groups-rings-fields

[limpid edge]

but where does the notation originate

and how is it used

[chilly ocean]

do you know about quotient groups?

[rustic crown]

(or quotient rings)

[chilly ocean]

(yea but if they know that they prob do q geoup too lol)

[limpid edge]

i think so

[chilly ocean]

what makes you think that

[limpid edge]

ok I know what they are now

[chilly ocean]

Z/nZ is just that

[limpid edge]

I see

ok I'm looking at the wiki page and it's explaining well

thanks

[latent stratus]

any idea how to find ker f? I am not too sure about the answer. Mine came +-1

[oblique river]

You are right

The kernel is the set of elements x such that f(x) is the identity element

in this case, the identity element of the group C* is 1

(because 1*x = x for any complex number x)

and so to find the kernel we need to answer the question "what are the complex numbers x such that f(x) = 1"

i.e.

"what are the complex numbers whose square is 1?"

and the only numbers x with x^2 = 1 are x = 1 and x = -1

[chilly ocean]

cuz C* is an integral domain and x²=1 <=> x²-1=0 <=> (x+1)(x-1)=0

[latent stratus]

ah you are right, thanks!! :D

[oblique river]

Carla I'm not sure if they know what an integral domain is

given that they're asking a qusetion about the kernel of a group homomorphism

[old lava]

is this C* algebra

:^)

[chilly ocean]

[oblique river]

C-star algebra, more like Beastars algebra amirite @chilly ocean

(sorry if this joke has already been made i'm behind on the memes )

[unique juniper]

[past temple]

if I have a bijection f

and I know that A is contained in it's image f(A)

does this imply that A = f(A)

?

[chilly ocean]

f(x) = 2x; f((0, 1)) = (0, 2) contains (0, 1) but is not equal to it

[past temple]

ah

okay

ty

[past temple]

wait

C isn't a Galois extension of Q right

because the algebraic closure of Q is a proper subset of C

and the algebraic closure is the splitting field of every irreducible polynomial with coefficients in Q

[oblique river]

correct

[chrome mirage]

Let R = {a, b, c, d}. Define a suitable addition + and a suitable multiplication · so
that (R, +, ·) becomes a field.

I have no idea how to do this....

[final pasture]

so first you need to chose some element to be the additive identity

and also choose an element to be the multiplicative identity

right ?

[chrome mirage]

choose one of the elements from R?

[final pasture]

yes

[chrome mirage]

okay so I choose a for additive and d for multiplicative

[final pasture]

right

so now you now that you need to have:
a + a = a
a + b = b
a + c = c
a + d = d
and also the reverse, i.e:
b + a = b
c + a = c
d + a = d

same thing for the multiplication

[chrome mirage]

hmm yes

[final pasture]

ok now you also need to choose additive and multiplicative inverses

(note that there's a faster way to do this, but I think doing it by hand is a bit more interesting to start with )

So a is it's own additive inverse, 'cause it's the zero of the field

[chrome mirage]

i like doing things the slow way :

[chrome mirage]

yup

[final pasture]

now what could be the additive inverse of d ?

[chrome mirage]

it would be a, yes?

[final pasture]

hmm, no ? the additive inverse of is some element x such that d + x = a

but d + a = d

so x can't be equal to a

so x € {b, c, d}

[chrome mirage]

so x can be any one of a or b or d?

[final pasture]

sorry {b,c, d}, not {a, b, d}

we just said x can't be a

[chrome mirage]

so it shouuuuuld be d

[final pasture]

Can you explain why ? (it is true )

[chrome mirage]

hmm i said d because we didn't get into b or c yet so it couldn't be one of them

[final pasture]

Fair enough lol.
Ok so let's assume d + d = a, and see how things goes

let's recap what we know for the moment

$R = {a, b, c d}$.\
We said that $0_R = a$ and $1_R = d$.\
\begin{align*}
a + a &= a\
a + b &= b\
a + c &= c\
a + d &= d\
d + d &= a\
d \cdot a &= a \
d \cdot b &= b \
d \cdot c &= c \
d \cdot d &= d \
a \cdot a &= a \text{ Because multiplying by $0$ gives zero in a ring, and a fortiori in a field}\
a \cdot b &= a\
a \cdot c &= a\
a \cdot d &= a\
\end{align*}
And all of that commutes

[chrome mirage]

that's understandable

[final pasture]

some part is missing

[chrome mirage]

that's alright!!

