groups-rings-fields

[delicate orchid]

this really is like nothing I've ever seen before - I shall adjust my mindset accordingly

I thought self teaching galois theory would kinda be like self teaching representation theory but turns out the former is much harder

[chilly ocean]

on the contrary I feel

[delicate orchid]

idk representation theory is just "oohhh use the inner product on the charaters funny direct sum wheeee"

[coral shale]

my mind still dazed about z^5-2

im convinced this thing is irreducible but it apparently isnt 🤔

[delicate orchid]

irreducible over what field

nvm it's a z clearly it's over C

[chilly ocean]

[coral shale]

huh were we not doing Q

[hidden haven]

It is

Eisenstein

[coral shale]

wat

scam?

[hidden haven]

And Gauss

[chilly ocean]

It is Einstein... 🤓

[coral shale]

and common sense

[hidden haven]

Nope

[delicate orchid]

Eisenstein and gauss sitting in a tree...

[coral shale]

ok so the degree of the extension is 5!

maybe

[delicate orchid]

120 degree extenstion no way

[chilly ocean]

120?

[hidden haven]

The issue was that splitting field is not Q[fifth root of 2]

It's not 120

[coral shale]

am i getting warmer

boi whatm i doing

[chilly ocean]

you want to find the splitting field or what?

[hidden haven]

Yes

[coral shale]

im thinking what we need to adjoin

[hidden haven]

Galois group

[coral shale]

not all 5 apparently

[hidden haven]

And I've said start with splitting field

And now I sleep

[coral shale]

ive only just done splitting fields

[hidden haven]

[delicate orchid]

hold on a moment

[coral shale]

thats probably my problem

like the definition, thats it.

[chilly ocean]

well you need other roots for a slpitting field

what are the roots of x^5 - 2?

[coral shale]

the 5th roots of unity times

2^(1/5)

[chilly ocean]

Q(2^(1/5)) doesnt have roots of unity

[delicate orchid]

you gotta get da complex ones too

?

[coral shale]

yes

ik theres 5 of them

[chilly ocean]

but adding just one complex one is enough

[delicate orchid]

yeah but as I've learnt that doesn't mean the degree is 5

[coral shale]

it is?

[chilly ocean]

yeah the roots are of the form 2^1/5 w^k

degree will be higher

[coral shale]

https://media.discordapp.net/attachments/496784958430380033/945025914742374470/Screenshot_20220220-183513.jpg

So if adding one is enough then degree of extension is 5

[chilly ocean]

yeah but it isnt enough

Q(2^(1/5)) is a subfield of reals

[coral shale]

huh i thought u said adding one complex one is enough

[chilly ocean]

oh yeah if you add complex one then you will have the splitting field but the degree won't be 5 over Q

[coral shale]

im uh...

[chilly ocean]

you need a complex root of unity + fifth root of 2

to get all roots

[coral shale]

Ok ima have to think about this carefully after we do this section

guess it was too early for me - but enlightens what i didnt know

thanks

[chilly ocean]

Okay so do you see why Q(2^1/5, w) is a splitting field?

[coral shale]

I understand now yes

but it wasnt obvious at the start to me

[chilly ocean]

yeah its not

[lapis trail]

Is there a result that says if you divide two polynomials that are relatively prime, the remainder is a unit?

[chilly ocean]

not think about the degree over Q in your free time

hhint: tower theorem

[lapis trail]

By which I mean gcd 1

[coral shale]

The degree should be 4x5

I believe

[chilly ocean]

Yep

yeah finding galois group is harder

[sharp sonnet]

have you tried looking at char>0?

[upbeat swift]

Quick question: If I have a function f:S_3 → Z_2, how do I identify the kernel and it's cosets? In the problem I'm solving, we are using tables. And I have the identity of S being ε and the identity of Z being 0.

[kind temple]

kernel is the set of all permutations which get mapped to zero by f. and do you mean the cosets of S_3/ker(f) ?

[upbeat swift]

It just says to list the cosets if K

Of*

[kind temple]

i’m assuming K is the kernel of f

[upbeat swift]

Yes

[coral shale]

👀

So do you have a specific f given

Or just any f.

[delicate orchid]

also is f a homomorphism or just a function?

[upbeat swift]

Just says f: S_3 → Z_2

[coral shale]

🤔

[upbeat swift]

if you wanna see

[coral shale]

Oh so you are given f

[upbeat swift]

yes

Using tables, i showed it was homomorphism

[coral shale]

Ok

You showed
f(a)f(b) = f(ab)

[upbeat swift]

im just confused on kernel and cosets

mhm

[coral shale]

Is there any extra info

From previous questions

like what those 6 greek letters are

[upbeat swift]

Oh yes

[coral shale]

yh... I think all of this missing info is needed

[upbeat swift]

Let me find the chapter and shows it's chart

[coral shale]

Well that aside - tell us what the kernel is, as in the definition

[upbeat swift]

K = {g in G: f(g) = e in H}

[coral shale]

Exactly.

Now identify what G and H are in that definition

[upbeat swift]

G would be S_3 and H would be Z_2

[coral shale]

Ok.

so what is 'e in H'

[upbeat swift]

The identity in Z_2 is 0

[coral shale]

K = {g in S_3: f(g) = 0 in Z_2}

So we have this.

So you are looking for this set K

everything in S_3 which f maps to 0

Does that make sense?

[upbeat swift]

Oh it does

[coral shale]

Utilise this

[upbeat swift]

Which is epsilon, beta, delta

[coral shale]

ok.

And then what definition of cosets are you given

[upbeat swift]

Left coset: aH = {ah:h in H}
Right coset: Ha = {ha: h in H}
(In this def, you have a subgroup H and group G)

a is an element in G

[coral shale]

Right.

So you find those

for every possible a

[upbeat swift]

Both left and right?

[coral shale]

uhhhhhhhhhhhhhh

Since the question doesn't specify, I suppose so.

So from the definition alone

Since there are 6 elements in G

It sounds like there are 6 left cosets

and 6 right cosets

But you should know some of these will be the same (there will be less)

[upbeat swift]

I'm no sure how to find these cosets

[coral shale]

Just utilise the definition

So go through each possible a

starting with epsilon, say.

eH = {eh : h in H}

Identify what this set is. (list its members)

And then do the same for the rest.

It looks like a lot of work, but its less than you think.

[upbeat swift]

I'm just trying to think of it all

[coral shale]

Write it down, do it step by step

eH = {eh : h in H}

oh wait a second, I wasn't paying attention

[upbeat swift]

👀

[coral shale]

@upbeat swift read the question carefully

What is H here?

