groups-rings-fields

[solemn rain]

f.v = fv?

[latent anvil]

I mean, what does fv mean?

[solemn rain]

okay whats even f

f is an n-tuple ig

[latent anvil]

You tell me?

You wrote it?

[solemn rain]

(a_1,a_2.....,a_n)

[latent anvil]

What?

[solemn rain]

nvm boys

idk what is f

[latent anvil]

you have two operations on V

• and *

where * : F×V -> V

And a*(v+w) = a*v + a*w and (a+b)*v = a*v + b*v

[stone fulcrum]

F is the set that contains your scalars

[latent anvil]

further, a*(b*v) = (ab)*v

and 1*v = v

@solemn rain so that's a vector space

[solemn rain]

over a field F

[latent anvil]

Yes

[solemn rain]

okay

[latent anvil]

An important family of examples:

[ripe crest]

I highly, highly recommend these videos when you learn linear algebra: https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[latent anvil]

If S is a set, let F(S) be the set of all formal sums a1 s1 + … + an sn where a1,...,an in F and s1,...,sn in S

So the weird thing is that those symbols don't mean anything

• is not an actual operation yet

I'm just defining what the elements of the set look like

[solemn rain]

sum ( a_nf_n) n = 1?

to k?

[latent anvil]

What?

[solemn rain]

can i write that with simga notation/

sigma*

but okay i get it

[latent anvil]

Why the = 1 though?

Anyways, we can define an actual + operation by adding coefficient by coefficient

So if we have formal sums v = as + bt and w = cs + dr, v + w = (a+c)s + bt + dr

and our scalar multiplication works by multiplying coefficient by coefficient. So a*(a1 s1 + … + an sn) = (a a1)s1 + … + (a an)sn

[solemn rain]

cool

[latent anvil]

Now if V is a vector space, a basis for V is a subset B of V such that every v in V has a unique representation of the form v = a1 e1 + … + an en for a1,...,an in F and e1,...,en in B

If V has basis B, then we get an isomorphism F(B) -> V by sending a formal sum to its actual sum

The amazing thing about vector spaces is that any two bases for the same vector space are actually in bijection, so they have the same cardianltiy

We define the dimension of V, dim V, to be this cardinality

Finally, two vector spaces V and W turn out to be isomorphic if and only if dim V = dim W

Oh, a homomorphism of F-vector spaces is a map L : V -> W satisfying L(v+w) = L(v) + L(w) (so it's a group homomorphsim) and L(av) = a L(v) for any scalar a in F and v in V. An isomorphism is an invertible homomorphism

[chilly ocean]

Now if V is a vector space, a basis for V is a subset B of V such that every v in V has a unique representation of the form v = a1 e1 + … + an en for a1,...,an in F and e1,...,en in B
@latent anvil don't they need to be linearly independent? Otherwise this would just be a generator

[latent anvil]

No, I said unique representation

take v = 0 to get linear independence

[solemn rain]

okay

i get all of htis

this * ( i hope )

[latent anvil]

Oh I guess I'm actually lying a little

You should have the ai be nonzero

Otherwise it's easy to get non-unique representations

By just adding in more terms

[solemn rain]

okay

now what else do i need to prove R is isomorphic to R^2

[chilly ocean]

An isomorphism 👀

[latent anvil]

So if R and R^2 as isomorphic as vector spaces over some field, they're isomorphic as groups

They're definitely not isomorphic over R, since they have dimension 1 and 2 respectively

But over the field of rational numbers, they actually do have the same dimension

[woven delta]

They are as Q vector spaces

Lol

[solemn rain]

how

[latent anvil]

fam I am aware

[chilly ocean]

Uncountable basis

[latent anvil]

So prove that R is not a finite dimensional vector space

Show that if V is an F-vector space, there's a natural vector space structure on the abelian group V×V

Show that if V is infinite dimensional over V, then V ≈ V×V

[solemn rain]

i will try that

[latent anvil]

By using a basis for V to get a basis of the same cardinality for V×V

[solemn rain]

ty

sm

vector addition on R here

would be just

addition?

[latent anvil]

yes

[woven delta]

Lol

[solemn rain]

and same with scalar multiplaction

[woven delta]

Yes

[solemn rain]

cool

[latent anvil]

lol

[solemn rain]

stupid questions?

[latent anvil]

i gave a very fast introduction to linear algebra

[ripe crest]

stupid questions?
no

[solemn rain]

a basis B is a subset of V such that every vector in V can be written as combinations of vectors in B

am i right?

[woven delta]

Yes

Uniquely

[solemn rain]

what yes is this

yes to stupid

orr yes to me

[woven delta]

Yes

[solemn rain]

both?

now

what yes is this now

[chilly ocean]

Lol

[solemn rain]

im just oging to say yes to all

ty

[ripe crest]

A basis B is a linearly independent subset of V such that every vector in V can be written as linear combinations of vectors in B

[chilly ocean]

A minimal generator

[ripe crest]

A subset B of V is said to be a basis if:

1. It is linearly independent
2. span(B) = V

[woven delta]

A basis is an isomorphism to R^k for some cardinal k

[latent anvil]

imo

[solemn rain]

oh my god im so stupid

[latent anvil]

a basis is a spanning set

[solemn rain]

why wouldnt basis of R here be R as every vector can be written as just itself

??

[latent anvil]

if you need linear independence you're trying to be too fancy

[woven delta]

Lol

[latent anvil]

lmaoo that's literally the joke I was making

life imitates art

[chilly ocean]

Oof

[upper pivot]

lmao

[solemn rain]

?

[chilly ocean]

R is not linearly independent

[latent anvil]

Okay, linear algebra time is over

[solemn rain]

what is linear independance

[latent anvil]

Let A be the infinite direct sum of Z with itself

Or the infinite direct product if you swing that way

Show that A ≈ A × A

[solemn rain]

okay

[latent anvil]

Here is your example, you're welcome

[solemn rain]

yea

[latent anvil]

This is secretly the linear algebra example

[solemn rain]

phi((a_1,a_2,a_3,....a_n) = ((a_1,a_3...,a_(2k+1)),(a_2,a_4....,a_(2k))?

phi is isomoprhism?

