groups-rings-fields

[chilly radish]

Here is fine

[coral spindle]

Just don't interrupt someone else asking a beginner's question

[chilly radish]

You're more likely to have people who remember what a sylow is in this channel tbh

[toxic zephyr]

alright cool thanks

[delicate orchid]

who? me? noooo never

[rocky cloak]

A sylow (noun) :
(1) great mathematician who inexplicably works as a high school teacher most of their life
"Dude, I had this total sylow in the 10th grade, now I can finally shitpost on mathcord"

[chilly radish]

40 years as a highschool teacher damn. Dude had a calling

[hidden wind]

i’ve dug up some short early 20th century articles by thoralf skolem about sylow in norsk matematisk tidsskrift i’m excited to read when i find the time

[rocky cloak]

I just find it funny that three greatest Norwegian mathematicians of history all have families of groups named after them.

[hidden wind]

also i went i went to that hs sylow taught at… big mistake

[rocky cloak]

Did it go down hill after he died or something?

[hidden wind]

i probably should not have mentioned this when i do not want to elaborate

[rocky cloak]

Which school even is it. Wikipedia lists some school in Denmark, which isn't correct

[hidden wind]

hartvig nissen

[rocky cloak]

Right, he was there for like two years or so. And then he worked at Fredrikshalds lærde og realskole, according to snl

Nissen is nice though, not sure why that would be a mistake

[hidden wind]

personal circumstances

[hidden wind]

oh i wasn’t aware of this second place… i guess it’s probaly mentioned in skolem’s article i’ve yet to read

[warm ember]

is it true that a p-group has subgroups of order p^i for all i

as long as p^i<=order of group of course

im guessing not

[topaz solar]

Have you heard of Sylow

[warm ember]

yes

im reading that rn

nvm

[warm ember]

it was used in the proof of sylow

im dumb

wait no

[warm ember]

only if every proper subgroup has index divisible by p

if its abelian it must have all subgroups

p group has nontrivial center so G/Z is smaller. then induction and we done

this correct?

[toxic zephyr]

stupid question: what would be an example of a group G with order |G|>|R|? obviously, the additive group on R has an uncountable order, but are there any uncountable order groups with order aleph2 or alephomega?

[topaz solar]

This does reduce things

[topaz solar]

Ok so

Dumb answer: free group on X generators has cardinality |X| + aleph_0 with choice

[topaz solar]

I’m afraid I dont remember all the conditions and corollaries regarding Sylow offhand, but I thought it said we had a p^i order subgroup for each sane i?

[topaz solar]

Also, aleph_2 or aleph_omega need not be above |R| anyway

But that’s sorta irrelevant to the intention

[toxic zephyr]

hm? wdym?

[warm ember]

o theres probably some variations to tthe theorem

[toxic zephyr]

if that's true, i don't think i understand aleph notation lol

[warm ember]

thanks tho

[toxic zephyr]

i thought aleph1=|R|

[topaz solar]

No

That’s the continuum hypothesis

[topaz solar]

Probably some include some corollaries in the statement

[toxic zephyr]

oh shit. well if i say aleph1 to mean |R| is that generally well understood? i know the CH isn't decidable, but i would certainly prefer it be true lol

[topaz solar]

Not really but it’ll get the point across most of the time

And usually just being uncountable is enough

CH is independent of ZFC so it’s fine to slap on too, but not needing it is always better obviously

I’ve slapped it on for some elevator pitch sketching for small model saturation stuff in a PDE fever dream

[grizzled spindle]

I’m not sure I get the last 2 bijections

H mapped to (G/H, H)?

Is it to specify which coset is taken (in this case the one with identity?)

[mighty kiln]

A pointed G-Set is a G-Set with a chosen element

The chosen element of G/H is H in this case

[grizzled spindle]

So H is the coset with identity right

Just to make sure

[mighty kiln]

Yea

[grizzled spindle]

Alright thx

[lean sail]

reading through an abstract algebra book, i just reached the section talking about subgroups and cyclic groups. i found this proof. i didnt write it, i just rewrote it by hand so that i can understand the steps as i was reading. my question is, in the third highlighted step, why does the author point out that $a^{n - m} \in \left < a \right >$? is this because we don't automatically know that $a^n (a^m)^{-1} \in \left < a \right >$? in other words, we know that $a$ to some integer power is an element of $\left < a \right >$, but not two elements multiplied.

[delicate orchid]

Yeah

[cloud walrus]

proofman

[delicate orchid]

That condition is equivalent to <a> being a subgroup so if we were to assume it our argument would be circular

[lean sail]

that makes perfect sense

we don't just have closure on some random set

right?

[lean sail]

Trying to understand the notation in this cyclic subgroup. From what I understand the elements are getting added. So if 2(-1) = -1 + -1, what is -1(-1)? somehow $\left < -1 \right >$ generates the all of the integers. i am failing to see how that is possible.

[cloud walrus]

proofman

[delicate orchid]

-1 times -1 is 1

and then it's pretty clear that 1 generates all of the integers additively

[lean sail]

wait a second, so this is multiplication?

[delicate orchid]

how is writing multiplication as concatenation notation abuse

[lean sail]

i think you understand my question

[delicate orchid]

that's very silly

abelian groups are Z-modules there is nothing wrong with writing it multiplicatively

[hidden wind]

this is very nonstandard

[lean sail]

yes

[lean sail]

but i think this is what the author does in this example

but then how does (-1)(-1) work?

[hidden wind]

in an abelian group the notation nx is often used to mean the same thing that in nonabelian groups is written x^n

please stop writing addition multiplicatively

[lean sail]

@limber thorn @hidden wind does this seem right?

[hidden wind]

^

[lean sail]

the inverse of $-1$ should just be $1$ i think

[cloud walrus]

proofman

[delicate orchid]

^

[lean sail]

so when the author says $-2(-1)$ that's like saying... add the inverse of $-1$ twice?

[cloud walrus]

proofman

[lean sail]

author is Gallian btw

[lean sail]

does this make sense?

so in this m times n notation, or whatever he is using, which one is supposed to "be like" the exponent?

sure, i am open to that

[delicate orchid]

my brother in the heavenly lord you are multiplying integers

you did this when you were 5 years of age

I don't get what the confusion is

[glad osprey]

I think he's asking whether ab in additive notation is like a^b or b^a in multiplicative notation

[delicate orchid]

it doesn't matter

think of it as either if you want

you're just trying to show that every integer is a multiple of -1

[glad osprey]

btw, why did you ask him to go to DMs, then delete all your messages?

