groups-rings-fields

[woven delta]

sort of

[next obsidian]

I mean also you can compute the size of AB

But it requires knowing the intersection’s size

¯_(ツ)_/¯

[woven delta]

no it's more complicated because they aren't subgroups

I think

[next obsidian]

Oh

Ahhh

Dealing with products of subsets

Reminds me of S^|G|

Banger problem btw

[hidden haven]

For subgroups this would be done by Lagrange

[woven delta]

okay so if we look at the set of all products GxG, how many of those elements do you need to get a projection to the group via the product map

if that is something I can find and it is less than my lower bound then I am done

okay how many representatives can there be for a single element in my product

oh that's what's happening

lol

[woven delta]

I think this is the critical idea

there can be at most the smaller subset many instances of any given element in the product

wait that's not enough

but I know the right way to think about this now

bleh this is annoying

[desert dome]

Is it true that for any finite extension L/k, we can find E/L/k such that E/k is galois?

[lavish nexus]

how to map O(n) into SO(n+1) injectively
like take A and map it to
A 0
0 1/det(A) ?

is there a catch because I think this is too silly for a problem

[sturdy marsh]

no, but you can do it for separable extensions

[sturdy marsh]

I think it works

[desert dome]

We do have the assumption it is separable! Do you mind explaining why?

[sturdy marsh]

if the extension is separable, then it is generated by a single element

take the minimal poly

and take its splitting field

[lavish nexus]

oh wait this is not orthogonal

[sturdy marsh]

oh SO

not SL

hmm

[desert dome]

I see. It's because of the Primitive Element Theorem (forgot about this). Thank you!

[sturdy marsh]

A 0
0 sign of determinant ?

wait that's the same map

[lavish nexus]

no that wouldn't be S

[sturdy marsh]

det(A) is 1 or -1

[lavish nexus]

oof

[sturdy marsh]

yeah this works I think

[lavish nexus]

yes

[lavish nexus]

this is what I somehow forgot

[sturdy marsh]

yeah I missed that too for a second

[desert dome]

Hi, I am trying to figure out this problem: let f(x) = x^4 - 4x^2 - 1 ∈ Q[X] and determine the
splitting field. Could I just go ahead and calculate all the roots?

[hidden haven]

Yes

[desert dome]

sure, I just thought there might be some smart things I should notice

[hidden haven]

There are plenty of smart things you can do with finding roots too

The way I would do this would be to find the possible values of x^2 ie treat this as a quadratic in x^2, and one root comes out to be positive real and the other is negative real. This gives a degree 2 extension, and then you adjoin the square roots of both of those roots, and you can argue that each square root you adjoin has to give you a degree 2 extension so in all you have a degree 8 extension. The last part is a bit tricky because you have to argue that each extension is proper but you can do some nice stuff there.

@desert dome if you need a hint

[desert dome]

Thank you!

[steady axle]

I have a field k of char =p >0 and suppose a \in p such that a has a pth root b in k. then b, 2b,... , (p-1)b are also pth roots of a right?

[chilly ocean]

yea i think so, and more generally should work for any a \in k?

[next obsidian]

I think so because 1,…,p-1 live inside the prime field and those are perfect. Translated to not Hurb language it just says they have p-th roots

Oh wait

I read this wrong lol

No, kb will not be pth roots of a, because the p-th power on the elements k from 0 through p-1 is a bijection on that set of elements

So like, k^p = 1 iff k = 1

Then (kb)^p = k^p•a

[chilly ocean]

oh yea

fuck im dumb and i hate myself

[next obsidian]

I just misread the question at first

It’s true that ka will have p-th roots

[covert vector]

k is integer and the whole field

[next obsidian]

Whatever

[covert vector]

Lol

[coral shale]

I'm trying to make 'sense' of these definitions...

If anyone has any insight why they make sense? Just a quick line/idea/intuition that might save me some thinking would be appreciated 😅

[delicate orchid]

Reminds me of the definition of the trace and determinant of a matrix via the eigenvalues

[coral shale]

sum of eigen is tr
product of eigen is det

Right. (now that leaves me wondering as well 'norm', not det)

[delicate orchid]

The determinant is a norm on the general linears

I think

Yeahhhh it probably satisfies the triangle inequality

[coral shale]

That I was not aware of, will have a look.

[delicate orchid]

Sorry, absolute value of the determinant might be a norm

[coral shale]

So now I guess I have to convince myself the images of alpha in the embeddings correspond to the eigenvalues in some way

[delicate orchid]

Idk if it’s relevant

It was just an observation

[chilly ocean]

When we say f : X -> Y has a right inverse, do we mean that for all y in Y, there exists x in X such that f * f_r y = y?

[delicate orchid]

Yup

Well

[coral shale]

uhhh

[chilly ocean]

Or do we mean that there is at least one element for which this right inverse function exists?

[delicate orchid]

f_r is the right inverse of f there

[coral shale]

are u talking about the function f

having a right inverse

[chilly ocean]

yes

[coral shale]

exists g : Y -> X
f o g = id_Y

Its this right?

[delicate orchid]

Yur

[chilly ocean]

But say X = {x_1,x_2} and Y = {y_1,y_2} and f maps x_1 |-> y_1 and x_2 |-> y_1, then f_r could map y_1 to x_1 or x_2, so we have two right inverses

But f is not surjective

[delicate orchid]

You don’t have a right inverse there

[chilly ocean]

Yes I do

What do you call it then?

[delicate orchid]

No, you don’t

A map that maps one thing to multiple things is not a function

[chilly ocean]

I didn't say that it does

[coral shale]

I'm not sure about this definition. In the first place the usage of quantifiers theres looks odd to me. You havent said what f_r is, and haven't used x.

[chilly ocean]

I am saying we have f_r1 and f_r2

[delicate orchid]

Consider making a choice to map y_1 to either x_i, then consider the image of the other x_j with j not equal to i

[chilly ocean]

And that f_r1(y_1) = x _1 and f_r2(y_1) = x_2

i.e. we have two right inverses

[delicate orchid]

You’ll find that it maps to x_i, not x_j

So it’s not an inverse

[chilly ocean]

There just this theorem in my book that states that every function has at least one right inverse,

But it does not have to be unique

[coral shale]

$$(\exists g)fg = id_Y$$

[chilly ocean]

But I thought we needed for a function to be surjective for a right inverse

[cloud walrus]

Shuri2060

[coral shale]

All we need is for g to satisfy this

f : X -> Y
g : Y -> X

Let's see

[delicate orchid]

I think you need surjecivity for a right inverse yeah

[coral shale]

No.