[woven delta]

Yeah

Well no

[solemn rain]

okay boys

fuck

[chilly ocean]

That doesn't look injective

[solemn rain]

oh

it is

[woven delta]

Just make a bijection between N and N disjoint union N

[solemn rain]

if u go form

from

Z/2Z

njo

product(Z/2Z) to product(Z/2Z) x product(Z/2Z)

ayy

am i right boys?

[woven delta]

Actually what you were doing is kinda fine

[solemn rain]

i fucked it up didnt i

but whats wrong

@chilly ocean that guy shamrock had enough with me being brain damaged what does linear independant mean

u gotta help me

[woven delta]

Lol

[chilly ocean]

,w linearly independent

[cloud walrus]

Display results online and refine query

Requested by N/𝔄#0985

[solemn rain]

oh okay

wtf is a matrix

is 1+1+1 unique?

like

if i have an element in R v

basis of R is 1?

{1} ?

[chilly ocean]

With respect to the basis {1} the representation 1*3 for 3 is unique.

[solemn rain]

it is/

cool

so the vectors are linearly indepednant if u can write them as a unique linear ocmbinations with vectors from the basis

[chilly ocean]

Mb

[bleak abyss]

mo2men you need to like

Pick up a linear algebra book

And just work through it

[woven delta]

Yeah

[solemn rain]

i just wanted to learn linear algebra enough as like a

lemma or someshit for the problem i had

i never liked this linear algebra shit from high schol

[woven delta]

Why are you learning algebra if you don't know linear algebra

[solemn rain]

its not that big of a deal ig

the book literally has

[bleak abyss]

You're not gonna learn shit by just derping around on a discord server

[solemn rain]

part 3 ( linear algebra )

well im going through the text books im just asking for help with problems

[woven delta]

Why are you doing math if you don't like linear algebra

[solemn rain]

and its just a conincidence that i asked shamrock to teach me some new definitions

[latent anvil]

Dark secret: when I first learned group theory I couldn't do the matrix groups problems because I couldn't multiply matrices

[chilly ocean]

Algebra before linear algebra kinda sounds like first semester category hell

[solemn rain]

yea see at shamrock

and now that guy is teaching me algebra like hell

[latent anvil]

oh yeah it also had category theory in it @chilly ocean

I was a meme high schooler

[bleak abyss]

And turns out shamrock probably was disserviced by this lol

[latent anvil]

Yes

Absolutely

[solemn rain]

@woven delta idk i probably dont know linear algebra enough to hate it i jusut iddnt like playing around with matrices using gauss stuff

i know matrice operations forom high schoool

[bleak abyss]

Don't learn that type you can learn actually interesting stuff

If you wanna learn algebra along with linear algebra

[solemn rain]

and i think it helped me to understand group examples of matrices

yea rtin

artin

we talked about this

[bleak abyss]

Pick up a book called Algebra by Artin

[solemn rain]

b4

and i just hated matrices for whatever reason

why u thought i was derping around on discord tho?

[bleak abyss]

Turns out math isn't kind to people's desires, it's a very git gud or git fuckt subject

[solemn rain]

yea

i am in the git fuckt category

im into bdsm

..

[bleak abyss]

And you ask why I think you're derping

Anyway

[solemn rain]

i mean

wdym by that tho

[bleak abyss]

My point here is like, this isn't the way to learn stuff

[solemn rain]

'Turns out math isn't kind to people's desires, it's a very git gud or git fuckt subject'

oh

[bleak abyss]

If you don't wanna learn linear algebra you're just screwed

[solemn rain]

well i read the textbook , do some problems and post problems here that i have difficulty wiht

[bleak abyss]

Like this is a barrier of entry to the subject

[chilly ocean]

There is a reason why they don't start with algebra in uni

[solemn rain]

if you byw hat you mean by this is just

[bleak abyss]

You clearly didn't read much of Artin or even like, Hoffman and Kunze

[solemn rain]

that this way im not learning linear algebra then i totally agree

[bleak abyss]

If you're asking for definitions on discord

[solemn rain]

yea i agree with you obv

[bleak abyss]

This isn't the place to learn the foundations of the subject, that's a book or a class

[solemn rain]

i think the book has its fair share of algebra

[latent anvil]

yeah so I am still bad at linear algebra

[solemn rain]

and the book even sometimes imply that some readers would not be fmailiar with lienar algebraa

[latent anvil]

Because I didn't pay attention to it in high school

Because I thought it was boring

[bleak abyss]

And if you got quasi-ghosted it means you've probably used up his patience

[solemn rain]

so i think i will learn lin aglebr when i get to it in like chapter 15

[latent anvil]

And never took a second course

Learn linear algebra

[bleak abyss]

Which isn't something you wanna do

[latent anvil]

For real

[solemn rain]

whats quasi-ghosted

[bleak abyss]

Stopped responding for a bit

[tribal pasture]

yeah so I am still bad at linear algebra
@latent anvil same lmao...

[solemn rain]

i mean

[tribal pasture]

Like those basis change problems

[bleak abyss]

I am much better at later stuff than earlier stuff in LA

[solemn rain]

i wont mind much if i get quasi-ghosted

ik im stupid

[bleak abyss]

Since I never really had like, much practice with Gaussian elimination

[stone fulcrum]

Yeah change of base still throws me off

[tribal pasture]

Oh yeah... gaussian elimination the LU stuff

[bleak abyss]

But I think of everything in terms of minimal polynomials

[stone fulcrum]

I get the process but I can't always carry it out

[chilly ocean]

But I think of everything in terms of minimal polynomials
@bleak abyss wdym by that?