[upbeat swift]

Z_2?

[coral shale]

no

Verify that f is a homomorphism, find its kernel K, and list the cosets of K

[upbeat swift]

Oh K

[coral shale]

For a in G
Left coset of H: aH = {ah:h in H}
Right coset of H: Ha = {ha: h in H}

===

To clarify your definition above.

So for the first a

we chose e

eK = {eh : h in K}

ok?

[upbeat swift]

Mhm

[coral shale]

So you should hopefully notice eK = K

because we are left multiplying everything in K by e

which does nothing

To give yourself an easier time, did you prove somewhere that the Kernel of any homomorphism always forms a subgroup of the domain?

Maybe not which is fine also.

[coral shale]

Utilise this table to give yourself an easier time finding the cosets

[upbeat swift]

Wait so say I'm doing aplhaK = {alpha h : h in K}. What exactly am I doing here?

[coral shale]

use the cayley table

{ak1, ak2, ak3}

There are 3 elements in K, I will just call them k123

This is what aK is right?

Next figure out what each of those are from the table

Which is epsilon, beta, delta

^ you said this was K

So
aK = {ae, ab, ad}

[upbeat swift]

Oh okay

Thank you

[desert dome]

Hi, does anyone know why "G acts on this set by permutation"? I understand that galois group permutes the roots of an irreducible polynomial, but here we don't know if it's galois yet and we don't have a ground field... I would appreciate some explanation

[lapis trail]

Is it a fact that if gcd(n,k)=1, then r is a unit in $\mathbb{Z}\mod k$, where $n=qk+r?$

[cloud walrus]

seth.delacroix

[proud bear]

yes. if gcd(n,k)=1, then by bezout's lemma, there are integers a and b so that an+bk=1. taking that equation mod k, an=1 so n is a unit mod k. since n=r mod k, r is also a unit

[lapis trail]

Woahh thanks

[next obsidian]

This is a criterion. It’s iff

Proof: Bezout’s lemma

[lapis trail]

I was trying to proveF[x] where F is a field, if p(x) is irreducible then <p(x)> is a maximal ideal. So I wanted to know

[proud bear]

for $\sigma\in G$, the action of $G$ on the set should be $\sigma\cdot(\sigma_i(\alpha))=(\sigma\circ\sigma_i)(\alpha)$ where $\circ$ is composition of automorphisms. $(\sigma\circ\sigma_i)(\alpha)$ will equal one of the other $\sigma_j(\alpha)$ because if it wasn't, then you've found another conjugate of $\alpha$ that is not in that set.

[delicate orchid]

there's an easier way to do this that doesn't rely on it being a Euclidean domain btw

[desert dome]

ohh I see. Thank you!

[lapis trail]

What is it if ya don't mind me askin

$numbers$

[delicate orchid]

use the fact that it's a PID and so if a divides b if and only if (a) contains (b)

[next obsidian]

How tf you prove it’s a a PID without showing it’s a Euclidean Domain?

[delicate orchid]

it's more that this method works for a general PID

and they've assumed it's an ED anyway so you can assume it's a PID

[prisma shuttle]

for the second statement, when it uses the notation $\alpha^n$, ima used to that meaning $\alpha\cdot \alpha\cdot \cdots \cdot \alpha,$ where there are $n$ $\alpha$s. However, if $M_{\alpha}$ is an additive monoid, does that mean that $\alpha^n = \alpha+\alpha+\cdots +\alpha,$ where there are $n$ $\alpha$s? aka. is $\alpha^n=n\alpha$?

[cloud walrus]

JustKeepRunning

[chilly ocean]

can someone help me on this? idk what exactly i need to read in the book for this problem

i read direct products

and im kinda ehhh on correspondence theorem

this question isn't even algebra

it should be a review of basic facts about functions

that's where you'll want to read

[paper flint]

This seems to be an unusual problem for "homework 7" on (presumably) an abstract algebra class. This is basic set theory, #proofs-and-logic or #discrete-math are better suited.

[lavish nexus]

something-jective

[chilly ocean]

I agree it's definitely the wrong channel but I think it would be a good refresher for students to put such a problem on a hw especially to build intuition for something that uses the same logic, like the correspondence theorem which was mentioned

so personally I don't think it's too unusual :P

[paper flint]

Hmm, fair enough.

I'd rather drop this as a hint to the main problem. 😛

[hybrid island]

can someone help me out w these

[lavish nexus]

you can write out columns based on how T sends basis vectors

[hybrid island]

like [[0, 0, 1], [1, 0, 0], [0, 1, 0]]

?

[chilly ocean]

Can someone explain to me why the product group of two mod groups have to be coprime in order for it to be cyclic?

I must be misunderstanding how to generate cyclic groups, because in my mind you can generate the entire product group with the action +1 for both groups

[tawny pine]

@hybrid island yes

[chilly ocean]

Bijective is what i thought

[chilly ocean]

How would i prove a fill in the blank question

Should i just draw a diagram?

Like domain codomain

T’ etc

[terse crystal]

You consider the smith normal form of diag{a,b} when a and b are not coprime

[chilly ocean]

I'm a noob

[terse crystal]

?

[chilly ocean]

I dont know what a smith normal form of diag{a,b} is

Anyways

I somewhat figured it out

[terse crystal]

It is diag{gcd(a,b), lcm(a,b)}

Okay then

[lavish nexus]

[[0,1,0],[0,0,1]
you can figure out the last column

[lavish nexus]

one is injective the other is surjective it's 50/50

[chilly ocean]

do i draw

or like

how would i prove this

or show it to myself

[lavish nexus]

just write normal proofs

[chilly ocean]

you think you can walk me through parta then i can try to do part b myself?

