]

ig thats the standard like graduate level qualifying

[formal ermine]

you didn't not tell me you're not a math major

you told me you're not one

[void cosmos]

like thats what a normal person would expect haha

[void cosmos]

yea thats true im not

i am a cyber sec student

cs

[summer path]

that seems like standard undergrad algebra material though >.<

[void cosmos]

u think?

maybe ur only missing like

the module stuff

like this + modules would be definitely grad lvl for me atleast

[summer path]

oh

[formal ermine]

lmao this is basically free

[void cosmos]

yea at the end it depends on which uni

[summer path]

analysis quals meanwhile like

might just be that i haven't looked at analysis in a while though

[delicate orchid]

quals always seem biased towards analysis

[summer path]

:(

[void cosmos]

the problem isnt with the syllabus

its the damn problems in the exam itself are whats crazy for me

would 2 be done using orbit-stab?

[void cosmos]

.

or how

[delicate orchid]

2 is like modular rep theory lol

[formal ermine]

this looks so easy

how do you show that F(alpha, beta) : F(alpha) isn't 3 lmao

[delicate orchid]

I'm not looking at the galois ones lol

ok nvm even I could probs do 6

[elder wave]

these look quite doable

[delicate orchid]

2 looks the most fun

1 would be more fun if it was just straight up proving the pq-theorem

[void cosmos]

😠

is being able to solve qualifyings like this a good indicator that my self-studying bonafides are somewhat working?

or are those considered too easy? ( i dont think so )

i did other ones aswell

[delicate orchid]

I wouldn't call these easy but I wouldn't call them hard either

ok some of them I would DEFINITELY call easy

[void cosmos]

yea medium

i would call this medium

the one tabular post

[delicate orchid]

it's the easier side of medium imo

[void cosmos]

agreed

but its not that u cant know whats goign on

to do it

[delicate orchid]

also I did question 1 without realising I was breaking my streak of never using the sylow theorems

very sad!

[karmic moat]

hmm these are interesting

[void cosmos]

let F be a field and A,B be 3x3 non-singular matrices over F such that 2A=B^-1AB

find char(F)

any hint?

[lusty marlin]

Determinant maybe?

[void cosmos]

yea but how do i get to char

[delicate orchid]

I was thinking something to do with the eigenvectors yur

*values

fuck the vectors!!

[lusty marlin]

You'll get an expression in terms of det A

[void cosmos]

ok

[lusty marlin]

Which will give you a hint

Since det A is nonzero

[delicate orchid]

yeah ||det(2A) = det(A)|| is a bit of a hint lol

[void cosmos]

well i have det(2A) = det(B)det(B^-1)det(A)

so det(2A)=det(A)

[lusty marlin]

Yes

[void cosmos]

oh thats coo

ol

[lusty marlin]

And that essentially gives you your answer

[void cosmos]

yea im stupid

thank you

[lusty marlin]

Where is this question from btw?

[void cosmos]

qualifying exam i am just goign through qualis for fun

algebra

dk which uni

nvm

its uc irvine

😦

its not easy for everyone my man

[delicate orchid]

brutal

[void cosmos]

yea 😦

[formal ermine]

riemann hypothesis is trivial tfym

[karmic moat]

four color theorem is trivial

[barren sierra]

TFW thing easy for me isn't easy for everyone

[void cosmos]

i dont know how good every uni is

[karmic moat]

proof: give me four markers

i draw four colors

boom

[void cosmos]

i hear uc irvine is one of the good schools so

i jusut checked the qualis out for fun

maybe asia is much harder

idk much about asia either haha

just watched a documentary about taniyama

lmao

yea to each their own ig my man

u can check out the stanford qualis i linked above

i am sure only a small fraction of problems would be easy for a large fraction of people