[delicate orchid]

You posted a counter example for a non-surjective function

[coral shale]

g does not have to hit X everywhere for f o g to be defined right?

[delicate orchid]

For it to be well defined I’m pretty sure it does

[coral shale]

I think not but I am not 100% sure

[delicate orchid]

I’m going to google it

[coral shale]

But I dont see whats wrong with a non-surjective g there

The concept of composition doesn't break

[coral shale]

only if f is surjective, that I agree

[chilly ocean]

Okay, then what's wrong with my example?

[coral shale]

But g need not

[delicate orchid]

I’ve literally explained it

[coral shale]

Could u write it explicitly in numbers. I found it hard to read

[delicate orchid]

Consider mapping y_1 back to x_1 then consider the image of x_2 under ff_r1

It’s not mapped back to x_2

It’s mapped first to y_1 and then back to x_1

So the composition is not the identity

By symmetry this is also true of the image of x_1 under ff_r2

[coral shale]

We require for all y in Y
f(g(y)) = y

I agree there could be multiple right inverses

=======
For example f maps from integers to evens.
f(2n) = 2n
f(2n+1) = 2n

g then needs to maps from evens to integers.
g(2n) = 2n + 1
h(2n) = 2n

Both are right inverses

oops

[delicate orchid]

I really don’t see how either of those are right inverses

Ah ok

Yeah I see it now

[chilly ocean]

I am confused now

So every function does have at least one right inverse

But not every function is surjective

So how can that be?

[delicate orchid]

Surjective functions have a right inverse

[chilly ocean]

That's not what this says

[coral shale]

You can make any function surjective by restricting codomain

thats what it means i believe

[chilly ocean]

No, we're not looking at restrictions of the codomain

[coral shale]

Then its clearly wrong lol

[chilly ocean]

Why?

[coral shale]

A non-surjective does not have a right inverse

[sharp sonnet]

what's right inverse of a constant function?

[coral shale]

Unless you say the domain of the right inverse is the range of f?

And so we only care about id being id on the range

[sharp sonnet]

this talks only about the image

[delicate orchid]

I hate set theory so much

[sharp sonnet]

if f is constant, then so is f \circ g for any g

and constant functions are in general not the identity

[chilly ocean]

So the statement is wrong?

[coral shale]

It depends on the defn of right inverse

[chilly ocean]

One second

[coral shale]

the typical one --- yes, it would be wrong, technically

although it doesnt take much to amend that statement

[chilly ocean]

A function $R: T(V) \to V$ is called a right inverse of $T$ if $T[R(y)] = y$ for all $y$ in $T(V)$. That is if $TR = I_{T(V)}$, where $I_{T(V)}$ is the identity transformation on $T(V)$.

[cloud walrus]

n/c

[coral shale]

T(V), what is this set?

[hidden haven]

If you are talking about sets then it is true that surjective is equivalent to right invertible which is equivalent to right cancellable

[chilly ocean]

The image of T on V

[chilly ocean]

What else were you typing?

[coral shale]

Then indeed, this defn is different from the typical one

Because you essentially restrict the codomain of T

and make it surjective

[hidden haven]

The right cancellability is not true for rings but I realised you are talking about invertibility not cancellability

[hidden haven]

Yeah usually right inverse is defined from the codomain not the image

[chilly ocean]

Okay I see

[coral shale]

The typical definition has for f : X -> Y, a right inverse must be a function g : Y -> X and fg = id_Y

But that is not the case here.

[coral shale]

The modulus of det of a matrix is a norm on GL right? (if GL over number field)

The above definition of norm corresponds to the determinant? This is surely not necessarily non-negative

[delicate orchid]

Yeah hence why I quickly amended my statement to “abs value of the determinant”

Although I think that it’s just completely wrong now

[coral shale]

Yeah but I mean this defn given

@hidden haven come bak 🙏 👀

[sharp sonnet]

the field norm is not a norm

[coral shale]

oh.

They couldnt just have called it determinant, now could they

[sharp sonnet]

at least not a vector space norm

[hidden haven]

I don't really have any intuition for trace and norm beyond that they allow proving sometimes that certain extensions are proper

Det could probably help

@rustic crown

[rustic crown]

i'm in a lecture rn >.<

[hidden haven]

Cringe

What lecture at 6pm

[coral shale]

so am i lul

[delicate orchid]

I am as well

[hidden haven]

English?

[sharp sonnet]

you use the norm to extend complete valuations uniquely, right?

i mean they are maps from the extension field into the base field with nice properties i guess

norm is a hom of multiplicative groups, trace of additive groups

[coral shale]

this was the original Q btw. Just wanted to make sense of this

[sharp sonnet]

so it makes sense to use them to extend stuff from the base field to the field extension

i think proving that the definition of norm/trace given in this image follows from the linear algebra definiton is a bit involved

[delicate orchid]

Oh there actually is a connection

Cool

[coral shale]

The linear map I am investigating is multiplication by alpha right?

Hunting for eigen:

Given a in L,

Solve ab = Lb for L in K and b in L

Then

(a - L)b = 0

^ this is wrong right? Need to treat it as vector space, not field

[sharp sonnet]

actually if you take a primitive element a and basis {1, a, ..., a^{n-1}} then you can write down the matrix of the linear transformation quite easily
then you can write down the minimal polynomial and read off the trace and determinant and connect them with coefficients of the minimal polynomial, which is given by galois conjugates

and then vieta gives you the result

then the hard part is dealing with the case when a is not primitive

but then you consider the tower of fields L/K(a)/K

and show that the trace/norm of L/K is the composition of the one of L/K(a) and K(a)/K

one of those things is easy, the other hard

this is also how you compute trace and norm in practice

[coral shale]

I will try, thanks

[sharp sonnet]

so yeah, if you want to write this down start with a in L and consider the cases a in K and L = K(a) and write down the matrix of the multiplication by a map

in the first case (a in K) multiplication by a is just diag(a)

and in the second case you can write down the matrix in terms of the coefficients of the minimal polynomial of a

if you choose the basis {1, a, ..., a^{n-1}}

[subtle ivy]

whats $\mathbb{Z}/a_i$ here? ($a_i$ an integer)

[cloud walrus]

shortcut

[subtle ivy]

is it supposed to be $\mathbb{Z}/a_i\mathbb{Z}$?