[lavish nexus]

maybe do contradiction and conjure up a counter example

well let's say f(S) is not all of T

take T' in T\f(S)

what do you get

[chilly ocean]

one sec

well you get everything In T that is not mapped

[lavish nexus]

I mean you can proceed directly from there

you're looking at f(f^(-1)(T'))

[chilly ocean]

yeah

how would i check for injective or surjective

[lavish nexus]

write out what can be in f(f^(-1)(T'))

[hybrid island]

i think we have the same answer, urs is in terms of columns and mine is in terms of rows

[lavish nexus]

like what even is f^(-1)(T')

if T' is in T\f(S)

[chilly ocean]

isnt it just f(S)

[lavish nexus]

what's the preimage

[chilly ocean]

like f^-1(T') is is just f(S)

[lavish nexus]

where is f(S)

[chilly ocean]

in T

[lavish nexus]

how can a preimage of a subset in T still be in T

you're mapping from S to T

[chilly ocean]

im so confused lol

okay

lets go step by step

on the left side

isnt that just f(f(s))

[lavish nexus]

f^(-1)(T') = {x in S| f(x) in T'}

[chilly ocean]

yeah

[lavish nexus]

meanwhile T' in T\f(S)

so what is f^(-1)(T')

is there anything in S that maps to something in T'

[chilly ocean]

there can be

[lavish nexus]

what is T\f(S)

[chilly ocean]

everytihng in T after f(S) is removed

[lavish nexus]

ok the entire image has been removed

we're picking T' from what's left

is there anything in S that maps to something in T'

[chilly ocean]

not sure

well we removed

everything in T

[lavish nexus]

not everything because assumed f is not onto

[chilly ocean]

i honestly dont know bro

im confused

this shouldnt even be hard

[lavish nexus]

nothing can map to T'

f^(-1)(T') is empty set

[chilly ocean]

because we removed f(s)

[lavish nexus]

f(f^(-1)(T')) is f(empty set) which is nothing

[chilly ocean]

so it has to be surjective?

[lavish nexus]

it proves one direction

-> direction

so go ahead and prove surjective leads to f(f^(-1)(T'))=T'

[chilly ocean]

can you write how we proved -> in one line

i cant really see how you did that

we kinda just

[chilly ocean]

i didnt get this part tbh

[lavish nexus]

suppose not surjective, then statement is not true
so if statement is true it must be surjective

T' is any nonempty subset of T removing f(S)

the statement has to hold for any subset of T
since f not surjective, f(S) is not all of T, so T\f(S) contains something other than empty set, and we can take T' nonempty
then f(f^(-1)(T')) = f(empty set) = empty set != T'

[chilly ocean]

so this is a proof by contradiction

and we are supposing that the function f is not surjective

[lavish nexus]

more like contrapositive

[chilly ocean]

how is T' any nonempty subset of T removing f(S)

[lavish nexus]

just take any

[chilly ocean]

or are we letting that happen

[lavish nexus]

yes because it is possible by assumption of f not onto

[chilly ocean]

so we first suppose f is not surjective. Let T' be any nonempty subset of T removing f(s)

yeah i get what onto is

basically we are saying not all of codomain has something in the domain that maps to it

[lavish nexus]

yes

[chilly ocean]

gimme a moment let me write down the t-> direction

and see if understand it clearlt

since f not surjective, f(S) is not all of T, so T\f(S) contains something other than empty set, and we can take T' nonempty " in this sentence isnt T\f(S) same thing as T'

[lavish nexus]

might be if there's only one element in T\f(S)

[chilly ocean]

so its more like T' cannnot be the empty set

if thats what you mean

[lavish nexus]

no it's we can take T' nonempty

[chilly ocean]

then why did we take it empty in the first placE?

[lavish nexus]

we didn't

we're only justifying that we can take it nonempty

because otherwise if T' must be empty then f(f^(-1)(T')) = f(empty set) = empty set = T'

we're not proving anything

[chilly ocean]

bro

this is what i understood

[lavish nexus]

yes this finishes ->

[chilly ocean]

imi dont understand the last line

it seems extra

idk how the f(empty set) takes play

[lavish nexus]

it shows the statement is false

[chilly ocean]

we basically said in the thing above it that T' is not empty

[lavish nexus]

that f(f^(-1)(T')) != T'

that's the point

T' is nonempty but f(f^(-1)(T')) is empty
they can't be equal
this proves contrapositive for ->

[chilly ocean]

i didnt even talk about the empty set

right here

so first thing equals f(empty set)

[lavish nexus]

yeah

cuz nothing maps to T'

[chilly ocean]

i didnt even say that in my proof

[lavish nexus]

it's obvious

T' is not in the image

obviously its preimage is empty

[chilly ocean]

one sec

i think i get it

dont we have to say let T' not be in f(S)

[lavish nexus]

yeah

[chilly ocean]

i didnt say that

like read my proof

its hella unclear

[lavish nexus]

take T' nonempty in T\f(S)

[chilly ocean]

so i wouuld state that in my proof right?

[lavish nexus]

yes

[chilly ocean]

this is clearer right?

[lavish nexus]

yes

[chilly ocean]

it makes ense

so basicallty

this is what is happening

since we removed f(S)

and we put T' in T/f(S)

[chilly ocean]

it is extra

[lavish nexus]

pf: obvious

[chilly ocean]

the takeaway is that T’=T/f(s) contradicts the hypothesis since f^-1(T’) is empty

the otherway is definition of surjection, right inverses, or you can use another definition and show its equivalent

@lavish nexus now i would do <=

assume surjective

[chilly ocean]

yes

read my previous message

[acoustic pine]

I'm trying to come up with a quadratic form that is compatible with the dual numbers over R, that respects multiplication if at all possible. Is such a thing possible? I'd rather not just use the vacuous q(a+b\eps)=\lambda a^2, but If thats all I have to work with thats fine

[terse crystal]

I don’t understand what you mean by saying “dual numbers over R”, anyway, any quadratic form of R^n over R has the form x|—> x^tAx where A is a real symmetric matrix.

=Σx_k^2 (k from 1 to p) -Σx_k^2 (k from p+1 to p+q) after applying a linear isomorphism on R^n

[next obsidian]

Dual numbers are R[x]/(x^2)

[chilly ocean]

im kinda stuck on the <=

[terse crystal]

I see, thanks. And what does compatible mean?

[acoustic pine]

Thats a fair question. I'm trying to think of what I mean myself. My initial thought was "it respects multiplication in R[x]/(x^2)", but I'm really not sure if that would be of any use, or even possible

I'm trying to draw a parallel between complex/hypercomplex and dual numbers, and work with it geometrically

But I need a quadratic form, because one thing I want to explore is if I can do anything like geometric algebra using clifford algebras over R[x]/(x^2)

[terse crystal]

I don’t even know you are considering quadratic form on what… on a k-vector space? What field k?

