groups-rings-fields

[next obsidian]

I totally forgot oof

But if you’ve seen the concept of an associated prime

An associated prime of A/I is called a prime divisor of I

And you can see that this is exactly equal to (I:a) for some a not in I

There’s also things about the products of these ideals and stuff which become useful, I kind of view them as a like, compact way to write stuff which also makes them computationally efficient

If you also extend away from just ideals and define it for modules, if (A,m) is a local ring the socle of M is the module (0:m)_M

Which is the elements of M which are annihilated by m

[chilly ocean]

its called socle?

how do you pronounce that

[next obsidian]

This matters for eg defining Gorenstein rings

So-cull

Like I’m

Umm

If you know the word “focal”

Just replace the f sound with an s

[chilly ocean]

oh okay

[next obsidian]

Overall though, they’re kind of niche

[chilly ocean]

the anihilator of a module is called a socle

[next obsidian]

Not quite

[chilly ocean]

that looks like anhillator though

[south patrol]

That sounds so cool.

[next obsidian]

It’s specifically the stuff killed by the maximal ideal

Altho if everything else is invertible…

It’s supposed to be different

[chilly ocean]

oh lol

i didnt notice little m is maximal ideal

also slightly off topic

but do you know any resources for drawing / representing modules/rings in general

[next obsidian]

Oh yeah, that’s right. the socle is inside of M

It’s a submodule

[chilly ocean]

or is it one of those things i pick up along the way

[next obsidian]

The annihilator is an ideal

[chilly ocean]

ok

an example ill give

[next obsidian]

Undergraduate commutative algebra has some pictures, I think the first page has a visualization of a module as a bundle over the ring

But for me

I don’t really visualize the stuff

[chilly ocean]

yeah thats a cool conceptualization of it

i realized that in diff top

[next obsidian]

Rings have become geometric to me, but I don’t really run off of visual intuitions

So I really don’t, no

[chilly ocean]

im wishing i had visualization but i probably should ask for it in this case

in my alg nt class we are talking about p-adics alot

and my intuition is very bad for these things

we kinda quickly ran through the construction of Q_p

and then he mentions the closure of Q_p and then the completion

[next obsidian]

Yeah idk, I think you can try to visualize the p-adics specifically

[chilly ocean]

i have no clue what these objects look like to start with

[next obsidian]

But I don’t run spatial intuition for algebraic things

I have never tried to visualize them and don’t care to and don’t know how I would

[chilly ocean]

im really shaky foundationally with this stuff

my prof mentioned these things called character groups and he kinda wanted us to visualize what they are

but i felt burnout by definitions

ig this goes back to having to struggle with understanding to finally get something after rereading

[next obsidian]

Yeah, idk, I am content with viewing these as algebraic objects and just like, sets of stuff with operations

[chilly ocean]

I have to prove A-algebra A[x] is flat. Afaik this means two equal things. The first is that tensor products preserve exact sequences(phrased weirdly). The second is that f:B->C injective linear A-mod homomorphism implies f’:B tensor A[x] -> C tensor A[x] is injective.

This means im trying to show that f’(b tensor g)=f(b) tensor g 0 implies b tensor g is 0. for b in B, g in A[x].
But I thought that a tensor b = 0 implies a is 0 or b is 0. At same time I have the feeling that Im only believing this because I am not writing the tensor product as the proper formal sum that it is

Also made a mistake, everything here is an A-algebra

nvm

I just need to use two facts that A[x] is cnt inf direct sum of A -algebras

and that if A=(A_i) is flat as A-module iff every A_i is flat as A-module

i do want to show the direct way though

[hidden haven]

A[x] is not a direct sum of A-algebras

For that matter, direct sums of algebras are not a thing, unless you mean direct sum as A-modules

Which should work yeah

What you said about a tensor b = 0 is not true, for example, 2 tensor 2 in Z/4Z tensor Z/4Z

And else, not every element of a tensor product is of the form a tensor b

Just that elements of that form span the tensor product

One way to do this is to show that A[x] tensor B is isomorphic to B[x]

And this should be an isomorphism of functors

And taking polynomial rings does take injections to injections

Direct sum of modules should be easier though since the underlying module of the tensor product of algebras is the tensor product of underlying modules of the algebras

And injectivity only depends on the underlying module

[chilly ocean]

holy mackrel thank you

I should work on showing A[x] tensor B is isomorphic to B[x]

oh shit

there is a canonical iso from Hom(M tensor N,P) to Hom(M,Hom(N,P))

if I let P be A and N be A[x]

nah nvm idk where im getting at yet

at this point im curious what universal property is for polynomial rings

[hidden haven]

It is the free pointed ring on any given ring

ie there is a forgetful functor from the category of pointed rings (pairs (R, r) where r is an element of R, which we call the basepoint of this pointed ring) which has morphisms as the homomorphisms that map basepoint to basepoint to the category of rings

The polynomial ring functor is left adjoint to this forgetful if you view it as R maps to (R[x], x)

You can also view it as the free commutative R-algebra on the singleton

[chilly ocean]

atm im trying to think of example of why direct sums of algebras arent a thing . I would just assume that the multiplication defined over the original R-algebra works component wise

maybe you are saying the inf sum isnt well defined?

[hidden haven]

Yeah if you have infinitely many algebras, then the direct sum under pointwise multiplication won't have an identity

The identity in the case finite direct product is (1,...,1)

No such element in the infinite direct sum of the underlying modules

The coproduct of R-algebras is tensor product over R, rather than direct sum

[chilly ocean]

oh wow

that is fucking cool

when people talk about infinite direct sum do you need to take limits?

[hidden haven]

I'm not sure what you mean

[chilly ocean]

in category theory why arent infinite direct sums or products different objects with their own universal property? why do product and coproduct have to be compatible with infinite versions of their construction? im sorry but it is sort of hard for me to phrase.

[hidden haven]

They have their own universal properties

Product of a collection {A_i} is an object A with projections p_i: A → A_i such that whenever you have some B with f_i: B → A_i for all i, there is a unique f' such that p_i f' = f_i for all i

Coproduct is the dual of this

In R-mod, the product is the direct product and the coproduct is the direct sum

I'm not sure if this was your question

[chilly ocean]

Yea sort of. I guess I was wondering why there arent two definitions, one for finite collections and another for infinite, but i guess its for convenience for the most part to have one compatible definition

[lethal cipher]

Okay, so my prof is working on proving this in a lecture and I am a bit hesitant. Why are we allowed to assume that J is a principal ideal?

[hidden haven]

F[x] is a PID for any field F

It's a Euclidean domain with Euclidean function given by degree

[lethal cipher]

What does Euclidean domain/function mean?