[cloud walrus]

shortcut

[hidden haven]

Yes

[lethal dune]

is this ftofgmopid

[hidden haven]

bruh I can't believe I understood that abbreviation

[unreal portal]

I see two ways to do this, but I get stuck on both of them because I don't see any obvious algebraic relationship between $\mathfrak p$ and $\mathfrak q$. First is by supposing $R_{\mathfrak q}$ is reduced, so $f^n\neq 0$ for all $f\in R_{\mathfrak q}, n\in\mathbb N$. Through the inclusion map $R_{\mathfrak q}\to R_{\mathfrak p}$, I think we can conclude that those same elements are nonzero in $R_{\mathfrak p}$, but then there are elements $a/b\in R_{\mathfrak p}, a\in R, b\in (R\setminus \mathfrak p)\cap \mathfrak q$ that could still be a nilpotent.

The other method, trying to do it directly, $f=a/b\in R_{\mathfrak p}$ such that $f^n=0$, runs into the same problem where $b\notin \mathfrak p$, but $b$ could be in $\mathfrak q$. Is there some additional structure I need to realize to solve this? I don't see any way using the nilradical definition may help, but maybe that's the right way to go

[cloud walrus]

cgodfrey

[hidden haven]

There is an obvious set theoretic relationship between p and q

Did you use that I can't tell

[unreal portal]

q supset p, I sort of use that implicitly that $(R\setminus\mathfrak q)\subset (R\setminus\mathfrak p)$

wait that's a supset

[hidden haven]

Isn't the inclusion the other way

Ye

[cloud walrus]

cgodfrey

[hidden haven]

I see

So you want to show that a localisation of a reduced ring at a prime is reduced

Maybe use the fact that nilradical is the intersection of all primes, and the correspondence between the primes in R and the primes in a localisation of R

[unreal portal]

I dont' think there's any assumptions on R, just that it's noetherian

[hidden haven]

If they have a trivial intersection in R, their images should also have trivial intersection

[hidden haven]

Here I took R = R_q

Wait

You have to show the other direction

Interesting

oh wait

Yes this is fine

I am showing contrapositive

R_q is reduced implies R_p is too

[hidden haven]

Then use this reasoning

Also there is an #algebraic-geometry channel now

[unreal portal]

Oh wtf

I thought i postes this there

Mybad

[hidden haven]

F

[unreal portal]

I'll read on that correspondence of prime ideals in localization

[hidden haven]

Proposition 3.11 in Atiyah Macdonald

[unreal portal]

just to confirm, if you're taking R = R_q, is (R_q)_p = R_p? At least, localization at the image of p in R_q?

[hidden haven]

Yes, localisation at the image of T under localisation at S is the same as localisation at T

Same proof as quotienting the quotient by I by the image of J is the same as quotienting by J

Modulo some containment conditions

I think I subset J and S subset T

[unreal portal]

okay, thanks! Algebraic geometry is hard when you never took a course on commutative algebra lol

[spice whale]

is the characteristic of a ring related to topological characteristic

because i can't exactly see the relationship

[hidden haven]

I don't think it is

[spice whale]

oh lol

I'm sure you could force a connection sonehow

[chilly ocean]

what is topological characterisitc

oh euler characterisit

[spice whale]

euler characteristic

$\chi = \sum\limits_{i=0}^\infty (-1)^i k_i$

[chilly ocean]

what are k_i

[cloud walrus]

Alison40

[spice whale]

k_n is the number of cells of dimension n

[chilly ocean]

n-cells are images of [0,1]^n right

[spice whale]

yeah

[chilly ocean]

this doesn't make sense if there's infinitely many does it

[spice whale]

uhh

[hidden haven]

Ye this is only for finite cell complexes

[spice whale]

i put the infinity to generalise to any dimension

[chilly ocean]

i mean

[hidden haven]

You can define it using homologies

[chilly ocean]

infinitely many cells

[hidden haven]

Then it generalises

[chilly ocean]

for example whats the Oyler char of R^2?

[hidden haven]

To some degree

[hidden haven]

Same as that of I^2

[chilly ocean]

why

[hidden haven]

because it happens to only depend on homotopy type

[chilly ocean]

hmm

but not every topological spce is homotopy equivalent to an image of some finite n-complex is it

[hidden haven]

No

Not even weakly equivalent I think

[chilly ocean]

how do you define wheeler char for those then

[hidden haven]

Then you use this definition

Replace each k_i with rank H_i

Which is defined as long as each rank H_i is finite

[chilly ocean]

what if is infinite

[spice whale]

what if they're not

[hidden haven]

Whomst've been studying infinite rank homologies

[spice whale]

uhhh

i mean

[hidden haven]

Lol it is not always defined

[spice whale]

idk what homologies are really so

[chilly ocean]

this is so sad. you hate to see this kind of thing

[hidden haven]

[chilly ocean]

i kinda have idea of what homologies are

but idk well yet

ik its related to sequences of modules s.t. image contained in next kernel

[hidden haven]

homology is study of homo, of course 😌

[spice whale]

algebraic topology sounds very based

[spice whale]

🏳️‍🌈

[chilly ocean]

you sound very based

[spice whale]

[hidden haven]

It is a measure of how far such a sequence is from being exact

[chilly ocean]

oh yea

[next obsidian]

Actually it’s a measure of how far a sequence is from being a good boy

[chilly ocean]

yea it's the kernels right

or is it the cokernels

[hidden haven]

kernel/image at each stage

[next obsidian]

It’s about the kernels and then the cokernels of the kernels which is the same as the kernels of the cokernels (the image)

[hidden haven]

[chilly ocean]

yea i heard of cokernels of kernels

and kernels of cokernels

somewhere

[next obsidian]

Yeh those are the image and coimage respectively

But the 1st iso says those are the same

[chilly ocean]

ah

up to iso

[next obsidian]

Via a very natural map

There’s an induced map between image and cokernels

And that one is an iso

[chilly ocean]

up to canonical iso

[hidden haven]

up to iso is not needed here

[next obsidian]

I suppose so yeah

[hidden haven]

image is anything that satisfies uni prop

same for coimage

You can just say they are equal

You can make the same object satisfy both universal properties

[next obsidian]

I kinda disagree?