[cloud walrus]

Shika-Blyat

[final pasture]

ok that's good enough lol

So we already know some stuff

[chrome mirage]

yup! where do we go now

[final pasture]

we're gonna look at the additive inverses of b and c, which are the only inverses that we don't know yet

the nice thing is that we know that the multiplication is distributive over the addition

we know that the inverse of b is either b itself or c

[chrome mirage]

Can you tell me why exactly was d the additive inverse of d?

[final pasture]

Just because you said so, we're just trying stuff for the moment

if we ended up with something absurd in our laws, we would just have shown that d can't be it's own additive inverse and we would've restarted with another additive inverse for d

[chrome mirage]

Could you help me understand...the faster method

[final pasture]

Yeah

do you know what the characteristic of a field is (or more generally the caracteristic of a ring) ?

[chrome mirage]

smallest number of times i have to use the multiplicative identity to get the additive one, yes?

[final pasture]

yes

well the characteristic of a finite field is always a prime, do you know why ?

[chrome mirage]

hmm i don't

[final pasture]

a finite field, sorry*

[chrome mirage]

oh wait

a field has no zero-divisors

so it makes sense the char is prime

[final pasture]

yes

that's exactly that

so now do you also see why the characteristic has to divide the field's cardinal ?

[chrome mirage]

i can't remember the cardinal term

[final pasture]

the number of elements in the set

[old lava]

all fields*, (0) is a prime ideal, hence 0 is a prime

[chrome mirage]

oh yes

[final pasture]

[chrome mirage]

i can see why it would have to divide the field's cardinal

[final pasture]

why ?

[chrome mirage]

because for each time i use the mult iden to get to the additive one, i am moving through each element

[final pasture]

🤔

no, not necessarily ? the characteristic can be smaller than the cardinal

[chrome mirage]

it made sense in my head, i figure it should give 1 upon division, no?

oh

okay then why

[final pasture]

I could give you an example, but the simplest one is exactly the field your exercise is asking you to define, so that'll wait lol

[final pasture]

check that {n 1_F, n € N} is an additive subgroup of (F, +) (for (F, +, .) a field)

and then use lagrange's theorem

[chrome mirage]

i never thought of using Lagrange smh

[final pasture]

Anyway so, I'll leave checking that to you, but the important part is that in our case, R = {a, b, c, d} contains 4 elements, so there isn't a lot of prime that could divide 4, in fact only one

so we know that the characteristic must be 2

and thus that 1_R + 1_R = 0_R, or in our notations, d + d = a, meaning that d is its own inverse

[chrome mirage]

yup, the char is 2

[final pasture]

[chrome mirage]

ah

yup i can now see that

[final pasture]

Ok so continuing with the less painful way of solving this, there's also something you can think of, if you already know it:
there's only 2 groups of cardinal 4 up to isomorphism, so our field must be one of them

[chrome mirage]

what are these groups

[final pasture]

[chilly ocean]

wdym by cardinal 4?

Oh I thought there are more lmao

[chrome mirage]

the set in my exercise has 4 elements

[chilly ocean]

its klein four and Z_4

[final pasture]

yes

so $\bZ/4\bZ$ and $(\bZ/2\bZ) \times (\bZ/2\bZ)$

[cloud walrus]

Shika-Blyat

[final pasture]

[chrome mirage]

Z_4 it is then because we never did klein four in class

[final pasture]

klein four is just a fancy name for (Z/2Z)²

Now you can probably try to define a multiplication on both groups

and see for which one it works

Hint: Z/2Z must be a subfield of your field

because it has characteristic 2

and the {n1_R, n € N} is not only a subgroup but a subfield, isomorphic to Z/2Z

[old lava]

it's physically impossible to have done basically any group theory and not have come across Z/2Z x Z2/Z

[chrome mirage]

the thing is, i am not exactly sure what i am supposed to do when the question asks me to define a multiplication or an addition

[chrome mirage]

i did come across it, i just didn't know that specific name

[final pasture]

just try defining a law that respects the field structure

like, because of the field structure you already know what a x b must be equal to for some a's and b's, like you know that a x 0 must be equal to 0 etc

[chrome mirage]

does defining the cayley table count as defining the multiplication

[old lava]

an insanely tedious way of doing it, sure

you can technically define the multiplication by writing out the entire cayley table (making sure it's all correct and works)

[final pasture]

that's pretty much what I was trying to do initially

(by writing everything and deducing stuff step by step)

that's a bit tedious but that works and it's probably the easiest way if you don't see any slick way of doing it (and you can do it faster by thinking of all these few things, like characteristic etc, that enforces some identities, like 1+1 = 0 in our case)

[chrome mirage]

i thiiiink

this should be it

for the multiplication

is this correct Shika

[final pasture]

lemme check

[chrome mirage]

[final pasture]

I think there's something wrong

[chilly ocean]

All good 😎

[final pasture]

you sure ?