[solemn rain]

u change basis if u conjugate ig?

[tribal pasture]

Same, I understand what happens but like if somebody brings me a question, I am like now waaitt a min

[solemn rain]

dk nvm

[bleak abyss]

Like idk char and min polys of matrices are how I solve a lot of problems

[stone fulcrum]

That's pretty much right mo

[solemn rain]

@kaaynex

cool

isnt that rep theory

lmao

[chilly ocean]

I guess that .. works

[woven delta]

[solemn rain]

lmaaao rep theory is turning algebra to linear algebraa

xD

ig im screwed

[upper pivot]

lol does anyone remember inverting matrices or calculating determinants even

[tribal pasture]

linear rep theory i fink

[bleak abyss]

Just read Artin

That's your assignment

[stone fulcrum]

Rep theory would say that a conjugation by an element is iso to a conjugation by a matrix

[bleak abyss]

Or like idk Hoffman-Kunze

[chilly ocean]

lol does anyone remember inverting matrices or calculating determinants even
@upper pivot i do

[woven delta]

Even Axler

[solemn rain]

i will do that after i finish this chapter on group theory

[bleak abyss]

No not Axler

[solemn rain]

that last one

[tribal pasture]

Apparently there is a big formula for determinants using cofactors and shiz

[bleak abyss]

Fuck Axler

[woven delta]

Axler is alright

[bleak abyss]

Not really

[tribal pasture]

axler has nice colours

[bleak abyss]

He thinks about determinants like an actual dumbass

[stone fulcrum]

I wish more books had nice colors

[upper pivot]

lol N/mathfrak(u) i never remember those of the top of my head, always have to look up the algos

[tribal pasture]

same imagine rudin colourised

[solemn rain]

the parity of a permutation is the determinant of the peermutation written in array form

boom

[bleak abyss]

Which is fine if you're doing functional analysis because determinants aren't nearly much of a thing

But if you're doing anything else and your definition of the characteristic polynomial is

[chilly ocean]

Mathfrak A!

[stone fulcrum]

Like, easiest way to make a book more readable, put definitions and theorems in colored boxes

[chilly ocean]

Close enough

[bleak abyss]

"Complexify the matrix, upper triangularize that, and then take the product of t-diagonals

[solemn rain]

@stone fulcrum topology without tears right?!

[latent anvil]

I feel like I still don't really understand determinants

[tribal pasture]

Axler did that with his FA book as well

[bleak abyss]

Like if that's how you define a characteristic polynomial you need help

[chilly ocean]

That's a really good book

[bleak abyss]

And you also probably need to lose tenure

[latent anvil]

Dear God

That's awful

[stone fulcrum]

Yes. Call me inspired. I honestly want to write a similar book for abstract

[chilly ocean]

You think?

[woven delta]

Yeah that's bad

[latent anvil]

Like, at least define it as like product of (x-λ)

[solemn rain]

@stone fulcrum first buyer

[tribal pasture]

Determinant something something volume

[solemn rain]

is the characteristic polynomial

[latent anvil]

Err I guess how do you talk about multiplicity?

[solemn rain]

the triangle thiny?

[chilly ocean]

Why do you believe that?

[solemn rain]

thingy?

[stone fulcrum]

But I've tried and it comes out sucky

[bleak abyss]

Char poly of a matrix A is det(tI-A)

[solemn rain]

i think i saw this when defining partiy of permutatioln

[bleak abyss]

-_-

[tribal pasture]

Char poly of a matrix A is det(tI-A)
@bleak abyss ad-bc ftw

[solemn rain]

ok boys anyways

thanks all for the help

and making me sure im stupid

[bleak abyss]

You can fix that by actually learning linear algebra

[latent anvil]

I feel like this devolved into everybody saying they're bad at linear algebra

[bleak abyss]

Nah I'm good

[woven delta]

I'm okay at linear algebra

[tribal pasture]

You can fix that by actually learning linear algebra
@bleak abyss o.O or.....

[latent anvil]

You said you were much better at the later stuff dami

ergo

Proctor hawk

[chilly ocean]

I'm not bad at linear algebra

[bleak abyss]

x < y doesn't imply x is small

[tribal pasture]

I'm not bad at linear algebra
@chilly ocean oh yeah? give me the formula for the determinant of a 7x7 matrix

[latent anvil]

Yeah duh it implies y is small

[bleak abyss]

Hmm?

[chilly ocean]

Bruh wolfram alpha

[latent anvil]

Oh sorry lol

[tribal pasture]

o smort

[latent anvil]

< not >

I'm bad at analysis sorry dami

[bleak abyss]

So it seems

[tribal pasture]

i wish there was a field of math i was good at ):

[latent anvil]

And reading

[chilly ocean]

I mean you can compute it using gauss

[bleak abyss]

Wait hold up it actually worked

I actually got it wrong the first time and edited the message

Master gaslighter

[latent anvil]

I've been hoodwinked!!!!

[woven delta]

Nice

Anyway the point is that if people try to do more advanced stuff and ask dumb questions because they don't know essential stuff they may get btfo'd

[solemn rain]

idk why

[woven delta]

When no one is getting paid to answer questions there are in fact dumb questions

[latent anvil]

yeah who knows

[solemn rain]

but the textbook assumes some1 is not fmailiar with lienar algebraa

and i was not supposed to know linear algebra in solving the problem

[latent anvil]

Anyways why is a^2 - b^2 = (a-b)(a+b)?

[solemn rain]

i was just extra curious to learrn how linear algebra solves this problem

thats it

[tribal pasture]

well i tried with different values of a,b and it seemed to work so it must be true

[woven delta]

Btfo shamrock

[latent anvil]

Owned

lol idiot just look at the zero sets and apply the nullstellensatz

Z(a^2 - b^2) = Z((a-b)(a+b)) and they're both clearly radical

[chilly ocean]

Who hurt you

[tribal pasture]

LA

[ripe crest]

@latent anvil , there's actually a really nice geometric proof

[latent anvil]

oh?

