groups-rings-fields

[warm ember]

but by definition v1^2 is not necessarily in the extension

if you want it to be in the extension you need to seperatly add v1^2 to the basis

the basis dont interact with each other or itself

[chilly radish]

$v_1^2$ is absolutely in $K(v_1,\ldots,v_{n-1})$

[cloud walrus]

ShiN

[chilly radish]

i'm not taking a span here

[warm ember]

yes

[chilly radish]

this is notation for the field generated by these elements

[warm ember]

but that means v1^2 is also part of the basis

[chilly radish]

it might be

it might not be

[warm ember]

if its not then its generated by some other linear combination of basis

but you cant say v1 x v1 =v1^2 so therefore v1^2 is in the ring

remember the field extension is just a vector space

[delicate bloom]

that's not a basis, it's linearly dependent, 1+w+w^2=0

[warm ember]

o

my bad

1,i,w,iw

sorry

[chilly radish]

yes you can!!

it's a vector space, but it's also a field...

[warm ember]

bruh

reread definition of basis

[chilly radish]

bruh

do you know what an algebra is

[warm ember]

no

[chilly radish]

[warm ember]

i know what group ring field vector spaces are

i mean i know what an algebra is but ive never done anything with it

[chilly radish]

a field extension $L/K$ is also a $K$-algebra naturally, so thereiis an internal notion of multiplication. In particular, if $v_1,\ldots,v_n$ is a basis of $L/K$ as a vector space, this does not contradict the fact that $v_1^2\in L$, simply because multiplication is defined

[cloud walrus]

ShiN

[chilly radish]

I am not sure what you are talking about

anyways, you can always find a basis element that if you exclude, the degree goes down, and conclude by degree considerations

[warm ember]

ok idk what you mean but in linear algebra basis dont interact with each other or itself and the same is true in field extensions

[chilly radish]

you have two DIFFERENT structures here

that interact

[warm ember]

they dont

[chilly radish]

you have multiplication which interacts with the vector space structure

[warm ember]

its a matter of definition not a big deal

reread the definition

its important for defining the degree of an extension

[chilly radish]

kevin, I think you need to review the definition of an algebra over a field

[warm ember]

ok

idc about algebras

im too young for algebra

[brazen slate]

wait til you learn what channel you're in

[delicate bloom]

I think this should be an example, take the degree 2 extension {1, sqrt(2)} then remove sqrt(2), now you can't generate sqrt(2)

[chilly radish]

For posterity: I am assuming $L/K$ is seprable, so I can choose the basis to be powers of a primitive element, then I can just choose e.g. the square of the primitive element

[cloud walrus]

ShiN

[chilly radish]

or sorry, I want to remove the primitive element itself

i.e. if $L=K(a)$ then $a\notin K(a^2,a^3,\ldots,a^{n-1})$

[cloud walrus]

ShiN

[warm ember]

ok consider

Q(cbrt(2))

it is deg 3 extension

meaning it has basis of size 3

and 1,cbrt3 isnt a basis since cbrt9 isnt in Q[1,cbrt3]

the basis is 1,cbrt3,cbrt9

[chilly radish]

uhh, it is

cbrt(9) = cbrt(3)^2

but it is not a basis

you are confusing notation here

square brackets mean the ring generated by Q and those elements

this include linear combinations of PRODUCTS of those elements

this is different than the Q-linear span of these elements

which is smaller, and also lacks a field or ring structrue

hmmm taking a primitive element doesn't work a priori either I think

[warm ember]

yes

basis are just like span

and rings do just that

[delicate bloom]

I think it's false if you have a primitive element like that

suppose $a$ satisfies the minimal polynomial $$\sum_{k=0}^nc_ka^k=0$$ Then we can pull out $c_1a+a^n$ and factor out $a$ to get, $$a=\frac{-1}{c_1+a^{n-1}}\sum_{k=0}^nc_ka^k$$

[cloud walrus]

Merosity

[delicate bloom]

sum should have k != 1 or n

[chilly radish]

hmmmm

[delicate bloom]

so we can make a from what you listed, a^2 to a^{n-1}

[chilly radish]

then i'm not sure why this proof i'm looking at is true

[warm ember]

the word basis in extensions is called basis cuz its the same as the word basis in vector spaces

[delicate bloom]

ok and

[warm ember]

they dont interact

am i missing something??

[delicate bloom]

well c_1 + a^{n-1} is invertible so it just boils out to a linear combination

whwen you multiply it out together

[chilly radish]

mero, i'm trying to show that finite extensions of complete discretely valued fields are complete discretely valued. On completeness we want to show that if you take a basis for the extensions, then a cauchy sequence gives cauchy component functions. at one point we derive a contradiction by taking that the n+1-st element of the basis is not in the extension generated by the first n basis vectors

[delicate bloom]

idk, is that not invertible here cause we don't have a

[chilly radish]

my question is essentially why can you always find such a basis element

[warm ember]

just because im high schooler doesnt mean everything that i say is bs 😭

[chilly radish]

or is this proof wrong

it's not because you're a highschooler, it's because you're fundamentally misunderstanding how different algebraic structures can interact

[warm ember]

no bro

you misunderstand

[chilly radish]

it's true that 1,cbrt3 is not a basis of Q[1,cbrt3], but it still generates Q[cbrt3] AS A RING, which is what the square brackets mean

[warm ember]

degree of extension is same as dim of vector space over base field

which is size of basis

[chilly radish]

you can have generating sets that are smaller than bases

K[x] is generated by a single element, but it is infinite dimensional over K, where x is some indeterminate

[warm ember]

we are looking at generating rings here

like generating vector spaces

[chilly radish]

you might be

i'm not, the notation of square brackets has a very specific meaning

[warm ember]

yes i know

but basis dont interact so its like generating a ring

but if the basis is correct then the ring is actually a field

[chilly radish]

this makes no sense

[warm ember]

ok

[chilly radish]

@rocky cloak I see you lurking, any idea?

[warm ember]

https://en.m.wikipedia.org/wiki/Field_extension

Given a field extension L / K, the larger field L is a K-vector space. The dimension of this vector space is called the degree of the extension and is denoted by [L : K].

[chilly radish]

yes, that is a true statement.

[rocky cloak]

What are cauchy components? Like you express each element of the sequence in the basis?

[chilly radish]

I meant like

yea

expand out the sequence w.r.t the basis

and consider the components of each of the basis elements

these should be cauchy sequences

I mean that's just background

[rocky cloak]

Alright, and then since the components converge the whole thing does, yeah seems reasonable

[chilly radish]

nono

that's not the question haha

there's a detail in the proof i'm not sure about

that's what i'm asking about

[chilly radish]

this is true just by the defn of convergence in valued fields

I'm asking if for a finite separable extension $L/K$ of degree $n+1$, can you always find a basis $v_1,\ldots,v_{n+1}$ such that $v_{n+1}\notin K(v_1,\ldots,v_n)$

[cloud walrus]

ShiN

[rocky cloak]

I would expect you can basically never do that...

[chilly radish]

then i'm not sure why this proof we gave is true at all

I think maybe i'm misunderstanding the notation my prof is using

nono

something is wrong here

[rocky cloak]

What is the proof, or what are you proving

[chilly radish]

this is the theorem, specifically the part of proving the extension is complete

this is the proof

[rocky cloak]

I don't see the problem

This is just what you said before right. That a the components become Cauchy sequences and then it's complete

[chilly radish]

the problem is the contradiction

to prove that the compoennts are actually cauchy

specifically this part

the contradiction is only given if L(b_1,...,b_n) has strictly lower degree

[rocky cloak]

But L<...> is just the linear span right?

[chilly radish]

then why would the induction hypothesis hold

we are inducting over degree of extension

so in particular we are inducting on field extensions

[chilly radish]

that's what I thought

[rocky cloak]

I think they're just inducting over the size of the basis (which equals the degree of the field extension)

[chilly radish]

how would you phrase the induction hypothesis in this case, since E is fixed

[rocky cloak]

I guess it would be a little cleaner to say that you're proving that L<b1, ..., bn> is complete, then do induction on n

[chilly radish]

saying L<b_1,..,b_n> is complete is a little murky since we can't define a valuation on a vs

I guess this is a semantic matter moreso

if we consider L<b_1,...,b_i> as subspaces of E I guess that's fine

[rocky cloak]

Yeah it would be a subspace of E

[ivory trail]

are you confusing generating an algebra with generating (spanning) a vector space

basis elements don't "interact" in the sense that they are linearly independent, but you often have algebraic relations between these basis elements in a field extension, meaning they satisfy some nontrivial polynomial (or multiple)

[tardy hedge]

If im given some subset of a group, what are quick ways or faster tricks of checking its a subgroup or not?