[delicate orchid]

which is trivial because (-n)*(-1) = n

[lean sail]

yes, but more generally, i just dont understand what is going on

[glad osprey]

<a> = {..., a^(-2), a^(-1), a^0, a^1, a^2, ...}

set a = -1

[glad osprey]

Note: I've written the group operation multiplicatively, and repeated application of the group operation as exponentiation. When you're in Z the group operation is addition, and repeated addition is multiplication. This can be confusing, because when we write a+b we could be talking about a generic group operation in an abelian group, or we could be talking about literally adding together a and b

So it's not really a matter of multiplicative vs additive notation, it's just that when you're in Z, the group operation is addition

[lean sail]

makes sense, thank you!

[winged flicker]

🫣

[grizzled spindle]

I think my brain was not working cause I wake up and this makes sense

The pointed G-sets represent the same G-set and they map to conjugate subgroups of G, but if we take the operation in reverse the need for a pointer is removed if we instead remove all conjugates of subgroups

[grizzled spindle]

Well I need to also exclude the empty set cause it never works

[winged void]

I want a hint if someone can help Suppose R is a ring of 10 elements. Prove that R is commutative.

[white oxide]

i'm not entirely sure what I'm supposed to show here, that the prime ideals act as a basis for the jacobson topology on prime(z)?

[noble lynx]

if I have a homomorphism R[x, y] -> R[x] that sends x to x and y to x^2 how do I show the kernel is exactly (y-x^2). Having trouble showing the kernel doesnt contain any more elements

[coral spindle]

So one way to do this is to try and look at an arbitrary element of the kernel and then reduce it in some way

So we have some polynomial f(x, y) = whatever

Now suppose it has some power of x in it somewhere, x^n

If f(x, y) is in the kernel, I can substitute something else I already know is in the kernel and get another thing in the kernel

So what I mean for example is that x^4 is in the kernel iff y^2 is

So I can 'reduce' this thing to include only single powers of x

WLOG I can assume that f(x,y) = g(y) + xh(y) for some polynomials g and h of y

And now I can think about when these would be in the kernel

If I can prove that this is in the kernel iff it's in (y-x^2) then, since I have only used manipulations due to (y-x^2), I have shown that the kernel is (y-x^2)

[winged void]

@coral spindle can you maybe give me a hint for the the question of 10 elements in a ring

[coral spindle]

I have no idea. Also please don't ping random people.

[noble lynx]

can you explain this part more carefully? I get that the reduction thing works for monomials, but why do we expect it to work in general, like maybe the sum of monomials is in the kernel even if each monomial is not

[coral spindle]

But that's not what I've claimed

[rocky cloak]

Hint: What can the characteristic of such a ring be?

[coral spindle]

Let me give a tiny example

[winged void]

the thing is we did not yet learn over the characteristic of a ring

[winged void]

so i tried to prove it using contradiction

[next obsidian]

The underlying abelian group is cyclic

[delicate orchid]

oh I know! pick me pick me!

[winged void]

my idea was we know that two of the 10 elements are {0,1} and with contradiction we know that ab \neq ba for all elements in R and plus a \neq 0 \neq 1 same for b

[next obsidian]

Does that finish it?

I feel like that + the unit of the ring commuting finishes it

Idk tho

Maybe I’m wrong

I feel like the unit has to be a generator of the abelian group and then you go woohoo

[rocky cloak]

I feel like I would write out this argument as well, but yeah that's basically it

[coral spindle]

Let's say we're trying to discover if f(x, y) = x^2 + 2yx + 1 is in the kernel. Now we know already that (y - x^2) is contained in the kernel -- this is obvious. Therefore f(x,y) is in the kernel iff f(x, y) + (-1)(y-x^2) = y + 2yx + 1 is in the kernel. In particular, f(x,y) is in (y - x^2) iff that latter sum is in (y - x^2).

[winged void]

Idk i still find it difficult to be fair

[rocky cloak]

Yeah, that's okay

[noble lynx]

thanks got it!

[winged void]

i know but can you give me like what you meant with the chracterstic so i know where to go

i know what a unit is

[rocky cloak]

So 1 commutes with everything. So also 1+1 commutes with everything, and 1+1+1, etc.

The characteristic is the smallest n such that 1+1+1... (n times) equals 0.

Notice that 0, 1, 1+1, ... forms an additive subgroup of your ring, so this limits what the characteristic can be (by Lagrange)

Notice also that x+x+... = (1+1+...)*x, so if the characteristic of your ring is n, x+x+... (n times) is 0

And what can you say about abelian groups where all elements have order 2 for example?

[winged void]

a^2 = 1 but we have that a + a = 0

am i right ?

[winged void]

i do not know if im right with what i mentioned

[kind temple]

an abelian group whose elements are all order two is a Z/2Z vector space
make a cardinality argument

[rocky cloak]

I'm not totally sure what you mean here...

[winged void]

i mean because they are of order 2 right

[rocky cloak]

Yeah, so that means a+a=0

[winged void]

exactly but that means as well that a^2 = 1

because a ring has two operations right

like + and multiplication

[rocky cloak]

Sure, but we're just talking about addition right now

[winged void]

sure agree so this mean that a has chracterstic of 2

but how that is going to help us with proving the statment that 10 elements implies commutative

[kind temple]

im actually unsure if levy is a good reaction sticker or not

[rocky cloak]

Well, it will tell you what the only ring with 10 elements is

[winged void]

Sure but I mean I do not know what are the elements of that ring so how come I’m going for example to guess that each element has order of 10

In such a way that a+a 10 times equal = 0

[winged void]

I know I’m asking stupid question sorry but I do not see it yet

[rocky cloak]

So do you know what the finite abelian groups are?

[winged void]

Yes that is when a+b = b+a

For all a,b in R

Right

That is for addition

[rocky cloak]

Yeah, but I mean do you know which groups that is?

Like the classification for example

Like do you know what the abelian groups of order 10 are?

[winged void]

Not really

But I guess it’s going to be

Module 10

Because they are always ableian

In addition

And multiplication

[rocky cloak]

Hmmm, okay, so what tools do you have? Like do you know much about groups and rings besides the definitions?

[kind temple]

the characteristic determines the order of all of your non-zero elements, and the characteristic can only be 1, 2, 5, or 10

they can't all be order 1
it can't be 2, since R would be a Z/pZ vector space for p = 2
it can't be 5, since again, R would be a Z/pZ vector space for p = 5
so all the elements have to be order 10

the rest follows from the classification of finite abelian groups

[rocky cloak]

Or maybe you can just argue like this:

The characteristic must be either, 2, 5 or 10 (from Lagrange, clear?)

If it's 2, then take x not 0 or 1. Then {0, 1, x, x+1} is an additive subgroup of order 4, is this possible? (Think Lagrange)

If it's 5, {0, 1, 2, 3, 4, x, x+1, x+2, x+3, x+4} are 10 elements that all commute.

Similarly if it's 10, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} all commute.