[chrome mirage]

[chilly ocean]

No 😎

[chrome mirage]

which part makes you sus

[final pasture]

iirc in F_4, the 2 nonidentities elements aren't their own inverses

did you check that the distributivity works with that law ?

[chrome mirage]

i haven't defined the addition yet so i can't

[final pasture]

[old lava]

ya, this would be the F_4 multiplication table

[final pasture]

yeah ok I'm not crazy

[viscid pewter]

fields are weird ngl

they're like a group, and then another group in that group with one less element

[chrome mirage]

interesting

so that's half the thing done

[old lava]

F_4 addition is really obvious

[chrome mirage]

now on towards the addition table

[old lava]

just treat \alpha as a variable

add things normally

if there's a coefficient of 2 on anything

remove that term

it's that easy

[final pasture]

Ok so that conversation leads me to asking myself something

if I have a field (not necessarily finite) $(\mathbb{K}, +, \cdot)$

[cloud walrus]

Shika-Blyat

[final pasture]

And another field $(\mathbb{F}, +', \star)$

[cloud walrus]

Shika-Blyat

[final pasture]

if $(\mathbb{K}, +)$ and $(\mathbb{F}, +')$ are group-isomorphic, are $(\mathbb{K}, +, \cdot)$ and $(\mathbb{F}, +', \star)$ field-isomorphic ?

[cloud walrus]

Shika-Blyat

[old lava]

if it's finite field, then I'm 99% sure it's true

[chrome mirage]

when you said remove the term if 2, what would i be putting in the box of (1,1)?

[final pasture]

for finite fields it's obvious yeah

[chrome mirage]

for addition

[viscid pewter]

i bet it breaks down for infinite fields tho

[old lava]

since group isomorphic implies same number of elements implies isomorphic field

[viscid pewter]

complete guess

[final pasture]

since there's only field of card p^n up to iso

[old lava]

for infinite fields, ya, there's probably some shittery happening

[final pasture]

Yeah I think it breaks down too, but I'm looking for an example

[old lava]

like Q(a) and Q(b) are in general not isomorphic as fields for different a and b, but I think they are group isomorphic

if you map a to b

[final pasture]

oh yeah

[old lava]

(looking at the additive group)

[final pasture]

if the degree of a and b are different

🤔

ig

[scarlet estuary]

there are deep theorems which you dont have to know to solve these intro-finite-field-theory questions

but they make them really easy

[old lava]

I mean even if it's same degree, like sqrt(2), sqrt(3)

don't work

[scarlet estuary]

theorem 1: there exists a finite field Z/pZ for every prime p

[chrome mirage]

would I be putting 0 in the (1,1) box for the addition table?

for F_4?

[scarlet estuary]

theorem 2: any other finite fields are constructed by "stapling together" copies of Z/pZ

so the field of 4 elements is Z/2Z "stapled to" another copy of Z/2Z

[old lava]

didn't know this was a theorem, I always visualized it this way

[scarlet estuary]

corollary: all finite fields have order p^k where p is prime, and finite fields of the same order are isomorphic.

[final pasture]

^

[viscid pewter]

por que no los dos

[scarlet estuary]

(here, the field of 1 element is not considered "a field")

[old lava]

sadge

[scarlet estuary]

(we usually count it as a ring but not a field)

[final pasture]

you don't need to know how to prove that @chrome mirage, but just knowing it's true really helps

also you can add that every field of order p^n has Z/pZ as a subfield

[old lava]

just 1, 1+1, 1+1+1, etc

[final pasture]

^

[old lava]

until you get 0

and it'll be an isomorphic copy of F_p for some prime p

[final pasture]

the {n1_R, n € N} subgroup/subfield I was speaking about

[final pasture]

well it is a ring but not a field

[scarlet estuary]

yes

[final pasture]

like that's not quite a matter of convention, it's more like by definition, no ?