[hidden haven]

It's a ring with a function that works like absolute value for ℤ or degree for polynomials (because it is degree for polynomials lol)

[delicate orchid]

gonna try and re-prove this, seems fun

[hidden haven]

The important property is that for any pair a, b, you can write a = bq + r

And the remainder r should be smaller than b

This smallness is given by the Euclidean function

ie euc function of r is smaller than euc function on b

[hidden haven]

Anyway if you need a hint for this

||take the least degree element of J||

[delicate orchid]

ha so I am on the right track

always nice to know

[hidden haven]

Wew I was waiting for you to come online and be devastated by all the universal property yeah that was happening

But you never came

[delicate orchid]

sorry moldi I was probably... busy...

[chilly radish]

Try proving the converse is true too!

If F[x] is a PID, then F is a field

Where we assume F is integral

[delicate orchid]

I feel like the converse is easier

[chilly radish]

I.e. commutative with identity and no zero divisors

It's actually quite harder imo

F[x] is no longer assumed to be euclidean

Only principal

[hidden haven]

Do we need this

[delicate orchid]

hmm good point

[chilly radish]

Well it's implicit fron F[x] being a Pid

By the canonical embedding

[hidden haven]

Right

Lul

[chilly radish]

I just wanted to point it out

Same thing with assuming commutative and has 1, all implicit gy F[x] being a PID

Basically all that you have to prove is that F has all inverses

There's 2 proofs of this I know. A quick and clever proof and a more hands-on proof

[hidden haven]

Or that it has no non trivial ideals

[chilly radish]

True

[delicate orchid]

this proof is way harder than I thought it was gonna be I'm ngl

[chilly radish]

Quick and clever: ||consider the evaluation map f -> f(0), use 1st iso and maximality of <x> (equivalent to primality) to conclude F is a field||

Want a hint?

[delicate orchid]

I know that I need to take the minimum degree polynomial, and then use the fact that every coefficient has an inverse to simplify it down into some nice form that collapses the entire ideal

[delicate orchid]

but what that exact simplification looks like I cannot visualise

oh!

[hidden haven]

Hint

[chilly radish]

Note that if F isn't a field degree isn't necessarily a euclidean function anymore

So you can't assume division with remainder holds

For the converse that is

Wait wew which part r u doing

[hidden haven]

You would know if you knew how to read shin

[delicate orchid]

the non converse

i.e. the proof you said was easier

[chilly radish]

Ah ok my b

[hidden haven]

I would not say either is easy

After all, there are no obvious universal properties here

[chilly radish]

I didn't say easy, I said easier

[delicate orchid]

consider the variety generated by the ideal (p_1, \cdots, p_n) this is clearly generated by something something direct sum of the varieties generated by the ideals (p_i)

[hidden haven]

Sir this is the shitposting channel why are you taking me seriously

[chilly radish]

Wew you're really close

[delicate orchid]

I feel like I'm missing just one puzzle piece

the problem is there might be other polynomials of the same degree with weird coefficients

[chilly radish]

Why are you assuming a priori all ideals are finitely generated

[delicate orchid]

oh I mean

[hidden haven]

Do they actually use the normal subgroup notation for ideals

[delicate orchid]

good point but I think this still holds

[hidden haven]

I've never seen that before

[chilly radish]

Yes moldi

[next obsidian]

Hurb

[hidden haven]

But that's pretty based

[next obsidian]

You can show F[x] is a Euclidean domain

[delicate orchid]

I do because I'll take any excuse to write "trianglelefteq"

[chilly radish]

Try assuming you have an element in the ideal that's not a multiple of the minimum degree poly and see what goes wrong

[delicate orchid]

yeah that's what I've been trying to do in my head for 5 mins

I'll keep thinking though

[chilly radish]

Also you will then see that any other poly of the same degree is an associate

In the ideal thantis

[delicate orchid]

ok lemme consider an example of what's throwing me off

I'm thinking of something like

(2x-1, x-2)

[chilly radish]

Over which field

Q?

[delicate orchid]

just R I guess

ah the difference is in the ideal

there we go

[chilly radish]

Yea, the key point here is that the explicit generators you take might not actually generate the ideal principally

[delicate orchid]

yeah so the ideal of polynomials contains the differences of all the polynomials as well, right ok

so if there's two "minimal degree" polynomials, we can multiply them by the inverse of the leading coefficient to get two minimum degree polynomials with leading coefficient 1

so the difference will be a lower degree

contradiction! 🚨⚠️⚠️⚠️⚠️

[chilly radish]

Well, a lower degree or 0

[delicate orchid]

yeah and if it's degree 0 it's just the whole ring

[chilly radish]

if it's 0 you're good, but you've shown they're associates

Nono

It just means those two polynomials differed by a unit

I meant the 0 polynomial

[delicate orchid]

oh ok yes

[chilly radish]

So either way, you get that if there's a unique minimal degree monic polynomial, or equivalently unique minimal degree polynomial up to units

Now you just need to show it generates the ideal

[delicate orchid]

you could iteratively generate the higher dimensional p_is by multiplying the minimal degree by some x^k for some k and some unit a_0, and then adding on the minimal degree times x^{k-1} and some unit a_1, and so on

it's got a kind of Fourier transformation flavour to it

[chilly radish]

You're being too constructive about this

I mean, that might work idk

I'm just saying there's an easier way

[delicate orchid]

well if we're able to form all the original (p_1, p_2, ...)s out of the minimal degree m then we get (m) = (p_1, p_2, ...)

seems like the most obvious way to me

[chilly radish]

Well yes, but then to prove the full claim you'll still have to prove every ideal is finitely generated

[delicate orchid]

yeah I've realised my method won't work even for finitely generated ideal

if m is the minimum degree you could have something like ax^{m+1}+bx^{m-1} and there might not be a way to construct this as you can't multiply by x^{-1} to get the lower coefficients to line up

[chilly radish]

That's also true. Well, obviously that couldn't happen if you prove the claim but a priori you don't know that

So, up until now you haven't really used the Euclidean properties of the degree

I'd refer back to moldi's hint

[delicate orchid]

probably because I have absolutely no experience with euclidean domains

[chilly radish]

The key point with euclidean domains is that there is division in remainder

[delicate orchid]

although thinking about it

[chilly radish]

Which works exactly as in the integers

[delicate orchid]

you can write the higher degree polynomials as a quotient and a remainder

and the remainder will be a nice low degree, lower than the minimal though?

cause if it's lower than the minimal then it has to be zero

[chilly radish]

NOW YOU'RE COOKING

Exactly, the division algorithm promises you a remainder that is either.0.or.lower than the divisor

[delicate orchid]

oh so it's guaranteed to be lower than the divisor ok

oh wait yeah obviously

that's the definition isn't it

[chilly radish]

Yes

[delicate orchid]

ah ok so the remainder is 0 so all the other polynomials must be a multiple of the minimal degree one

hence the ideals are the same

[chilly radish]

Yup

So the minimal degree poly generates the ideal

And you're done

[delicate orchid]

wow that was WAY harder than I expected

[chilly radish]

This also works if you're not considering finitely generated ideals, because the natural numbers are well-ordered

[chilly radish]

The good thing is this is basically how you prove most things about euclidean domains

Degree consideration and division with remainder

[delicate orchid]

I can very easily see how this generalises to the proof "Euclidean domain => PID"

[lethal cipher]

So we are currently constructing a field containing Q which has the root of x^2-2. However, my professor made an expansion mistake (a+bx)^2=/= a^2+2abx+x^2, and the rest of the proof kind of rides on that. I was wondering if someone can help me figure out how to do this correctly, because I am having a hard time seeing it.