Oh sorry I thought you meant about R and R^2 being isomorphic as groups

Yeah no I was meming

[oblique river]

R and R^2 are isomorphic as groups

[latent anvil]

yes

we were talking about it earlier

[oblique river]

oh, sorry, I misunderstood then

[latent anvil]

Someone had a problem to find a group G iso to G×G

They said they tried R and it didn't work

And I was like "well it does"

[oblique river]

ohh I see

gotcha

[latent anvil]

But then they didn't know linear algebra

[oblique river]

there are easier solutions anyway

[latent anvil]

And the proof I know for that involves linear algebra

Yeah for sure

[oblique river]

yeah

gotcha 👍

[toxic zephyr]

let me see if I have this right: A set S with a law of composition is a group if there exists a surjective homomorphism from a group G to S?

[oblique river]

yep, think that's true

[toxic zephyr]

Okay cool.

Another question: when we talk about the quotient group $G=\bC^{\times}/U$ with U being the set of complex numbers on the unit circle, we're talking about all of cosets of U? So the elements of G are sets of the form $xe^{i\theta}$ for all theta and some $x\in \bC^{\times}$?

[cloud walrus]

nix:

[toxic zephyr]

With this group G being isomorphic to the positive real numbers under multiplication?

[oblique river]

yes

it's isomorphic to R^\times

[toxic zephyr]

So the map from the quotient group to R^\times is the absolute value map, and would the map from R^\times bring a real number r to the circle of radius r in C^\times?

[oblique river]

yes

that's right

[toxic zephyr]

Great thank you

[oblique river]

👍

[gritty fractal]

do ring homomorphisms preserve multiplicative identity?
wikipedia says yes, but my textbook says only if the homomorphism is onto

also which convention is more prevalent: rings with identity or rings without identity

[oblique river]

with

[gritty fractal]

ah thx

what about the homomorphism thing?

[oblique river]

the more common convention is that homs preserve 1

[gritty fractal]

thx

im glad lol i feel like itll make them easier to work with

[oblique river]

yep haha

just be careful if you're taking a class and your prof expects you to use your textbook's definition

I saw that :P

it's good to keep in midn that there are other conventions out ther

but the most common one is definitely the one with 1

being preserved

[chilly ocean]

I forgot what I posted lmao ok coolio

How would you use the division algorithm, I don't see it immediately

True!
Any other way of proving that anyone can see in the context of an intro analysis class tho? I'm trying to think of a really simple way to show at least that it is at most countable

without development of anything outside the context of set theory or reasonable prerequesite ideas

at that point in the chapter rudin just develops stuff regarding showing countability and the like

but does give a hefty amount of properties of complex numbers from the last chapter

[latent anvil]

Mega, what have you tried?

[chilly ocean]

let $p(x)$ be the minimal polynomial then $p(\alpha) = 0$. Now as F remains unchanged under $\sigma$ so we have $\sigma(p(\alpha)) = \sigma(0) = 0$, but $\sigma(p(\alpha)) = p(\sigma(\alpha))$ as coefficients of $p(x) \in F$.

I wanted to verify my proof

[latent anvil]

That's correct so far

[chilly ocean]

Isn't that enough ? Have I missed something ?

[latent anvil]

No

You showed that if α is a root of some polynomial p, then σ(α) is too

That doesn't say anything about the roots of the minimal polynomial, or bijectivity

[chilly ocean]

oh okay, so I also need to show that for all alpha, sigma(alpha) is unique ?

So that it is a complete cycle ?

[cloud walrus]

mega:

[chilly ocean]

Ah okay, it was my mistake not mentioning the minimal polynomial, and as sigma is automorphism, it's already unique

[steep hull]

Yep, you’re done after that

[chilly ocean]

I see, thanks!

[thorn flint]

Hi, can someone help with this question?

I'm trying to prove the hint part first

So I need to prove that it is a subgroup and it is cyclic

[mild laurel]

Yes

[thorn flint]

I think I can see how it is cyclic but how do I prove that it is a subgroup?

[hot lake]

just to be sure, you know that ${\pm 1}$ means ${-1;1}$, right ?

[cloud walrus]

Zef Klop 🍃 🌿 🌻:

[thorn flint]

The set which has -1 and 1 right?

[hot lake]

yes

[thorn flint]

Yup

[hot lake]

then I don't see what is stopping you from showing it is a subgroup

[thorn flint]

I don't understand how we can show that is a subgroup

[hot lake]

use the definition of subgroup

[thorn flint]

So it has to satisfy that G contains the identity and inverse of each element of GL_2(Q) right?

[hot lake]

no

and they meant subgroup of G, not subgroup of GL2(Q)

in the hint

[thorn flint]

Okay wait so what is the question really asking

[hot lake]

the hint is asking you to show that the set H = {(1 b // 0 1) with b in Z} is a subgroup of G, and is cyclic (generated by one element)

the question is really asking you to show that G can be generated by 3 elements

[thorn flint]

When they say "determine the elements x,y,z such that G=<x,y,z>", I need to find 3 matrices (x,y,z) that have elements cycles back to each other?

[hot lake]

they don't say "determine the elements x,y,z such that ..."

they say "determine some elements x,y,z such that ..."

you need to find 3 elements of G (matrices) such that those three generate G

[thorn flint]

Okay\

So G has 4 different matrices (1b 01), (1b 0-1), (-1b 01) and (-1b 0-1)

[hot lake]

G is infinite

but I guess it has 4 kinds of matrices

[thorn flint]

So i tried to compute (1b 01)^n and found that in general, it becomes (1nb 01)

And if i do the inverse, (1b 01)^-n, in general we get (1-nb 01)

[hot lake]

yes

[thorn flint]

But I don't see any patterns of them being cyclic?

except identity

[hot lake]

a cyclic group is a group that is generated by one element

you are thinking of torsion groups

not the same thing

[low surge]

i just started group theory at school so im still learning but cant u say that the one in the hint is a cyclic subgroup bc from the form uve found, its isormorphic to a subgroup of integers, and subgroups of cyclic groups are always cyclic?