[winged void]

I see let me try that approach

[rocky cloak]

(the characteristic 5 example doesn't actually exist, but I guess that's no hinder to the argument)

[winged void]

I see but

Just one question Lagrange has to do with cosets

Right

[delicate orchid]

yes that's how you prove it

[topaz solar]

It has to do with sizes of subgroups

[winged void]

i see

[delicate orchid]

oh yeah? watch THIS

nvm they're right

[topaz solar]

Wew when 5^2 \neq 10

[delicate orchid]

it was earlier today

[long obsidian]

Can someone show me how to use gauss' lemma about polynomials to show y^2-x^3+x is irreducible in k[x,y]?

[rotund aurora]

why do you think you need Gauss' lemma?

this is a quadratic in y

[long obsidian]

I thought that was the standard method to show this. Why is it being quadratic obviously imply it's irr?

[rotund aurora]

if it were reducible then x^3-x would be a square in k[x]

[long obsidian]

Why would x^3-x be the square of something?

Does this have to do with there being no mixed terms like xy?

[rotund aurora]

If it were irreducible it would factor as (y-a)(y-b), where a,b are polynomials in x

right?

[long obsidian]

Right I agree with that. But then ab= x^3-x right? Where does the square come from? I can find an example of a and b e.g. a=x and b=x^2-1.

[rocky cloak]

I guess you can use gauss lemma to see this. Like gauss lemma tells you that this is irreducible in k[x, y] iff it is irreducible in k(x)[y].

And in k(x)[y] you know how to factor quadratics

[rotund aurora]

If it is irreducible in k(x)[y] it is trivially irreducible in k[x][y], it is the other direction that is not trivial (which is not needed here)

[rocky cloak]

You can also just sort of write it out explicitly though

[rotund aurora]

the point is that the coefficient of y is zero. So a+b=0

[long obsidian]

Ohh I see thank you that makes sense! And then squares of polynomials have even degree so this can't happen and it follows y^2-x^3+x is irreducible in k[x,y]. Thanks again!

[west sinew]

Is this seemingly intuitive statement true: If we have two modules $M\cong M'$ and submodules respectively such that $N\cong N'$, then $M/N\cong M'/N'$

[cloud walrus]

adi

[west sinew]

how do you cook up subspaces which make the quotients non-isomorphic?

[chilly ocean]

There's also a very simple counterexample with small finite abelian groups

[west sinew]

not sure i'm seeing the full picture

OK so broadly, you're suggesting an infinite-dimensional vector space H with a proper subspace $S\cong H$, so on side you have nothing and on the other you have whatever $H/S$ is?

[cloud walrus]

adi

[topaz solar]

Yes

[west sinew]

thanks! yeah it didn't strike me that you could come up with such subspaces in an infinite-dimensional setting

[west sinew]

just out of curiosity, what's this one?

[chilly ocean]

Z/2Z x Z/4Z

[tropic spade]

Where is my thinking going wrong with this? I take the automorphism of Q that maps x to the multiplication by 2 map on Aut(Q). I take the semidirect product of Q and Z with respect to this automorphism. I'm supposed to still have Z as a subgroup of Q in this semidirect product.

But I'm sort of lost about how this works out with the actual definition of the semidirect product and the identifications used for Q and Z.

In the product we identify Q with {(q,1) : q in Q}.

We also identify {(1,z) : z in Z} with Z.

But {(q,1) : q in Q} has a second subgroup iso to Z, it is {(z,1) : z in Z} (and is a distinct set from the other Z, but by our identification we ought to technically also call this Z too).

So, isn't the claim that in my semidirect product Z <= Q sort of ambiguous?

Here are pictures of the section from df I'm talking about:

G is the semidirect product of H and K

[delicate orchid]

it's this one

if it was the other Z then xZx^-1 = Z because x is a generator for that Z

[tropic spade]

Okay, I see it now. This is some fucked up naming/identification lmao.

[delicate orchid]

it was kind of obvious from context

they really did just have to write Z < Q < G though

[tropic spade]

Seems like they ought to point out that Z here refers to two totally different subsets of G though.

[topaz solar]

The <x> thing is called iso to Z but not identified 😔

[tropic spade]

Ah yeah that's true. Didn't realize that technicality.

[topaz solar]

Anyway, they also point out Z < Q, and Q has a pretty uhhhh distinct choice

[delicate orchid]

yes.... {x/101, x in Z} is the clear canonical choice

[topaz solar]

If we callin that Q maybe I should just Q-uit

[tropic spade]

I have changed my mind. I hate Z iso to K. Z=K as god intended.

[delicate orchid]

All categories are skeletal

[topaz solar]

Remove my skeleton

[autumn flower]

WAIT REALLY

YOU SHOULD

I REALLY LIKE THE PSETS IN JACOBSON

[warm dove]

no I'm a lazy fuck

[autumn flower]

NO MARLINS

[warm dove]

I don't have time for that crap with physics

[autumn flower]

wait should i apply to ross this year?

[warm dove]

I'm doing a lot of the basic super easy exercises in judson

[warm dove]

yes

ross was best 6 weeks of my life by a very high margin

[autumn flower]

i think i can get in

but hopefully

[warm dove]

you probably could have gotten in last year if you got your rec letter

lol

[autumn flower]

dude my teacher

fucvkkk

[warm dove]

there was exactly one kid at ohio who had done abstract algebra

and he tied for most sets finished along with some other ppl

[autumn flower]

I LOVE ARTIN DUDE

[warm dove]

🤓

just use judson

[autumn flower]

no im doing artin

but jacosbnon has some nice exercises

i do

[warm dove]

I'm planning on doing this great (mostly easy) problem soon

[autumn flower]

wait this is

[warm dove]

for fun

[autumn flower]

highly trivial i fear

[warm dove]

very trivial

but fun

[autumn flower]

dude ive been stuck on this one analysis problem

about fixed poins

[warm dove]

💀

idk analysis

[autumn flower]

i cant do any math until i solve it

[warm dove]

go torment amukh or smth

[autumn flower]

perhaps

asking for help is cheating so

i suffer in silence

[warm dove]

there's a complex analysis indy study at my school rn actually

but I have no time to join it

[autumn flower]

o_o complex before real?

[warm dove]

idk

I could lol

but I have no time

I'll just do real analysis next fal

[autumn flower]

idts

[warm dove]

I still need to take linalg in the spring after absalg this semester

[autumn flower]

any good comple text assumes some familaiority with pst

[warm dove]

🤡

[warm dove]

idk lol

[autumn flower]

U HAVENT DONT LA YET MARLINS

omgh

[warm dove]

I have a math phd to explain it to me

[warm dove]

I've taught myself a lot of LA

for physics

[autumn flower]

oh good

:)

read lang, its super boring but super interesting

if that makes snse

[warm dove]

I mean I still want to take the class ofc so I can turn the "jumbled picture in my head" into an actual understanding

[warm dove]

joever

too lazy though the textbook for my class in Strang (sadly)

[autumn flower]

for those of us without such a privelage, we have to follow hte prereqs

[autumn flower]

erm ... not slay at all !