[chilly radish]

That's basically what you did. Nothing special about polynomials really

Well, eventually

[chilly radish]

Can you show the rest of the proof

Are you just trying to find what polynomial satisfies p^2=2 in the quotient?

[lethal cipher]

Yes

He uses the fact that x^2 is congruent to 2 mod(x^2-2), which then means we are just trying to see when a^2+2abx is congeuent to 0 mod(x^2-2).

But of course, this won't work since properly expanding (a+bx) gives b^2x^2 at the end. So really we should be looking at when (a^2+2abx)=(2-2b^2)mod(x^2-2)

Which is different

[chilly radish]

Well but isn't that it then? the polynomial g(x)=x satisfies x^2=2

[lethal cipher]

-_-, yeah... that's it

[spice whale]

just started knapp

it's all integer divisors

[chilly radish]

Idk why your prof wanted to like, explicitly show that like that. Generally, when you quotient by an irreducible polynomial, x will become a root of it

well mod the ideal

Related question, if I quotient by an irreducible poly that has multiple roots in some extension, I know that this is isomorphic to a simple extension by one of the roots. A priori, x doesn't 'become' any specific root right? Like, all the extensions are isomorphic to this quotient, so I can just choose which root I wanna treat x as.
In that sense, it's still an indeterminate

[neat pollen]

is this true: Let I be an ideal in F[x]. If f(x) is an irreducible polynomial in I, then I = <f(x)> ?

F[x] is a pid so it is generated by some polynomial, say g(x). Then f(x) = a(x)g(x) for some a(x) in F[x]

so we must have a(x) in F right

[chilly radish]

Well, there are 2 options here

Either a is a unit, in which case f and g are associates so <f>=I, or g is a unit, in which case I=F[x]

If you require that I be a proper ideal then what you said is 100% correct though

[neat pollen]

since a in F i think it must be a unit

because its a field

but if we had some ring then it wouldnt always be a unit

[chilly radish]

f being irreducible just tells you either a or g are units

Which in F[x] is the same as being in F

But it could be that I=F[x], in which case f can't generate it as irreducibles are non-units, so we must have that g is a unit

[neat pollen]

so either g is a unit or not a unit. If g is a unit then I = F[x]. Otherwise f(x) = ag(x) for some nonzero a in F which means that I = <f(x)>

[chilly radish]

Exactly

[neat pollen]

thanks!

[chilly radish]

Np!

[chilly radish]

Bump. Am I right here? It feels weird because in the actual simple extensions there is an actual element that satisfies some identity that says it's a specific root, but in the quotient there isn't necessarily? I don't think

Basically you can't canonically embed the quotient into the larger field from whence the roots came is what i'm asking

[waxen hedge]

Let's say you have an irreducible polynomial $P\in\mathbb{Q}[X]$.
You have $\mathrm{deg}(P)$ embeddings from $\mathbb{Q}[X]/P$ to $\mathbb{C}$, one for each root of $P$ in $\mathbb{C}$.
So yes, there isn't a canonical embedding from the quotient to $\mathbb{C}$.
Actually, the point of Galois theory is precisely to describe this ambiguity.

[cloud walrus]

Adrien

[chilly radish]

I see. Yea that's what I figured after a bit of thought. Still weird to think about. Taking Galois theory next sem so I guess i'll see that

Thanks!

[waxen hedge]

Yes these are quite subtle questions

[chilly radish]

Earlier in this course I proved that the only automorphism of Q[sqrt 2] is the one sending sqrt 2 to - sqrt 2. I now see why that's the case

Besides the identity*

[waxen hedge]

I would say this is not true but I don't know how to prove it, I need to think about it

[chilly radish]

I think it's not true i.e. in the example Q[x]/x^2-2

x is distinctly sqrt 2

Well, x and -x

[waxen hedge]

Yeah, you can't distinguish sqrt(2) and -sqrt(2)

[chilly radish]

Generally if you quotient by an even degree polynomial you're gonna get conjugate roots

But canonically x isn't + or - any of them

Right

All of them isn't necessarily true

[waxen hedge]

I think the good viewpoint of this is considering embeddings

Because Q[X]/P doesn't depend of anything

On the other hand, if you wanna talk about "the roots of P", you need to choose a field in which P has roots

ie an embedding

[chilly radish]

Yes of course

I agree

[waxen hedge]

I'm not sure about that

Well, depends what you call conjugate roots

Do you know about normal extensions ?

[chilly radish]

Not yet

[waxen hedge]

Ok so, let's stay in the case of Q (or any perfect field if you wish)
Let's pick K an extension of Q
We say K is a normal extension of Q if every embedding of K into C has exactly the same image

It's equivalent to ask that for every polynomial P in K[X], if P has a root then P splits in K

[chilly radish]

I see. Yes that makes sense

[waxen hedge]

(since we are working over Q, normal extension are exactly the Galois extension)

So these are extensions which behave well

In K is a finite extension of Q, then it happens that there exists a polynomial P in Q[X] such that K is the smallest field in C containing every roots of P in C

Basically, take a polynomial, add every roots of the polynomial and you get a Galois extension

(you probably need to assume that the polynomial is irreducible in Q[X])

[chilly radish]

So basically every number field is the splitting field of some polynomial?

Or of some irreducible polynomial at least

[next obsidian]

It isn’t a splitting field I think

[delicate bloom]

Q(cbrt(2)) is a number field that's not the splitting field of x^3-2

[next obsidian]

A splitting field would always be Galois

[chilly radish]

So is this not true? Or am I misinterpreting

Oh sorry

I think adrien waa talking about normal extensions specifically

I misinterpreted

[waxen hedge]

Finite Galois extensions (over a perfect field) are exactly the splitting fields yes

And the Galois group tells you how much you can swap roots without messing things up

[chilly radish]

I see

Galois theory seems really cool

[waxen hedge]

Yes
It's quite complex tbh

It's also a geometric theory

But that's for later

It's related to covers/fundamental group

[chilly radish]

I've heard of galois correspondence

Really cool stuff

[waxen hedge]

In the context of Riemann surfaces, the analogy between the Galois group and covers become concrete

But I don't know the details

[chilly radish]

John explained it to me once a while ago

I can probably find the convo if you want

[waxen hedge]

Don't bother, I have a book on it

I just need to read it 💩

[chilly radish]

Lol ok

[upbeat swift]

Two quick questions for clarity: how would you describe the cyclic subgroup <1/2> in R* and the cyclic groups <1/2> in R

[next obsidian]

Multiples versus powers (including negative powers) of 1/2

[delicate orchid]

<1/2> = {1/2^n such that n is an integer} for R* and <1/2> = {n/2 such that n is an integer} respectively

[upbeat swift]

Ah okay, thank you

[next obsidian]

That’s what I said wtf!