I know by lagrange thm, if number of elements doesnt divide order of G then automatically it cant be subgroup

[abstract rock]

a(b^(-1)) is in the set for all a b in the set

[tardy hedge]

The eg i have rn is this

Group is units of Z21 and subset is {1,4,10,13,16,19}

So do i just have to manually check that condition

For each element?

[delicate orchid]

each pair of elements, although you can infer some pairs from others

[tardy hedge]

Yeah

So basically, u do kind of have to manually “check”

[abstract rock]

there isnt another elementary algorithm thats faster, so yes you do have to check

[low wyvern]

I had this question...

[tardy hedge]

I see

[abstract rock]

this requires you generate the inverses first ofcourse

[low wyvern]

It seems like you have to look for like inverses and that

[tardy hedge]

Seems like it could take a min

[abstract rock]

whats the question

[abstract rock]

most things with groups do

[tardy hedge]

I see, thats good to know tbh

Know what to “expect”

[low wyvern]

quick way to find 'all' subgroups

but it looks like you just have to check systematically

[abstract rock]

there are some algorithms that require highly non trivial results

that are faster than this

[low wyvern]

I see

[tardy hedge]

Random thought but classes really do go so quickly thru material

Its like u just have time to have some sort of grasp of the topic before moving on

[low wyvern]

Same with mine, and they leave out a lot of detail...

[tardy hedge]

Lol my prof doesnt even give accurate definitions

Its so dumb

[low wyvern]

are you UK?

[tardy hedge]

Canada

Hahaha

[regal star]

Hey Guys I am new here. I am following Dummit and Foote's book on abstract algebra and I am on Class Equations. Solving questions takes a lot of time? Any advices in general?

I am stuck at this question Q19 now. Any hints would be much appreciated

[tardy hedge]

Someone else also thought he knew me

[low wyvern]

We all just struggling with groups...

[abstract rock]

is it a big uni? i'd be surprised

there's a lot of good algebraists at the big unis

[abstract rock]

Its generally gonna take a lot of time until you develop the intuitions that work for you. How far into 19 are you? Have you built up some examples, (say from S4 or S5?)

[regal star]

For now I am trying to prove the first part of the problem. I am not able to figure out how to calculate k, I know x is in K and If I could calculate the centraliser of this then I think maybe able to work out the number of conjugacy classes

[grim hollow]

Hey all, assuming this is the abstract algebra chat. I'm currently taking this in uni with Paulo Aluffi's Notes from underground textbook.
I would like to master this topic. Not sure how popular this text is, but is anyone interested in doing every problem and writing a solutions manual together?

[abstract rock]

got a link?

[abstract rock]

I think that hint is plenty useful if you remember some stuff about inner products

whats getting you stuck with the hint in mind?

[abstract rock]

cool! would it be alright if i could help with some specific chapters?

[regal star]

Inner products? I am not able to find the centraliser

[grim hollow]

Yea! Which ones are you interested in?

My semester covers Chapter 1-5, 10, 11, and 12. So I would like to prioritize those

[abstract rock]

HC_G(x) is the innerproduct of H and C_G(x) no?

[regal star]

Oh yes but how is it related to that I don't see

[abstract rock]

Well the hint talks about how the intersection of those two is non trivial, so that innerproduct wouldnt be isomorphic to the direct product, maybe im wrong but perhaps that could tell you about the size of its index with respect to G

[abstract rock]

I can certainly help with those, was hoping to recall some stuff doing chapters 7-9 and 15 as well

[regal star]

Thanks I will give it a shot

[abstract rock]

bon chance!

[warm ember]

yes i know

im not confusing anything with anything

im not an algebrist or whatever and i dont do stuff with algebras

[ivory trail]

I just don't know why you were so persistent then

[warm ember]

ok

they dont know how basis of field extension work

[ivory trail]

you are confused

[warm ember]

ok idc

i was trying to help someone now im getting bullied

[terse crystal]

Use H CG(x) / CG(x) being isomorphic to H/ (H cap CG(x))

[abstract rock]

guy asked for a hint not the answer :(

[terse crystal]

Editing

[warm ember]

just becuas wim a highschooler doesnt mean i have to get bullied

but i do understand cuz i bully elementary schoolers who claim to know calculus

[abstract rock]

if you cant help, dont help friend

[warm ember]

Xooks

[terse crystal]

Who was helping whom? Checking previous messages…

[abstract rock]

the comment easnt for you mr.mind, it was for yang

[warm ember]

.

[terse crystal]

I see, thought it was resolved long ago

[warm ember]

not really but ok

[cursive spindle]

Is it a requirement that every irreducible polynomial must be monic?

[crystal turtle]

No. Consider 2x over Q

[delicate orchid]

nope

minimal polynomials have to be monic though

maybe that's where you're getting confused

[rocky cloak]

If you're working over a field then every polynomial is associate to a monic one, so you might as well consider monic polynomials. In general you can have irreducibles not associate to a monic (like 2x+1 over Z)

[abstract rock]

there's also the cyclotomic polynomials over Q or R, no?

[rocky cloak]

What about them?

[abstract rock]

they're not reducible to monics

[rocky cloak]

They are monic, or what do you mean?

[abstract rock]

oh woops, my bad!

[cursive spindle]

Atm I don't care about Q

I'm working with Z

[crystal turtle]

still no.

And why don't you care about Q we love Q here

[cursive spindle]

I don't

I'm working with Z/nZ

and the special case F_p

[delicate bloom]

you're in a field so that's all that matters

a different prime field no less, basically Q

[cursive spindle]

Is it better to only care about monic cases?

Well I'm counting both cases and see what I get

I had another question but I forgor

[delicate bloom]

your coefficients are in F_p?

[cursive spindle]

My coeffs are all Z

I mean Z_n

[rocky cloak]

n prime or not?

[cursive spindle]

iirc Z_p = F_p

[delicate bloom]

"yes"

[cursive spindle]

not necessarily prime

Z_n where some stuff are rings and others are fields

[next obsidian]

[rocky cloak]

If n isn't prime then you should also consider non monic polynomials

[delicate bloom]

use CRT to separate into powers of primes, then work in F_p and Hensel lift back later

[cursive spindle]

right

[delicate bloom]

Hensel lift from F_p up to Z_p so you're in Q_p, so you have Q hanging around too why not

[cursive spindle]

There aren't many "non-monic" poly over Z_n so I thought why not include them

Oh yeah I now remember what I was gonna say

is there a special notation where you only denote linear polynomials or quad or whatever in F[x]?

[next obsidian]

F[x]_n may work for degree n polys, this is notation used for graded rings. In some sense F[x]/(x^n) is degree < n polynomials

But you should just say how you’re gonna denote it tbh

[cursive spindle]

also fair

Yeah I'm just gonna create some fancy notation for that

[cursive spindle]

Is there a way to count how many irreducible monic polynomials are there given a degree n and Z_m?

[rocky cloak]

If m=p is prime, you can use the fact that there's a unique field of order p^n , then use a little Möbius inversion to get that the number of minimal polynomials of degree n is

Sum μ(d) p^(n/d)

If m is square free you can use Chinese remainder theorem, and reduce the problem to the primes that divide m.

If m is not square free like m=p^k it's a little more complicated. Like you can account for most of the polynomials using Hensels lemma, but you have stuff like x^2 + 1 modulo 4, which I don't see any obvious way to account for.

[cursive spindle]

Can you show me an example of how to do the second case?

[delicate bloom]

I guess the differentiator here is unramified vs ramified extensions of Q_p

like, all the irreducible polynomials of unramified extensions of Q_p end up reducing down to irreducible polynomials of F_p, and then the extra junk lifted up with them with Hensel in Z/p^kZ

[rocky cloak]

Hmm, that's interesting.

So say m=6 = 2*3. Then Chinese remainder theorem gives us that

Z/6 [x] = Z/3[x] \times Z/2[x]

A polynomial of the form (f, g) can be factored as (f, 1)*(1, g), and (f, 0) is factored as (f, 1)*(1,0). So the irreducibles are on the form (f, s) or (s, g) with s a unit in Z/2 or Z/3 respectively. But such polynomials will never be monic, so I guess there are no irreducible monics in these cases. For example

x + 1 = (3x+1)(4x+1)

[cursive spindle]

Thanks

[boreal inlet]

I am for the 2nd part, I am unable to find proper submodules so that they are disjoint, hence I can't express C^3 as the required direct sum.