Never mind.I barely know anything about Clifford algebras or other stuffs anyway

[acoustic pine]

The R-algebra R^2, with bilinear operator ((a,b) X (c,d)) = (ac,ad+bc)

[terse crystal]

I see, a quadratic form on R-linear space R[x]/(x^2)

[acoustic pine]

Exactly

[terse crystal]

All possible forms are Q(a+bx)=a^2

And zero

Since Q doesn’t equal zero and you want it to satisfy Q(a+bx)Q(c+dx)=Q((a+bx)(c+dx)),then Q(1)=1 Q(x)=0, Q(1+x)=1

[acoustic pine]

I figured it would be something along the lines of thi

[cyan raft]

shameless plug: hi can anybody help me in #help-9? i have an abstract algebra question and i dont wanna interrupt the conversation

[terse crystal]

It won’t

Question solved we ended already

[cyan raft]

oh okay

here goes

can anybody check my proof (polynomial ring over field is PID)

(integral domain: trivial)
call the field and the polynomial ring F and F[x]
first i'll prove that in F, multiplication is a bijective map.
then, i'll prove that the ideal generated by elements ax^n, bx^m(n<m) contains all polynomials with the last n-th terms = 0.
then, i'll prove that gcd(ax^n, bx^m) = zx^n (still, n<m) where z is an element in F
and concluded that (gcd(ax^n, bx^m)) = (ax^n, bx^m).

i've heard that this proof requires bezout's lemma, so i'm pretty skeptical about my attempt. can anybody tell me if this is correct?

[terse crystal]

What do you mean bijective? Any b, a|—> ab is bijective you mean?

it’s only injective

I believe the standard proof is proving it’s an E.D

[cyan raft]

shouldn't it be surjective as well? for the map x->ax with nonzero x, for every value n you have nx^-1x=n

[terse crystal]

If b has positive degree

Then ab has clearly higher degree than a

for example b=x clearly not all polynomials can be divided by x

[cyan raft]

it's bijective in F though, right?

[terse crystal]

In other word x is not invertible

You mean b is from F? Sure

[cyan raft]

yeah

[terse crystal]

Non-zero also

[sharp sonnet]

i think the point of this in the proof is only to reduce the case ax^n in the ideal to x^n in the ideal

[cyan raft]

yeah, it makes the rest much easier

[sharp sonnet]

anyways, you probably implicitly already use that F[x] is an euclidean domain when computing the gcd

the thing i would be worried about is how what you have shows its PID

do you want to inductively reduce the generators of an ideal?

then you need to show noetherian as well

[cyan raft]

arent fields always noetherian?

[sharp sonnet]

well, yes

so is F[x]

but you need this

and then this works i think

you implicitly use heavier tools though when computing the gcd

this proof sketch also gives you the "correct" idea

you show that you can ignore higher degree terms in an ideal in some sense

so pick an element in the ideal with lowest degree

[terse crystal]

Then he returns to the standard proof actually 😂

[sharp sonnet]

yes

the gcd computation in your proof should more or less have all the meat already

(this is where you use the fact that F[x] is an euclidean domain)

[cyan raft]

hmmm
does the standard proof require this step?

[sharp sonnet]

yes

you can do it implicitly without stating it

but its what makes this thing easy to show

its also what lets you compute the gcd of two polynomials easily

[cyan raft]

got it, thanks

[bleak abyss]

I'll talk here

Some more algebraic groups stuff

So take the subgroups U(n) and B of GL(n)

Think of them as stabilizers for natural actions of GL(n)

Let's find something

So GL(n) acts on A^n, A^n - {0}, P^{n-1}, Gr(n,k)

I guess we want GL(n) to act on "the space of inner products"

But I wanna think of that linear algebraically because I want to think of it as a scheme

I'll worry about details later

GL(n,C), which I guess we think of as a real group for now

Acts on the space of Hermitian inner products

g.(x,y) = (gx,gy)

And stabilizer of the usual one is U(n)

As for B, I guess we think of GL(n) as acting on the space of flags

In some way

I mean the way you expect probably

And then B is the stabilizer of the flag {0, {e_1}, {e_1,e_2},...}

[terse crystal]

No need to view it as a scheme, it is a group scheme, I believe it’s discussed in detail in that AMS book which I didn’t read

Representation of algebraic groups

[desert dome]

Hi, I am so frustrated with finite fields.. Does anyone know why F has characteristic p if F is the set of all the roots of x^p^n - x over F_p? (I know F has to be a field)

[next obsidian]

F_p < F

Char p would mean that 1 + 1 + … + 1 p-times is 0

But 1 as an element of F lives inside F_p

[hidden haven]

Set of all roots of a polynomial over F_p is kinda weird to say though

[next obsidian]

And you know that p•1 = 0 inside F_0

[desert dome]

I read this from dummit. What's the better way?

by F_0, do you mean F?

[desert dome]

Why F_p < F? sorry I am so confused...

[next obsidian]

I mean… F_p is roots to linear polynomials

Or well

[hidden haven]

You should specify what the ambient space is when you say set of all roots. If you take the set of all the roots in F_p, you get just F_p itself. If you take set of all roots inside some ring, you can get a lot more roots then you want

[next obsidian]

That won’t show it but

x^p - x

[hidden haven]

F_p

[next obsidian]

Is satisfied by every element of F_p

[hidden haven]

It's like "take a union of ℝ and F_p" and if you say this without context anyone's gonna be confused about where this union is happening

[desert dome]

I see. yeah I dont quite understand why element in F_p satisfy x^p - x but I can look it up

[hidden haven]

Lagrange

On the multiplicative group

[desert dome]

I see. Thank you chmon and moldi

[hidden haven]

✓

[coral shale]

Yesterday I was shown we need to adjoin 2^(1/5) and a 5th root of unity to split z^5-2, so the degree of the extension is 20.

I am convinced this is definitely sufficient to split the polynomial.

[cloud walrus]

Shuri2060

[coral shale]

I am currently consider the same but for z^6-2, and am stuck for this reason

[coral shale]

typo, missing i in the exponent of e. So

$$2^{1/5}e^\frac{2\pi i}{5}$$

[cloud walrus]

Shuri2060

[hidden haven]

Are you convinced that adjoining 2^(1/5) is not enough?

[coral shale]

yes

[hidden haven]

Try proving that adjoining 2^(1/5) or the element you said, both give isomorphic fields

[coral shale]

ahh so thats the way

ty

[hidden haven]

In general, an element behaves exactly like any of its conjugates

[coral shale]

For the minimal polynomial only?

when u say conjugate

[hidden haven]

Yes

[coral shale]

ie. for roots of unity, 1 doesnt count

[hidden haven]

Yes

Conjugate always means roots of minimal polynomial

This is only true of extensions when you view them as sitting inside their algebraic closure though

For example, in ℝ, √2 and -√2 can be algebraically distinguished

Since one has a square root and the other doesn't

But over ℂ, these are indistinguishable

Algebraically

In the sense that there is an automorphism of ℂ

That maps one to the other

[coral shale]

uhhh dont we always talk about splitting fields within algebraic closures

otherwise you cant always split

[hidden haven]

Yeah you can take that approach

But not necessary

[coral shale]

So splitting fields can be broadened to this

So a splitting field of Q in R doesnt have to split to linear factors

but just as much as possible??