Idk

[hidden haven]

Maybe because we don't say that any 2 images are equal

[next obsidian]

I’m gonna take the kernel of the cokernel of your kernel

[hidden haven]

Ye idk it just seems stupid to me to have to say that 2 things are isomorphic when they are only defined up to isomorphism

[next obsidian]

Bitch

[chilly ocean]

what about homology but with nonabelian groups

[next obsidian]

Do not ask such questions

You stray too far from the light

✝️

In general tho you can’t define it

Because the image need not be normal

[hidden haven]

There actually is stuff there

[next obsidian]

Yeah that’s why I said my shit at first

[hidden haven]

Weibel does some things on it

[chilly ocean]

i know there is

[hidden haven]

But yeah you need this

[next obsidian]

Oh are you talking about group cohomology?

[chilly ocean]

that's why i asked

[hidden haven]

no

[chilly ocean]

Grp is a homological category

[next obsidian]

Okay yeah

[hidden haven]

This is in the simplicial methods chapter

[next obsidian]

It’s not nice

Yeah I was about to say

For nonabelian homology stuff you usually have to go to simplicial stuff

For eg rings

[hidden haven]

Right

[next obsidian]

You can define andré-Quillen homology

[chilly ocean]

what is simplicial stuff

[next obsidian]

And you do it by taking simplicial resolutions and then tensoring with, or homming into stuff

And then you can take homology as modules

Simplicial in the sense of simplicial objects

[chilly ocean]

idk what are those

[next obsidian]

They’re functors from the simplex category

You can think of it as like

A sequence of elements indexed by n

With these maps going forward and backwards

[chilly ocean]

why are those functors important

[next obsidian]

Which satisfy some equations mimicking the inclusion of n-simplified into n+1-simplicies

[chilly ocean]

ah okey

[next obsidian]

So here’s why they might make sense to be “chain complexes”

[hidden haven]

Let me just link a clerk rant from 4 months ago

[next obsidian]

Dold-Kan actually says simplicial objects in A where A is an abelian group

[chilly ocean]

oh i heard of dold-kan

[next obsidian]

Are equivalent to chain complexes in A which are concentrated in nonnegative degree

So you could think of simplicial objects in any category as sorta hinting at that

You can also do homotopy on these things because of the simplicial identities blah blah

[hidden haven]

#point-set-topology message

[chilly ocean]

concentrated?

[next obsidian]

Simplicial sets for example I think are a model of infinity categories

It just means for all negative degrees, the objects are 0

[hidden haven]

This describes the universal property of the simplex category

[chilly ocean]

also why are the inclusions backwards

[next obsidian]

Wdym?

What inclusions

[chilly ocean]

isn't n-simplex subset n+1-simplex

[next obsidian]

Yeah

So the maps are about those inclusions

Including an n-simplex as a face

But there’s also degeneracy maps

[chilly ocean]

so why is it contravariant

when n<n+1

[next obsidian]

Which collapse an n-simplex to an n-1

It’s contravariant

[hidden haven]

basically whenever you have a monoid like thing you have a simplicial object, and monoid like things arise as soon as you have a universal property, and these simplicial objects can be used to construct resolutions that give cohomology theories

[chilly ocean]

i don't get why contravariant

n+1 simplices are subsetted by n simplices

and n+1>n

shouldn't it be covariant?

[next obsidian]

It just be like that

You get maps going both directions

Because you can project from an n+1 simplex onto one of the edges

[hidden haven]

In the simplex category the face maps go from the higher degree to lower degree

Because rather than inclusions

They are face assignments

[chilly ocean]

hmm

[hidden haven]

And when you contravariantly map this to say Top

You realise those face maps as inclusions

Hence contravariant

[next obsidian]

I did this shit a year ago for a homological algebra class so I forgor some of this stuff

💀

[chilly ocean]

why are you using emojis on a serious topic

[next obsidian]

Cuz I forgor 💀

[hidden haven]

This is because the objects in the simplex cat are formal

Not actual top spaces

[chilly ocean]

hmm so

[hidden haven]

So you can't really include them

[next obsidian]

Smfh what construction are you using?

I used the one with [n]

[chilly ocean]

i mean wasnt simplex cat just category of finite ordinals

[hidden haven]

same

It is finite ordinals

I mean the numbers themselves are formal

Each number is supposed to be thought of as a simplex

but it is just a point

[chilly ocean]

are they dressed formally

in a suit

or what

[hidden haven]

Yes

[chilly ocean]

anyway

[pastel cliff]

just spent an entire algebra lecture proving the most mundane statements

[chilly ocean]

good

i mean bad

[pastel cliff]

this is supposed to be my fun class this semester

[chilly ocean]

what's a simple example of a contravariant functor from simplex cat to Top

[chilly ocean]

file a complaint

[hidden haven]

Tell your prof that I said he is bad

[next obsidian]

Any simplicial complex carla

[chilly ocean]

sue them

[next obsidian]

You can model them as a simplicial object in Top

[chilly ocean]

hmm how

[pastel cliff]

surely he'll cower in fear of the great moldilocks of the math server

[next obsidian]

I forgor

💀

I just remember this

[pastel cliff]

actually yk what's funny, one of my old professors is in here, but has never sent a message

[next obsidian]

There’s a nice paper on some of the basics plus Dold-Kan by Akhil Matthew

If you want to look into this stuff

[hidden haven]

You can make any CW complex into a simplicial set

[chilly ocean]

i want

[next obsidian]

Just google “Akhil Matthew Dold-Kan”

[chilly ocean]

but like

[spice whale]

ping them

[hidden haven]

In a way that you can reconstruct the homotopy type of the original

[chilly ocean]

there's only 1 n-ordinal in simplicial cat

[spice whale]

time to do some more knapp

[next obsidian]

Geometric realization so true

[chilly ocean]

but there can be various n-simplexes?

in a complex

[next obsidian]

Idk maybe I misremembering

Like I said, I forgor 💀

[chilly ocean]

do you want to relearn it together

[spice whale]

all the number theory is doing my head in

[chilly ocean]

maybe it's time to review the basics then, Alison

have you tried Weil's Basic Number Theory?