If someone has any hints for this it would be appreciated, as I am stuck at this for hours

[boreal inlet]

<@&286206848099549185>

[cobalt heath]

How do I avoid common blunders like thinking there is no surjective homomorphism from free group of 2 generators to Z/2Z * Z/5Z * Z/10Z

[terse crystal]

That matrix has three distinct complex eigenvalues

[terse crystal]

By not missing Z/2Z times Z/5Z = Z/10Z next time

[cobalt heath]

How

How do I keep myself from missing them?

[terse crystal]

Idk

[cobalt heath]

😭 I guess I cannot avoid some deductions then..

[open sluice]

practice more?

[cobalt heath]

Hmmm

[boreal inlet]

OH WAIT SO

It's all about the eigenvalues

I just did the R^3 case

e1 and e3 are enough to generate it

[terse crystal]

Looks irrelevant

[boreal inlet]

Because (2A + I)e_1 = sqrt(3)e_2

[terse crystal]

What I wanted to say was
Ax=λx gives you a cyclic F[T] module Fx, which is isomorphic to F[T]/(T-λ)

[boreal inlet]

Or should I say 2x + 1

[terse crystal]

My bad, three cyclic R[T] modules not three cyclic C[T] modules, rethinking

[cobalt heath]

Maybe the mistake of mine was just because I was out of time solving the problem

[boreal inlet]

@terse crystal figured it out

(1,i,0), (0,0,i), (0,0,1) are the generators of the each of the cyclic R[x] modules

And as C[x] modules, we can safely omit the 2nd one as that's generated by the third

[terse crystal]

I was thinking four , (1,i,0),(1,-i,0), (0,0,1), (0,0,i)

[boreal inlet]

Do you need that one?

[terse crystal]

Looks like if you remove the one generated by (1,-i,0), it’s not going to cover C^3

[boreal inlet]

Wait..

wait how

[terse crystal]

Either i am being dumb, or this question might be wrong. Cause you view C^3 as R^6, three real three imaginary, then T acts on R^6 by matrix diag{A,A}. So since you have written R^3 as direct sum of two cyclic submodules , then R^6 here is direct sum of four cyclic submodules

And four many add up to dimension 6: two copies of R[T]/(T^2-2cos(2π/3)+1) , two copies of R[T]/(T-1). 2+2+1+1=6

[coral steeple]

Is there a name for monic quintics whose quadratic term has coefficient 1?

[tiny jolt]

Any ideas for how to prove this? We defined the determinant as a wedge product and I'm not sure what I should be taking the wedge product over. (k is a field)

[terse crystal]

If A ^ otimes n you mean n many A taking tensor product, and A otimes B you mean the blocked matrix (aijB), then for m times m A, n times n B, we have det(A otimes B)=det(det(A)B^m)=det(A)^n det(B)^m. From this you can easily generalize it to det of multiple otimes . Second looks trivial

[coral spindle]

Can someone give me a quick hand with magma? I'm wanting to describe a subgroup of GL_8( Q(i) ) generated by a handful of matrices. I realise this is probably very simple, but I would be very appreciative of a code example as I've not used magma before.

A simple example of a matrix group over Q(i) with any number of generators would be very helpful.

[delicate orchid]

if G = GL_8(Q(i)) then you can make subgroups like sub<G | g_1, g_2, ..., g_n> where ur g_i are your generators

[coral spindle]

Ah ok so sub is the generator notation

So how can I describe Q(i)? I'm working through the docs slowly and haven't got there yet

I won't ask much more lol this is the wrong channel really

[delicate orchid]

#computing-software if you want but I feel like it's fine here

I've answered magma questions in here and #advanced-algebra before

[coral spindle]

I fear people will start saying "just use matlab 🤪" if I go to that channel

[delicate orchid]

ext<Q | x^2+1> should work

[coral spindle]

Oh very nice

This thing is p good huh

[delicate orchid]

yeah you'll say that until you try and do anything even remotely weird

like, oh idk, try and work with fusion systems?

[coral spindle]

Sounds like a you problem

[delicate orchid]

my supervisor made a library for them but the uni will not buy me the full magma version so I can't use it

[untold cloud]

Hi, guys, let k be a field, i know that k(a_1,...,a_n) = {f(a_1,...,a_n)/g(a_1,...,a_n) : f,g\in k[x_1,...,x_n]}. But Let A be a set, what does the element in k(A) look like? A could be uncoutable

[delicate orchid]

the same? I don't see why A being uncountable would affect anything

you just couldn't write it using that notation

[untold cloud]

i am thinking if A is uncountable, I write element in k(A) as f(a_1,...,a_n)/g(b_1,...,b_m) where f,g\in k[x_1,x_2,...], a_i,b_j\in A. But doesn't k[x_1,x_2,...] give a way to count elements in A?

[delicate orchid]

_ _

[untold cloud]

But what notation should i use

[delicate orchid]

I'd just use f(a)/g(b)

with a,b being understood to be elements of $\prod_{i \in \mathbb{R}} A$ if you want to be explicit

[cloud walrus]

WewGhostTbh

[untold cloud]

Oh, that makes a lot of sense

thank you!

[delicate orchid]

np

[south patrol]

Why R lol

[delicate orchid]

first uncountable set that came to mind

[south patrol]

Sure just seemed a lil misleading

[delicate orchid]

I will mislead and you are powerless to stop me

[crystal turtle]

Idk, potato has some power now

[surreal acorn]

I'm looking for people for a server that deals with nature that helps not contaminate anyone wants :))

@here

[open sluice]

I will contaminate this channel with analysis

[crystal turtle]

Nooooo

[cursive spindle]

Algebraists gonna hate us

[rustic crown]

.<

[open sluice]

e

[verbal tulip]

Is there a systematic way to find which functions belong to certain irreps of a group?

I'm struggling to find functions that belong to the one dimensional irreps of the tetrahedral group and I can't find any intuitive way to tackle this issue.

[delicate orchid]

what do you mean by "functions belong"?

if you mean class functions on G then you would take inner products with the irreducible characters to find the decomposition coefficients

[verbal tulip]

I mean that for each irrep of a group there is a set of functions, usually called partners, that transform among themselves when operated on by the matrices that form the irrep. Finding these functions has been very difficult for me.

I think I have an idea of what you're trying to say. If I understand correctly, for each irrep there is a projector operator that depends on the irreducible characters that will project an arbitrary function to the function that belongs to the said irrep. But in practice, these are difficult to implement, even for relatively small groups like the tetrahedral.

[delicate orchid]

they're not difficult to implement, you just do the inner product. Explicitly multiplying by the projection idempotent inside the group algebra is unwieldy I'll agree there but it would be possible for a group of order 12

haven't heard of these "partners" things before, I'll have a google

[delicate orchid]

there seems to be next to no information on them, could you link a source?

[verbal tulip]

Thank you for your help. Wigner calls them partners in "Group theory and its application to the quantum mechanics of atomic spectra", chapter 12.

And so does Hamermesh and Tinkham.

[delicate orchid]

ah it's applied that'll be why I haven't heard of them!

[torn warren]

I want to construct this field by following this instruction, if I set p=3, k=2, then the order of F is 9, but what is F?

F can't be Z9

[delicate orchid]

F_3[x]/(x^2+1) as the definition says

[torn warren]

is F_3[x]=Z_3[x] ?

[delicate orchid]

do you understand the definition that you've posted

[torn warren]

it is to construct the coset of the maximal ideal, right?

[delicate orchid]

yeah but you're asking what F_3 means when the thing you've posted uses F_p so

[torn warren]

I don't understand what is F_3 is, because I need to choose coefficient from F_3 when construct the polynomial

if I don't know what F_3 is, how can I choose coefficients...

[winter shore]

In an ordered ring, does it follow that a<b implies b=a+e for some e>0?

seems not but can't figure any counterexample

[lapis latch]

How about e=b-a

[delicate orchid]

you can think of it as Z/3Z, yes

[winter shore]

damn I'm very dumb sometimes ty

[torn warren]

but I am confused, let K=Fq, and if the char of K is p, can I say 1_K is the generator of K?

I mean to treak K as a group

[main needle]

Working on a proof that required to show this : if $a_1, \cdots, a_{2g} (g > 1)$ are complex numbers (if it helps they are eigenvalues of an invertible matrix with integer coefficients). Want to show that I can find an integer $n >1$ such that $\sum a_i^n \neq \sum a_i$. This feels very easy but I don't have any immediate ideas. I am thinking about how characterstic polynomial change when you raise the matrix to a power but I am not getting it

[cloud walrus]

ru0xffian

[rocky cloak]

Since K has characteristic p, the subgroup generated by 1 will have p elements

So it's not all q of them (unless p=q)

[torn warren]

If K has dimension n, then let k1, k2, ... kn to be the basis, (where k1=1_K), can I say k1 generate a subgroup of K and this subgroup is F_p which has p elements? and similarly, k2 generate a subgroup with p elements (is not equal to F_p) ... kn generate a subgroup with p elements (is not equal to F_p).