[elder wave]

what an odd thing to say

[void cosmos]

yea im more of a poet myself

[elder wave]

not you

[void cosmos]

idk how rank corelates with this

i would tihnk it depends on like

[karmic moat]

so much emphasis is placed on uni rank

[void cosmos]

which subfield of math is the department more focused at

like for example university X is like more analysis-based

[karmic moat]

i feel like for grad school uni rank matters less than the specific department for ur study field

yea

[void cosmos]

so u would expect a hrader analysis quali .. something like that

i hear for example uwaterloo are like insane at discrete math stuff

[karmic moat]

school rank is silly

[lusty marlin]

Mine is an engineering school that doesn't do much math

[karmic moat]

it's just a way for us schools to justify one kajillion dollars per credit hour

[lusty marlin]

I happen to be a math major there though

[void cosmos]

is it important for a career in academia?

[karmic moat]

been thinking about that, no idea

[formal ermine]

uhhhhh

[karmic moat]

i wouldn't think so?

hopefully it comes down to your research you did in grad school/postdocs

[void cosmos]

i would think how good is your own research/advisor is more importnat

[karmic moat]

i'm also an undergrad so take my words with more than a grain of salt

[void cosmos]

like an extreme case u prove twin prime at university of moamen

[karmic moat]

like a tbsp of salt

[lusty marlin]

You guys pay per credit hour?

[karmic moat]

i think that's how it usually is here in the states

[lusty marlin]

Sad

[karmic moat]

pay per credit hour + additional fees and whatnot

[void cosmos]

anyways back to algebra:

its not a quali problem

but its just something thats been bugging me

now suppose R is a commutative ring

with identity

now by zorn's lemma (M3 in stanford quali lol) there exists a maximal ideal

call it m

R --> R/m by natural map

R/m is a field , so it has the invariant dimension property

but we know that if the homomoprhic image of a ring has the property then ring itself does

so now we know that all commutative rings have the invariant dimension property

can someone give me an example of a ring that does not have the invariant dimension property

[elder wave]

what is invariant dimension property

invariant basis number?

[void cosmos]

all basis over all free R-modules would have the same cardinality

specific

like inside a free-R module 2 basis would have the same cardinality

[lusty marlin]

Could you please elaborate on this?

[formal ermine]

consider Z --> Z/3Z

latter is a field former not

or Z --> Z/4Z lol

[lusty marlin]

Yeah, that statement is false

[formal ermine]

BRO I KEEP TYPING PRIMES WHEN I'M TRYING TO TYPE COMPOSITE NUMBERS

[void cosmos]

i think i need surjection.

yes i do

[formal ermine]

huh?

[void cosmos]

sorry boys and girls

okay okay

let f: R-->S be an epimorphism of rings with identity

then S has property --> R does

[elder wave]

okay yeah this is the same as IBN

[void cosmos]

okay

[elder wave]

the endomorphism ring of K^\mathbb{N} doesn't have it

[void cosmos]

now is the proof of any commie ring has IBN true?

and can anyone

give me an example of a ring that does not have IBN

cuz i am struggling

[devout crow]

look right above you buddy

[void cosmos]

the ceiling?

oh

whats K^N?

[devout crow]

sequences in a field K

[void cosmos]

ok

can you prove it?

[elder wave]

your proof is correct though

[void cosmos]

cool

[elder wave]

if R/I has IBN then R has it

[void cosmos]

yea exactly

but R/I is S by iso

so yea

but can u prove that this counterexample works?

would appreciate it

[devout crow]

why is this? I don't see it

[void cosmos]

consider the free R/I- module F/IF

this is an S-module by first iso

so it has IBM stuff or whatever

IBN

so like the "pull backs" of any two basis in this module would be the same in the R-modujles

[elder wave]

oof it's been a while

this also uses a different version

[void cosmos]

okay so suppose X and Y are basis of some free R-module ( call it F )

let pi be the canonical epi from F--> F/IF

then pi(X) would be a basis of F/IF

and so as pi(Y)