[warm dove]

hehe

:DD

[autumn flower]

i miss zorn

[warm dove]

yeah idk lol

[autumn flower]

:(

bring them back they didnt do NUN wrong

[warm dove]

are you not still on analysis study

like the server

[warm dove]

joever

[autumn flower]

i am, but im not doing the measure theory seminar

[warm dove]

huh

[autumn flower]

i have too much schoolwork, i would drag them all down

[warm dove]

is the RA thing officially over

[warm dove]

this is so real

[autumn flower]

im following folland at a slower pace

[warm dove]

my research project is eating my time

[autumn flower]

and i have a friend who is vetting my proofs

[warm dove]

unironically 20 hours a week

[warm dove]

wait this is goated

[autumn flower]

yeah its nice

anwway, this is BRUTALLY off topic

[warm ember]

is that why i didnt get in

cuz ross thought i would have know everything already

[warm dove]

not really

they did accept ppl with a decent amount of background

[warm ember]

what about this

[warm dove]

they don't accept everyone with huge amounts of background

idk

maybe you just got unlucky

[warm ember]

i had huge amountts of background bruh

[warm dove]

there's a lot of people who I think should have gotten in that I know that didn't

and some people that I met that maybe weren't taking full advantage of it

[warm ember]

is it better to show not a lot of background on psets

cuz i used field extensions on pset

i think that may be reason for no accpet

[warm dove]

I mean you also have to explain where that background and knowledge came from

if you just randomly used field extensions

it probably looks like you're cheating

💀

[warm ember]

wtf

[warm dove]

high schoolers don't typically know what a field extension is

💀

[warm ember]

🤷‍♂️

[warm dove]

and just pulling out a lot of high powered theory is pretty cooked, especially if the other problems have more elementary solutions

[warm ember]

high schoolers typically dont know what a derivative is

[autumn flower]

the demographic that applies to ross prob does tho

[warm dove]

no

the majority of ross application pool is comp math kids

[autumn flower]

oh god

ew

[warm dove]

no no

[warm ember]

i am comp math kid ...

[warm dove]

that's one of the things about ross

it exposes them to higher math

[autumn flower]

comp math "people" when they see showers

[warm dove]

I convered some comp math kids ik into pure math

converted

at ross

[autumn flower]

my entire school is comp matth people

its a struggle every day

[warm dove]

joever lol

[warm ember]

isnt ross mosty elementary nt tho

[warm dove]

yeah

that's one of the nice things

[autumn flower]

i tried to show them how important de morgan's theorem was

to ... no avail

[warm dove]

even if you have high powered theory

[autumn flower]

😔

[warm ember]

comp math kids know that

[warm dove]

elementary number theory has hard problems

[warm ember]

i meant

[autumn flower]

wait im actually ass at nt

[warm ember]

the camp

[warm dove]

do they know the uniqueness of F_p^k?

[warm ember]

like promys

i do

[warm dove]

most comp math kids don't

💀

[warm ember]

but that isnt elementary nt

[autumn flower]

ur not like tese other girls kevin

[warm dove]

do they know hensel's lemma, quadratic reciprocity, pell's equation, etc.

[autumn flower]

doesnt this follow frm teh fact that any two fields with the same # of elements are iso?

[warm ember]

most oly kids know qr

some know pell

[warm dove]

you're confusing the Ross application demographic with like the application pool for PRIMES

lol

[warm ember]

some know hensel corollory

[autumn flower]

wait u shoul;d apply to primes

i feel like u would get in

[warm dove]

I'm not good enough for primes

[autumn flower]

arti could prob do it

[warm ember]

o i have that book

[warm dove]

the pset is very oly styled (and they weigh oly, which I'm crap at), there's graph theory that I don't know, etc

[warm ember]

its strange

[autumn flower]

oh i like graph theory

[warm ember]

read diestel

[warm dove]

yeah I get told this a lot

I'm a little busy with physics

💀

[warm ember]

o ive told you before

[autumn flower]

isnt that a gtm

[warm ember]

ye

[warm dove]

yes

[warm ember]

but its not hard

[warm dove]

idk some GTMs are accessible

[warm ember]

you dont need gt knowledge

[warm dove]

ireland and rosen is accessible after one UG absalg book

I'm planning to take an indy study out of ireland and rosen with my math teacher if he approves

next year

[autumn flower]

what is ireland and rosen lol

[warm dove]

but he says he doesn't actually understand the book

[warm dove]

the de facto "intro to advanced NT book"

[warm ember]

whats advanced nt

alg nt?

or harder elementary nt

[warm dove]

well both alg nt and analNT

[autumn flower]

well it appears to have less numbers

[warm dove]

ireland and rosen does some o fboth

[warm ember]

dont you need commutative algebra and gal theory for those

[autumn flower]

dude holy fuck

i love gal theory

[warm dove]

ireland and rosen is an intro

💀

[warm ember]

o

[warm dove]

it teaches some of the basic comm. alg iirc

and presupposes some of the gal theory

2 semesters of UG algebra hopefully covers galois theory (if it doesn't, wtf are you doing with your life lmao)

[autumn flower]

marlins making my inferiority complex grow by the second

[warm dove]

idk any of this crap

I'm 3 weeks into absalg right now buddy

we proved Lagrange yesterday in class lmao

[autumn flower]

💀

[warm ember]

3 weeks for lagrange

[warm dove]

well 1.5 weeks was on prereq set theory and ENT and stuff

[autumn flower]

fair enough

[warm ember]

oh

[warm dove]

for people that hadn't taken number theory/intro proofs beforehand

[autumn flower]

grop theory as an intro to proofs is weird

[warm dove]

well it's not meant to be an intro to proofs

[autumn flower]

oh but i go to a sweat school

[warm dove]

some people just registered for the class without the prereqs

lol

[autumn flower]

so everyone just learned prof based LA in the womb

[warm ember]

intro to proofs just do oly lol

[warm dove]

🤓

[warm dove]

🤓

[autumn flower]

some of us, marlins, go to public schools that cut their funding for toilet paper last year

[warm ember]

only benefit of oly is knowing how to write proofs and college apps

[warm dove]

I live in a residence hall that was scheduled to be torn down 10 years ago

[autumn flower]

oly teaches proof wrirting?