[scarlet estuary]

definitions are conventions

[final pasture]

fair

[chrome mirage]

i am really confused from when you said if there's a coefficient of 2 on anything, to remove that term 😭

[final pasture]

our field R has 4 elements, so it is of char 2

[old lava]

2=0

in F_2

[final pasture]

in particular 1 + 1 = 0

so 2 = 0

[old lava]

so replace every instance of 2 with 0

rip, my internet is lagging

(1+a)+a = 1+2a = 1
a+1+1 = a+2=a

[prisma ibex]

0 days since someone has mentioned the field with one element

[old lava]

for example

[scarlet estuary]

well THE field with one element is a different thing

~~ ~~

[old lava]

[final pasture]

how do you show they aren't isomorphic ?

[old lava]

sqrt(2) satisfies x^2-2, no element of Q(sqrt(3)) does that

can't map sqrt(2) to anything in Q(sqrt(3)) in a field homomorphism

[chilly ocean]

Also, can you jsut show they arent isomorphic by showing sqrt2 mapsto sqrt3 isnt isomorphic?

oh ye sniped

[old lava]

no, you have to show sqrt(2) can't map to anything

[chilly ocean]

but obviously cant map to anything in Q right

[old lava]

there may theoretically be some random element that might work (there isn't, but there might until you prove otherwise)

why not

I mean it feels obvious

still gotta prove it

[chilly ocean]

cause that would be weird

[old lava]

that's not a proof technique

[chilly ocean]

it is for me

[old lava]

[chilly ocean]

I mean ye, same contradiction applies for any sqrt2 mapsto q in Q(sqrt3) Im pretty sure

Oh wait maybe I was thinking something of the way: only ismorphism of Q is identity and I guess that implies the only chance for those to be iso is sqrt2 mapsto sqrt3 no?

[final pasture]

oh

thanks

[old lava]

no, please leave this mindset

of thinking that the natural option for the map is the only map that could work

[chilly ocean]

But can't this arguemnt work: assume phi iso of those fields, then phi_Q is iso to phi(Q)=Q since only automorphism of Q is identity?

Asking cuz not sure maybe I dont get this

[final pasture]

the field isomorphism necessarily fixes Q, yes

but that's not enough, ig ?

[chilly ocean]

But that I think implies sqrt2 has to map to sqrt3 for it to be surjective then

[oblique river]

that's just not true

[chrome mirage]

okay my final question for the day i don't see them having multiplicative inverses...isn't that necessary for a field

[oblique river]

why don't you think the function which sends sqrt(2) to 1 + sqrt(3) is surjective?

[chilly ocean]

Yeah I just realized

[old lava]

according to your logic, we might conclude that $\bQ(\sqrt[8]{2}, \sqrt[8]{2} \exp(2 i \pi /8))$ is not isomorphic to $\bQ(\sqrt[8]{2}, i)$ because the second generators cannot map to each other

[cloud walrus]

F[x]-module

[old lava]

which would be very wrong, since they're the exact same field

[chilly ocean]

ye

[old lava]

you cannot just look at the generators and conclude the possible maps that work

that's just completely faulty logic

don't try to use it like ever again

for your own good

[old lava]

learn to read multiplication tables I guess lol

[chilly ocean]

So in field iso generators dont have to map to generators?

[old lava]

(hint: if a b = 1, then a and b are inverses)

there's multiple choices of generators

[oblique river]

I mean, the image of a set of generators will be a set of generators

[old lava]

in general

[oblique river]

but maybe not the set that you had already chosen

[old lava]

like yes, generators will map to generators, but a specific fixed set of generators won't map to another specific fixed set of generators

[chilly ocean]

So they do map generataors to generators

[delicate bloom]

each nonzero row will have a 1 in it, that tells you the thing in that column is its inverse

[chilly ocean]

but not necessarily the ones I thought about

[old lava]

every single element of Q(\sqrt(3)) that is not in Q is a generator for Q(\sqrt(3))

so there's a lot of generators, which is why your logic of only looking at the map to sqrt(3) was faulty

[chilly ocean]

I see, thanks for correcting.

[balmy flower]

for a metric space (X,d) if we know supX and infX exist, is saying for all x,y in X infX <= x < y <= supX implies d(x,y) < d(infX,supX) sufficient?