[spice whale]

i haven't done NT before

[chilly ocean]

exactly

[spice whale]

i wanted to do aa

[spice whale]

I'm not at a computer to sail across the ocean rn anyway

[chilly ocean]

you poor poor thing

[spice whale]

ok I'm just going to skim it

I'm sure not every single proposition there is incredibly important

[next obsidian]

Don’t read this book for an intro to NT

Lmfao

This Shit is hard

[spice whale]

nono I don't mean this book

I'm just skimming the ch1 nt section in knapp

[barren sierra]

Does anyone have a simple example of a finitely generated group with a subgroup that is NOT finitely generated?

The example my prof gave in class is kinda 💀

[hot lake]

probably something in the free group on 2 elements

[barren sierra]

That is the example he gave

Is there nothing simpler? I don't think I would have come up with that myself

[chilly ocean]

how is the free group on 2 elements not a simple example

[barren sierra]

hm fair enough

[hot lake]

I don't think there are examples with abelian groups

[barren sierra]

idk I just would have never come up with it

[hot lake]

so free group on 2 elements is the next best thing

[chilly ocean]

free group on x elements is the universal groups generated by x elements

[hot lake]

you can probably take the subgroup generated by ab,aab,aaab,aaaab, etc

[chilly ocean]

so its a pretty natural example

[hot lake]

wait uuh

well something looking like that

[chilly ocean]

uuh

[hot lake]

yeah not exactly those generators

[chilly ocean]

?

[spice whale]

why wouldn't that work

ohh

inverses

[hot lake]

if you contain ab and aab you have a and b

[spice whale]

yeah

because aab * (ab)^-1 = a

[chilly ocean]

the free group on countably many elements is a subgroup of the free group on 2 elements

hint: think of parentheses

[spice whale]

for the free monoid on 2 elements this does work

[chilly ocean]

dont give up

[spice whale]

aa aba abba abbba ... maybe

not sure if this works

[chilly ocean]

haha

[spice whale]

yeah i think it does

lol

[chilly ocean]

exactly

i think of the a's

as separators

the bs in middle tell the character

so u get free group on countable generators

🍾

[spice whale]

and you can even define a concatenation operation x | y = x * (aa)^-1 * y

[chilly ocean]

i dont understand that

[spice whale]

basically

the first bit

you have x

let's say abbbbaabba

and y

[chilly ocean]

haha

[spice whale]

let's say abba

[chilly ocean]

shouldnt concat be just abbbbaabbaabba

[spice whale]

nono

by concat

i mean concat the bs

sorry

less like concat

[chilly ocean]

sum of characters maybe

[spice whale]

yeah

let's see

what properties does this operation hold

it's closed

[chilly ocean]

all operations are closed

[spice whale]

ok yeah

it's definitely associative

[chilly ocean]

if you restrict it to generators

you just rediscovered succ

i mean

[spice whale]

yep

[chilly ocean]

natural number addition

[spice whale]

but if you don't

it's weirder

it's associative

not commutative

has the identity aa

oh wait the inverses make this weird

[chilly ocean]

theres no inverses in general

[spice whale]

no i mean

the inverses in the base group

[chilly ocean]

oh

[spice whale]

I'll represent a^-1 as z and b^-1 as d

so like

hm

the inverse of aba under addition would be adz

wait

no

because the identity is aa

it's ada

not adz

aba zz ada = aa
ada zz aba = aa

nice it's a group

i think

what about an asymmetric one

like abaabba

abaabba zz addzzda = aa

addzzda zz abaabba = aa

[chilly ocean]

hahah

that looks so funny

[spice whale]

I'm pretty sure it's a group

[spare raptor]

Let $X = {a,b,c}$ and $\mathcal{P}(X) = {\emptyset, {a}, {b},{c},{a,b},{a,c},{b,c},X}.$ $\mathcal{P}(X)$ together with operations $A + B = (A\setminus B)\cup(B\setminus A)$ and $A\cdot B = A\cap B$ define a commutative ring.

Show that there are precisely two subrings of $\mathcal{P}(X)$ that contain ${a}$, namely ${\emptyset,{a},{b,c},X}$ and the entire ring $\mathcal{P}(X)$.

[cloud walrus]

sam

[spare raptor]

Is there a non-bruteforce way to show this?

And if so could I get a hint? ^-^

[chilly ocean]

Show that if a subring contains {a} then it has to conatin {b,c} and if it has any other element then it generates the whole ring

[spare raptor]

yh

but that last step

should I just go through it case-by-case

[chilly ocean]

yeah there are few cases to check

[spare raptor]

kk

yeah thats easy enough

just figured there was maybe a more elegant way

[chilly ocean]

I dont see any faster method

[spare raptor]

oki

thx

[chilly ocean]

only 2 cases to check actually I guess

what happens when theres {b} or {a,b}

other cases are symmetrical

[spare raptor]

also there was a sidenote to the question that read "Don't forget to show that the sets are ideal." with no further specification

cant figure out what they mean by that

maybe they meant to say that the sets are rings? or subrings?

[chilly ocean]

Yeah hmm actually idk what the sidenote means

[spare raptor]

aha ye

im just gonna

ignore it

:D

[spice whale]

sorry if this is a stupid question but
are polynomial fields possible (excluding the trivial case of 0 indeterminates)

[chilly ocean]

hmm?

what do you mean by a polynomial field?