[warm dove]

yes

lol

[warm ember]

and clout in math community

[warm dove]

all oly problems are proofs

[autumn flower]

oh yeah

[warm dove]

lock in

[autumn flower]

but the proofs i wrote for usajmo and the proofs i write now are

[warm dove]

imagine making usamo

[autumn flower]

VERY different

[warm dove]

🤓

[autumn flower]

in terms of rigour and generality

[warm ember]

wait

[warm dove]

college apps is cooked

[warm ember]

when was usamo for you

what year

[autumn flower]

jmo

sorry

[warm ember]

oh

what year

jmo becomming xooked

[warm dove]

yeah you could have gotten into ross this year lmao blackbeard

only 5/60 people at ohio were JMO/AMO quals

and they were mostly more cracked than the rest of us

I barely kept up with them

[warm ember]

i am also jmo qual

[warm dove]

🤓

[warm ember]

but i didnt get in ross

[autumn flower]

2022

i was in 8th

[warm ember]

orz

[autumn flower]

wait no

[warm dove]

orz wtf

8th is crazy

[autumn flower]

dude i am 11th grade now and its 2024?

[warm ember]

2022 jmo is hard

[autumn flower]

what uear was i in 8th graade for?

[warm dove]

I got a 4 on AIME this year lmao

[autumn flower]

i literally cant remmever

sorry

[warm ember]

2021

[warm dove]

bc I sillied 3 questions

[autumn flower]

ok like numbers are hard

[warm dove]

still would have only gotten like a 7 💀

no I'm just incompetent

[autumn flower]

i got above the median score, but i didnt place in the top 35 or whatever

[warm dove]

I don't practice comp math like at all

so it's really my own fault

[autumn flower]

i did alright but not enough to justify the amount of time i spent studying

[warm dove]

my roommate at ross missed MOP by 2 points

from JMO

[autumn flower]

motherfuckkkk

[warm dove]

it's ok I pure math pilled him

[autumn flower]

good.

[warm dove]

he's applying to primes now :DD

[autumn flower]

gooood.

[warm ember]

who

[warm dove]

damn the ross memories are hitting hard lol

[autumn flower]

preach the gospel of rudin

[warm dove]

Aiden Jeong

[autumn flower]

🙏

[warm ember]

o

orz

[warm dove]

he's grinding dummit and foote

[warm dove]

you know aiden?

based

[warm ember]

id ont know

[warm dove]

sully

[autumn flower]

wait fukcing ew

tell him about artin or jacobson

[warm ember]

what

[autumn flower]

cmonn

[warm ember]

dummit and foot is good

better than artin

[autumn flower]

HWAT THE UFKC

JAIL

RIGHT TO HELL WITH YOU BYE

[warm ember]

artin is so bad bruh

[autumn flower]

WHAT

[warm ember]

its only about matrices

[autumn flower]

MARLINS

marlins bro

[warm dove]

joever

[autumn flower]

IM CRTINHG

I JUS SHED A TEAR

[warm ember]

artin called a monoid a semigroup

and they use 1 for group identity

[warm dove]

dummit and foote is fine lol

idk how he stands it it's such a dry text

[autumn flower]

its good for a reference but

its sooooo boring

[warm dove]

but it covers everything comprehensively

[autumn flower]

jacobson is moer interesting than artin

artin is more beginner friendly

[warm ember]

dummit and foote just skim it

to learn

[autumn flower]

d&f problemsets are boring as all hell

[warm dove]

idk

my class uses fucking judson which is a trimmed down book for students without much mathematical maturity

[autumn flower]

judson sounds like a hockey player

[warm dove]

it's a pretty lightweight book

it's a lot lighter than artin

lol

[autumn flower]

not my best take

[warm dove]

artin is designed for honors UG classes

lock tf in

[autumn flower]

im lockded

bro

im doing artin and AoM at the same time

theres always some interesting parallels

[warm dove]

yeah ur cooked

[autumn flower]

nah

throw a course of folland in there

and ap art history

NOW im cooked

[warm dove]

joever

modern physics is cooking me right now

[autumn flower]

apush is cooking me rn

we fr turned this chnanel into discussion-1 😭

[warm dove]

🤫

nobody cares smh

[warm dove]

I have creative writing next semester it's so joever

[autumn flower]

oh i rly like writing

[warm dove]

I like it a lot

and I have my favorite english teacher teaching it

but it's gonna be so hard

lol

[autumn flower]

marlins

how do i convert my yucky classmates into pure math

i was reading munkres in class and

some kid gets on my ass for messing up a weird computation in multi 😭

[mighty kiln]

Put them in the pure math grinder

[autumn flower]

they are so mean to me

"whats the point of proofs"

"put munkres away, this is english class"

[mighty kiln]

Munkres is an English text

[long obsidian]

Let k be a field and consider the integral domain A=k[x,y]/(y^2-x^3+x) and it's field of fractions Q. Even more consider [x]/1 in Q with [x] the equivalence class of x in A.

then how can I show [x]/1 is transcendental over k?

Suppose kn( [x]/1)^n+...+k0= [0]/1. I think it follows that kn x^n+...+k0 in (y^2-x^3+x). But since the only non zero elements of this ideal has a term with a y factor maybe it follows that kn x^n+...+k0=0. But we know x is transcendental in k(x,y) so this can't happen.

Does this argue that [x]/1 is transcendental?

[warm dove]

I dobut they actually believe this

or say this

lol

[warm dove]

just don't sell in multil ol

[warm dove]

you have to give them interesting but simple questions

so they can think deeply abt simple things

I'm wearing a tshirt with that on the back rn

[autumn flower]

i gave one quesiton ab thomae's fucntion once

its been a year

[autumn flower]

i learned multi alr so we are chilling

[warm dove]

more elementary

[autumn flower]

i should show ss if it didnt have their names

[warm dove]

give them number theory

💀

[autumn flower]

yeah i did

i asked them to prove the uniqueness of exponentiation

apparently thats so "pointless" and "trivial" and i am just being overly pedantic

[warm dove]

don't just straight up give them boring ass proof problems like this

[autumn flower]

its interesting to me

[warm dove]

it is worth doing the exercise

but you have to motivate even "why we want to study the integers so rigorously" first

[autumn flower]

i constructed R

[warm dove]

by explaining why the integers are interesting

[warm dove]

this is also a highly technical argument

[autumn flower]

i used cauchy

[warm dove]

well

not that technical

but still

[autumn flower]

marlins i tried

[warm dove]

you need to give them something intuitive that they can play with

lol

[autumn flower]

bruh

[warm dove]

give them like legendre symbols

[autumn flower]

they think i speak in tongues bruh

its so annoying

[warm dove]

or talk about sums of squares

[autumn flower]

cuz like ill be doing a basic anal problem

[warm dove]

or sumes of cubes

[autumn flower]

and they look over and see a few symbols

[warm dove]

💀

[autumn flower]

like shut the fuck up

goddamn

[autumn flower]

... they are comp math people, not 4th graders man

[warm dove]

sums of squares is a great theorem

if you want

[autumn flower]

its naslo fairly trivial

to prove

[warm dove]

give them the proof using minkowski

or give them the proof with gaussian integers

[autumn flower]

nah they just dont care abt that i think

[warm dove]

and explain how that gives you Jacobi's number of sum of squares representations result

idk

sometimes there's nothing you can do

[autumn flower]

beyond savibng

btw we burried someone's quesrion .