[cyan marten]

What do sup and inf mean?

These are defined for R because it's ordered.

[balmy flower]

we also know that (X,d) is compact so I think we can assume inf and sup exist/are defined since its bounded?

nevermind i see what youre saying, i'll have to rethink how im doing this problem then

[latent anvil]

Are you trying to show X has finite diameter, ie d(x, y) is bounded among x, y in X?

Also this seems more appropriate for #advanced-analysis

or maybe #point-set-topology

[balmy flower]

yes, given that (X,d) is compact. and okay i'll keep that in mind, its a metric space topology class so im not quite sure where it'd fit in since its pretty entry level

thanks to both of you

[chilly ocean]

metric space topology class
not sure where to post it
there is literally a channel called topology

[old lava]

[proud bear]

[sinful mirage]

my prof said that $(\alpha,\beta)=\frac{1}{|G|} \sum_{g \in G} \limits \bar{\alpha}(g) \beta(g)$ is an inner product on the space of class functions on a finite group

[cloud walrus]

ProphetX

[sinful mirage]

I am trying to prove that it really is,and I do not seem to use anywhere that $\alpha,\beta$ are class functions. Is this an inner product also on the space of functions on a finite group?

[cloud walrus]

ProphetX

[compact needle]

This works for the set of all complex-valued functions on G. It does not even require that G be a group, just a finite set.

[sinful mirage]

thanks @compact needle

[sinful mirage]

this might be trivial, but then: it is true that the functions on a set ( to C) form a vector space using pointwise multiplicaiton and addition, right?

[old lava]

just test the vector space axioms

[compact needle]

Yup

[cloud walrus]

ProphetX

ProphetX

[sinful mirage]

how can I conclude that the character is a root of unity aswell?

the character would be the sum of the eigenvalues of the rep matrix,but the sum of roots of unity is not always a root of unity

[sturdy marsh]

@sinful mirage the matrix corresponding to \rho(g) has finite order and therefore is diagonalizable

\rho(g^{-1}) is diagonal in the same basis

[sinful mirage]

ohhhhh

and then the eigvals of $\rho(g^{-1})$ are $\lambda_i^{-1}$ and I can use $tr(rho(g^-1))=sum_i(\lambda_i^{-1})=\sum_i \bar \lambda_{i}=\bar \chi(g), right?$

[sturdy marsh]

yes

[cloud walrus]

ProphetX

[sinful mirage]

why can one think of the group algebra as linear combinations $\sum_{g \in G} \limits f(g) g$?

[cloud walrus]

ProphetX

[sinful mirage]

in the group algebra there should be functions,no?

[sturdy marsh]

what is f?

[sinful mirage]

my prof defined it like this: $f \in \mathcal{C} G$(the group algebra), f=\sum_{g \in G} \limits f(g) g$

[cloud walrus]

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

[sturdy marsh]

the coefficients are functions

[sinful mirage]

i'm super confused cause I understood the algebra of functions on the group before

and i don't understand what the definition of the 'group algebra' is

[sinful mirage]

from this

[sturdy marsh]

that's the definition of the group algebra

formal linear combinations

of elements of G

[sinful mirage]

I defined the group algebra differently actually

[sturdy marsh]

how

[sinful mirage]

and I managed to prove 1) and 2)

is this not the group algebra?

[sturdy marsh]

I dont think so

[hidden haven]

for finite groups it will I think

Because a function G -> C can be veiwed as a C-linear combination of elements of G

[sinful mirage]

if the algebra of functions on a finite group and the group algebra are two totally different concepts then it's ok

[sturdy marsh]

sure

[sinful mirage]

but i can't see how they are related

[sturdy marsh]

but is the multiplication the same

[sinful mirage]

how?

[sturdy marsh]

send f to sum_g f(g) g

[sinful mirage]

a function G->C needs to eat an element in G and return a number

[sturdy marsh]

okay I think it works

[sinful mirage]

ahh f(g)=\sum_{g} f(g) g?

[hidden haven]

yeah multiplication is the same actually

[sturdy marsh]

ye

[sinful mirage]

why could I not just say f(g)=\sum_{g} f(g)?