[rocky cloak]

Yes, as a group K will look like (Z/p)^n

[torn warren]

is Z/p= Z/pZ ?

or do you mean Z_p?

[summer path]

Z/p is standard notation for Z/pZ

Z_p is p-adic integers

[torn warren]

i see, first time to see this notation :_)

[prisma ibex]

You can always identify F_p with Z/pZ, take 1 as the generator

for something like F_9 the underlying Abelian group is going to be Z/3Z x Z/3Z, and the multiplication is going to be determined by the polynomial you used in the construction

[torn warren]

so if treat K as a group, then it is like a n-tuple (a1, a2,..., an) and each ai is selected from Zp, hence there are totally p^n elements, and each slot _ of (_, _, ..., _) is the basis vectors ki, and ki is in K, if write it in vector form we get a1 k1+...+ an kn, is this correct?

[prisma ibex]

yes you can always write field extensions as vector spaces over the base field

[rocky cloak]

Yes, allthough writing it this way is perhaps not the most helpful if you want to understand the field structure

[delicate orchid]

important to note that even though the multiplication is different, the fields will still be isomorphic regardless of what polynomial you chose

[torn warren]

what is the best way to write this?

I just learn field and still used to the group format...

[delicate orchid]

as polynomials

for example I would write the elements of F_4 = F_2[x]/(x^2+x+1) as 0, 1, x, 1+x

[torn warren]

wow!

how do you choose the coeffficients to be 0 and 1, do you treat F_2 as Z_2?

[delicate orchid]

yes

[torn warren]

thank you!

[delicate orchid]

np

[hazy flax]

I have a small question regarding the proof of this

[cloud walrus]

PROnoob

[wraith cargo]

so n/d is the maximal number that divides both n and m
Now if we divide one of these by the gcd, they no longer have any common divisor apart from 1 since we just removed the one they had

[hazy flax]

oh i see

is there a direct way to go from the second last step to the last one?

[wraith cargo]

you can probably argue by like prime factorization

[hazy flax]

hmm

i'll have to think about it

thank you!

[hazy flax]

i just realized that for your argument to work we need to divide both n and m by the gcd.

if we divide just one of them, there may still be common factors

so like if m=n^2

[torn warren]

should I use choice 1 or choice 2?

the irreducible poly is not unique...

[delicate orchid]

They produce isomorphic fields it doesn’t matter

Pick either or (I recommend x^2+1)

[summer path]

or do it both ways and verify that they are isomorphic

[prisma ibex]

Are you sure this first polynomial is irreducible?

[delicate orchid]

Or prove there’s only one dude of each order

[delicate orchid]

Wait is that an x^3 or an x^2

[prisma ibex]

Yeah I’m assuming this is a mistake

[delicate orchid]

I presumed x^3 but then theyre not the same order so yeah

[summer path]

i don't read things that aren't texed

[karmic moat]

i close my eyes and compile the tex in my brain

[hazy flax]

here's what i came up with but i think i might be making some mistake

[cloud walrus]

PROnoob

[white oxide]

how exactly does this follow?

i understand that p can't contain any units

but i don't understand the converse

[karmic moat]

maximality

any non-unit is contained in some maximal ideal

(shoutout to zorn's lemma)

[white oxide]

oh is this a theorem

[karmic moat]

yep, prove it :)

[white oxide]

classic lang

[karmic moat]

anyone know if this can be proved without zorn's lemma (or some equivalent)? or if this is equivalent to zorn's

[delicate orchid]

No need for zorns here

If x is some non-unit element outside of p then p U (x) is a strictly bigger proper ideal, contradiction :uponthewitnessing:

[delicate orchid]

Showing this requires zorns tho

Can’t see anyway of doing it otherwise

[karmic moat]

oh yeah that's what i meant

[white oxide]

does anyone know if there's a copy of solutions to lang's algebra

[summer path]

if you aren't familiar with the kind of zorn's argument, do it as an exercise

[next obsidian]

You can prove existence of primes from the ultrafilter lemma which is weaker. Existence of maximal ideals in this form requires Zorn’s I believe

[torn warren]

it is x^3, but the book says the irreducible ploy is unique (up to a constant factor) and is a minimal degree greater or equal to 1

[torn warren]

it is x^3 , so it is irreducible

[delicate orchid]

the minimal polynomial should be unique, but... those polynomials are different degrees?

why would they be correlated?

the quotients get you two completely different fields

F_27 and F_9

[torn warren]

why can't I choose x^3+2x+1 as the irr poly?

[delicate orchid]

you can?

you just get F_27, because it's degree 3

[torn warren]

I know it is degree 3, I don't konw why can't I choose degree 3?

is that due to degree 3 is not the minimal poly?

[delicate orchid]

... because you want to get F_9...

so you need a degree 2... because 9 is 3^2, not 3^3

[torn warren]

ah, i got it...

sorry i am confused just now, the book says it is unique, so do you mean if I fixed the degree of irr poly first, then the irr poly is unique?

in this prob, it is to ask to construct the 3^2, so the degree must be 2

[delicate orchid]

no, the minimal polynomial is unique

there can be many irreducible polynomials of a given degree (even up to scalar multiples)

[torn warren]

is the book wrong?

[delicate orchid]

for example, x^2+4x+1 and x^2+x+1 are both irreducible in F_5

[delicate orchid]

this is the minimal polynomial of alpha, as I said

and as the book says, it just says it's irreducible not that it is the irreducible

[torn warren]

do you mean, after I plug in alpha, since alpha^2 will be expressed by the linear combination of 1 and alpha, then the two ploy alpha^2+4alpha+1 = alpha^2+alpha+1 ?

[delicate orchid]

the polynomial p(x) is the polynomial of minimal degree that has alpha as a root

it should be obvious why it has to be irreducible

[torn warren]

I understand it is irr, because <p(x)> is the max ideal, then implies F[x]/<p(x)> is a field, I don't understand why it says unique but your counter example shows not uniqe?

[delicate orchid]

father give me strength

YOU claimed the "irreducible polynomial was unique", my post was a counter example to that claim

the theorem you posted, makes a different (true) claim

[torn warren]

oh, it says alpha is a root, so given this fact, then the irr is unique, right?

i miss the alpha as a root

[delicate orchid]

the corrisponding irreducible (called the minimal polynomial - third time now) is unique yes

[torn warren]

your example shows if the alpha is not first given, then the irr ploy is not unique, right?

[terse crystal]

First we assume |aj|<=1 otherwise n big enough it can be proved. Now if we really have tr(A^n)=tr(A) for any n.
Then recall that there is its characteristic polynomial b0+…+bm x^m (m=2g)
So b0+…+bmA^m=0
Multiplied by A both sides, then take trace , we have (b0+…+bm)tr(A)=0
We also can take trace directly of b0+…+bmA^m=0
(b0)m+(b1+…+bm)tr(A)=0
We have b0(m-tr(A))=0.
Since A invertible, b0 non-zero
tr(A)=m. And their length<=1, so either all aj are 1, or all aj are -1. Latter case take n=2 ,m=-m, contradiction, done. So unless A is identity matrix , it’s true

[summer path]

my eyes

[torn warren]

I fix the steps, f(x) is a coset, not a poly. Is this time right?... @delicate orchid

[summer path]

@main needle since i think they pinged texit and not you

[white oxide]

what does lang mean by "set of representatives for the irreducible elements"

we're not talking about equivalence classes or anything here right

he doesn't mean collection of irreducible elements of right

because then why would he say "representatives"

[terse crystal]

Oh. Didn’t know that. Thank you. I have pinged other’s tex lots of times😂

[white oxide]

here this decomposition cx/b + cy/a doesn't satisfy the theorem right, since we could choose c to be not coprime to b

[next obsidian]

Pick one under each equivalence class where the relation is differing by a unit

[jovial bough]

Sup gang

anyone able to give me a hand with this?