[serene dune]

what is this glorified chat here

[trail cave]

Any recommendations for a rigorous/higher level counterpart to my current first semester of undergrad algebra course to read after class? Would Milne's Group Theory work?

[serene dune]

naw they need defeated sanity

[serene dune]

umm artin ?

dummit foote

nathan Jacobson

[trail cave]

do you feel like giving pros/cons of these

our current textbook is Fundamental of Abstract Algebra by Malik/Mordeson/Sen

[crystal vale]

#book-recommendations

[serene dune]

oh chatgpt does a great job of tha

[trail cave]

there's so many channels in this server I've yet to discover

[serene dune]

take yo time

idk this seems nice too

[trail cave]

ty

[serene dune]

https://www.math.clemson.edu/~macaule/classes/s24_math4120/index.html

these notes are well made too

richard borcherds in his channel did a great job altho its not proof based and u dont get the notes

i have it all handwritten by myself

1 was great problem

need help with 2a

i mean is trial and error the only way or is there other way around

[autumn flower]

thats how i would do it

[serene dune]

idempotent doesnt work either, its at exact 7 its becoming identity

[serene dune]

joever

512 of em

[autumn flower]

construct a GL(3, F_2) matrix and jus check if M^7 = I_3

😭

[serene dune]

and 3x3 is hard to see if its invertible or not

yeah joever

GL(2, F_2) very good

its D3

🤫

[autumn flower]

doesnt this group have order 168, meaning a matrix of order 7 is possible by lagrange's theorem since 7 divides 168

or am i tweaking

[serene dune]

ermm

there is a nice way of counting

[autumn flower]

,calc (2^3 - 1)(2^3 - 2)(2^3 - 4)

[cloud walrus]

Result:

168

[serene dune]

oh yeah

thats it

by the possibilty of existence its conjectured

$\text{GL}_3(\mathbb{F}_2)\cong\text{PSL}_2(\mathbb{F}_7)$

[cloud walrus]

yeshua

[warm ember]

cauchys theorem

[serene dune]

oh yeah u are correct

how fool i am 🤦‍♂️

7 a freaking prime, should have striked earlier

[crystal vale]

let aut(G) be cyclic group then G is abelian group. Since G/Z(G) is isomorphic to inn(G) and inn(G) is cyclic so G /Z(G) is cyclic thus G is abelian.
Is it correct?

[warm ember]

yes thats correct

[crystal vale]

okay thank you

[old hollow]

Is $\text{Aut}_{\mathbb Q},\left(\mathbb Q(\sqrt[3]2)\right)$ trivial?

[cloud walrus]

a.b.s._.0.

[old hollow]

[automorphisms fixing Q]

[rocky cloak]

Yes

[old hollow]

💵

[tawdry herald]

To show that $\mathfrak{S}_n$ is $\mathfrak{A}_n \rtimes <(1 2)>$, Does it suffice to say that the subgroup on the right is of order 2, and why ?

[cloud walrus]

rraanto

[tawdry herald]

Oh the intersection is trivial and the order of $\mathfrak{A}_n$ is half that of $\mathfrak{S}_n$

[cloud walrus]

rraanto

[cobalt heath]

Don't you also need to specify the action

[crystal vale]

To determine the centre of ring M_n(R), may it be enough to determine the centre of {e_ij | i,j= 1,2,..,n}.

I don't know how to write correct proof but my guess is a scalar matrix will work.

I think we need R to be commutative

If R is not commutative then the centre of M_n(R) is identity.

Is it correct?

[inner harbor]

Guys

Sry it's in french

I have X³-bX-b

Which "cancel" a lmao what the word in eng

Like P(a) = 0

P ... a

Anyway, how do I prove that P is irreducible in k(b)

a is transcendant btw on k

[tender jay]

rj

[charred iris]

<@&268886789983436800>

[rocky cloak]

The center of Mn(R) is Z(R)*I

To determine the center it's enough to determine what commutes with re_ij

[tropic spade]

Muted for spam.

[crystal vale]

Yes

And to determine centre of e_ij then? I tried so I get scalar matrices

[rocky cloak]

Yeah, that sounds right

[crystal vale]

But I am not sure how to write properly, should I worry about it or just skip it ?

[serene dune]

its scalar matrix
a.In
where a belongs to the field

[serene dune]

wdym by R is commutative
its a field right?

[rocky cloak]

R is a ring

R for ring you know

Not that the argument is any different

[rotund aurora]

What does M_N/{+-1} mean?

[coral spindle]

Surely it means modulo the diagonal embedding of {+1, -1}?

Strange thing to write though

[rotund aurora]

but (a,b) could have order N while (a,-b) no

[coral spindle]

But they're not equivalent modulo the diagonal embedding

(a, b) and (-a, -b) are

[delicate orchid]

real but they don't have order N unless N = 2. This is dumb

[coral spindle]

Uh

[delicate orchid]

they should've just put \pm m and \pm n

[coral spindle]

Am I just missing something

I'm pretty sure (1, 1) and (-1, -1) have order N in (Z/NZ)^2

[rotund aurora]

oh oops I read what you wrote as (1,-1). But still, (a,b) can have order N while (-a,-b) not. In any case, M_N is not a group or anything. Well I suppose you have to include some factor depending on whether N is even or not

[delicate orchid]

it has to be the group generated by elements of order n or something

[rotund aurora]

I understood multiplicative order

[coral spindle]

OH multiplicative order

Then I have no idea what they mean

[delicate orchid]

yur

[rotund aurora]

because the condition (m,n)=1 already gives (m,n) has additive order N

[coral spindle]

Might be time to inspect the context 🙏

[delicate orchid]

woke

[coral spindle]

Ur right the context is filled with snowflake liberals

We should apply facts and logic to it

[knotty frigate]

lmao what

[coral spindle]

Fun fact: you are allowed to make jokes while doing math

[delicate orchid]

jokes?

wokes...

[coral spindle]

So true.....

[rotund aurora]

okay this is pretty horrible notation. The order is actually additive order. They are just picking a set of representatives. Like the lifts of an element (m',n') of M_N/{+-1} to elements m/n in Q with m,n coprime forms an equivalence class

[coral spindle]

Grody

[knotty frigate]

proof?

[coral spindle]

See above

[crystal vale]

Suppose n is a homomorphism of R into R' and n is epimorphism. Does this imply that n is an epimorphism of the groups of units of R onto the group of units of R' ?.

No because we can take n:Z -> Z/pZ then it is not true when p >2 and p is prime number.

Is it correct?

[delicate orchid]

the prophecy is true

[coral spindle]

Yes

[crystal vale]

You mean it implies that n is an epimorphism of the groups of units of R onto the group of units of R'?