[sturdy marsh]

f(g) is an element of C

[sinful mirage]

right,but that's what we want to get

and f eats an element in g

[sturdy marsh]

the RHS is a formal linear combo

{functions on G} <-----> {formal linear combinations}

that's the correspondence we're talking about

[hidden haven]

not as written

[sturdy marsh]

oh

[hidden haven]

tho whats written is also wrong

[sturdy marsh]

sorry

I misread

[sinful mirage]

yes this is what i'm confused a bit about

can't see this correspondence(it was stated in the lecture)

[sturdy marsh]

$f \mapsto \sum_g f(g) \cdot g$

[sinful mirage]

so a function on G is just a map G->C

[cloud walrus]

Brofibration

[sinful mirage]

what does it mean to eat a function and send out a c number?

or is f not a function here?

[sturdy marsh]

f is a function

[hidden haven]

RHS is not a complex number

[sinful mirage]

rhs is an element in G, sorry

[sturdy marsh]

no

[hidden haven]

nope

[sturdy marsh]

C[G]

[sinful mirage]

[sturdy marsh]

or CG

whatever you called it

[hidden haven]

formal C-linear combination of elements of G

[sinful mirage]

f(g) is a complex number right? and g is an element in G

[sinful mirage]

ah yes

yes

[hidden haven]

you cant multiply elements of g by complex numbers

[sinful mirage]

ok so i understood this,but why can this viewed as a function?

here we map a function to a linear combination

[sturdy marsh]

not with that attitude you cant /s

[hidden haven]

You can view a formal linear combination as a function on G, it eats g and gives the coefficient of g

easy to see that this is the inverse of map from functions to linear combinations

[sinful mirage]

ohh ok

it makes sense now

sorry i was mega confused

pure math is new to me im a physicist

[hidden haven]

physicist

[sturdy marsh]

mirror symmetrist

[sinful mirage]

but with interest in mathematical physics

[sturdy marsh]

@chilly ocean

eisenstein coordinates

[chilly ocean]

who ping

oh hi prophetx

[tawny latch]

Why can't you grow corn in Z/6Z?

[proud bear]

it's not a field? idk

[tawny latch]

top notch comedy

[next obsidian]

This is the type of content we were missing

[sturdy marsh]

nathannerd for honorable

[chilly ocean]

brofibration for honorable

[sturdy marsh]

brofibration for honorable

,iam honorable

[cloud walrus]

No selfroles matching honorable.
See ,selfroles --list for the list of valid selfroles.

May 13, 2021, 5:34 AM

[sturdy marsh]

wha

that worked

whoa

[chilly ocean]

I thought that was going to have something to do with kernels

[chilly ocean]

this is wrong for noncommutative rings right? it doesn't seem like it is in general closed for right multiplication by elements of the ring but seems tedious to construct a counterexample

[next obsidian]

what do you mean wrong for commutative rings?

Do you maybe mean noncommutative rings?

In a general context this is a left ideal generated by S, but for commutative rings left and right don't matter

[chilly ocean]

yea i mean noncomutative

makes sense if it's the left ideal thank

[minor badger]

Okey so Im dumb, but I really dont understand the proof of Nakayamas lemma.

[golden pasture]

could send screenshot of the proof

which step

[minor badger]

Well Im quite stressed so everything is just a mess in my head.

But first the existence of the monic polynomial

[hidden haven]

Existence of p? That is proved later

The first part is giving the idea of the proof

[cloud walrus]

bim

[hidden haven]

you see how $\sum_{j=1}^n(\varphi\delta_{ij} - a_{ij}) x_j = 0$?

[cloud walrus]

Moldilocks

[hidden haven]

@minor badger

This is a system of n linear equations in n variables

so you can turn this into a matrix equation,

$A \cdot [x_1 \quad ... \quad x_n]^T = 0$ where $T$ means transpose and $A$ is the $n\times n$ matrix with $ij$-th entry being $\phi\delta_{ij} - a_{ij}$

[cloud walrus]

Moldilocks

[hidden haven]

multiply this equation by Adj A on the left, and for a commutative ring, (Adj A) A = det A so you get (det A) [x_1 ... x_n] = 0

so you get the equation from this since this gives det A = 0 (or M = 0, but if M is the zero module then the whole thing is trivial anyway)

[sinful mirage]

ok so I managed to prove that there is an algebra isomorphism between the functions on a finite group and the formal linear combinations CG

however,I still would like to ask again what do we mean by formal linear combinations(if the multiplication between c numbers and group elements is not defined)