[spice whale]

a polynomial ring that is a field

[chilly ocean]

if it has x it also needs to have 1/x

[spice whale]

oh yeah

so

[chilly ocean]

K(x) is the smallest field containing K and x and its basically p(x)/q(x) where p and q=/=0 polyonomials in K

[spice whale]

hm

[next obsidian]

If you take a polynomial ring over a field in 0 variables sure

[spice whale]

i mentioned this

[next obsidian]

Oh

[spice whale]

what's the difference between a polynomial field and power series field

is it just that a power series field can have infinite terms

[next obsidian]

Neither are fields

But a power series ring is just formal power series

You just have infinitely many terms yeah

And you add them term by term and product is given by convolution stuff

[spice whale]

i mean ring

ok nice

[spare raptor]

Having some trouble with another basic proof. Let a and b be associates in some integral domain. Show that if a is irreducible then so is b.

Think I shouldn't be struggling so much with this..

[next obsidian]

You ought to be able to just take a decomposition for one

And transfer it to the other by multiplying by a unit

[tribal moss]

In other words, prove the contrapositive: If b is reducible, then so is a.

[prisma shuttle]

can someone give me a hint for this problem

Suppose $S$ is a multiplicative set and $I$ is an ideal of $R$ such that $I\cap S = \phi$. Show that $I$ is contained in a prime ideal $P$ such that $P\cap S = \phi$.

[cloud walrus]

JustKeepRunning

[next obsidian]

So hmm

Also

$\varnothing$

Don’t use phi hahaha

[cloud walrus]

Chmonkey

[spare raptor]

or \emptyset

:)

[next obsidian]

So there’s kinda two hints

I could give

So uh heads or tails?

[prisma shuttle]

ok lol

[prisma shuttle]

heads

[next obsidian]

Often times you have a collection of ideals

And you can show maximal elements are prime

In fact you can describe this very very generally, there’s something on the stacks project about it

But you could try to consider collections of ideals here and show maximal@ones are prime

That won’t finish it, but there’s pretty standard ways to finish from there

[next obsidian]

Idk how, but I found varnothing first, and so I’ve always used that

Lmao

[pastel cliff]

sorry for interrupting but this is one of my hw questions for my alg class and ???

not asking for help just ???????'ing

[spare raptor]

Let $a$ and $b$ be associated in some integral domain. This gives $a = bc$ for some invertible $c$. Assume that $a$ is reducible, so that $a = xy$, where $x, y$ are non-invertible.

$$xy = bc$$
$$b = c^{-1}xy$$

From this point stackexchange tells me $(c^{-1}x)$ is non-invertible? (which would mean that b is reducible, woho) But why is it?

[cloud walrus]

sam

[spare raptor]

and I think \emptyset is nicer :D

[next obsidian]

If it was invertible, so is x

Usually properties of elements are preserved under multiplication by units

[spare raptor]

ah ok

so a non-invertible * invertible is always non-invertible?

[next obsidian]

Right

Go try to prove it

:)

You could do it manually

Or…

The set of non invertible units is the union of all proper ideals (you only have to look at maximal ones even)

Because you’re a unit iff you’re in no proper ideals

So you can use absorptive properties of ideals to see the product is also in a proper ideal

[spice whale]

how tf

[pastel cliff]

i am not capping

[spice whale]

how do you even approach this

[pastel cliff]

rotman, introduction to abstract algebra, chapter 1 section 1

[spice whale]

is it like
pedantic ambiguity of "two different ways"

[pastel cliff]

his eminence moldilocks surely knows the answer

[waxen hedge]

Maybe for (iii), it's something trivial like 0 = 0^3 + 0^3 = 2^3 + (-2)^3

[hidden haven]

My eminence has short circuited upon seeing this question

bruh

That is most likely it lol

Once you allow negative numbers it becomes false

[pastel cliff]

i emailed because it's horseshit but it's mostly just funny

[hidden haven]

Might as well allow complex numbers while we are at it

[tribal moss]

(Once I was running late for work and had to call a cab. The one that arrived had a cab license numbered 1-1729 proudly displayed in front of the passenger seat, but the driver was woefully unaware of its mathematical significance).

[hidden haven]

Then it is trivial

[pastel cliff]

"hardy was a crank" - ryc

[chilly ocean]

There is a funny book called

Mathematics made difficult

One of the jokes goes roughly as follows

A prime number is a number who’s only factors are 1 and itself

[pastel cliff]

YOUR MOM

https://tenor.com/view/barrack-obama-obama-out-peace-bye-goodbye-gif-8239799

[chilly ocean]

Many high school mathematics teachers incorrectly assume that 7 is a prime number

Clearly 7 has 4 factors

1, 7, -1, -7

[pastel cliff]

groethendiek prime

[hidden haven]

Are you not counting i

[chilly ocean]

The keen student should note that then -1 is the only prime number

[hidden haven]

[chilly ocean]

And then it goes on and continues this for another few pages doing stuff like

[tribal moss]

If we want to be pedantic, -1 has plenty more factors than that -- for example -pi and 1/pi.

My answer would be 9 = 1³+2³ = 2³+1³.

[chilly ocean]

Clearly what is meant by such a statement that is that up to units a the only factors of a prime number are 1 and p. However if this is the case, the then the text should state this. The text should say something like the following: Needless to say, we only consider factors unique up to units.

[hidden haven]

This is the most based solution

[spice whale]

at that point you have to ask which element of the equivalence class do you use

[potent briar]

im being asked to see if vector space direct sum is associative

[next obsidian]

Oh boy…

So there’s a natural isomorphism

But like as actual objects they aren’t equal if you build them up as pairs

So its “associative in any way we care”

[potent briar]

yeah thats what i wrote

[delicate orchid]

same goes for the tensor product
stay tuned for more fun tensor product facts

[next obsidian]

Yup

But it isn’t strictly associative

You have to qualify it with “up to isomorphism”

And it ends up being natural in 3-components

And blah blah blah

Don’t worry about that, just say “associative up to isomorphism”

[potent briar]

👍

[barren sierra]

The first part of this

doesn't have anything to do with P being a Sylow subgroup right?

that's just a property of normal subgroups in general?

nvm it isn't

damn

[lethal dune]

yes, normality is not transitive

[barren sierra]

yea

Say n_p = number of Sylow p-subgroups of G

n_p = 1 => P, the Sylow p-subgroup of G, is normal in G

but does the converse hold?

it does right?