[delicate orchid]

no the exact opposite

[crystal vale]

My counter example?

[coral spindle]

[delicate orchid]

I got the clout soz boss. It's like top trumps

[crystal vale]

Sorry I don't get it what is correct or what is not

[coral spindle]

I should start speaking in riddles instead of saying things plainly

[delicate orchid]

it's obviously correct

have some confidence in yourself

[crystal vale]

When I feel confident in my theory then it will be found wrong 😭

[delicate orchid]

it doesn't even descend to a group homomorphism in your case

[coral spindle]

Thy proclamation... its verity is plain as the day is long; blatant as the grass is green

[delicate orchid]

wait yes it does

[crystal vale]

Yes

Actually they define n is homomorphism when n(1) = 1

Oh this is true

In any ring which has unit element

But no we need a surjective for that

So they define n(1) = 1

[coral spindle]

Calling a homomorphism n is the most fucked up thing

[crystal vale]

\n 🥲

I don't know how to use that symbol

[coral spindle]

Do you mean eta

The greek letter eta

$\eta$

[cloud walrus]

Boytjie

[crystal vale]

Thanks

[rocky cloak]

I like that when Latin and Greek made their miniscule H, they both took n and made one part a little longer

Just a different part

[delicate orchid]

my dumbass was literally about to ask what the latin one was

[crystal vale]

If R is a commutative ring with prime characteristic p then a -> a^p is Endomorphism of R. But it is not necessarily automorphism, right?

Because take R = (Z/2Z)[x] then it has characteristic 2 but it is not surjective mapping.

Yes if it is a finite field then it will be true.

[delicate orchid]

yup, good counter example

[coral spindle]

Me when perfect fields

[delicate orchid]

dunno what that is but I agree

don't tell me I don't want to know

[coral spindle]

OK

lol

[delicate orchid]

some things are best left to erode in the sands of time

[crystal vale]

What is that?

[coral spindle]

Fields being perfect is rly important.......

[coral spindle]

The tl;dr is that a field of characteristic p is perfect if x |-> x^p is a surjection

[crystal vale]

Thank you

[white oxide]

what's an example of a subset that's not closed under this topology on Prime(Z)? for instance, if I take any subset of prime ideals of Z and consider their intersection, it's just going to be their product - but the prime ideals which contain this product are only going to be the prime ideals which are in the intersection originally, no? Hence, it looks as if every subset is closed

[chilly ocean]

There's a prime ideal that is special, different than the others..

[white oxide]

oh 0

assuming that R has no zero divisors ig tho

[chilly ocean]

You asked about Z

[white oxide]

oh yeah

so a subset won't be closed if it contains 0

[chilly ocean]

It can be closed

[white oxide]

well if it's just 0 right

[chilly ocean]

If it's just (0) then it is not closed

There are nonclosed subsets that don't contain (0) though

They might not be finite though..

[white oxide]

what if i forgot all my algebra

[chilly ocean]

?

[white oxide]

does this require some theorem from abstract algebra

or number theory

[chilly ocean]

Shouldn't require much more than knowing what the prime ideals are

{(0)} is not closed because its closure is not {(0)}

[white oxide]

oh yeah oops

it's Prime(Z) right

[chilly ocean]

the closure of {(0)} is Prime(Z) yes, since every prime ideal of Z contains (0)

[white oxide]

i really have no clue about the infinite case lol

that is if we take an infinite number of prime numbers and consider their product

wow i just turned blue

[chilly ocean]

What is the intersection of infinitely many different prime ideals in Z?

[white oxide]

well if you have some integer in this intersection, then all the primes divide Z, of which there are an infinite number of them

[chilly ocean]

There's no such thing as infinite product of ideals, but you want intersection not product

[white oxide]

yeah, but something tells me that the intersection of a number of prime ideals will be equal to (product)Z

[chilly ocean]

that only makes sense when there are only finitely many

[chilly ocean]

What integers have infinitely many (prime) divisors?

[crystal vale]

If the ring has a prime characteristic then not necessarily the domain ring.

Let Z/2Z such that addition under modulo 2 and ab = 0 for all a,b in Z/2Z.

Right?

[rocky cloak]

Are you asking if any ring of prime characteristic is a domain?

Then that is indeed false, and you've given a non-unital example. You can also find unital examples.

[crystal vale]

Sorry but not necessarily unity in my ring

[crystal vale]

{0,2,4} under modulo 6, it is a unity ring with characteristic 3, right?

Wait

[rocky cloak]

Yes, though it's better known as Z/3

[crystal vale]

But I missed it not be a domain

Let me find

Z/2Z × Z/2Z, right?

[chilly ocean]

Yes, that works

[crystal vale]

Thank you

If the ring is a domain, non - commutative ring with unity then characteristic is prime, is there any need R to be commutative?

[chilly ocean]

There are noncommutative domains with unity whose characteristic is 0 like the quaternions

But I don't understand what you are saying

[crystal vale]

There is a result that if R is an integral domain then R has characteristic 0 or prime

And they define integral domain as commutative ring with unity

But it seems we don't need commutativeity

[chilly ocean]

Do you know how to prove it?

[crystal vale]

Yes

[coral spindle]

Indeed

[chilly ocean]

The proof should be the same, yes

[crystal vale]

Yes

[chilly ocean]

I think we don't need unity either

[crystal vale]

I used that but maybe we don't need

[chilly ocean]

I have a question too. If R is a ring with unity of nonzero characteristic n. Then the cardinality of R is a multiple of n, right? Because we can do an equivalence relation x~y iff x=y+k for some k=1,...,n. And each equivalence class has n elements. Is this still true for rings without unity? (define characteristic as the smallest n such that nx=0 for all x)

[crystal vale]

But not necessarily R is finite

Take a polynomial ring over Z/2Z

[chilly ocean]

Doesn't matter, you can still define multiples for infinite cardinals. But if that confuses you you can imagine I added the finite condition to my question

[coral spindle]

Then {0, 1, ..., n-1} is a subgroup of order n

Additive, ofc

I don't know what a characteristic of a nonunital ring means exactly

[chilly ocean]

Infimum of all n's such that nx=0 wrt divisibility order on N (N includes 0)

[coral spindle]

If the infimum is achieved by some x then the same argument works

But if it is not, I have no idea.

[chilly ocean]

Yeah and if it's nonzero then it must be achieved by some x because this poset is uh locally finite or whatever its called when you remove 0

Because given a nonzero n, there are only finitely many things smaller than it

[crystal vale]

If the ring has zero divisor can I say there exists an ideal which is not {0} and ≠ R ?

[coral spindle]

As ever, I will ignore nonunital rings.

Yes

[chilly ocean]

Otherwise {0,2} mod 4 is a counterexample

[crystal vale]

Any hint to show that?