[delicate orchid]

yes but I said it better

[next obsidian]

You said it worse

[delicate orchid]

nuh uh

[upbeat swift]

[next obsidian]

You used set builder notation

I used English

[delicate orchid]

explicit constructions are always "swagger"

[next obsidian]

You have no swag

[delicate orchid]

yeah cause I put it all into this statement

[next obsidian]

Get kobenaenae’d

[delicate orchid]

[next obsidian]

Bruh

Okay so it’s a choice of 3 elements in Z

Wait

Fuck you

Lmfao

[chilly ocean]

I got another banger

[delicate orchid]

wait a minuite is that fermats last fucking theorem

[next obsidian]

Yeah

Lmao

[delicate orchid]

LMFAOOOOO

that is a good one

[next obsidian]

I mean there are maps

They just are trivial in a sense

[delicate orchid]

trivial homomorphism

[next obsidian]

They won’t be actually trivial, but it’s solutions to that equation which we know

[chilly ocean]

For when is x in (2)[(p)+(q)] x in Z

[delicate orchid]

goldbach

I think

[chilly ocean]

What about x in Z_p @-@

[delicate orchid]

[chilly ocean]

Are there any resources that teach basic number theory but from algebraic perspective?

Or like some intro to algebraic number theory books?

Nothing too hard though

[upbeat swift]

Is P_2 abelian?

[tribal moss]

What is P_2?

[upbeat swift]

A group

A group of subsets of two element sets

[tribal moss]

Under which operation?

[dusty sapphire]

symmetric difference?

[tribal moss]

That sounds like a good guess.

[dusty sapphire]

yep btw order of P_2 will be 4, hence every group of order 4 is abelian. If P_2 is group then it will abelian.

[upbeat swift]

I'm not sure

I'm using Cayley's theorem

[tribal moss]

BL is right that all groups of order 4 are abelian, so in that sense it settles your question. But you really shouldn't be asking the question without first being sure what the group you're asking about is.

[upbeat swift]

Well I know it could be (identity, a, b, (a,b))

[dusty sapphire]

still operation is not clear.

[upbeat swift]

That's why I'm asking cuz I'm unsure of the operation

Whether it's addition or multiplication

I can show you guys the whole thing, I may be reading this entire thing wrong

[proud bear]

what does chapter 3 exercise C say

[upbeat swift]

Let me show that too

Oh!! It's addition

That's the operation

[tribal moss]

How does addition of subsets of a size-2 set work???

[tribal moss]

Okay, symmetric difference. It's rather uncommon to notate that with a +, but let's roll with it.

Symmetric difference is definitely a commutative operation (which should be clear from its definition).

Therefore every group where the group operation is symmetric difference is abelian.

[upbeat swift]

Oh alright

Thank you

[woven delta]

Boolean rings

You are asking about the additive group of the prototypical boolean ring

[tribal moss]

Okay yeah, so it's not unprecedented to write it with a +.

[chilly ocean]

I was working on this last week and did the whole pset except for 1.6(b), tried writing some stuff out like setting cos(nt)sin(nt) = a sin(nt) + b cos(nt) + c to see if we can get closure but I was not sure how to proceed

from Artin btw

[tribal moss]

The definition is written a bit sloppily, but I think the intention is that S can contain integer combinations of the trig functions with different n, so the base set is really {1, sin t, cos t, sin 2t, cos 2t, sin 3t, cos 3t, ...}.

[chilly ocean]

That weakens the condition, but I still have no clue where to go from there lol

[tribal moss]

Have you looked up a set of trigonometric product formulas?

[chilly ocean]

I mean they usually involve higher powers or products of sin/cos right

[tribal moss]

Just google and see.

[chilly ocean]

ahh I see the ones you mean

[tribal moss]

There are some pesky factors of 1/2 in the relevant formulas, which mean that the set of integer combinations is not closed under multiplication. However making a completely foolproof argument out of that takes a bit more footwork than the problem setter might have realized ...

[chilly ocean]

yeah cause it looks pretty unintuitive right now at least 😢

oop cropped out = sign but top row and bottom row should be equal by assumption

kinda looks like some fourier analysis shit but idk much about that lol

I might be able to get some intuition for how to show that such an equality is impossible by actually trying the case we kinda ruled out cause it probably was worded weirdly, and seeing how to argue that too it is impossible

[tribal moss]

I think the easiest is to pick a single example and argue that for example sin²(t) is not in the set.

Which would be enough to establish non-closure.

That could be based on the convenient property that $\int_0^{2\pi} f(t),dt = 0$ for all of the base functions \emph{other than} $1$.

[chilly ocean]

wait what is the relevance of sin²(t)

[cloud walrus]

Troposphere

[tribal moss]

sin²(t) is the product of sin(t) with sin(t).

[chilly ocean]

wait I'm buggin

lmao

if you think of me as a bumbling idiot then you have the perfect image lol 👍

they really handin out math degrees to just anyone these days

[sturdy marsh]

[chilly ocean]

boutta adjoin myself to < , rope>

[sturdy marsh]

[chilly ocean]

I just joking 😉

[sturdy marsh]

:petthecat:

pretty neat problem

[chilly ocean]

tbh my head still whizzing from it

[sturdy marsh]

did you figure it out

[chilly ocean]

no lul

[sturdy marsh]

use troposphere's trick

what can the integral of some Z-linear combo of those functions over 0 to 2pi be

[chilly ocean]

only the term that was originally constant

[sturdy marsh]

right, so the integral is always an integer times 2pi

what is \int_0^2pi sin^2(x) dx

[chilly ocean]

^

[sturdy marsh]

yup

[chilly ocean]

nice

would never have thought to introduce calculus into this problem lmao

[sturdy marsh]

lol

[chilly ocean]

that's a good angle

[sturdy marsh]

yup it's a cool observation

[chilly ocean]

"what's the relevance of sin^2(x)" in this context will go on my list of idiotic things I've said that replay in my brain before I sleep, along with "Italy is in northern Europe" 💀

[sturdy marsh]

[chilly ocean]

take my degree away someone pls

[sturdy marsh]

weil asked if there can be infinitely many finite groups of fixed order @chilly ocean

[chilly ocean]

yeah I'm just being irrational, I shouldn't speak that poorly of myself
thanks for being a homie ❤️

hard to stop thinking like that sometimes though

it's not all so serious and we can laugh at silly mistakes and know they are of no consequence to our real worth as people

[sturdy marsh]

as long as it's in jest, that's okay

[chilly ocean]

moments of weakness make it tough doe

nah it wasn't but you cracked me out of that thinking pattern <3 ty

[sturdy marsh]

:petthecat:

[chilly ocean]

Funny question

The question sounds like a contradiction though/malformed question

[chilly ocean]

So is S for part b supposed to be S:={sum_i z_ix_i where z_i is an integer and x_i is 1,sin it, cos it }?