( this just follows directly from linear independnace of X )

but those are equal ( cardinality-wise )

and pi is a surjection

so X would have same card as Y

[void cosmos]

is there no easier example

oh and sorry I is the kernel here

from the first iso

[elder wave]

IBN is pretty weak

not easy to come up with counterexamples

[void cosmos]

hmmm 😠

speaking of exams

how hard is this exam

@elder wave solve number 1 haha

[karmic moat]

this is interesting question, if someone solves it tag me please

[wooden ember]

Lol I didn’t see the « answer ONE of the following » at first for I and I thought to myself the exam looks hella long

[karmic moat]

only thing i've been able to muster is to break down [K : F] = [F(a,b) : F] = [F(a,b) : F(a)] * [F(a) : F]

[void cosmos]

hahaha yea

anyways can u help with a counterexample

[formal ermine]

yeah

[void cosmos]

or just an example of a ring that doesnt have IBN

wiht a proof

[formal ermine]

and the last thing is = 5

the first prodcut is either 4, 3, 2

[karmic moat]

yeah im trying to figure out how to conclude [F(a) : F] is 5

[formal ermine]

can't really see why it can't be 3

[formal ermine]

minimal polynomial

[karmic moat]

oh duh right

[formal ermine]

we probably need to use the fact the polynomial is separable

[karmic moat]

if K/F is the splitting field of f, then would K/F be the algebraic closure of F

since f is irred and separable

[wooden ember]

Honestly idk I’ve never heard of this concept before

[formal ermine]

heh?

[wooden ember]

After a quick search on Wikipedia though there’s a counterexample there

[void cosmos]

which concept

IBN?

[wooden ember]

But apparently you need to go through non commutative rings and me no like that

[wooden ember]

Yeah

[void cosmos]

yea any comm ring has IBN

[wooden ember]

It’s surprising there are rings for which it isn’t true

Kinda gross

Anyways yeah Wikipedia says column matrices of countable dimension

[formal ermine]

@karmic moat I'm thinking like

if it were 3

then its galois subgroup would be cyclic or something

this would be overkill but then uh

the subextension on the left would be simple

and so = 1

which is not possible

[karmic moat]

haven’t gotten that far into alg :^)

[formal ermine]

yeah it's really nieche

@karmic moat lol the solution is easier than I thought

ok so

one property of the minimal polynomial is that if some other polynomial has that element as a root then the min poly divides it

right

so m_beta | f/(x - alpha)

[karmic moat]

whats m beta

[formal ermine]

minimal polynomial of beta

[karmic moat]

ah

[formal ermine]

hmmm now I'm thinking

why is m_beta now just 1

cuz like m_beta | f

and it obviously can't be 5 right

[south patrol]

invariant cringe number

[formal ermine]

hi potato

[south patrol]

Hi

What are we doing rn

[formal ermine]

basic field theory

this question

[south patrol]

hm which one

ah

[formal ermine]

tower law

right part is just 5

because minimal polynomial the left part is either 1, 2, 4

now I'm buffled as to why it isn't 1 for the same reason

like given [F(alpha, beta) : F(alpha)]

we need to find the minimal polynomial of beta over F(alpha)

we know that f(beta) = 0

so minimal polynomial of beta = m_beta divides f right

but f has degree 5

m_beta can't be degree 5

so why is it not just degree 1

LOL ignore all of that

I'm stupid

polynomial degree is additive not multiplicative 🤦

[south patrol]

hm imma think about this, this is interesing

i think I did essentially thsi question

[south patrol]