[gritty sparrow]

yes the converse holds, every sylow p group is conjugate to each other

[barren sierra]

since Sylow p-groups conjugate

yea

cool cool

[rapid slate]

Is there a name for this result?

I couldn't find a derivation on the net

Any book that has a proof?

[terse crystal]

Jacobson

Basic algebra volume 1 chapter 4

[rapid slate]

Thanks!

[lavish nexus]

man tf is a conjugate subfield

should just fix gK

[rapid slate]

The proof is not at all Elementary though 😂

[lavish nexus]

ah yes BA

is about very basic algebra

every proof is written in giant paragraphs

[rapid slate]

😂

[lavish nexus]

wait what is U

[rapid slate]

Group of units

Modulo n if its U(n)

[lapis trail]

Is someone willing to explain to me what the algebraic closure of a finite field is?

[delicate bloom]

not really different than an algebraic closure of any other kind of field

[chilly ocean]

Are inverses in groups unique? If we have element g∈G then would g^-1 be unique to g?

Sorry if this is a simple question just double checking my knowledge

[chilly ocean]

Good question

What do the elements look like

Think about algebraic extensions of finite fields

Once you think of one example you can extrapolate to figure out what algebraic closure looks like

[delicate bloom]

try to prove it

show what you come up with

[chilly ocean]

So then it is true? Hm I’ll have to think about it

[delicate bloom]

I will neither confirm nor deny

might also help as a hint to think about if a left inverse is equal to a right inverse too

[chilly ocean]

Okay just did some multiplication tables it seems to be true

[chilly ocean]

Idk any examples of groups that dont have unique inverses

I think we can use contradiction and suppose that there are two inverses then use the associative law to deduce they’re equal and thus they’re unique?

also it really comes down to fact that ab=ac implies b=c from cancellation rules

[chilly ocean]

Yeah that’s what I thought

very useful trick you should always keep in mind

[next obsidian]

Why do this as contradiction :)

You could just take two inverses and then use this to show they’re equal

No contradiction needed

[chilly ocean]

Oh fair point

Hi Chmonkey

[next obsidian]

Hello tocyem

[chilly ocean]

Thanks for the help

[next obsidian]

Swagger

[chilly ocean]

I'm a little confused. If $T \in \mathcal{L}(V,W)$ is a 1-1 function, then $T(x) = 0 \implies x = 0$. The proof of this is simple, so $T(x) = 0$ and so $T^{-1}Tx = T^{-1}0$ implies that $x = T^{-1}0$. And $T^{-1}0 = T^{-1}(0 + 0)$ so $T^{-1}(0) = 0$. Therefore, $x = 0$. However looking at the example $(x,y,z) \stackrel{T}\mapsto (x-1,y-1,z+1)$, we don't have that $T(0) = 0$, yet $T$ is still 1-1?

[cloud walrus]

n/c

[gloomy night]

Hey there. It's because the stated result only holds for linear maps, and the example you give is not linear.

[chilly ocean]

Oh I guess you're right T wasn't mentioned in L(V,W). I must have hallucinated... thanks anyway

[gloomy night]

Np

L(V,W) usually means the space of linear maps from V to W.

[chilly ocean]

Yeah I know

[olive mirage]

I'm looking at the ring of invariants of the polynomial ring C[x,y] under the group of order 4 generated by (x,y) \mapsto (-y,x). The only ones I'm finding are x^2*y^2 and x^2+y^2 but I'm pretty sure I am missing another one. Does anyone see it? or is anyone confident I'm not missing one?

(That is: Is C[x,y]^G = C[x^2+y^2, x^2*y^2] ? )

[delicate orchid]

oooh it's like a matrix meme

[olive mirage]

[The motivation is I'm trying to develop some intuition for surface singularities generated by cyclic quotients. And the intuition I do have says there should be a singularity, but if the answer to my question is yes, then the fixed point of G does not become singular in the quotient]

[But I'm guessing there is a thing involving the number i that I'm not seeing]

ah, xy and (x-y) are both multiplied by -1, so their product xy(x^2-y^2) is also invariant

[barren sierra]

is the reason we use <= for subgroup because subgroup defines a partial order?

so the notation makes sense in that regard?

[delicate orchid]

yeah I suppose

[cloud walrus]

Moldilocks1337 ✓

[delicate orchid]

Moldi say it ain't so

[hidden haven]

I ain't saying so

[delicate orchid]

[cloud walrus]

Moldilocks1337 ✓

[hidden haven]

😌

[delicate orchid]

moldi look in off topic voice and scroll up a bit

is this you?

[chilly ocean]

What is U(8)

What does it mean

I know it’s all the coprimes to 8

But like I want to check if it’s cyclic

[tribal moss]

It's the multiplicative group of units in Z/8Z -- in other words {1,3,5,7} under multiplication modulo 8.

Since the group has order 4, it is cyclic iff you can find an element of order 4. Checking all of them won't take long ...

[chilly ocean]

Hmmm,

How do u get 1 3 5 7 exactly?

So from all integers

[delicate orchid]

consider what happens when you choose a number not coprime to 8

[chilly ocean]

Idk what the first part of his sentence means tbh

It's the multiplicative group of units in Z/8Z

[chilly ocean]

polynomials in polynomial rings, f(x+1) irreducible implies f(x) irreducible

how?

[delicate orchid]

consider the function from the polynomial ring to itself given by f(x+1) -> f(x)

[next obsidian]

Just like

You can turn an expression

Making one of them reducible

And translate that

And get one for the other one

You might have to assume A is an integral domain

[chilly ocean]

oh yeah lol

a is a ufd

mb

[delicate orchid]

that's essentially what you do yeah, but idk I feel like you should show that the translation preserves irreducibility

[next obsidian]

Yeah

[delicate orchid]

it obviously does

but you should still show it

[next obsidian]

I mean this is where you Lowkey might need integral domain

I think I wrote my proof using that

[delicate orchid]

mm

yeah I can see if failing if x is a non-zero divisor but x+1 is a zero divisor

[next obsidian]

Wtf lmao

[delicate orchid]

beat me to it

[next obsidian]

What happened to catfan1336?