[coral spindle]

Look at the ideal generated by the zero divisor

[crystal vale]

Yes but there is a possibility that the zero divisor has one sided inverse

Or you mean ideal generated by zero-divisors

[chilly ocean]

I think Boytije was thinking about commutative rings

[coral spindle]

there is a possibility that the zero divisor has one sided inverse
Is there?

Think this through.

[chilly ocean]

There is if the ring isn't commutative

[coral spindle]

Really?

[crystal vale]

Yes

[chilly ocean]

In fact rings of matrices over a field are simple

Despite having zerodivisors (for n>=2)

[coral spindle]

Right, right, you need some sidedness on the zero divisor's complement. I see.

[crystal vale]

Take (a_1,...,a_n,..) maps to (0,a_1,...,a_n,..) it has left inverse and also left zero-divisor

[chilly ocean]

You can if R is commutative (and unital). If it isn't, you can't

[crystal vale]

There is a question that if R is simple then it's characteristic is 0 or prime

[chilly ocean]

Does R have unit?

[crystal vale]

Yes

[chilly ocean]

There should be an obvious choice of an ideal then

Well, more than one

[crystal vale]

Hint?

[chilly ocean]

Suppose the characteristic of R isn't 0 or prime (or 1)

[coral spindle]

There's already a hint given

[chilly ocean]

What must the characteristic be like then?

[crystal vale]

Composite

And then R has non-zero zero divisor

[chilly ocean]

Yeah but it's more than just a non-zero zero divisor

Which non-zero zero divisor are you thinking of?

[crystal vale]

If n = ab, then I take a as zero divisor

[chilly ocean]

Good

But a is more than just a zero-divisor

Can you see why (a) is not R?

[crystal vale]

Wait

No

[chilly ocean]

What would need to happen for (a) to equal R?

[crystal vale]

a has right inverse or left inverse

So a has right inverse

[chilly ocean]

Does it?

[crystal vale]

We are not sure

[chilly ocean]

A priori it's more complicated than that, but thankfully a commutes with every element of the ring

[crystal vale]

I see

[chilly ocean]

Suppose ax=1. Can you reach a contradiction from here?

[crystal vale]

a commute with every element

Yes

[chilly ocean]

How?

[crystal vale]

a commutes with every element so xa=1 implies b = 0

[chilly ocean]

Yeah

[crystal vale]

Thank you ❤️

[chilly ocean]

Alternatively, a commutes with every element so if it had a right inverse it would also have a left inverse

[crystal vale]

Yes

[chilly ocean]

But zero-divisors cannot be invertible

[crystal vale]

Yes

I don't know why I didn't see a commutes with every element 😭

[chilly ocean]

It's okay, I also only noticed after you said this because I noticed this wouldn't be enough to prove what we want

Now I wonder if there can be a noncommutative nonunital simple ring of composite characteristic

[crystal vale]

To show if I and J are relatively prime ideals then R/(IJ) is isomorphic to R/I × R/J by x + IJ -> (x+I, x+J) it is surjective by chineese remainder theorem and if x-y in I intersection J then x-y in IJ, so it is injective.

Right?

[dull ginkgo]

Times Miz thinks simple rings means no nontrivial proper sided ideals instead of two sided and confuses the fuck out of himself: 4

[rocky cloak]

Yes, consider the additive group of R and use the classification of finite abelian groups

[chilly ocean]

Overkill as we can still prove it without choice (there can be infinite cardinals which are not multiples of n without choice)

[west sinew]

can I ask a problem from an assignment or is that frowned upon

okay cool. So I've been asked to give an example of a simple module wherein two (nonzero) elements have different annihilators.

to begin with the only easily constructed simple modules which come to mind are Z_p, but those won't work here because that's just a field

but uhh yeah if anyone has an idea of the approach here i'd appreciate a nudge in the right direction

[rocky cloak]

One thing you might notice (or try to prove yourself) is that if the ring is commutative this never happens

Another thing that could be useful is that any simple module is isomorphic to R/m where m is a maximal left ideal

[grizzled spindle]

Is there a way to find what this is without doing it by hand (here D8 is generated by (13) and (1234))

[delicate orchid]

if each of those are actually different copies of D_8, which I'm 90% sure they are, then this is the largest normal 2-subgroup of S_4

which is <(12)(34), (13)(24)>

[grizzled spindle]

That does make sense

Kinda not sure

[delicate orchid]

the intersection of all Sylow p-subgroups of a group is the largest normal p-subgroup, if you haven't seen this before it's a good, fairly easy, exercise

[grizzled spindle]

I haven’t seen sylow subgroups yet

I got it explained here tho

[grizzled spindle]

But yeah these are the Sylow 2-subgroups of S4 as u probably guessed

It’s obviously normal because it’s the kernel of an homomorphism

[delicate orchid]

they're also the Sylow 2-subgroups of A6

[grizzled spindle]

That’s cool

[chilly radish]

Le Sylow theorem 2 has arrived

[warm ember]

||if thats H then H<=P for each sylow p-subgroup P. So conjugation fixes H because conjugation permutes the sylow p-subgroup. If H is a normal p-subgroup then H<=P for some sylow p-subgroup P. conjugating this we get H is subgroup of all sylow p-subgroups||

kewl

[icy totem]

A ring, I, J ideals. Let v(I) denote the set of prime ideals that contain I. Find an ideal M of A such that v(I)Uv(J)=v(M).

I found M=IJ, but the solution tells me M=IΛJ. Are both valid?

[topaz solar]

If you contain I, or you contain J, you contain the intersection

[icy totem]

Yea but I also contains IJ, cause IJ is contained in both I and J no?

[topaz solar]

Yep, so then the question is if there can be a prime p which can contain IJ but not I n J?

[icy totem]

My answer is no, cause if P doesn't contain a j€J then still it contains ij for every i€I, and being prime it contains every i€I.

[topaz solar]

So once you contain IJ you contain one of I or J, hence I n J, and IJ < I n J, so….

It doesn’t seem like it matters right?

Products of ideals are nice, however, so IJ would be the suggested answer

[icy totem]

Ok thanks

[void cosmos]

is the set {x^3,y^3,xy^2,x^2y}?

[topaz solar]

Pretty sure it should be

[void cosmos]

yeah thank you!

is S_4 generated by (1234) and (1243)

[delicate orchid]

yup

[void cosmos]

how the fuck

i tried to write (12) in this set but i couldnt

as like generated by this set

can u do that

[void cosmos]

yeah ig i believe u bro but like

how to prove it

[delicate orchid]

just did

called "computer assisted proof" it's all the rage

[void cosmos]

bro im doing this qualifying exam

they dont even allow books in this shit

XD

i can't just bring a computer (or can i)

practicing

[delicate orchid]

idk there's at most 50 different combinations you can try so I'd suggest trying all of them