[brisk granite]

I need a bit of help with one of the problems I'm working on

[cloud walrus]

PolyBeanDip:

[brisk granite]

This the problem and part of my solution. I'm not really sure how to proceed from here

Also, I only included problem 1 so that problem 2 makes sense

[leaden finch]

can someoneone help me with a

[brisk granite]

use rational root test to note -3 and 1/2 are roots

then you have a quadratic

by the rational root test, you can show it's irreducible

(x+3)(2x+1)(x^2 + 1)

should be the factorization

<@&286206848099549185> could I have help with the question I have above?

[oblique river]

what do you mean?

your solution is complete

[brisk granite]

why is that a contradiction tho?

[oblique river]

just say "therefore the galois group can't have an odd permutation"

[brisk granite]

can't a = -a?

for example a = 0 or elements of F_2

[oblique river]

the discriminant is nonzero because you've assumed the poly is separable.

[brisk granite]

I didn't assume that

I only did that for the first but

*bit

[oblique river]

alright, one sec

so this theorem is just false if the characteristic is 2

because of the issue you noticed

[brisk granite]

oh dang

did you find a counterexample?

[oblique river]

sure, take x^2 + x + 1 over F_2. The discriminant is 1 but the galois group is Z/2Z = S_2

whereas A_2 would just be the trivial subgroup of S_2

you can kind of see this directly. let's think of quadratic polynomials. if the discriminant is a square, then the quadratic formula tells you the roots are in the base field

but the quadratic formula fails in charactersitic 2 because it requires dividing by 2

[brisk granite]

ooh, cool

[oblique river]

as for the nonseparability issue, take the polynomial f(x) = x^2(x^2 + 1) over Q. it's not separable and therefore its discriminant is 0 (so, a square in Q). Label the roots in order 0,0,i,-i. As as subgroup of S_4, the galois group is {e, (34)}

which isn't a subgroup of A_4 as it contains an odd permutation

it's not a difficult example to construct but it did take me a sec -- this is why I had assumed that your assumption of separability was for the whole problem, but you're right that the problem doesn't really add that as an assumption

[brisk granite]

ok thanks

[oblique river]

for this example, you can basically take any poly that has a nonsquare discriminant (and therefore a galois group not contained in A_n)

and then force it to have discriminant 0 by multiplying by x^2

and then the galois group will now be a subgroup of S_(n+2) since we've added 2 roots

but it won't be contained in A_(n+2)

sorry for not reading your question closely enough at first

[solemn rain]

u done here boys?

i just wanna make a concept check not a problem

okay so this says that if G is a finitely generated abelian group then G is isomorphic to Z^r * Z_n_1 X Z_n_2 X .... X Z_n_k such that n_i >= 2 for all i , n_(i+1) divides n_i and r>=0

now the problem we are doing now is trying to find the abelian groups of a certain order

doing this with using what i just said up would be 'ad hoc' and would be hard to find such ns which are now called the invariant factors

so there is an identical theorem

for finite abelian groups

let G be of order n = (p_1)^a(p_2)^a_1..... G is isomorphic to some groups A_1 X A_2 X... X A_k such that :

|A_i| = p^(a_i) and

is isomoprhic to Z_p^b_1 X Z_p^b_2 .... such that b_1 + b_2 +....b_j = a_i

am i doing everything right here?>

i know i butchered the notation cuz idk latex but is there somethging wrong

[oblique river]

you haven't done much yet

[solemn rain]

yea

im jst stating what i think i know

[oblique river]

but you haven't said anything false

[solemn rain]

okay cool

so now we call the p^b_is the elementary divisors of G ( dk why )

now this is what i know

plus i proved that Z_n X Z_m is isomorphic to Z_nm iff (n,m) = 1

now i still dont know how i would find abelian groups of certain orders

[oblique river]

yes, so that fact is how you go between the two different theorems

basically just factor the order

[solemn rain]

okay

say

[oblique river]

if you get a p then that factor must be Z/pZ

if you get a p^2 then it's either Z/p^2 Z or Z/p x Z/p

etc

[solemn rain]

i want to find all abelian groups up to isomoprhism of order 180

cool

[oblique river]

I have to go now but hopefully that will help you start

[solemn rain]

so factorization 180 would be 2^2(3^2)(5)

okay

G ~= A_1 x A_2 x A_3

|A_1| = 2^2 , |A_2| = 3^2 , |A_3| =5

A_1 ~= Z_2^b1 x Z_2^b2 .....

such that b1 + b2 + ... bn = 2

and bi >= b_i+1

now this is a partition of 2

now bs would all have to be one here

2 bs

so A_1 ~= Z_2 x Z_2

am i doing this all right ?

same shit with A_2 and A_3 ?

and i think i find other nonisomoprhic abelian groups

when i play around with the partitions

if i change the b_s ( not for this case cuz 2 has only this partition ig ) i get new As hence news Gs

hope im right

ig the rest of the section is trying to explain how to go from the invariant factors to the elementary divisors and the othe way around

[toxic zephyr]

Since multiplication must be commutative, is the general linear group not a field? It seems to satisfy the necessary properties of every element having an inverse for both addition of multiplication, so if it's not a field, what is it?