[barren sierra]

doesn't this need that H is normal in G?

not just that H is a subgroup of G?

Say P_H is a Sylow p-subgroup of H. I have by Sylow 2 that P intersect H is a subgroup of P_H

I think

so then I need the other direction of inclusion

and that's where I got stuck

cause I need that P intersect H is normal

for it to be a unique Sylow p-subgroup also

[proud bear]

for P\cap H to be a normal subgroup of H, no

[barren sierra]

oh yea

cause P is normal

ok yea then I'm stuck

[next obsidian]

Normal p-subgroup is unique

[chilly ocean]

nice problem

[next obsidian]

Show that P\capH is a Sylow-p subgroup of H

[barren sierra]

yea that's where I'm stuck

[barren sierra]

I showed this

or well I realized it

cause it's known fact

[next obsidian]

It has a p-Sylow, call if F

For F = P\cap H it’s equivalent that F < P

[barren sierra]

it is?

lemme try to parse that

ok so F < H clearly, how does F < P imply F = P cap H

doesn't F < P only then show F < P cap H

[barren sierra]

nvm I got it

cause sylow p-subgroups are maximal in size in a sense

[lavish nexus]

usually you can consider |PH|

as to normality we just want P\cap H to be normal in H

so conjugate by any element in H and see what happens

[barren sierra]

you can't tho

cause H isn't normal

so I don't know if that cardinality argument can be made?

[lavish nexus]

you can

P is normal

in fact you don't need normal here

|PH|=|P||H|/|P\capH|

so |P\capH|=|P||H|/|PH|

[barren sierra]

|PH|=|P||H|/|P\capH|
I thought this only held if both were normal?

if not then this problem would have been easier 💀

[lavish nexus]

one normal is enough for PH to be a subgroup

but the order formula works for any groups

I think this is in dummit 3.2

basically if A is normal in G and B is any subgroup
A\cap B is normal in B
then aba'b'=aa'a'^(-1)ba'b' is in AB
so it is a subgroup

hmm also this is directly second iso thm
I'm kind of going in circles

[next obsidian]

You only even need one of them to be a subgroup of the normalizer of the other for it to be a subgroup

[prisma shuttle]

can someone explain wut this notation means

[proud bear]

$k^{\oplus n}=\underbrace{k\oplus k\cdots\oplus k}_{\text{$n$ times}}$

[prisma shuttle]

wait so like wut does that mean

i am a bit confused what k represents here

is k a basis?

[thorn delta]

k is a field

[prisma shuttle]

oh i see ok thx

also is anyone applying to virginia reu (or any other program through the mathprogram.org application)

because basically i had a teacher who wrote a rec for another program through mathprograms.org and they uploaded a generic letter and now for the virginia reu program it also has that letter uploaded

and even when i press the "x" button when i refresh the check mark comes back, meaning it'll be sent

[wise igloo]

can someone explain to me what this means?

if you need more context lmk

im confused on what is unique exactly

[barren sierra]

So (1 2 3) can be written as (1 2) (2 3)

But it can also be written as (1 2) (2 3) (1 3) (1 3)

Cause those last two permutations cancel

[wise igloo]

mhm

[barren sierra]

And 2 and 4 (the lengths of those chains of transpositions)

Are even numbers

And that theorem says that (1 2 3) can NEVER be written as an odd number of those transpositions

[wise igloo]

oh so for that cycle it will never have an odd number of permutations basically?

alright

[barren sierra]

There's no way

Ye

[wise igloo]

so same works if you had an odd number then

alright that clears it up

tysm

[barren sierra]

Ok so now I got a question

This seems obvious but there may be a dumb exception

If every subset S of P has a maximal element of S in S

Must P have a maximal element of P in P?

it must right? Cause a set is a subset of itself

[lethal dune]

zorn's lemma?

[patent crescent]

My class defined a right coset of H with respect to b as the set of all elements in G which when you multiply by inverse of b on the right, it is in H

But online I see that it is just multiplying on the right by b instead of inverse of b. Is there a reason for this difference?

[lethal dune]

honestly i've never seen a right coset being defined like that

[barren sierra]

Idk?

I hope not

[lethal dune]

any chain of P is a subset so it has an upper bound (by your assumption)

[lavish nexus]

it is

[patent crescent]

damn my professor making it confusing then :((

Anyways, it's still basically the same thing since inverse of b is also in G right?

[lavish nexus]

this is basically Hb

I mean what is Hb in the usual definition

{hb|h in H}

[barren sierra]

It is? #5

[lavish nexus]

so for any hb in Hb, multiply on the right by b^(-1) you get h in H

[lavish nexus]

every non empty subset is still a poset

so treat as an ascending chain

[barren sierra]

Ok so S has a maximal element in it right?

Or I mean

P

[lavish nexus]

this is definition pushing this is not Zorn's

[barren sierra]

hm

Ok

[lavish nexus]

every S has a maximal element

so every ascending chain has a maximal element

[next obsidian]

One cannot fear Zorn’s

Zorn’s is your friend

[chilly ocean]

hey guys if a group has a prime order how would it have 1 generator?

or 0

[lethal dune]

it has ϕ (p) generators

[chilly ocean]

what does that symbol men

mean

[lethal dune]

Euler phi function

[chilly ocean]

so how many does it have

[lethal dune]

ϕ(p)

[lavish nexus]

lol

[chilly ocean]

bruh

ive never seen that in this class

[lavish nexus]

what can the order of an element be?

What does Lagrange say

[chilly ocean]

uh

i dont think this has anything to do with lagrange tho

the order of order(h) | order(g)

to be a subgroup

[lavish nexus]

Sure it does

yes

[chilly ocean]

thats what it says

[lavish nexus]

but every element generates a cyclic subgroup

and the order of that subgroup is the order of the element

so anyways what can the order of the cyclic subgroup be?

[chilly ocean]

how are you getting cyclic subgroups from this

all i said was that a group has prime order

[lavish nexus]

take any element g

[chilly ocean]

oay

ok

[lavish nexus]

{g^0=e, g, g^2,…}

is a subgroup

because the whole group G is finite

This also has to be finite