[sturdy marsh]

wdym

the answer is no

[chilly ocean]

Well yea

But the question if posed as if there is an answer other than no

It isnt a bad question to ask though, makes you think about fundementals

speaking of

[sturdy marsh]

I mean it's pretty easy to realize that there can only be finitely many as there can only be finitely many multiplication tables

[chilly ocean]

what are first examples of non concrete categories

[sturdy marsh]

you could come up with a stupid bound

there are at most |G| choices for each entry in the multiplication table

[chilly ocean]

the way i interpret it is how many possible functions from f:A x A -> A, which corresponds to tables in general

when you restrict with group structure then it becomes less

[sturdy marsh]

sure, that's why it's a stupid bound

[chilly ocean]

but i mean if we are being pedantic there are infinite

[sturdy marsh]

but it still tells you that there can be only finitely many

[chilly ocean]

up to isomorphism there are finite

[sturdy marsh]

uhhh

sure

but that is definitely not what weil meant lmao

[chilly ocean]

lol

i wish there was initiative to make math words shorter

ive always hated morphism ending

[sturdy marsh]

schemes

[chilly ocean]

actually?

i have no clue of general definition

i only know maybe what affine schemes are

or only a specific type

[sturdy marsh]

an easier example is the homotopy cat of topological spaces

but proving that a category cannot be concretized is probably very hard lol

[chilly ocean]

yeah

im assuming that

i dont really know where i would start either

its probably super logic based

[sturdy marsh]

you need to show that there cannot be a faithful functor U: C ----> Set

[chilly ocean]

yeah

idk the objects of homotopy category either, i remember thinking its just homotopies and topological spaces but then my prof said its more complicated

[sturdy marsh]

it's more or less that

[chilly ocean]

because idk where to fit in the continuous maps either

[sturdy marsh]

not homotopies of top spaces

but homotopy classes of maps between top spaces

some people call this the naive homotopy cat

[barren sierra]

I am struggling to show the second condition for the subgroup closure

I showed the identity was in the group, that was chill

and then after that I have absolutely no idea how the hint helps

[hot lake]

I think the hint is for determining the size of the subgroup

[barren sierra]

ok yea I got this

[barren sierra]

I mean the p-sylow subgroup of Sym(p^2) would be p^(p +1) right?

[hot lake]

Yeah

[barren sierra]

so I think the hint is more about describing the operations that such phi in the group do

But then I'm lost about that lol

[hot lake]

You have to show that H has size p^(p+1)

[barren sierra]

yea

Hm my first thought is some sort of counting argument but we'll see

[fossil shuttle]

nice

[coral shale]

https://i.imgur.com/USMSBR6.png

I've seen proofs that use properties of field extensions (show $K(a,b)$ finite). My idea, though:\

Let $s$ be the field extension map. Take $f, g \in K[x]$ with $(sf)(a) = (sg)(b) = 0$.\

Claim: $(s(gf))(a+b) = ((sg)(sf))(a+b) = 0$.\

We have

$$(sf)(a + b) \equiv 0\pmod b$$
$$(sg)(kb) = s(k)(sg)(b) = 0$$\

Hence $(s(gf))(a+b) = ((sg)(sf))(a+b) = 0$.

[cloud walrus]

Shuri2060

[coral shale]

I feel this idea works, although the usage of 'mod' is a bit sketchy? 🤔

[chilly ocean]

why do we have 0 mod b

and like what's s?

[coral shale]

If I write down the general form of the polynomials, it's clear (sf)(a+b) will turn into a polynomial in b (linear combination of powers of b). And then by linearity (sgf)(a+b) will be 0

f and g are in K[x]. s maps from K to L (and induces a map from K[x] to L[x]). So technically I'm looking at f and g in L[x] which are s(f) and s(g)

[coral shale]

I wanted to go with this argument but wondering if there's a more concise/elegant way to put it

I'm not 100% it works ig

[coral shale]

Ohhhh wait I see, I guess not

No wait yh, it should be 😵

When you expand powers of (a+b) using binomial expansion

You collect all the powers of a (and the constant) together

[chilly ocean]

Yeah

[coral shale]

That will make 0. Everything that's left is a multiple of b

[sharp sonnet]

what about the constant term of f

[coral shale]

That gets collected with the powers of a

[chilly ocean]

yeah looks suspicious imo

[coral shale]

I'll write it out in full

Let $f(x) = \sum^m_{i=0}k_ix^i$.\

Then $f(a+b) = \sum^m_{i=0}k_i(a+b)^i$.

$f(a) = \sum^m_{i=0}k_ia^i = 0$.

Then $f(a+b) = f(a+b) - f(a) = \sum^m_{i=0}k_i((a+b)^i - a^i)$.

[sharp sonnet]

how about you try an example

[cloud walrus]

Shuri2060

[sharp sonnet]

(x^2 - 2)(x^2 -3) for x = sqrt(2) + sqrt(3)

then your claim is ?

this is zero?

i mean the overall claim

[coral shale]

uh yh

wait no

[chilly ocean]

[sharp sonnet]

yeah, that is not true

[coral shale]

I take 2 polynomials which make rt2 and rt3 0

And compose them

not multiply

[sharp sonnet]

so ((x^2 -3)^2 - 2) for x = sqrt(2) + sqrt(3) ?

[coral shale]

,w ((sqrt(2) + sqrt(3))^2 - 3)^2 - 2

not 0 🤔

[sharp sonnet]

indeed

[chilly ocean]

[sharp sonnet]

the construction is more complicated than what you are trying

[coral shale]

what goes wrong then 🤔

This shows f(a + b) is a multiple of b

but then g(kb) =/= kg(b) is my problem I think

ah.

[sharp sonnet]

there is a constant term for one

[coral shale]

No there isn't

k_0(1 - 1) for i = 0

[chilly ocean]

there are a^kb^n-k terms

[foggy merlin]

for showing that the sum and product are algebraic, trying to actually construct the polynomials for them is absurdely tiresome, like don't even try for your own sake

[coral shale]

yh ok, didn't realise

[foggy merlin]

you're better off using a different argument

there is a way to construct them actually, but for the love of god, don't do it to yourself

[coral shale]

Thanks everyone 😅

[foggy merlin]

do you want a hint for the actual argument?

[coral shale]

I've seen it

(unfortunately)

[foggy merlin]

oh ok

well do you want to know how you would construct it?

I'm not giving a proof of why it works tho

[coral shale]

the polynomial?