[cloud walrus]

JoelWantsCoffee

[jovial bough]

a paper I'm looking at has this line:

[cloud walrus]

JoelWantsCoffee

[jovial bough]

any insights much appreciated 🙏 🙏 🙏 🙏

[terse crystal]

y^p-y=Π(y-s)

Replace y with g(x)

Polynomials y^p-y and Π(y-s) equal because Fp is made up with p roots of y^p-y

[winter shore]

Any justification on why the definition of ordered ring is as such? It gives the consequences I'd expect, but the conditions seems very arbitrary

looking for some abstract nonsense that justifies it

or something like that

[next obsidian]

I don’t see how it seems arbitrary

You have an order that works with addition and multiplication as you’d expect

[open sluice]

it means that addition/multiplying by a positive preserves order
is there something else you expected

[winter shore]

hmmmmmm

not really sure, like I guess I'm expecting something like a topological group has continous operations because that's what a group object in Top implies.
Sum with one parameter fixed preserving order while somewhat natural feels like its missing something

[open sluice]

if you want I think you could also develop the same thing using the idea of a positive subset
either x is positive, x is 0, or -x is positive
and positive plus/times positive = positive

[winter shore]

now that feels better to me I think

but still somewhat hmmmm

like consider the theory of some n-ary operator. The 'natural' way to impose an order is to require that the n-ary operator preserves all orders in each coordinate

i.e. ai<bi implies f(a1,...,an)<f(b1,...,bn)

what general nonsense tells me that this is natural indeed

or maybe it's just an illusion that this is natural and is actually somewhat arbitrary

[open sluice]

a < b implies f(a) < f(b) is pretty natural to me
it’s an iff statement in the case of a total order
maybe you can read it as saying that f distributes over <

[graceful dome]

is this equvalent to asking how many rotations by 28/37 degrees until I get R_360 (the 2x2 Identity)?

Cause it would take 475.7142857 rotations but does it make sense to say a^475.7142857 and that the order is 475.7142857

or am I not making any sense

[primal beacon]

the order is either an integer or infinty, your first sentence is correct the rest doesnt make sense

you dont have an integer cause
$475/37\approx13$ and 1328=364 so your close but you dont do exactly an exact full turn. This means you have to do multiple turns, (2 full turns are equivalent to 1 full turn) in fact you know that doing 36037 turns would be a full turn so the order is finite

[cloud walrus]

rayane

[primal beacon]

rotate 37 times you get 28 degrees ( 37 is prime so you have to rotate multiples of 37) then 360 and 28 have gcd 4 so that $37\alpha$ has order 90 (prove it) and so $\alpha$ has order 37*90

[cloud walrus]

rayane

[graceful dome]

Ohhh

that makese sense, it most certainly didnt feel right putting a non integer as my order

[primal beacon]

yeah but to get a feel for it try that : given a cyclic group of order n and a generator g. compute the order of $g^r$

[cloud walrus]

rayane

[primal beacon]

then suppose your cyclic group is contained in another cyclic group of order m and generator a such that $a^k=g$. Compute the order of $a^r$

[cloud walrus]

rayane

[primal beacon]

I promise this will be useful, try to link that to your original question and prove it

[graceful dome]

will do!

@primal beacon does this make sense while we're at it?

I'm assuming we cant tell because we dont know if the cycles are disjoint

since this theorem only holds for disjoint cycles

or is there another way around it that im missing

[primal beacon]

you are right non commutativity breaks everything

consider $S=\begin{pmatrix} 1&1\ 0&1 \end{pmatrix}$ $T=\begin{pmatrix} 0&-1\ 1&0 \end{pmatrix}$

[cloud walrus]

rayane

[primal beacon]

then $ST$ has infinite order while $TS$ has order 3

[cloud walrus]

rayane

[primal beacon]

which means any commutative expression of the product order from the orders will fail

[graceful dome]

ohh okay

[primal beacon]

im not completely honest because here T has infinite order which cannot happen in $S_n$ but you get the point

[cloud walrus]

rayane

[wraith cargo]

There wouldn't be

If there would be then what we divided by wouldn't be the gcd

The idea is that by dividing one of them by the gcd you remove all the common factors from one of them

So that means those factors are no longer common between them

So gcd = 1

[sly rain]

If im working with two finite k (=some field) vector spaces e.g. k^2,k^3 and im given two explicit matrices that represent an endomorphism from k^2->k^2 k^3-k^3 and I shall compute Hom_kX.
I look which k-homomorphisms f respect "y * (f(v1,v2)) = f(a * (v1,v2))" where "y" lies in k[X], keeping in mind that different matrices act on X depending on where im multiplicating?
Also I can assume that f is k-linear because otherwise it would certainly not be k[X]-linear?

Im roughly aware of the very basic definitions of module stuff.

[dim widget]

That can’t really be true since they are conjugate by T

I think they both have order 6

[dim widget]

S has infinite order not T

[primal beacon]

oh yeah you’re right I exchanged the names and I said some nonsense

I dont what I was thinking about

[dim widget]

I think the interesting thing is that T and ST have finite order and they generate a group of infinite order

[primal beacon]

yeah SL2

[dim widget]

In particular that ST*T^{-1} has infinite order

[primal beacon]

thats what I should have said

[dim widget]

Which was probably what you were going for

[primal beacon]

check this

[primal beacon]

something like that but it turned out bad ahah thanks for the correction

[terse crystal]

We can just say option 2,3,4 are false. But we can’t say option 1 is true. You can’t know for sure, if there is some one who has some secret formula order(ab)=f(order(a), order(b)), but that some one hasn’t told anyone else. Also “can’t be determined by…” looks not very formal.

[rocky cloak]

You can just give an example where a and a' have the same order and b and b' have the same order, but ab has a different order to a'b'

[terse crystal]

Oh that’s right

[regal star]

@terse crystal I managed to solve it using the ismorphism theorem. I am just not able to prove that all conjugacy classes inside K are of the equal size

[terse crystal]

Let me recall what was discussed last time.
Take any x in K.
|K|=[G: CG(x)|
The conjugacy class under H containing x, is of the size [H:CH(x)] so k=[G:CG(x)]/[H:CH(x)]=[G:CG(x)][H:(H cap CG(x)]=[G:CG(x)]/[HCG(x):CG(x)]=[G:HCG(x)]
This is what was discussed last time right, if all conjugacy classes are of the same size then k=[G:HCG(x)]

This time you want to know how to obtain that all conjugacy classes are of the same size, I see

[regal star]

Yes exactly I assumed that they have the same size and applied the ismorphism theorem

[terse crystal]

[H:CH(gxg^-1)]=[H: gCH(x)g^-1]=|H|/|gCH(x)g^-1|=|H|/|CH(x)|=[H: CH(x)]

[regal star]

Thank you. I got it now. 👍

[regal star]

@terse crystal for Q35 I don't understand the question. Is it asking to find the number of conjugacy classes which have elements of order p in them? I am not able to construct an example

[terse crystal]

Elements of order p form a subset of Sn

It’s asking you how many conjugacy classes this subset is divided into

(So essentially how many p-cycles can a permutation has, since two permutations are conjugate iff they have the same shape)

[delicate orchid]

strange question

[terse crystal]

Without the thing in the parentheses it seems fine. What in the parentheses looks strange. I didn’t use it anyway

[rocky cloak]

I'm guessing you're supposed to think about the order of a product of disjoint cycles

[terse crystal]

I see. Though this case cycles are of the same length

[rocky cloak]

Sure, but to know that you have to first think about what the order of a product of cycles is

[delicate orchid]

I solved this question for all p-elements a few months back it was a mess

[terse crystal]

Why, isn’t it just ||[n/p]|| shapes

[rocky cloak]

I guess p-element means element of order p^k

[delicate orchid]

it does indeed

[terse crystal]

Oh my bad

[delicate orchid]

no worries

[rocky cloak]

Still doesn't seem so bad though...

[delicate orchid]

so it's the number of ways you can partition n where each part of the partition is a power of p

[terse crystal]

Yeah… present n in p-adic form

[halcyon hull]

(p!)^k?

[regal star]

Since p is prime there will be one conjugacy class containing all elements of order p? Am I right?

[terse crystal]

No

[halcyon hull]

Oh nvm not permutations

[delicate orchid]

these are permutations

we're talking about the group of permutations

[halcyon hull]

I see

[delicate orchid]

anyway surprise surprise, it's iterative

[regal star]

Oh so there will be conjugacy classes with elements of order p^k as well?

[rocky cloak]

No

[terse crystal]

No, number of p-cycles change

[regal star]

Sorry I meant p^k cycles

[terse crystal]

We are discussing things beyond the case of your problem. Not relevant to you solving your problem I mean

[terse crystal]

p elements have ||n|| conjugacy classes I think , right?
Since n= am … a0 in p-adic. for a permutation having bj many p^j-cycles, it’s just numbers (in p-adic) … b1b0 <= am…a0

[halcyon hull]

What channel do I go to for lower level abstract algebra questions/discussions?

[delicate orchid]

you're in it

[rocky cloak]

Isn't this it?

[terse crystal]

Oh and minus 1

[rocky cloak]

Or what does "lower level" mean?