[solemn rain]

lmao

[latent anvil]

@toxic zephyr it's not closed under addition

I + (-I) = 0 is not in GL, but I and -I are

or even just it doesn't have a 0 lol

[toxic zephyr]

oh shit youre right haha my b

[latent anvil]

np

[oblique river]

@solemn rain did you forget about Z/4Z

[solemn rain]

is Z/nZ the cyclic group of order n for you?

anyways

no? idk

im still finding out first group

so now i have G = Z_2 x Z_2 x A_2 x A_3

[oblique river]

jyes

[solemn rain]

cool

so im doing all right now?

i have this weird feeling that i dont understand it it sjust

[oblique river]

no I was responding to what you said about Z/nZ

[solemn rain]

confusing lmaom

okay

[oblique river]

Z_n is nonstandard notation

[solemn rain]

yea

okay so

no i did not?

is htere another way i can write 2

[oblique river]

yes

[solemn rain]

as a sum of positive integers?

[oblique river]

2

[solemn rain]

?!

[oblique river]

there it is

2

[solemn rain]

is that one

[oblique river]

2 = 2

and 2 = 1+1

are the two ways

[solemn rain]

okay so i have Z_4

and also Z_2 x Z_2

which are non isomorphic

thats just for A_1

oh boy

[oblique river]

why is that an "oh boy"?

it'sliterally the same for A_2

and it's trivial for A_3

[solemn rain]

okay i m going to work that out jusut to check

my understanding

dk how u got that that fast xd

[oblique river]

because you already factored 180

[solemn rain]

|A_2| = 3^2

[oblique river]

the exponent is the same

and it literally only depends on the exponent

what are the options for A_2

[solemn rain]

A_2 ~= Z_3^b_1 x Z_3^b_2 .... x Z_3^bn

so i can write 3 as 3

2+1

1+1+1

[oblique river]

no no no no no

[solemn rain]

what

[oblique river]

check the theorem again

the b_i don't sum to 3

they sum to the exponent on the 3

[solemn rain]

fuck

[oblique river]

that's why this is so easy

[solemn rain]

yea they ssum to 2

oh

the same thing

oh thats why

[oblique river]

yes

[solemn rain]

i got you

and A_3 sums to 1

[oblique river]

wrong channel

[solemn rain]

okay so i think i can get the handle of this

[oblique river]

also you're interrupting another conversation

gtfo

[solemn rain]

oh fuck

yo chill man

[oblique river]

someone already told you to read #❓how-to-get-help

just follow the rules

[solemn rain]

chill boys

oh fuck

put that on me hyper

my bad

so anyways

is that it for this section?

[oblique river]

yep

[solemn rain]

okay but

look at this

how can i find

the b's

if i have the ns from the first theorem

the general one

and vice versa?

oh sorry

how can i find elementary divisors

from invariant factors

[oblique river]

uhh which is which

[solemn rain]

the invariant factors are the (n_1,n_2,n_3...)

ssuch that G = Z^r x Z_n_1 x Z_n_2 ...

Z^r = Z x Z x Z ( r times )

[oblique river]

basically you have to factor each of the n_i

into primes

then split each Z_{n_i} into its prime pieces

and then combine all the pieces prime by prime

so for example if you had Z_6 x Z_2

Z_2 is already its own prime piece

and Z_6 = Z_3 x Z_2

[solemn rain]

yea

from chinese

[oblique river]

so once you regroup

[solemn rain]

thati proved

cool

[oblique river]

you get Z_3 x (Z_2 x Z_2)

the two theorems are just two ways or organizing the same data

the first way is nice because you can write any group as aproduct of cyclic groups, which is nice

[solemn rain]

what first way

the invariant factors one?

[oblique river]

(and the n_i divisible by n_(i+1) is just a little "bonus")

yes

and the other way is nice because sometimes its useful to decompose a group prime-by-prime

[solemn rain]

you mean write any abelian group?

[oblique river]

yes

[solemn rain]

any finitelyu generated abelian group

ig

[oblique river]

yes

sorry I'm being a bit loose

[solemn rain]

okaay one more thing

no np thank you for even having patience with me

why is the number of abelian groups up to isomorphism

is the number of partitions of beta ( which somehow is not rleated to the prime its being raised by )

( beta is the elementary divisor )

is it because you have ddifferent As

[oblique river]

what do you mean the elementary divisor

[solemn rain]

let G be of order n = (p_1)^a(p_2)^a_1..... G is isomorphic to some groups A_1 X A_2 X... X A_k such that :
|A_i| = p^(a_i) and
is isomoprhic to Z_p^b_1 X Z_p^b_2 .... such that b_1 + b_2 +....b_j = a_i

the elementary divisors are b_1 , b_2 ...

[oblique river]

so it's the number of partitions of the a_i

[solemn rain]

fuck

yea

XD

sorr

y

[oblique river]

you can just see that from the statement

[solemn rain]

is it because

when you change the bs

[oblique river]

and also we talked earlier about why the prime doens't actually matter

[solemn rain]

you get new A_s hence new Gs?

[oblique river]

it doens't matter if it's 2^2 or 3^2 or 11^2

there are 2 partitions of 2

[solemn rain]

lmao

cool

yea

[oblique river]

corresponding to Z_p x Z_p

and Z_(p^2)

now you take the product of hte number of partitions of all of the a_i

and that's your answer

[solemn rain]

yea i got that

okay cool

thats it for the section?

i thank god that id ont have to prove this shit

its too big boy for me apparentrly

[gilded trellis]

erm... I'm a bit confused about : "What if the condition 1 != 0 is omitted?" part

[bleak abyss]

Ah yes the legendary Definition (2.3)

[gilded trellis]

F^{x] doesn't even contain 0

lol

imma share it

just thought I'd post what I'm confused about first then follow with it :"3

[scarlet estuary]

What do you think would happen if 1 = 0?

[gilded trellis]

but in the scope of the axiom (ii) 0 isn't in F^{x}

that's what I find confusing ...

[scarlet estuary]

0 is in F

it's defined by how it acts on F+

but you can still talk about multiplying by it

for example, the real numbers form a field

[bleak abyss]

I think his confusion is that F^{\times} = F - {0} is now empty

[scarlet estuary]

so?

[bleak abyss]

So how can it be an abelian group if 1\ne 0?