[foggy merlin]

yeah

[coral shale]

I'd be interested 😄

[foggy merlin]

I'll give it for the rationals here, but this should be applicable to any field unless I misunderstood sth really badly

[coral shale]

Thank you

[foggy merlin]

it uses some funky properties of the tensor product of linear maps

[sharp sonnet]

the funky property is bilinearity

[coral shale]

The circled x is tensor product or?

[sharp sonnet]

ye

the general idea isnt that bad

you want to construct a matrix with eigenvalue alpha + beta

and use the connection with the minimal/characteristic polynomial

[valid basin]

the mentioned corollary states that if $a$ is in a group $G$, then $a^{|G|}=e$

[cloud walrus]

loozoo

[valid basin]

i don't understand how $x=x^{sm}x^{tp^n}$ implies $x^{sm}\in H$ and $x^{tp^n}\in K$

[cloud walrus]

loozoo

[valid basin]

to me the argument seems to hinge on the fact that either $x\in H$ or $x\in K$

[cloud walrus]

loozoo

[valid basin]

since we can show that $x^{sm}\notin K$ by definition of $K$, and similarly so for $x^{tp^n}$ that $x^{tp^n}\notin H$

[cloud walrus]

loozoo

[coral shale]

By the corollary, for any y in G

[cloud walrus]

Shuri2060

Shuri2060

Shuri2060

[coral shale]

Similarly for the other case.

@valid basin ^

[valid basin]

ohhh right

[coral shale]

urgh typo

[valid basin]

that clears things up

[coral shale]

missing s, but yes.

[valid basin]

cheers, thanks so much

that's all good

[coral shale]

If I in J in R and I, J are ideals of R. Then JI is an ideal of R/I.

Then I believe you have something like this? One of the isomorphism thms?

Revising/rediscovering stuff from a while back

[rustic crown]

do you mean J/I? because JI is an ideal of R, which is very different from R/I

[coral shale]

I think I meant what I wrote (which may be wrong)

[rustic crown]

what you're doing kinda seems like the third iso theorem

[coral shale]

'JI is an ideal of R'

What is this in set notation?

{ij : i in I, j in J} ?

[rustic crown]

the ideal generated by that

[coral shale]

ok... uh what I meant by JI is

[rustic crown]

so the set of
{i_1j_1 + ... + i_nj_n | i_r in I, j_r in J}

[coral shale]

Ok, backtracking

I meant JI

But what I meant by that was

{jI : j in J} which is an ideal of R/I

[rustic crown]

Oh

that's denoted J/I

wait no

you probably mean {j + I : j in J} right?

[coral shale]

no, I meant jI

D:

Maybe I do mean that (probably remembered the quotient ring wrong, lemme check)

Ah yes.

I've written everywhere R/I = {rI : r in R}

Undoing this mistake, I will probably end up with {j + I} my ideal...

Thanks for clearing that up at least, brb 😄

So the idea of J/I := {j + I : j in J} is well defined?

J is an ideal, not a ring. But we take the quotient in the same way

And uh... I guess J is like a 'subideal' of I?

[rustic crown]

yea, but people would prefer the word "R-submodule"

[coral shale]

How do modules (I remember them as vector spaces, but for rings) and ideals relate?

[rustic crown]

Right... so ideals of a ring R are precisely the R-submodules of the R-module R

submodule is an abelian subgroup and it should be closed under the 'scalar' multiplication

[coral shale]

submodule vs module
subspace vs vector space

same analogy?

Ok, ima have a big think. Thanks

[rustic crown]

and recall that ideals are also subgroups which "absorb" the multiplication by 'scalars' from R

the category of Rings isn't the best one out there, but the category of R-modules is much well behaved

[coral shale]

Is the 'ideal quotient' (I : J) := {r in R : rJ sub I} for any I, J ideals of R related to any of this

Seems not, but asking in case.

[spice whale]

quotient of a ring by an ideal is effectively setting all the members of that ideal equal to 0

right?

[prisma shuttle]

yes basically

this is the reason that any ideal must always contain 0

and also the reason why saying an ideal is closed under subtraction implies that it must also be closed under addition (because it must contain additive inverses)

however note that if being closed under addition is not sufficient, because $0 + (-a) = -a$

[cloud walrus]

JustKeepRunning

[rustic crown]

i haven't seen much of commutative alg, to be able to use this definition beyond just the immediate consequences, so ig it's not related to this general stuff unless you're talking a lot about ideal products.

did honorable yellow get darker?

or is it my night light

[gritty sparrow]

Dehydrated honorable

[coral shale]

#changelog changes 👀

[hidden haven]

I disagree

[coral shale]

Me tryna prove 3rd isomorphism thm for myself >.<

There's this set in a set in a set in a set

[hidden haven]

Is third the one that says (G/H)/(N/H) = G/N?

[coral shale]

yh

doing it for rings

[hidden haven]

[coral shale]

cus idk

[rustic crown]

Let G be a ring and H < N be ideals >.<

[hidden haven]

Keep the arrows as they are and apply yoneda 🙈

[coral shale]

im tryna make a commutative diagram for 3rd iso

(after im done)

[delicate orchid]

The commutative diagram for the 3rd iso, or as I like to call it: The commutative diagram for the 1st iso

[rustic crown]

or as i like to call: universal property of quotients

[hidden haven]

Weird, I just call it Discussion's Greatest Moment

[rustic crown]

why add extra letters >.<

the I/J is def wrong

[coral shale]

J/I

gg

[rustic crown]

wait, did you just do that?

(R/I)/(J/I) = (R/I)(I/J)

[coral shale]

I mean this
It's right, right?
THIS

[rustic crown]

yea looks right...
just add a / between R/I and J/I

[coral shale]

aaaaaaaa

yes ty

Lines from R/I -> R/J or S -> R/J uhhh

There should be a natural homomorphism right?

[delicate orchid]

Oh is this the correspondence theorem

[cloud walrus]

det

[rustic crown]

well you can always use the isomorphisms to get those arrows, no need to write the extra letters and make life more complicated >.<

[coral shale]

Well why I'm doing all this ---

idk, just convincing myself R/(maximal ideal) === field

[chilly ocean]

[coral shale]

And then I remembered there was a lot of things I forgot

[rustic crown]

relatable 🙈

[coral shale]

In particular if S is the field, it will have a proper non-trivial ideal which is the kernel of the map from S to T (If I in J in R are proper)

[rustic crown]

usually it's phrased using the correspondence theorem, but really all these 3-4 theorems are just the same thing >.<
like ideal of R containing I are in bijection with ideals of R/I
so
R/I is a field <=> 2 ideals in R/I <=> 2 ideals in R containing I <=> I maximal

[coral shale]

👌 Thanks

[next obsidian]

No one should have all that power

No one should have all that power

[rustic crown]

[coral shale]

I was reading this earlier and went ❓❓❓ 😂
https://i.imgur.com/l7BSKZw.png

[next obsidian]

It’s so true

[delicate orchid]

One day I will be able to remember the difference between [a] and just adjoining a

Today is not that day

[hidden haven]

What do you mean by that

What is [a]

Do you mean like ℚ[✓2] vs ℚ[x]?