[delicate orchid]

for S_n? definitely not

[halcyon hull]

Undergrad lol

[delicate orchid]

but which elements are you missing

[halcyon hull]

Ah I see

[delicate orchid]

this is all very much undergrad dw about it

[delicate orchid]

for example for S_{p^2} you should have p+2 conjugacy classes of p-elements, the trivial, the p^2 cycles, and k p-cycles for 1 \leq k \leq p

[terse crystal]

Thinking…

[rocky cloak]

Surely it's not just a simple formula in n and p

[delicate orchid]

as I said before

I didn't consider the question for n not a power of p btw

the p-adic expansion is a good idea

gives you an upper bound on the number of p^k-cycles you can have in any cycle decomposition

[terse crystal]

n= am… a0, just b= … b1b0<=am… a0, so n, then except those b such that all bj=0 for j>=1, so ||n-p+1||

I am finding error of this…

Nvm I am dumb

This is completely wrong…

[delicate orchid]

n = 6, p = 2, our partitions would be 1+1+1+1+1+1, 1+1+1+1+2, 1+1+4, 2+4, 1+1+2+2
so there are 5 different conjugacy classes of p-elements
for example

[terse crystal]

I see

[delicate orchid]

let me find what I came up with

it was this apparently, I adapted the standard recursive formula for the number of partitions

[regal star]

Does this look right?

[terse crystal]

Pp(n) number of partitions of n into sum of powers of p?

[delicate orchid]

correct

[terse crystal]

The answer is right

I see. No wonder Jagr said no direct expression…

[delicate orchid]

yeah I asked some other nerds in this server about it - you can adapt the generating function for the partition numbers to get it as well

$\sum_{n=0}^\infty P_p(n)x^n = \prod_{k=1}^{\infty} \frac1{1-x^{p^k}}$ or something? I forget

[cloud walrus]

WewGhostTbh

[rocky cloak]

,w oeis A8642

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jagr2808

[delicate orchid]

yeah as expected the p'-part of the number doesn't affect anything

[rocky cloak]

This seems to match for p=2 and the first 10 ns

[delicate orchid]

I did prove it works but

[terse crystal]

Looks reasonable , but I guess from 0 to infinity?

[delicate orchid]

I don't trust my own proofs

[delicate orchid]

the standard partition product goes from k = 1 so

I don't know anything about generating functions

[rocky cloak]

Ahh, says "number of partitions of n into 1, 2 or 4"

So not quite right

[delicate orchid]

ah so we're missing the ol 8+1, 8+1+1

[rocky cloak]

,w oeis A18819

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jagr2808

[rocky cloak]

There we go

[torn warren]

I revised the work, is this right?

[delicate orchid]

are you just trying to write the multiplication in F_9

[torn warren]

yes

fix yesterday's work

[delicate orchid]

yeah so it's standard polynomial multiplication except we set x^2 to be -1 = 2

so this looks right

[delicate orchid]

(which is what it means to quotient by x^2+1)

[torn warren]

the step on alpha^2=2 shows all higher order terms can be downgrade to combination of 1 and alpha

[delicate orchid]

correct, hence why you get a field of order 9 and nothing bigger

[torn warren]

yesterday you wrote as the combination of 1 and alpha, and ignore the <p(x)>, I know in the quotient ring the ideal <p(x)> serves as 0, so we ignore it just for simple in writing?

[delicate orchid]

well if you use alpha you automatically have alpha^2+1 = 0

so no need to write the ideal

but yes it's incredibly common to skip writing out the +ideal when working in quotient rings

[torn warren]

i see, thank you very much!!

[low wyvern]

Guys for the dihedral groups, when there is R^2, does that mean R * R or R+R. I only ask because technically you are carrying out successive rotations (which would imply you add them), but supposing you were given R = 2pi / 3 or 2pi / n for that matter , if also looks like you should to R * R ...

[delicate orchid]

there is only one operation in a group

I'm concerned that this is even a question

it corresponds to applying the same rotation R twice, as you said

[low wyvern]

Ye so R+R

[delicate orchid]

no? I'm presuming you have a matrix representation here which is why you're getting confused

[coral spindle]

What does + mean to you in that case.

What is R, and what is +

[low wyvern]

R is a rotation

[coral spindle]

Is it a matrix or is it just a symbol representing the idea of a rotation.

[low wyvern]

symbol

[coral spindle]

OK

[low wyvern]

So if you carry out 2 rotations it should be R+R, but the R^2 notation is throwing me a bit

[coral spindle]

Please answer my second question: what does + mean to you here.

[delicate orchid]

D_n only has one operation

[low wyvern]

I haven't studied dihedral groups yet

[delicate orchid]

and using additive notation for non-abelian groups is a no-go

[delicate orchid]

it's... the definition of a group?

[low wyvern]

it means carry out successive rotations

[delicate orchid]

do you know what a group is

[low wyvern]

yes

[coral spindle]

OK then there is no difference between these notations

It is just very unusual to use + for it.

[low wyvern]

ok so if R = 2pi / n then R^2 = 4pi/n

[coral spindle]

Hold on, you said earlier that R is just a symbol representing the idea

[delicate orchid]

so you are thinking about them as matrices

[coral spindle]

but in fact you're saying now that this is equal to a number?

[delicate orchid]

just 1x1 complex matrices

[low wyvern]

well a rotation through an angle

[coral spindle]

So is R a symbol that represents the rotation through 2pi/n or whatever, or is it actually being set equal to 2pi/n

[low wyvern]

Is there a difference?

[coral spindle]

Yes

[low wyvern]

mm... Ill go and think about it. Thanks anyways

[coral spindle]

[delicate orchid]

the difference is why we care about groups at all

[coral spindle]

It being a number in particular would also mean something that is highly unusual in the study of dihedral groups: that the turn by 360 degrees is not the same as the turn by 0 degrees.

[low wyvern]

The reason I asked was more because I was confused at the construction of the Cayley table of a dihedral group

I see

[coral spindle]

Why don't you ask about that directly. What's confusing you.

[low wyvern]

Ok hold on

ill get an example

Ok so the table of the D3 dihedral group has like R * R = S where R is a rotation 2pi / 3 clockwise and S is the same anti clockwise. My question is specifically, in the table where it gives the result R * R = S it is simply an addition right? If this is correct then does that carry for all dihedral groups?

[coral spindle]

Again, it is very rare and strange to write this as addition. I am going to say that we are going to only use 'multiplicative' notation in the style of R^2 or R * R from now on to avoid confusion.

[tardy hedge]

Thx ppl for the help on my permutation homework last week

I did well on it

[coral spindle]

Yes, this expression R * R should (in your mind) mean the rotation that we get by rotating 2/3 of the way around a circle, clockwise.

It is not addition; it is just an abstract operation which we think of as doing one action after another.

And it is most certainly not true that R * R = S for all dihedral groups, because they often do not contain the elements R and S, and in any case the comparable elements in the other dihedral groups behave differently.

[low wyvern]

Ok, so say I was given another dihedral group = { pi /3 , pi/5 , pi/4, ....}, when setting this up in a Cayley table how would I perform the operation between the elements?

[coral spindle]

That's not a dihedral group.

[low wyvern]

well say a group of rotations then

[coral spindle]

That's also not a group of rotations, if by pi/n you mean rotating clockwise by 1/2n of a full circle.

[low wyvern]

On my tutorial sheet it uses hexagon

[coral spindle]

We can do the example for hexagons, sure

[delicate orchid]

this smells like chemistry, are you a chemist

[low wyvern]

it uses specifically { pi/3 , 2pi /3 , pi , 4pi/3 , 5pi/3}... How would I form a group table ?

[coral spindle]

That's still not a group

Anyway, we will fix this and discuss it

[low wyvern]

they are anticlockwise rotations

[rocky cloak]

You're missing one rotation

[delicate orchid]

D_6 is the set {1, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5}

[coral spindle]

I am not going to use the notation you've been using, because it is easy to confuse the number pi/3 with the rotation that is anticlockwise by pi/3 radians. I am going to write R for the rotation clockwise by pi/3 radians.

[low wyvern]

That's all I have

[delicate orchid]

r being rotation and s being a reflection

[low wyvern]

Oh and the identity

[coral spindle]

The the group you're actually talking about is {I, R, R^2, R^3, ..., R^5} where I is the identity (hence capital i) which is the rotation by zero degrees.

R^n means 'repeat R n times'

[low wyvern]

But my point is that's technically adding them?

[delicate orchid]

no

[coral spindle]

I'm not adding anything. These aren't numbers.

[low wyvern]

you carry out rotation 1... then you carry out rotation 2

[delicate orchid]

which is function composition if anything

[coral spindle]

That's right. We call that composition in mathematics, not addition.