I can give CG an algebra structure by just defining addition and multiplications between elements of the type $\sum_{g \in G} f(g) g$, but I do not know how to interpret these elements,since f(g) g is not defined

[cloud walrus]

ProphetX

[sinful mirage]

the algebra structure of CG is inherited from C and G,but I do not see what do we mean by its elements,even though I proved there is an iso between this and the algebra of functions on the group

this would be the proof

but I can't interpret the elements of CG because in the strict sense there is no multiplication between f(g) and g,why would this not be an issue?

so i see that the algebra CG and the algebra of functions on a finite group are isomorphic,therefore I Can think of them in either way. I just don't comprehend/see what would be the power of doing it in the CG picture or how CG elements "exist at all"

[cloud walrus]

bdobba

[untold sapphire]

@sinful mirage

[sinful mirage]

I didn't use the dual of the group algebra at all

[untold sapphire]

are you not talking about linear functionals from k[G] to k here?

if not, where are your functions on $G$ mapping to?

[sinful mirage]

I meant the functions from G to C, i.e. f:G->C

[cloud walrus]

bdobba

[sinful mirage]

in order to have a dual,I need a vector space structure on G

I do not have that is my issue

[untold sapphire]

functions from $G$ to $C$ form a basis for the dual

[cloud walrus]

bdobba

[untold sapphire]

more specifically, the coordinate functions $\delta_g$

[cloud walrus]

bdobba

[sinful mirage]

can you elaborate what do you mean by dual please?

for me a dual is a map from a vector space into the field it is defined over

[untold sapphire]

sure, so are you familiar with duals in vector spaces?

[sinful mirage]

but I do not have vector space structure on G

[untold sapphire]

right, yes you do - because you have an algebra

[sinful mirage]

the algebra is not on G,rather the functions on G,or?

the functions on G form an algebra,not G itself

[untold sapphire]

an algebra in your case is just a vector space equipped with a multiplication map

hold on, you can also form an algebra from $G$ itself

[cloud walrus]

bdobba

[sinful mirage]

and in order to have a dual in this sense,i'd need a map,which eats a function on the group and spits out a number

[untold sapphire]

ok slow down

[sinful mirage]

this is what i don't see

[cloud walrus]

bdobba

[sinful mirage]

yes,I think I made it

[sinful mirage]

here

but i can't interpret the word 'formal linear combination' because my issue is that there is no multiplication or anything defined between f(g) and g, so I do not know how to interpret these elements

but defining the above addition and multiplication,I can make these elements into an algebra,yes

[untold sapphire]

are these your notes or course notes?

[sinful mirage]

my notes

I struggled with the proof using the help of math encyclopedia

[untold sapphire]

because we can make a vector space $k[G]$ with basis ${g_1, g_2, \dots \g_n }$, which is just ${ z \cdot g_i \colon g_i \in G, z \in \mathbb{C} }$,

[cloud walrus]

bdobba
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ProphetX

[sinful mirage]

there is no multiplication defined between complex numbers and group elements

[untold sapphire]

that's it, that's all you need

you're just treating it as a set

let me give you a different (silly example)

Given a set $S = {a,b}$ , I can make a vector space out of $S$ just by declaring elements to be of the form $z_1 \cdot a + z_2 \cdot b$

[cloud walrus]

bdobba

[sinful mirage]

what is z_1*a here?

[untold sapphire]

$z_1 \cdot a$ doesn't have to be evaluated

[sinful mirage]

I can't see what is $\cdot$

[cloud walrus]

bdobba

ProphetX

ProphetX

[sinful mirage]

why would we not need to evaluate it?

[untold sapphire]

if $z_1$ were a real number, it's just the same of saying we have $z_1$ copies of a vector

[cloud walrus]

bdobba

[untold sapphire]

because the structure of a vector space says we don't need to

it just happens that in a Euclidean vector space, say $\mathbb{R}^n$, you can

[sinful mirage]

ohhhhhh shoottt

[cloud walrus]

bdobba

[sinful mirage]

a vector space is simply a set equipped with 2 laws

and if these 2 laws are ok,it is a vector spcae

[untold sapphire]

yeah it's an incredibly simple structure

[sinful mirage]

oh lol

[cloud walrus]

ProphetX

[untold sapphire]

you can put a bunch of stuff on it like multiplication, inner product etc but you don't have to

[untold sapphire]

right, it's just saying that you have $f(g)$ copies of your vector $g$

[cloud walrus]

bdobba

[sinful mirage]

explicitly $g,g,g,\cdots g$ f(g) times?