One is abuse of notation, while one is the polynomial ring

Sometimes abuse of notation, sometimes adjoining in a subring

[delicate orchid]

The check mark for square root

I mean more

Q[sqrt 2] vs Q(sqrt 2)

Although I think these are the same

Q[2^1/3] and Q(2^1/3)

[hidden haven]

The first means smallest ring generated by these elements, and the second one is the smallest field

Just that if the element you are adjoining to a field is algebraic over that field, then both are isomorphic

but Q[x] and Q(x) are not isomorphic

[delicate orchid]

Ah ok I see

Q(x) would contain x^-1?

[hidden haven]

Yes

[delicate orchid]

Got it

[hidden haven]

Also 1/p(x) for any polynomial p

It will be the fraction field of Q[x]

[delicate orchid]

Well yeah it’s gotta have all the inverses to be a field

[hidden haven]

Yeah, assuming that it contains Q[x]

[barren sierra]

I am having major dumb moment

How do I show Aut(D_8) = D_8

I have shown the first part

I have shown also that D_8 is normal in D_16

where do I go from there

my prof gave a hint which makes no sense

[delicate orchid]

What was the hint

[barren sierra]

The fact that D_8 is isomorphic to a normal subgroup N of D_{16} means you can get a homomorphism D_{16}/N --> Aut(N)=Aut(D_8), which you can show is injective. I described how this works during lectures on Wednesday.

but like

idk what to do with that

since D_16 /N has index 2

which isn't what we want?

[delicate orchid]

I’m thinking

Yeah I think we definitely want to quotient by C_2

[barren sierra]

yea

which is {e, r^4}

cause then you get index 8 which is the size of D_8

cause I can show D_8 = D_16 / C_2

I think

[delicate orchid]

Yeah but then the proof doesn’t follow

[barren sierra]

well my thought was

let D_16 act on D_8

you get a homomorphism from D_16 to Aut(D_8) right?

[delicate orchid]

Ah ok that’s smart

Yeah you get a homomorphism

[barren sierra]

yea cause every group action defines a homomorphism

[delicate orchid]

Ok then if we can find the kernel of this homomorphism we can use first iso

And I think the kernel will be iso to C_2 just from an indexing argument

[barren sierra]

yea

so then I have to show that this homomorphism is surjective right?

[delicate orchid]

Yes

[barren sierra]

and I get D_16 / C_2 = Aut(D_8)

and then D_8 = D_16 / C_2

and that works?

[delicate orchid]

I believe so

[barren sierra]

and I'm also right that my prof's hint doesn't really make sense?

that would be the 2nd time this week a prof of mine confused me with a hint

[delicate orchid]

Although I think you already have that the homomorphism is surjective from the fact that D_8 is a subgroup

And yeah

It was kinda odd

But it led you to thinking about D16 acting on D8

[barren sierra]

yea

I only thought about that because inner automorphisms are just conjugations

and group actions define conjugation actions iirc

or well

conjugation is a group action

[delicate orchid]

Gonna add group actions to “list of things I need to revise over the summer”

[barren sierra]

this class is moving so fast

week 1 of class

week 2

week 3

and now we're on week 4

it's crazy, I feel so out of my wheelhouse

[delicate orchid]

Jesus that is fast

[barren sierra]

I'm just lucky that I don't need this class for quals like the other students in the class (since I'm an undergrad)

[barren sierra]

yea and the HWs are brutal

https://faculty.math.illinois.edu/~rezk/500-sp22/ps-03-500-sp22.pdf
is this link accessible to you

I'm just like 💀
every week

[delicate orchid]

Oh the last one looks fun

[barren sierra]

it was nice

[delicate orchid]

Equivalence class type beat

[barren sierra]

ya

LONG proof tho lol

[delicate orchid]

I can imagine

[barren sierra]

or at least for me it was, I'm sure i can trim it in a bit

8 was cute

9-11 kinda tedious but good practice

now to finish 6 and 7 (got stuck in a similar spot for 7 but I think I can solve it in a similar way)

[delicate orchid]

6 and 7 do look very similar

[barren sierra]

they are

[wooden ember]

I think here they mightve meant D_16 / C(N)

with C(N) the centralizer of N

[barren sierra]

ah

right cause C(N) = C_2

[wooden ember]

which is C_2 if im not wrong

yeah

[barren sierra]

that makes much more sense

[wooden ember]

cause i was struggling to see how you get a homomorphism from D_16/N to Aut(N) otherwise lol

[barren sierra]

yea

if this class doesn't have a curve, imma cry

[wooden ember]

you can dew it

[barren sierra]

💀

Hopefully grad schools are a little lenient

[wooden ember]

AA 1?

[barren sierra]

[wooden ember]

but it starts with review of groups

wtf

[barren sierra]

500 = grad level

[wooden ember]

how can it be I if you start by "reviewing" groups

[delicate orchid]

Abstract Algebra 1: Review of Abelian Categories

[barren sierra]

week 1 was a review of group theory and all the isomorphism theorems which I supposedly learned in undergrad algebra

however I only learned first iso theorem in undergrad

[wooden ember]

Oh so you had an AA class in undergrad okay

[delicate orchid]

(it's because they're all the first iso theorem)

[wooden ember]

makes more sense

[barren sierra]

undergrad being last fall 😭

and I'm still an undergrad

but yea

[wooden ember]

oof

[barren sierra]

I'm happy my other two classes are easy

but my research is going much slower than I want

[wooden ember]

why are you taking grad classes?

[delicate orchid]

why not

[barren sierra]

^

[wooden ember]

fair enough

[barren sierra]

minor regret but also

¯_(ツ)_/¯

[wooden ember]

research in undergrad, sounds pretty dope 👌

[barren sierra]

ya

[wooden ember]

we get to do that third year i think

[barren sierra]

it's algos research, not math research

hopefully in the future I do more math research

but algorithms are fun

[wooden ember]

yeah algorithms are cool

[barren sierra]

and I'm TAing two courses which takes up a good amount of time but I'm honestly super happy to sink time into that because I love it

but thank god number theory is a snooze fest so far

[wooden ember]

hell yeah i hope i get the grades to TA

seems like so much fun

[barren sierra]

it is

can't TA math courses since only grad students can

[wooden ember]

bruh what

[barren sierra]

so I TA the discrete math class for CS majors

and I TA the algos class

[wooden ember]

still pretty cool

[barren sierra]

math department has enough grad students that they can afford to make that restriction

[wooden ember]

fair enough

[barren sierra]

CS department has much higher demand since they want a much lower student : TA ratio

Math TAs really only grade, rarely teach

[wooden ember]

same honestly our TAs only really supervise exercises sessions and grade

but some take the effort to try and deepen understanding

[barren sierra]

yea for discrete math I help in an exercise session and hold office hours

and then for algos I hold office hours and grade

[barren sierra]

ya I'm the annoying TA who does stuff the socratic method

cause it's more fun for me that way than just regurgitating the steps

[wise igloo]

I don't wanna interrupt this conversation so i'll make a thread, but does anyone have a simple, fun proof directly involving field axioms? I proved (-a)(-b)=ab and I wanna have other exercises similar to this

[wooden ember]

gimme a sec i might have a few

[barren sierra]

Ok I'm having trouble showing this

what would be a map from D_16 to D_8?