[hidden haven]

You can view adjoining of roots as an abstract process

In which case you don't need a larger ambient field at all

If you have a field F, and an irreducible polynomial p(x), then adjoining a root of p(x) is taking the field F[x]/(p(x))

Then p(x) factors in this larger field, but now you instead write p(y) for a new variable y since you've added this element x to your field anyway

Ok that might be a very confusing way to phrase it

You have p(y) a polynomial with coefficients in F, then F[x]/(p(x)) has a root of p(y)

[coral shale]

(F[x]/(p))[y]

Like this or am i misreading

[hidden haven]

But not necessarily all roots, so you do this process on the new irreducible factors of p(y)

[hidden haven]

Yeah

But F embeds into this larger field

So there's a natural way to view p(y) as a polynomial with coefficients in this larger field

Maybe take some simple example and work it out, it should make things clear

Like p(y) = y² - 2 or y³ - 2

You construct the splitting field by adjoining one root at a time like this

And this construction is then isomorphic to say the splitting field as a subfield of ℂ

You can prove that all splitting fields of p(y) over F are isomorphic

Did any of that make sense

[coral shale]

Thanks, I will have a think on this...

not everything is clear yet

[hidden haven]

Cool

[coral shale]

F[x]/(p)

This adjoins some roots of p

but not all

[hidden haven]

This adjoins exactly 1 root

Try to find it

[coral shale]

for a general p, i am asking

oh

uh

[hidden haven]

Other roots may have come along, but there's only one root that it attempts to adjoin

[coral shale]

Ok ill figure this out first then i think everything else will make sense

ty

[plush wasp]

A question !! Splitting fields are defined as the smallest field extension over which the poly. can be reduced into linear factors?
so say, Q(w) and Q(w²) are both splitting fields of p(x) = x^3 - 1 because they're both roots of the same minimal polynomial? [w is cube root of unity]

[chilly ocean]

ye

those are the same fields

[plush wasp]

kind of makes sense that the splitting fields must be isomorphic

[coral shale]

Q(w, ww) = Q(w) = Q(ww) uwu

its when p(x) = x^3 - 2 when crazy things start to happen

[chilly ocean]

Schuris Lemma

[plush wasp]

oh btw shuri, your explanation the other day is what was discussed today in class 👀

[coral shale]

what explanation

[plush wasp]

the proposition was: For any field F, there exists an algebraically closed field K over F.

And the construction went exactly how you explained the other day

[plush wasp]

considering F/(f) and the maximal ideal argument

[hidden haven]

How did you construct algebraic closure without adjoining 1 root

[chilly ocean]

Someone pls teach me Field Theory 😭 😭 😭 😭 😭

no

[plush wasp]

the argument was K = U (K_1, K_2, ..., K_n)

[coral shale]

I didnt explain it, my notes did

[chilly ocean]

I like Probability, Graph Theory, Applied Computational Methods and stuff But this...

[plush wasp]

but I still fail to understand what's supposed to be very trivial

why does F/(f) contain the root of "f"

._.

[coral shale]

so do i. z^5-2 is very strange

[hidden haven]

Appl*ed mathematician

[plush wasp]

I think you're probably overthinking this one

[coral shale]

i can show you my notes later

[chilly ocean]

look at f + x

[chilly ocean]

lol why censor??

[hidden haven]

Try to find it

[coral shale]

if moldi doesnt give some wonderful explanation

[hidden haven]

Maybe with explicit example

[coral shale]

ah thats beautiful

[hidden haven]

[coral shale]

we dont say the a word here

[hidden haven]

Why would I write bad words

[coral shale]

only a words we allow are abstract and algebra

[chilly ocean]

you just used "are" as well

[frail zealot]

what's the sort of process of finding symmetry groups of more complex objects. Like objects with infinite symmetries or higher dimensional polyhedra?

[coral shale]

I don't think that is 'easy' for the general case

https://gowers.wordpress.com/2011/11/25/group-actions-iii-whats-the-point-of-them/

I read this recently

and i believe this is relevant to your Q

he uses group actions to explore the symmetries of the cube

And I think he implies he is doing this because this isnt easy to do otherwise

[frail zealot]

thanks

[tawdry crystal]

Holy cow, your suggestions help me fix a problem I've been stuck on for two weeks! Many thanks!!!!!

[barren sierra]

I know we use $\leq$ and $\trianglelefteq$ for subgroup and normal. Is there anything analogous for subring and ideal?

[cloud walrus]

Spamakin🎷

[hidden haven]

I think the same 2 things

[barren sierra]

Hm ok I've never seen those used in that context but I guess that makes sense

[lethal dune]

invent your own notation $I \catthink R$ be ideal

[foggy merlin]

is there a field of study where one looks at a given function or a given set of functions and then determines what type of binary operations they are the morphisms of? kind of the opposite of what one usually does, usually we have the binary operations given, addition, multiplication, compostion in the given structures (group, ring, etc.) then figure out what the morphisms look like.

[pastel cliff]

https://tenor.com/view/run-forrest-puppet-panic-run-scream-gif-15317712

@hidden haven

[hidden haven]

[hidden haven]

Perhaps take the set of all binary operations that they preserve

[lapis trail]

In the algebraic closure of F, p(x) is reducible to linear factors where p(x) in F(x)

[hidden haven]

I was serious about this btw This should prove that any collection of functions X → Y is exactly a collection of homomorphisms between from X to Y equipped with certain binary operations

If instead of something as random as the set of all binary operations (or I guess in this case set of all pairs of binary operations, one on X and one on Y such that the collection of set functions respects each pair) you want some nice conditions like invertibility, associativity etc, I think this would fall under model theory/universal algebras

[lavish nexus]

what is the generating vector field for some η in Lie algebra

[foggy merlin]

well that would be nice, not to just have any type of binary operation, but more based on the properties of f and maybe other underlying structure on X and Y, what type of binary operations, (with properties like associativity etc.) preserving f exist.

actually I'd be more interested specifically in the case where Y = X, which would reduce arbitraryness by a lot (I think)

[hidden haven]

Why restrict to binary operations

Sounds a lot like model theory idk but I am sure ultra would know something lol

Curiosity or did this come up somewhere?