Yeah nice

imma think if there's any other particularly nice way to think about it lol

[formal ermine]

not nice lol

that was wrong

[south patrol]

oh maybe i misundrstood lol

well

uh

Yeah okay lol

SO

The galois group G acts on the 5 roots of the polynomial and each element of G is determined by its action on alpha and beta

so that gives a bound <= 20

like 5 choices of where to send alpha, then 4 choices of where to send beta

The divisibility by 5 is easy

So that is the partial credit done at least

[formal ermine]

yeah that immediately follows from min poly stuff

[south patrol]

tru fair lol

And then it's fine to rule out 15

i think

e.g. w galois theory ||if F(alpha) = F(beta) then we have K/F of degree 5, otherwise there are two subgroups of G = Gal(K/F) of index 5, so that the order cannot be 15 (or otherwise it's be Z/15Z)||

[formal ermine]

how does this rule out 15

like uh

[wooden ember]

What are the groups of order 15

[formal ermine]

cyclic

why can't it be Z15

[south patrol]

does Z15 have two subgroups of index 5

[wooden ember]

Z/15Z only has one sub of index 5

[south patrol]

😎

[formal ermine]

uh where does the other one come from

[south patrol]

this is uh

F(alpha) and F(beta)

[wooden ember]

F(alpha) and F(beta)

[south patrol]

distinct fields of degree 5 over F

[wooden ember]

Sorry lmao I keep getting sniped I’ll let potato speak considering it’s their solution

[south patrol]

(i'm assuming - otherwise we're done anyway)

aha lmao

sorry

Anyway so

[formal ermine]

I see

wlog right

[south patrol]

ye

well

not so much wlog as by assumption yes

[formal ermine]

yaya

okay yeah

[south patrol]

but i wonder if there is a way that doesn't use galois theory

I just saw "irreducible and separable" and assumed Galois theory for that extra structure

¯_(ツ)_/¯

[formal ermine]

now I can finally take a bath knowing that that question is solved

oh god it's been 1:30 hours

[south patrol]

Enjoy

130 hours

[kind jacinth]

i dont get why the oribt of 0 is {0, 1, 2, 3} shouldnt it just be {0, 1}?

[south patrol]

No

You can use p to send it to 2, then q to send it to 3, for example

[kind jacinth]

ah i can use p to send it to 2 by doing p^2? because thats when 0 goes to 2

right? cz currently 0->1->2

and 2 goes back to 0

[south patrol]

Yes

[kind jacinth]

and for this how did they get the value 12?

i am not sure what they mean by recursively computing

[formal ermine]

yeah lmao

I was gonna practice the drums but now it's too late

it's already 9 pm and my neighbors would shout at me

[elder wave]

10 Uhr Ruhezeit

😠

[formal ermine]

digga meine nachbarn sind rentner

letzte mal schon mad weil ich um 19 uhr geübt habe

[elder wave]

L

[south patrol]

Da hast du ja n großes L gekriegt digga, ziemlich kringelig

[south patrol]

Rip

[formal ermine]

potato abi 😭

[south patrol]

more like potato asi 😎

[sinful mirage]

Let $\rho:G \to \text{PGL}(V)$ be a group homomorphism. Let $s:GL(V) \to PGL(V)$ be a section. I have to show that $$ (s(\rho(g)) s(\rho(h)) (s(\rho(gh))^{-1}$$ is in the kernel of the quotient projection $\pi:GL(V) \to PGL(V)$

[cloud walrus]

ProphetX

[sinful mirage]

it is kind of straightforward to verify that after applying the projection $\pi$, we get $\rho(g) \rho(h)$ for the first 2 terms.

[cloud walrus]

ProphetX

[sinful mirage]

how do we evaluate the third term?

[valid night]

Is this assumption correct?

[delicate orchid]

no

oh wait, BOTH normal

then yes I think so

[valid night]

Alright thanks

Is the proof doable for a first years math student?

[delicate orchid]

if you know what an internal direct product is, the proof is very quick

but maybe you can do it regardless

[valid night]

Never heard of it but I'll try to find something

[cloud walrus]

marc

marc

marc

[delicate orchid]

they're defining the equivalence relation in such a way that the following 3 points hold - I don't really get what you're asking if I'm being honest

are you asking how point 3 holds or?