I'm having trouble visualizing this

[delicate orchid]

r -> r^2

[barren sierra]

oml

right

💀

and s -> s right right

[delicate orchid]

yeah

[delicate orchid]

so the kernel is clearly order 2

[barren sierra]

I literally already wrote this earlier 😭

[oblique leaf]

I essentially have to show that the tensor product of V with V is equal to the direct sum of S^2 and wedge^2. I know this is easy when F contains 1/2 because then we can use a variant of the proof that every function has an even and odd component. But the problem is that all we know about F is that it is not characteristic 2

V also doesn’t have to be finite dimensional

[delicate orchid]

throwback sundays

[tardy yacht]

Can you use the direct sum of eigenspaces is the original vector space?

[oblique leaf]

Does that hold for infinite dimensional V?

[hot lake]

but in characteristic 2 there is only one eigenvalue ?

[oblique leaf]

No our field is any field not characteristic 2

[hot lake]

so it contains 1/2 then ?

[oblique leaf]

I dont think so, there are fields not characteristic 2 that dont contain 1/2 right?

My argument currently relies on 1/2 being in the field

[hot lake]

is 2=0 in your field ?

[oblique leaf]

No

[hot lake]

then by the field axioms, 2 has an inverse

[oblique leaf]

But is 2 even in the field?

[hot lake]

2=1+1

[oblique leaf]

What about the field of 4 elements that are 0,1,x and x+1

[hot lake]

it has characteristic 2

there, 1+1=0

[oblique leaf]

I thought field with char 2 was just the field with 2 elements?

[hot lake]

well you just provided a counter-example

[delicate bloom]

characteristic is the number of times you add 1 to itself to make 0

[oblique leaf]

Ok sorry sorry

I forgot what characteristic was

[hot lake]

for any prime p and any exponent e>=1 there is a unique field of size p^e, with characteristic p

so that's infinitely many fields of characteristic p

not even going into the infinite ones

[oblique leaf]

Ok yeah that makes sense

Thank you so much!

[hot lake]

don't do an exercise without knowing what all the words mean oO

[chilly ocean]

pls someone give me funny simple algebra exercise

[lament dawn]

Prove or disprove the existence of a perfect cuboid

[chilly ocean]

idk whats a cuboid

[lament dawn]

It was a joke lol

[chilly ocean]

this doesnt sound like algebra

open up atiyah macdonald to a random page and prove the first exercise/theorem you see

i dont have this book

sad

[lament dawn]

Prove Z/nZ is cyclic

[chilly ocean]

thats trivial

Z -> Z/nZ is epi so sends generators to generators

[proud bear]

prove that if 1-ab is a unit in a noncommutative ring, then so is 1-ba

[lament dawn]

Prove all groups of order <60 are solvable

[chilly ocean]

oh i think i remember proving that long ago

[chilly ocean]

that sounds very difficult

[lament dawn]

It isn't

Use burnsides

[chilly ocean]

i dont know burnsides yet

therefore its difficult

[lament dawn]

Ok you need to be more specific then

I've given you something too easy and too hard so I'm not sure what you're looking for

[hidden haven]

Isn't it a non trivial fact that epimorphisms of groups are surjective

Prove the fact

[chilly ocean]

i should have said surjective

but i was lazy

[hidden haven]

[cursive temple]

this was pretty fun

not too difficult iirc

[chilly ocean]

seems unnecessary to restrict to finite groups

isnt this trivial for infinite

yes

also

what they mean by nonvacuous

[cursive temple]

nonempty

[chilly ocean]

thats also unnecessary?

[cursive temple]

its from jacobson so the language is weird

is the product AB well defined if B is empty

but sure it doesnt make much difference

[chilly ocean]

it doesnt need to be defined

since inequality wont hold

[cursive temple]

A isnt necessarily proper so the inequality would hold for A = G, B = empty set

[chilly ocean]

no

its > not >=

[cursive temple]

oh yeah cardinality of empty set is 0 lol

either way

not a very interesting case

[chilly ocean]

ik

but i can only do trivial stuff

okay i have idea

Let |G|=n

if |A|=n trivial

suppose it holds for |A|=k+1

let new A s.t. |A|=k

[hidden haven]

If AB is the set of pairwise products, then it will be empty if either set of empty

[chilly ocean]

and |B|>n-k

[woven delta]

carla if |A|+|B|>|G| what does that imply about one of |A| or |B|?

[chilly ocean]

ik what it implies i thought about that too

[woven delta]

hmm

[chilly ocean]

i think my idea wont work

[woven delta]

okay here's a question

suppose we have 2 elements a \neq b and we want to look at aA and bA

what's the most that aA and bA can intersect?

you can assume that a is the identity if you want

[delicate bloom]

I suppose it suffices to prove for the case |A|+|B|=|G|+1

[woven delta]

I mean sure

[chilly ocean]

i dont know

if |A|=|G|/2 then can have full intersection

or can it

[delicate bloom]

just allows you to focus on the fact that it only depends on when A and B have at most one element in common, dunno haven't caught up to what you've done yet

[chilly ocean]

yea

for example G = Z2xZ2, A=Z2x0 and a, b the elements of A

[hidden haven]

I can't understand this hint

[woven delta]

it's not a hint

[hidden haven]

oh

ohh a new question

[woven delta]

no, I was kind of just seeing if thinking about that would help

[hidden haven]

oh lol ok

[woven delta]

I don't know the answer to this question

I guess the tone of my first thing was hint like

but I am thinking about this question rn

[cursive temple]

For a pretty comprehensive hint ||Consider the set A_g = {ga^(-1) : a in A} for some fixed g in G||

[woven delta]

okay so if I consider the number of pairs that we are multiplying together, I get a pretty large number. If we assume as mero said that |A|+|B|=|G|+1, then is there a nice lower bound on the number of pairs |A||B|? Maybe this is not the direction to go in but I will try it

if you optimize x(a-x) you get a-2x=0 which gives you a=2x or x=a/2 as your minimum, so the worst case scenario here in terms of numbers of pairs is the case where |B|=|G|/2+1 and |A|=|G|/2 (approximately, odd and even make this approach a little annoying)

[next obsidian]

I feel like this is pigeonhole applied to the function b -> ab over all a in A, b in B

That’s my first instinct for this at least

[woven delta]

I mean that's kinda what I'm trying