[scarlet estuary]

well yeah

[bleak abyss]

Err

If 1=0

[scarlet estuary]

so you've found the problem

is what you're saying

[gilded trellis]

(I apologize beforehand for my stupidity)

[bleak abyss]

He has but he's not sure how to parse it

[scarlet estuary]

can F* be a group if 1 = 0?

[bleak abyss]

Basically the point is this, if 1 = 0, then x = x1 = x0 = 0

[gilded trellis]

I see...

erm

[bleak abyss]

So F^{\times} is empty if 1 = 0

[gilded trellis]

so it was that simple ?

[bleak abyss]

Empty sets can't be a grourp

[scarlet estuary]

essentially, yes

[gilded trellis]

:[

[scarlet estuary]

this is why we're forced to define 1 =/= 0

[gilded trellis]

wth

[bleak abyss]

Corollary it's fine to omit that condition

It's implied by the rest

[gilded trellis]

I was thinking that there was some important thing that has to do with the identity of (F,+) and (F,x) and distributivity

idk

[scarlet estuary]

since 0 isnt in F* by definition, it cant be the multiplicative identity

basically

so you MUST have 1 \neq 0

i.e. the condition can be omitted just fine

as long as you define F* = F \ {0}

[gilded trellis]

yeah I understand ...

[scarlet estuary]

anyway, there is a temptation to say "okay, but what if we pretended we could do this"

"and we had 1 = 0"

in that case, you'd have the field of 1 element

[gilded trellis]

so basically

[scarlet estuary]

as dami established

[which doesnt really exist]

[gilded trellis]

I've thought of F less abstractly

erm

with 1 he means the additive id and 0 the multiplicative

[scarlet estuary]

other way around

[gilded trellis]

but for some reason I associated them with numbers

[scarlet estuary]

0 is the additive identity, 1 the multiplicative

[gilded trellis]

so it didn't make sense to even pose such a question here

[scarlet estuary]

yeah sure, thats understandable

[gilded trellis]

but once you think of them abstractly as different identities

the question makes sense

yes

[scarlet estuary]

this is why in a lot of group theory settings, they use a symbol like "e" or "id" instead of a number like 0 or 1

[gilded trellis]

I see why such a situation is possible

EXACTLY !

(he's been using e for a while so I got used to the notation and 1 being associated with numbers because he worked in C alot)

[scarlet estuary]

it turns out that for fields specifically

no matter what sort of field we make

it behaves "a lot" like regular ol' arithmetic

so calling things 0, 1, etc isnt usually too problematic

[gilded trellis]

Yep yep

[scarlet estuary]

but there is some nuance, as established

[gilded trellis]

just notation

[scarlet estuary]

for example, in the field of 2 elements, 1 + 1 = 0

[gilded trellis]

hmm

so 1 is defined to be its own inverse (?)

[scarlet estuary]

it's not defined that way, but you can prove it

[gilded trellis]

ah

yep

F^{+} is a group

so it must be

[scarlet estuary]

oh you mean

in the field of 2 elements

yeah, 1 is its own inverse there by definition

[gilded trellis]

-_-

[scarlet estuary]

sorry i thought you meant mutliplicative inverse in a general field

lmao

[gilded trellis]

lol

[scarlet estuary]

my bad my bad

[gilded trellis]

it's alright :3

[scarlet estuary]

we usually say "negative" when talking about additive inverses

and yes, 1 is its own negative

[gilded trellis]

yep I know

[scarlet estuary]

in the field of 2 elements

anyway, the field of 2 elements is just what you get when you take the additive group $\bZ/2\bZ$ and define multiplication on it the ``obvious" way

[cloud walrus]

Namington:

[scarlet estuary]

ie 1*1 = 1, 0*x = 0

[gilded trellis]

Ah

I'm currently finishing up with chapter 3 (real vector spaces)

I've left some 3 sections out of 10 in the group theory part (chapter 2)

but I've just had a lot of trouble understanding them I just decided to go to the next chapter and go back

but now I don't feel like going back

:"3

anyways

Thanks A Waxwing Slain :3 !!!!!!!!!

[solemn rain]

what text are u using if i maay ask

[gilded trellis]

Algebra by Artin

[solemn rain]

ty\

[gilded trellis]

np :3

[ripe crest]

what's an intuitive explanation of characters?

I guess what I'm asking is - what are characters used for? What information do they encode?

[scarlet estuary]

considering characters from a group to the multiplicative group of a field gives us an abelian group of morphisms

in other words, we can study the properties of the multiplicative group of a field by looking at a regular old group and considering relations by characters

squint hard enough and you get statements about, say, galois theory, although usually the other direction (as in using galois theory to discuss characters) is more fruitful

[ripe crest]

hmmm

okay..

[solemn rain]

the abelian groups of order 100 are Z_2 x Z_2 x Z_2 x Z_2 or Z_2xZ_2 x Z_4 or Z_4 x Z_4 ( up to isomoprhism )

is that right boys?

and is there something like an online thing that would classify groups ?

so i could check?

[steep hull]

Um those groups have order 16 and there are P(4)=5 such groups (you missed Z_16 and Z_8 x Z_2)

[solemn rain]

oh fuck

well

Z_16 has order 16

why isnt it 100 then"?

stupid question

they are non isomoprhic?