[lavish nexus]

I know the basis but how is w^(\xi) defined

[chilly ocean]

the set of all functions from X to itself is big for arbitrary sets X

same with X x X -> X

[chilly ocean]

it's the infinitesimal generator of the flow (t, x) -> \exp(t\xi)x

just a general definition for lie group actions

ig it does make me wonder which X x X -> X are useful

but ppl study this

[lavish nexus]

d/dt \exp(t\xi)x at t=0?

[chilly ocean]

that'd be w^\xi at x, yeah.

maybe up to a sign depending on your conventions

[lavish nexus]

ok ty lemme write it out

[foggy merlin]

well but given some functions from X to X, the set of binary operations that are being preserved by that set might not be

[coral shale]

is moldi somewhere 👀

F[x]/(f)

[hidden haven]

[coral shale]

So I worked out some stuff

eg. f = x^2+x+1

You are now doing

h^2+h+1 === 0 (mod f)

And the solutions to this are h = x, x^2

But now I'm wondering what if f isn't minimal 🤔

[hidden haven]

For that you need to know what F is specifically

Over a general field, it need not be

[coral shale]

I mistakenly considered h^3 + 1 === 0 (mod x^3 + 1)

Before realising

[hidden haven]

If it is not minimal, then this quotient ring isn't a field

[coral shale]

Right. Is it worth studying/looking into at all?

[coral shale]

this equation

[coral shale]

Or would it be completely unrelated to this

[hidden haven]

Not as far as I know

[foggy merlin]

not really that it did come up anywhere relevant, but I was looking at some functions and this question kind of popped up, since they didn't preserve any structure but seemed like they could preserve some structure, and knowing what that structure could be might help studying the functions itself

[hidden haven]

Then you wouldn't be doing field theory

[hidden haven]

In model theory, we often talk about the properties that a function preserves

There we already have structures on both sets, ie binary (or n-ary) operations on both sets, and the structures are of the same kind in the sense that there are corresponding operations

And then you can talk about the properties of these operations that are being preserved by the function

For example, properties that go like "There exists x_1, ..., x_n such that [some equations]" are called positive existential properties

All structure homomorphisms preserve positive existential properties

Structure homomorphism is the obvious thing, a function that preserves all operations

[coral shale]

For the roots of unity, we will have powers of x, x^2, ...x^(n-1) as roots.
I can see they're roots. But it isn't obvious to me there aren't more possible h that could work.

1+h+h^2+...+h^(n-1) === 0 (mod 1+x+x^2+....+x^(n-1))

Do you justify with the fundamental thm of algebra there are at most n-1 solutions to this polynomial?????

But then what about
h^5 - 2 === 0 (mod x^5 - 2) 🤔. This is supposed to only have 1 root h = x, but I wouldn't know how to justify this

[hidden haven]

In general, x in F[x]/(f(x)) is the only element guaranteed to be a root of f(h)

[coral shale]

Agreed.

[hidden haven]

Over an integral domain, a polynomial can't have more roots than its degree

[coral shale]

(That's the fundamental thm of algebra right?)

[hidden haven]

No, that is a lot more specific lol

[coral shale]

This has no name?

[hidden haven]

FTA says that complex numbers are an algebraically closed field

[hidden haven]

None that I know of

[coral shale]

👌

So for a specific example such as h^5 - 2 === 0 (mod x^5 - 2) is there some way to justify there are no more roots, or do I require another approach

[hidden haven]

For x^5 - 2, the easiest way to justify that there are non other roots in that quotient ring is to first prove the isomorphism of this with Q[2^(1/5)] as a subfield of R

[foggy merlin]

seems very interesting, is this the type of area of math that tries unifying lots of proofs that are very similar, the unification of all these universal property/isomorphism theorems for rings, groups, modules etc. in one go and more?

[hidden haven]

There is a lot of stuff of that sort in model theory yes

[coral shale]

👌 I will try a few more examples, (sry dr j stockfish) 😄

[hidden haven]

The universal property stuff is there in a lot of universal algebra study

Like you can prove that there is always a free universal algebra on any given set

Groups, rings, modules are universal algebras

While fields aren't

[foggy merlin]

rip 😦

[hidden haven]

You can prove that every universal algebra is a quotient of a free one, so you can always do presentations with generators and relations

[hidden haven]

Yeah, fields are nice objects on their own, but the category of fields has very few universal constructions. No product fields, free fields, etc

While the other things do, which is because these other things are universal algebras

[foggy merlin]

makes sense

[coral shale]

Ah, so I probably want to show F[x]/(f) ~=== F(a) for any root a of f (probably in my notes, but I will try on my own)

[hidden haven]

Ye

Where a is a root of f in any ambient field K containing F

[coral shale]

hmmmmmmm is this necessary? 🤔

Ah yh I guess so nvm lel

otherwise what am i proving lel

[hidden haven]

Which part of it lol

[coral shale]

the ambient field K

[hidden haven]

Oh ye like a has to be somewhere right

[coral shale]

yh yh

[foggy merlin]

but if I remember what my commalg prof said some time ago correctly, there are still good analogues one can create for fields when studying for example transcendence bases of field extensions and bases of vector spaces

[hidden haven]

Ye there are transcendence bases, but as far as I am aware those don't have any nice universal properties

You define a dependence relation and you get bases

There is some general theory about such things

Wikipedia has a page called dependence relations

[foggy merlin]

this was what he mentioned

[hidden haven]

I see

Yeah he is talking more about an elementwise study sort of

Not universal properties

Like you can't specify a field homomorphism out of an extension by specifying it on the transcendence basis

[foggy merlin]

well yeah I was now talking about sth slightly different here

[hidden haven]

Right

The model theory part he mentions

I guess it makes sense but I haven't studied this specific kind of thing in model theory before

He might be talking about the dependence relation thing or maybe matroids

Idk what those are but I have heard that matroids are some generalisations of dependence and might be model theory

[foggy merlin]

well if it isn't an approach that gives more insight or makes things simpler it would make sense why it isn't that common to study I guess..?

we aren't talking about the matroids used in combinatorics or are we?

[hidden haven]

I am not sure what the ones in combinatorics are

But I am guessing these are different

https://en.wikipedia.org/wiki/Matroid

wait

[lapis trail]

What's it mean for a polynomial to be primitive

[hidden haven]

It says combinatorics lmao

[hidden haven]

gcd of coefficients is 1

[lapis trail]

Ok thanks

[foggy merlin]

well that's why I was asking, because most applications I've seen of matroids so far were in combinatorics related subjects

[hidden haven]

Ye idk anything about them

[lapis trail]

Is an irreducible polynomial primitive?