[delicate orchid]

not addition

[coral spindle]

jinx

[rocky cloak]

A rotation is different from a number even though one uses numbers to measure the angle of a rotation

[low wyvern]

But wouldn't the angle itself be the sum of them?

[coral spindle]

No.

[rocky cloak]

But yeah, composing rotations corresponds to adding the angles

[coral spindle]

That's not true, still

[delicate orchid]

yeah, adding in R/2piZ perhaps

whole other kettle of fish

[coral spindle]

because for example a rotation by pi/3 then a rotation by 5pi/3 is a rotation by......... 0

(in this group)

This isn't addition anymore because of course 2pi is not 0!

This is the definition of the group.

[delicate orchid]

because if you spin something all the way round... it hasn't changed...

[low wyvern]

oh right

Ye I get it lol

its 2pi

which is ADDITION

[coral spindle]

But 2pi is not 0, yes?

Sigh.

OK it's clear to me you're not interested in listening. I'll leave you alone now.

[delicate orchid]

pi+pi = 0 folks you heard it here first

I'm not even going to bother stepping on egg shells here

[coral spindle]

Please don't ping me, thanks

[delicate orchid]

we're defining a formal symbol r, such that r^6 is the same as the identity of the group
then we are taking the group generated by this formal symbol r

you can think of this as rotations if you want, the whole point is it's abstract

it doesn't a priori corrispond to anything

it can be used to model ANYTHING that when done 6 times is equivalent to nothing

hopefully it is is clear that rotations of a hexagon where one corner lands on another obeys this "6 times is nothing" rule

which is why the group of rotations of a hexagon is what it is

this isn't the full group of symmetry of a hexagon however, which is D_6

as you have to account for reflections too

well with D_6 we now have two symbols, s (for reflection... for some reason) and r (for rotation)

and just by thinking about how a hexagon flips around it should be clear that reflecting twice does nothing, so we write s^2 = 1, and rotating 6 times does nothing, so r^6 = 1

and now the interesting part, if we reflect, rotate, and then reflect again, we get a different rotation

which is a lot harder to see

[coral spindle]

I learned recently that the German word for reflection is Spiegelung, apparently that's why

[delicate orchid]

ofc it's german when isn't it german

[coral spindle]

Same for the integers ofc

[delicate orchid]

but we can encode this symmetry of "reflect, rotate, and then reflect = another rotation" as "srs = r^-1" (it turns out the rotation you get is the same as the rotation you start with but clockwise instead of anti clockwise)

yeah, the group is how its elements compose far FAR more than what those elements actually are

which is why we care - we can look at all objects with a certain structure at once

[white oxide]

dumb question but here is g(X) included in the summation or is it outside

[delicate orchid]

outside I presume

I agree though they should move it to come first

[white oxide]

hmm yeah ok

oh wait no i'm pretty sure it's inside, given the context from the last couple of pages that i just realized

[delicate orchid]

strange

[white oxide]

i also don't really understand how the terms in the summation are of the form given. If we divide each $\alpha_p$ by $p(X)^{j(p)}$, shoudn't each summand be for the form $\alpha_p = p(X)^{j(p)}h(X) + g(X)$?

[cloud walrus]

okeyokay

[white oxide]

or can the term $p(X)^{j(p)}h(X)$ be simplified to $\frac{f_p(X)}{p(X)^{j(p)}}$ somehow

[cloud walrus]

okeyokay

[white oxide]

OHHH

we divide both sides by $p(X)^{(j(p)}$ then

makes sense

[cloud walrus]

okeyokay

[rocky cloak]

There are infinitely many polynomials in P, so if g was on the inside this will not be a finite sum. Also g doesn't depend on p, so it would be weird to have it on the inside

[white oxide]

oh nvm then rip

i've been verifying everything onwards from thsi theorem wrong then 😭

ok now i'm a bit confused about the "trivial" existence

[rocky cloak]

I think the point is just that if you have something like f/pq, then you can rewrite it as f_p/p + f_q/q and reduce so numerator and denominator are relatively prime.

But if f_p is a multiple of p, then you get a polynomial out (which will be your g)

(p and q relatively prime)

So this is just partial fraction decomposition

[white oxide]

ohh, ok thank you!

yeah this is a section on partial fractions lol

oh so basically like if you have f_p being a multiple of p then you'll get a polynomial g_1(x), and then if this happens for multiple f_ps then their sum g_1(x) + g_2(x) + \dots + g_n(x) will be a polynomial which they take to be as g(x)?

[rocky cloak]

Yeah, exactly

[white oxide]

i see, thanks!!

[delicate orchid]

no problem!

[white oxide]

why is it that f_p = 0 if j(p) = 0? cuz then we get f_p(X)/1 = f_p(X) which doesn't necessarily imply that f_p(X) = 0 right?

or is it just saying that we don't include that term in the sum, and instead move it to be one of the summands of g(X)

[delicate orchid]

this

[white oxide]

omg this

This.

so when Lang says "state and prove the theorem for rational numbers" does he mean let A = Z, and prove the theorem for Frac(Z) = Q? because if we let A = Q[X] then there's nothing to prove lmfao

[next obsidian]

No

He wants A = Z[x]

Or wait uhhh

Yes A = Z

The idea is that there’s a lot of similarities between Z and k[x]. A lot of this boils down to the fact both are PIDs, but eg the classification of primes in Z[x] and k[x][y] are very similar (actually it just follows the form for any poly ring over a PID)

One interesting way they differ tho is that Z has no derivations, which means some things that are easy for k[x] are not easy for Z at all

[white oxide]

oh interesting

[delicate orchid]

le homology or somesuch

[next obsidian]

In slang you can find a proof (literally only uses high school calculus) for the abc conjecture over a poly ring

And Fermat’s Last Theorem for poly rings follows

Ofc these are suitable adaptations for k[x], but it follows basically because d/dx exists

Look at IV.7 the Mason-Stothers theorem

Afaik this is where the F_1 thing comes from, we want to view Z as a polynomial ring over some field, but such a field would need to “be a field with one element” or something. I don’t know much

[delicate orchid]

wait what exactly do you mean by derivation here chmonkey

[next obsidian]

Liebniz rule and linear

[delicate orchid]

this would line up with other shit about F_1

[next obsidian]

They all vanish on Z

From linearity + liebniz

[delicate orchid]

alright so the diff geo derivation

[next obsidian]

Is there a different definition?

And for me this is the algebra derivation

Lmfao

[delicate orchid]

they're probably equivalent I'm just skill issuing memory wise

but that makes sense

[next obsidian]

What’s your definition?

Is it about shit in the ring of differential operators

[delicate orchid]

D(ab) = aD(b)+D(a)b with an optional twist/grading

but I remember some connection to homology

[next obsidian]

Ah yeah that’s what I mean. No twist tho

Uhhhh idk there’s connections of this stuff to homological stuff but the stuff I know I imagine is not what you are thinking of

[delicate orchid]

yeah the connection I have in my mind is elementary - like the first homology is (isomorphic) to some set Dev which was short for derivations, as I said I can't remember

they were probably just 2-cycles

[next obsidian]

Makes sense

My view of it is that Der_A(B,M) = Hom_B(Omega_B/A,M)

But this leads to various homological crap

And like cotangent complex André-Quillen homology stuff

[delicate orchid]

very smelly

[white oxide]

$\emph{Let P be the set of primes in $\mathbb{Z}$. Then any element $r \in \mathbb{Q}$ has a unique expression}

[r = \sum_{p \in P} \frac{\alpha_p}{p^{j(p)}} + s]
$\emph{where $\alpha_p$, $s \in \mathbb{Z}$, $\alpha_p = 0$ if $j(p) = 0$, $\alpha_p$ is relatively prime to $p$ if $j(p) > 0$, and $\alpha_p < p^{j(p}$ if $j(p) > 0$.}

[cloud walrus]

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

[white oxide]

so would this be the analagous theorem?

i did italics cuz it looks sexy

[rocky cloak]

So is the point that there are |k| derivations on k[x], but for Z there is just the zero derivation?

[next obsidian]

I don’t know if it’s |k|, you might get more

I mean d/dx is a derivation on Z[x]

[delicate orchid]

there's definitely at least one though!

[rocky cloak]

Aren't they all just multiplies of d/dx

[next obsidian]

what you want is a derivation on Z

[next obsidian]

Well if k has its own derivations…

And also, not in char p

[delicate orchid]

map everything to zero ;3

[next obsidian]

Idk, maybe some of what I say is wrong

My point here was just that d/dx makes this tick for k[x]

And Z you got squat

[rocky cloak]

If k has characteristic 0, then |k| is infinite, so that's not very interesting

[delicate orchid]

found it

[rocky cloak]

What's with the tilde in the middle?