[steep hull]

Two groups of different orders can’t be isomorphic

So yes

[oblique river]

I'm also not sure where 16 is coming from here. 100 = 2^2 * 5^2

[solemn rain]

yea ^ same

but ig he is right

[oblique river]

so you should be looknig for groups of order 2^2 and groups of order 5^2

[solemn rain]

so yes what

[steep hull]

There are P(2)^2=4 such groups

[solemn rain]

arent there 2 partitions of 2

so there are (2)(2) abelian groups of order 100 ?

im sorry im bad

[steep hull]

Yep

You’re decomposing into groups of prime power order

Like writing 100 as the product of prime powers

[solemn rain]

lmao

i thought i was finding the abelian groups

themselves

damn how can i be doing somethign withotu even understandingf what im doing lmaao

[steep hull]

Well you find the orders of the cyclic groups and then take direct products

[solemn rain]

can you tel lme how do you get Z_16

what i did was this

G is isomorphic to A_1 x A_2 where |A_1| = 2^2 , |A_2|=5^2 and A_1 is isomorphic to Z_2^a_1 x Z_2^a_2 x ..... x Z_2^a_n where a_1+a_2+....a_n = 2

so there are two possibilities for the a

a's

a=2

or 1+1

so i get Z_2 x Z_2 or Z_4

the same for A_2

thats what i did

[steep hull]

Yes

[solemn rain]

hwo do you get Z_16

[oblique river]

you were the one who got Z_16

[solemn rain]

plot twist?

no?

he abelian groups of order 100 are Z_2 x Z_2 x Z_2 x Z_2 or Z_2xZ_2 x Z_4 or Z_4 x Z_4 ( up to isomoprhism )

where

[steep hull]

You would have the product of cardinalities equal to 16

[oblique river]

ok, you were the one who got 16

[steep hull]

This corresponds to partitions of 4

[oblique river]

16 is irrelevant here

[steep hull]

Z_16 is the {4} partition

[solemn rain]

okay so now

ddid i miss anything

[oblique river]

write out the abelian groups of order 100 and we'll tell you

[solemn rain]

the abelian groups of order 100 are Z_2 x Z_2 x Z_2 x Z_2 or Z_2xZ_2 x Z_4 or Z_4 x Z_4

[oblique river]

what???

those have order 16

[steep hull]

For 100, the groups are Z_25 x Z_4, Z_5 x Z_5 x Z_4, Z_25 x Z_2 x Z_2, and Z_5 x Z_5 x Z_2 x Z_2 up to isomorphism

[solemn rain]

how did you get these

[oblique river]

wait first I want to know how you got the ones that you listed

so that we can correct the mistake

[solemn rain]

G is isomorphic to A_1 x A_2 where |A_1| = 2^2 , |A_2|=5^2 and A_1 is isomorphic to Z_2^a_1 x Z_2^a_2 x ..... x Z_2^a_n where a_1+a_2+....a_n = 2
so there are two possibilities for the a
a's
a=2
or 1+1
so i get Z_2 x Z_2 or Z_4

[oblique river]

yes great

[solemn rain]

its not a mistake asmuch as i not understanding

i think

[oblique river]

that's correct

now what about A_2

[solemn rain]

A_2 is the same

as A_1

oh fcuk

fucking bitch

its 5's

lmao

sorry boys

[oblique river]

really? does Z_2 x Z_2 have order 25?

lol

[solemn rain]

my bad boys

i was too happy saying the yare same cuz they had same power

but no its Z_5^a_1

[steep hull]

Just notice that these {1,1} and {2} options are allowed for each prime.

[solemn rain]

boys boys im sorry

tysm

so im finding abelian groups

by decomposing them into cyclic groups

by using fundamental theorem

[steep hull]

We know

[oblique river]

I'd say first you are decomposing them into their "prime" pieces

[solemn rain]

no no i meant i know

opkay so for A_2

[oblique river]

and then within each prime piece

you are decomposing into cyclics

[solemn rain]

A_2 is ismoprhic to Z_5^a_1 x Z_5^a_2 x... Z_5^a_n

we have either A_2 IS ISMORHPC to Z_5 x Z_5 or Z_25

[steep hull]

Prime pieces are allowed by Chinese Remainder Theorem

[oblique river]

yeah we went over that earlier

[solemn rain]

cool

i got it now

okay here is the ifnal part

going from invariant factors to elementary divisors

suppose G is isomorphic to Z_2 x Z_2 x Z_25

what type of G is it?

(2,2,25)?

[oblique river]

if you're trying to write out the elementary divisors

that can't be it

[solemn rain]

no

thats

the invarant

[oblique river]

oh

[solemn rain]

the invariant factors

type

[oblique river]

ok yes

[solemn rain]

let me show u this

[oblique river]

to get the elementary divisors, basically just pick the largest part of each prime piece

[solemn rain]

so now by chinese remainder theorem

look down b4 exercsies

[oblique river]

and work your way down

[solemn rain]

no thats not what i wanted to show you lmao

1 sec

[oblique river]

oh sorry

oh wait a minute

[oblique river]

we already have the elementary divisors

[solemn rain]

yea

thats what i usued

the second 'theorem'

now the elementary divisors are 2 4 5 25

right?

NO WAIT

[oblique river]

I mean not all of them at once

[solemn rain]

suppose G is isomorphic to Z_2 x Z_2 x Z_25

[oblique river]

each group of order 100 has different elementary divisors

ok yep let's take that one

[solemn rain]

divisors here are 2,2 ,and 25

[oblique river]

ye

[solemn rain]

5*

[oblique river]

no

25 was right

[solemn rain]

fuck it im using definitions

yea ur right

yea ayea

divisors are the a_is here

so idfc about intuition

okay so now

how can i find the invaraint factors

cuz thats like half the section

[oblique river]

for each prime that shows up

take the largest elementary divisor

[solemn rain]

?

how

[oblique river]

the primes that show up are 2 and 5

the largest divisor for 2 is, well, 2

and the largest for 5 is 25

so let's take 2*25

=50

we started with 2,2,25 and we took out 2 and 5

so we're just left with 2

and that's it

2 and 50

[solemn rain]

ohh

[oblique river]

Or Z_2 x Z_50

[solemn rain]

i gotta ask

is gcd(2,25) = 1?

if yes then im done

[oblique river]

😓

[solemn rain]

i understand everything now

WHAT XD

[oblique river]

are you seriously asking that?

[solemn rain]

okay 1 sec

[oblique river]

gcd(2,25)?

[solemn rain]

fucking internet