[hidden haven]

Over what ring?

[prisma shuttle]

when someone says that smth is unique up to association that just means its unique up to multiplication by units right

[hidden haven]

Yes

[pastel cliff]

is a monoid with distributive property anything

actually nvm you need two operations for distributivity to be a thing lol

is it normal for my prof to be discussing SL_n GL_n and S_n all in terms of matrices

bc i do not like

[hidden haven]

S_n usually no, the other 2 yes

[pastel cliff]

that is the one i not like

i'll assimilate with the other two

im summarizing our material covered so far

should i also stop thinking about D_8 as matrices

[hidden haven]

bruh

probably

[coral shale]

Going back to what we were discussing

I can see this must be true

[pastel cliff]

then it's true

[coral shale]

But is there an elementary proof? 🤔

[delicate orchid]

this looks fun

[pastel cliff]

✓

[delicate orchid]

something to do with if x^a = 1 and x^b = 1 => (x^a)^b = x^ab = 1

moldi gradually makes his way down the posts leaving a breadcrumb trail of reacts

[coral shale]

Like its gotta be true right D:

cus Z[x]/(f) where the roots of f are the roots of unity

[delicate orchid]

moldi what's your thoughts on this approach

[coral shale]

gets u this

[hidden haven]

Idk what that means lol

[delicate orchid]

trying to factor the x^ab bad boy into the smaller ones

[hidden haven]

This feels like the kind of problem where you first assume a and b are coprime

And then show that the general case reduces to the coprime case

[coral shale]

$$1+h+h^2+\cdots+h^{n-1}\equiv 0 \pmod{1+x+x^2+\cdots+x^{n-1}}$$

[cloud walrus]

Shuri2060

[coral shale]

Well just to backtrack

[delicate orchid]

yeah if they're coprime you get this implication turns into an if and only if i.e. x is a root of the numerator if and only if it's a root of the denominator

[coral shale]

we had this before

And h(x) = x^k is always a solution for any k

Hence I make the above conclusion

[hidden haven]

For any k not divisible by n I think

And you also want n to be prime

[coral shale]

ohhhhh

[hidden haven]

Otherwise that polynomial is not irreducible

[delicate orchid]

he's right you know

[hidden haven]

✓

[coral shale]

$1+x+x^2+x^3$

[cloud walrus]

Shuri2060

[coral shale]

ah 1+x^2 goes into this

[delicate orchid]

(1+x)(1+x^2) my beloved

[coral shale]

Show for $k,p\in\bN$ with $\gcd(k, p) = 1$, $p$ prime

$$\left(\frac{(1-x^{kp})(1-x)}{(1-x^k)(1-x^p)}\in \bZ[x]\right)$$

So we have this 🤔

[delicate orchid]

ok now this looks rather funny, kek even

[cloud walrus]

Shuri2060

[hidden haven]

Where are the arrows

[delicate orchid]

so this can only be in $\bZ[x]4 if 4(1-x^{kp})(1-x) = p(x)(1-x^k)(1-x^p)$ for some $p(x) in \bZ[x]$, right?

[delicate orchid]

moldi when he sees a number

[coral shale]

yh . . .

$$1+h+h^2+\cdots+h^{p-1}\equiv 0 \pmod{1+x+x^2+\cdots+x^{p-1}}$$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

Shuri2060

[coral shale]

This does follow from this right

[delicate orchid]

I have no idea what that means and I choose to ignore it

[coral shale]

ehhhh we were discussing

F[x]/(f) adding roots to F basically

[delicate orchid]

quotient rings give me eye strain, especially when viewed as field extenstions

[coral shale]

If you have f irreducible

[hidden haven]

Me with no arrow

[coral shale]

That quotient adds roots to f basically

[delicate orchid]

we can just look at (1-x^{kp})(1-x) = p(x)(1-x^k)(1-x^p) I'm pretty sure and get it

[delicate orchid]

I know what it does I just don't like it with the mod notation, like at all

[coral shale]

not sure how else to write it

1 + h + ... + h^{p-1} \in (f)

[hidden haven]

The idea of this is that you are setting f(x) = 0 when you quotient by (f(x)), so x is now a root of f(y)

[coral shale]

yah. And the powers of x will tag along if f has prime roots of unity (how to phrase this better lul)

[delicate orchid]

I'm thinking that this can only hold if p(x) is of degree 1 by examining the degrees of both sides and setting them equal, right?

[coral shale]

trying that 👀

[hidden haven]

ye any power of a root of unity is a root of unity

I feel like you guys are forgetting something I explained to you very thoroughly a couple days ago

For this kind of problem

You just go

@rustic crown

And it is solved

[delicate orchid]

cheating

[coral shale]

well this sure as heck doesnt look like its gonna happen

[delicate orchid]

moldi I swear I'm nearly there

[coral shale]

whats going wrong 🤔

p has to be 1+x

[rustic crown]

I'm sweepy

[delicate orchid]

I... kind of agree... maybe?

[hidden haven]

Why do you wanna sweep at 1 am

[delicate orchid]

what makes you say that

[coral shale]

what did i think wrong here

$$1+h+h^2+\cdots+h^{p-1}\equiv 0 \pmod{1+x+x^2+\cdots+x^{p-1}}$$

[cloud walrus]

Shuri2060

[coral shale]

$$h(x) = x^k$$

[cloud walrus]

Shuri2060

[coral shale]

should solve tHis

[delicate orchid]

both sides must be 0 when x = 1 so p(1) = 0 => p(x) = a(x-1)?

[delicate orchid]

looking at this again

[coral shale]

yh i tried that, and the result is just plain wrong

wuts going on lul

[delicate orchid]

wonder where my thinking has gone wrong

the lhs is definitely degree kp+1

[coral shale]

its probably me

hmmmmmmmmmmm

[delicate orchid]

you wanna just do the euclidean algo until we get a zero remainder?

[coral shale]

OH WTH

Degree 1 my ass

x^kx^p = x^{k+p}

😂

[delicate orchid]

_ _

[coral shale]

ahahaha u got me good

[delicate orchid]

basic polynomial multiplication has foiled me once more

[coral shale]

ok ok ima have to go for now, will hack at this later

ty for your 🧠 s tho ppl 🙇

[delicate orchid]

I'm gonna keep thinking about it, cause we still know that p(x) must be a multiple of (x-1) regardless of it's degree

actually no, that's just wrong

we need to euclid algo this

which I cannot be bothered to do!