[delicate orchid]

turns out these are just 1-cocycles and not proper derivations

really cringe!

[delicate orchid]

there's a tilde map sending m -> (g -> mg-m)

[rocky cloak]

They named it tilde? Some people ey

[delicate orchid]

smells like the augmentation map to me

[delicate orchid]

oh yeah that's how they prove it lol

[rocky cloak]

When M is a right module you can naturally turn it into a bimodule by having g act on the left by the identity. Then this is an actual derivation, and you get that group cohomology is a special case of Hochshild cohomology

[delicate orchid]

interesting

maybe cohomology does have a use after all

is there a way we can view higher order cocycles as "higher order" derivations

and what I mean by that is a function D(ab) that satisfies the leibniz rule for the nth derivative

probably not actually, each term in the leibniz rule expansion has a positive coefficient which is not the case for 2-cocycles

[sonic coral]

is a subgroup H of order p is normal in G if and only if H is a subgroup of Z(G)?

[rocky cloak]

C3 is normal in S3, but S3 has trivial center

[coral spindle]

Perhaps you had some confusion regarding the definition of a normal subgroup. The condition that gNg^-1 = N for all g in G does not mean that gng^-1 = n for all n in N, because conjugation by g might send n to some other element of N.

So gNg^-1 = N is not saying that every element of the subgroup N is fixed under conjugation, but rather is saying that the whole subgroup remains invariant, viewed all together.

[white oxide]

i mean there's not that much to prove here right, given this theorem

and uniqueness just follows from Z being a UFD right?

[sonic coral]

would it just be an if and not an only if then?

i feel like there’s some truth to this just not as strong as i was thinking

[rocky cloak]

Sure, any subgroup of the center is normal

[rocky cloak]

Don't think you really need that Z is a UFD for that. You just multiply out the denominators, then reduce modulo p^j

[charred flume]

help me understand class equation, is my answer correct, the class equation of Z_9 is
1+1+1+1+1+1+1+1+1

[delicate orchid]

$|G| = |\mathcal{Z}(G)| + \sum [G : C_G(g_i)]$, here $\mathcal{Z}(G)$ so the class equation is just $|G| = |\mathcal{Z}(G)| = 9$

[cloud walrus]

WewGhostTbh

[delicate orchid]

if you're using the better but less common version that doesn't have the centre separate then you're correct

[rocky cloak]

Is that summing over the size of conjugacy classes or what?

[delicate orchid]

yeah

altenratively just orb-stab but yk

can't teach something TOO cool now can we

[charred flume]

damnn, i think my prof expect the center to be separate

[delicate orchid]

then u best make sure the centre is seperate!

[charred flume]

i misunderstood it the first time i read about it.

i have already passed my problem set 💀

well at least i know now, that i should add up all the centers and make it separate

[delicate orchid]

I know it seems weird and arbitary to do the centre seperate - and that's because it is! I will never understand it

$|\mathcal{Z}(G)| = \sum_{z \in \mathcal{Z}(G)} [G : C_G(z)]$ so we can just move it inside and write it all as one sum but nope

[cloud walrus]

WewGhostTbh

[daring nova]

I feel like it's to separate the ones from the rest

I did it this morning, so I think it's useful for the 1st Sylow theorem, for example

didn't reread that yet though

[white oxide]

ah ok

would it be fine to say we had to expressions of the form and then use the fact that p^j(p) divides \alpha_p - \beta_p, which implies that \alpha_p = \beta_p?

also i don't really understand how p^{j(p)} dividing fp - \varphip implies that it's zero... does it have to do with what's underlined in blue here?

[white oxide]

or is it saying this

[rocky cloak]

Yeah, the fact that f_p is 0 when j(p) is 0 is a restriction. Without this restriction you wouldn't have uniqueness

So yeah, move all the j(p)=0 stuff into g.

[white oxide]

hm ok i'll have to think about that

thanks

i still don't understand what there really is to prove, it seems like the proof is basically like the proof of theorem 5.2

this feels like a trap

[rocky cloak]

Yeah, the statements are basically identical. Just shifting around which things are 0

And 5.2 having a stronger uniqueness condition

[white oxide]

i see, thanks

[charred flume]

hi for clarifications, this is not a homomorphism right? because what does [a][b] even means ( [a] means class of a)

[coral spindle]

Please post more context...

[coral spindle]

But no typically this will be a homomorphism Z → Z_n where we consider the group operation to be addition.

[charred flume]

im currently practicing some problems

[coral spindle]

[a][b] is notation usually reserved for the product in Z_n.

In this case the group operation would be the sum, although they should have mentioned that.

[charred flume]

the book im reading defines it like this

[coral spindle]

I know.

[charred flume]

so im kinda confuse what [a] [b] means

[coral spindle]

Nobody has written [a][b] anywhere, wym

If you're asking what the group operation is, as I mentioned it would be the sum of two equivalence classes in Z_n.

[charred flume]

ohhh so [a] [b] means sum of the two equivalence classes

[coral spindle]

No, what

[charred flume]

in a and b

[coral spindle]

No we would never write [a] [b] for that

We would write [a] + [b]

The group operation here is +

[charred flume]

okay i mean [a] * [b] (supposedly we don't know the operation)

[coral spindle]

I'm not totally happy with this notation but it's clear at least you know what you mean here.

[charred flume]

but + is the operation. it is not mentioned in the problem, so is this like directly assumed?

or what?

[coral spindle]

That is presumably the only group operation on Z_n that you know about.

[charred flume]

alright

okay it is still not homomorphism

[coral spindle]

No, it's a homomorphism.

[charred flume]

o yeah wait

yeah it is

[barren sierra]

Say we have a linear map f: V -> W with {v_i} a basis for V and {w_j} a basis for W. What is det(f) and Tr(f) in terms of just bases / dual bases?

without making identifications of f with a matrix

[chilly ocean]

in terms of bases
without making identifications of f with a matrix
this is confusing to me

what do you need this for?

one way or another the matrix of your map in the bases is going to come up if you're doing anything with it and a basis

[barren sierra]

I originally answered this by just saying "let M be the matrix representation of A" and computed like that

and my prof said that I was making an identification and "not keeping track" of said identification

and that I shouldn't do such a identification of A with a matrix

I know what a basis for Hom(V, V) is given a basis for V

[chilly ocean]

by "keeping track" they probably just mean being explicit with your isomorphisms and what not, and not just outright assuming that f is a matrix

[barren sierra]

like explicitly keeping the isomorphism Hom(V, V) -> M_n(R) in the picture for all the computation?

[chilly ocean]

you're gonna have to pick a basis anyways. i can't imagine it meaning anything else

[barren sierra]

I had to pick a basis for such a matrix representation anyways

whatever

[rocky cloak]

I guess just compose f with the isomorphism W -> V that maps w_i to v_i, then use the usual definitions

[charred flume]

is X_G the same as Z_G it's just that Z_G is specifically when G acts itself by conjugation?

[delicate orchid]

indeed

[charred flume]

nice, thanks

[heavy verge]

Trying to prove this, cant seem to figure it out.As G is p-hypo from the definition there exists a P' normal in G st P' is a p-group, aka |P'| = p^m for some m in Z and G/P' is cyclic. So letting |G| = n, the order of P' must divide n by Lagrange so n=p^mq for some q in Z, which gives |G/P'| = q, however I dont know how to use the fact that its cyclic to force that the quotient by P' forces all of the factors of p in n to be divide out. Because q could still be written as p^a * p_1^a_1 * ... so not coprime to q. Any thoughts? Maybe im thinking about it wrong.

[rocky cloak]

So you know that there is a p-group such that G/P' is cyclic. Now it's not necessarily true that G/P' has order relatively prime to p, but maybe you can find a bigger p-group P such that G/P is cyclic and of order relatively prime to p?

Hint: ||second isomorphism theorem||

It can also be an idea to investigate the easiest cases first, for example what happens if G is cyclic?

[delicate orchid]

that hint jumpscared me

[heavy verge]

Thank you for the reply! I cant look at it currently but ill take a look later and let you guys know if I get anywhere.

[crystal turtle]

i always forget about that one

[rocky cloak]

Of course since there isn't any agreement of the order, timanaku will have to guess which theorem I mean

[crystal turtle]

going off of what wiki calls 2nd iso, I forget that one. But maybe you mean what they call 3rd iso, in which case I do remember that one.

[rocky cloak]

One pretty much implies the other anyway

Wikipedia sidesteps the issue by calling then theorem A, B, C and D

But yeah I mean ||C/D||

[alpine island]

discussion's second greatest moment