[valid night]

The relation ~ is defined as $a~b iff a^{-1}b \in H$

[cloud walrus]

Kroros

[valid night]

And the theorem says that 1: ~ is an equivalence relation

[delicate orchid]

as kroros said, $a \sim b \iff a^{-1}b \in H$

[cloud walrus]

Wew

[delicate orchid]

or are you confused about what an equivalence relation is

[cloud walrus]

marc

marc

[delicate orchid]

they're in the same coset of H by point 3

where this is going is that when we quotient G by this equivalence relation we get the same thing as G/H

that's very odd lol

tbh I'd define the equivalence relation as point 3 as that's the important bit

[cloud walrus]

marc

[delicate orchid]

$a \sim b \iff aH = b H$

[cloud walrus]

Wew

[delicate orchid]

this is all motivated by quotient groups

hell, cosets themselves are motivated by quotient groups

which is why it's incredibly weird they're not introduced until "much later"

no way you've done orb-stab before quotient groups this is wacky

kind of based albeit

fairs

take a subgroup H, multiply it by some element a

that's it

no, not subgroups

subsets

[delicate orchid]

and it does this precisely because this is an equivalence relation

[formal ermine]

[delicate orchid]

yes that theorem holds for every subgroup

[cloud walrus]

marc

[delicate orchid]

sure

I'll agree with you on that one

[cloud walrus]

marc

marc

[delicate orchid]

no, not at all

a and b are in the same coset of H

you can see the second equivalence via [a^{-1}b \in H \Rightarrow a^{-1}bH = H \Rightarrow aa^{-1}bH = aH \Rightarrow bH = aH]

[cloud walrus]

Wew

marc

[delicate orchid]

I don't see how you can't understand that, sorry

either a^-1b is in H or it isn't

it implies they're in the same coset of H

as I've said

this motivates quotient groups

G/\sim_H = G/H

[kind jacinth]

Anyome have any idea?

[prime sundial]

can the notation (S)^* be used for S \ {0} or is that not common?
basically want to write a modded vector space but
(S \ {0}) / ~ looks ugly

i also now realize how bad this notation is for a vector space as it could also mean its dual

[south patrol]

I have seen it used for that a lot for rings and stuff

But yeah for vector spaces could be ambiguous as you say

[prime sundial]

is the space given by identifying antipodal points on S^1 is RP^1? is this general to S^n

i want to say yes since this construction is basically the same as the usual R^{n+1} \ {0} modding out scalar multiples

[south patrol]

Yes

This is very useful because it shows that RP^n is compact, for example

[south patrol]

(Also note that S^1 mod antipodal points is also just S^1, which is a special case)

[prime sundial]

hadn't heard that one before. is there intuition to it

im working on hatcher problem 2.2.10, where we're asked to compute Hn(X) for X = S^2 with the equator S^1 under this antipodal mod
is there a way to construct this as a CW complex

[south patrol]

Oh like s1 mod antipodes is S^1?

Uhhh one way is to consider squaring map S^1 -> S^1

It is surjective and identifies points iff they are antipodal

So you get a homeo S^1/antipodal -> S^1

[smoky cypress]

How do you define tensor product over non commutative rings

[fervent rock]

i have a question

is showing that there is no quintic formula

the same as showing that group R_5[X] is irreducible \

i dont know weve gotten to the end of this algebra class, i hate it so much, much im seeing these polynomial ring stuff

[prisma ibex]

showing that there is no quintic formula in terms of radicals is the same as showing that the group S_5 is not a solvable group (i.e. it doesn't have a sequence of normal subgroups such that the successive quotients are finite cyclic groups)

[fervent rock]

meow meow meow thanks

that actually kind of makes sense

so whats a galois field cause the chapter is called galois fields but they dont use it anywhere

[prisma ibex]

yeah like the order of the cyclic groups corresponds to the order of radicals you're taking and so on