[hidden haven]

✓

[pastel cliff]

moldi how do you think about D_8

i wanna shed the matrices

[hidden haven]

(╯°□°）╯︵ ┻━┻

flipping square

and sometimes just rotating it

[delicate orchid]

wow I'm aktually an expert in rep theory and you assk moldi wow ok just because I don't know how exponents work wow ok I see wow ok

but yeah it's just spinning a square

[pastel cliff]

L + ratio

damn

we spent sm time running circles around it

for what feels like no reason lol

[delicate orchid]

no just kidding I actually visualise the semidirect product directly

[pastel cliff]

idk what that means

[hidden haven]

It means wew big brain

[delicate orchid]

$D_{2n} \cong C_n \rtimes C_2$ I hope

[next obsidian]

Yeh

[hidden haven]

Perhaps

Nice

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

i feel like semidirects were taught to me poorly

you assume they just use an inner automorphism right?

if it's not stated

[hidden haven]

I do not know of this convention

[delicate orchid]

so like

[cursive temple]

for a and b both prime you can note that this is the ab-th cyclotomic polynomial and so it has integer coefficients :D

[delicate orchid]

multiplication in the direct product looks like $(g_1, h_1) \cdot (g_2, h_2) = (g_1\phi_{h_1}(g_2), h_1h_2)$ for some automorphism $\phi$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

and if the automorphism is not given you just assume it's conjugation by h_1 iirc

[hidden haven]

oh ye that

Isn't it always inner aut

That's the only semi direct I know lol

[delicate orchid]

I think it's an inner/outer product type situation

[hidden haven]

I haven't learned it in a course properly

I only know internal semi direct which is always inner aut I think

[delicate orchid]

inner product yes it's always inner auto

I know it to be so

[hidden haven]

Shouldn't external also always be inner aut

Inner in the larger resulting group

[delicate orchid]

lemme wikipedia it

[hidden haven]

cheating

[delicate orchid]

yup it's always inner

[delicate orchid]

I mean ok fine

[hidden haven]

nice

[delicate orchid]

the definitions of inner and outer semi direct products produce the same object so if you use an inner auto for the inner product you must also have to use one for the outer

there, no cheating required

[hidden haven]

Ye

That's what I said

No external cheating required yes

Internal cheating product

[delicate orchid]

but I discovered it independently

liebniz be like

[hidden haven]

nice

[delicate orchid]

sorry we got a bit off track

[hidden haven]

abstract chillgebra

[delicate orchid]

[hidden haven]

Haha putting all the pieces together

[chilly ocean]

semidirect product should not exist

[delicate orchid]

cope

[chilly ocean]

how

I thought you pushin p man

i mean its lower

[chilly ocean]

its weird notation ill admit

notation is fine

i dont know why math people have completely adapted a functional notation for everything

less ambiguous

[delicate orchid]

it's like, the SECOND most natural way for groups to act on each other

[chilly ocean]

same thing for tensors

[chilly ocean]

?

[pastel cliff]

[chilly ocean]

second most

[pastel cliff]

im just

[delicate orchid]

responding to catfan

most natural is the direct sum obvs

[chilly ocean]

natural how

some notations are bad and they are not being changed because one mathematician used them and then everyone did because they werent creative enough

ifs not natural its just generalizing how it could act in general

[pastel cliff]

every day i wake up and im still a math major

[chilly ocean]

i want this image

[delicate orchid]

it's all yours my friend!

[chilly ocean]

NOOO ITS MY NFT

[pastel cliff]

guess who made it B)

[chilly ocean]

can i have it

post here ill save

[pastel cliff]

cmd shift 4 on mac

[chilly ocean]

im on iphone

[delicate orchid]

I ain't got no iphonneee

[chilly ocean]

#groups-rings-fields message

[pastel cliff]

kiss u thru da phone

https://tenor.com/view/cat-butt-wiggle-gif-22852678

[delicate orchid]

anyway chat, thoughts on the symmetries of a cube?

[chilly ocean]

why

[delicate orchid]

cat

[hidden haven]

Isn't that the rubik's cube permutation group

[delicate orchid]

yeah, the symmetries of a cube

[hidden haven]

Rather than symmetries of the cube

[delicate orchid]

did I say THE cube?

[hidden haven]

Bruh if you say symmetries of a cube I would think S_4

[delicate orchid]

S_4 my beloved

exercise!

[hidden haven]

Stop copying my hw wew

[delicate orchid]

prove that all rotations of the cube form the group S_4

and find what they permute

[hidden haven]

Is it only rotations

[delicate orchid]

yeah

[hidden haven]

oh

[delicate orchid]

reflections you get something a bit weirder

[hidden haven]

No what

[delicate orchid]

I swear it's just rotatoins

[hidden haven]

oh ye

it is

[delicate orchid]

I think the full thing is S_4xC_2

[chilly ocean]

<|

wait bruh

how can you rotate a cube 4 ways

aint it 3

oh

[delicate orchid]

[chilly ocean]

rip

[chilly ocean]

you can rotate it in 3 directions

[hidden haven]

But does that give you S_3

[chilly ocean]

moldi what time do you wake up on average

[hidden haven]

And are those directions independent

[delicate orchid]

you can rotate the cube in 24 different ways from its initial position

[pastel cliff]

is this right

[hidden haven]

noon

[chilly ocean]

dont you have classes

[hidden haven]

Lmao imagine attending

[chilly ocean]

presidential day

[pastel cliff]

imagine having holidays

[delicate orchid]

are you trying to show that the homomorphic image of the identity is the identtiy?

[chilly ocean]

moldi celebrating US

[pastel cliff]

thanks obama

[pastel cliff]

yes

[hidden haven]

celebrating what

[pastel cliff]

in general just to understand how homomorphisms preserve structure

[hidden haven]

oh

[pastel cliff]

dumb america holiday

[hidden haven]

Imagine being american

[chilly ocean]

(1337 🍓 )

[delicate orchid]

don't like the phi(ab) = phi(Id b) line, just write phi(b) = phi(Id b) = phi(Id) phi(b)

[pastel cliff]

good thing im not

[delicate orchid]

and they preserve structure because times of thing is thing of times

[chilly ocean]

america number 1

[pastel cliff]

cringe

[chilly ocean]

bruh

who is number one?

[hidden haven]

me

[chilly ocean]

what country

moldi country?

[pastel cliff]

why if i may ask

[delicate orchid]

moldi is the entire nation of norway

[pastel cliff]

why no like

[chilly ocean]

i guess thailand