[cloud walrus]

ProphetX

[untold sapphire]

sure, that's not a bad way to think about it

at it's most basic level, a vector space can just be a useful way to keep track of stuff

like how many of object $a$ I have, how many of object $b$ and how many of object $c$

[cloud walrus]

bdobba

[sinful mirage]

you mean the linear combinations are these?

[untold sapphire]

this is a bit wishy washy, but it's a helpful way to think about it if you're used to euclidean vector spaces

[untold sapphire]

yes

[sinful mirage]

i have x number of e_1,y number of e_2, etc.?

[untold sapphire]

right

[sinful mirage]

oh lol,this makes sense

[untold sapphire]

*formal linear combinations

[sinful mirage]

I guess not having a math linear algebra course hits back hard now

but now it makes sense

[untold sapphire]

oh lol, yeah that would help 😉

but glad I could help

[sinful mirage]

my linear algebra was taught by a physicist and not def theorem proof style but rather application oriented

and now i have a grad rep theory course with math people

[untold sapphire]

that's alright, the group algebra $k[G]$ is actually quite a complicated thing even though it might seem simple

[cloud walrus]

bdobba

[untold sapphire]

once you start putting extra structure on it

[sinful mirage]

sorry if this is a burden,but could you please check if the proof for injectivity is correct/written down okay in the proper math sense?

phi is this

I was trying to prove that the map phi is an algebra isomorphism between the functions on the group and the CG algebra, and I managed to show it is a homomorphism,surjective,but i'm not sure whether my proof for injectivity is ok

[untold sapphire]

hmm, what have you been told about the map $( \star )^{-1}$ ?

[cloud walrus]

bdobba

[sinful mirage]

nothing

[untold sapphire]

since obviously we have an inverse map on the group but that doesn't generalise in a very straightforward way to the algebra. It does exist but I won't talk about it if you haven't covered it

[sinful mirage]

this is what I said about the map inverse

(but this is me,nothing in the course)

[untold sapphire]

specifically, how the map $(\star)^{-1}$, which I'll call $S(g)$ works in $k[G]$

[cloud walrus]

bdobba

[sinful mirage]

but in order for this to hold,G has to be finite and bijective,which I'm not sure the injectivity proof is ok

[cloud walrus]

bdobba

[sinful mirage]

yes,exactly

that is my issue

this is how I tried to prove it but i'm not sure it works

or rather not sure I wrote it down in a proper math language

[untold sapphire]

$phi$ does have an inverse, but it actually turns out to be quite a technical thing

[cloud walrus]

bdobba

[untold sapphire]

basically on single elements $g$, $\phi$(g) = g^{-1}$

[cloud walrus]

bdobba
Compile Error! Click the reaction for more information.
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[sinful mirage]

phi eats functions,or?

it eats functions and spits out elements in CG

[untold sapphire]

oh wait hold on

is Fun_g the functions on your algebra?

i.e. the dual space

[sinful mirage]

I defined it as the algebra of functions on the group

[untold sapphire]

linear functionals $f \mapsto \mathbb{C}$, right?

[cloud walrus]

bdobba

[sinful mirage]

they are maps from G to C

from the group to the complex numbers

but the group itself doesnt have a vector space structure

[untold sapphire]

no

[sinful mirage]

i can only put a vector space structure on the functions on the group

[untold sapphire]

ok, so can you think of a way to extend your functions on $G$ to the vector space $\mathbb{C}[G]$?

[cloud walrus]

bdobba

[untold sapphire]

also, what are your functions on $G$?

[cloud walrus]

bdobba

[untold sapphire]

actually forget that for a second

[sinful mirage]

yes,this what I proved here that there is an algebra homomorphism between the space of functions on the group and CG

[sinful mirage]

any function f:G->C

[sinful mirage]

I want to prove this is an isomorphism but I'm not sure whether my argument for injectivity holds

[untold sapphire]

in order to have an algebra homomorphism you need a vector space first

[sinful mirage]

yes

[untold sapphire]

so how can you use the functions on $G$ in this way?

[cloud walrus]

bdobba

[sinful mirage]

I equipped the functions on the group with a vector space structure by pointwise addition and multiplication inherited from C

[untold sapphire]

ah ok

[sinful mirage]

and made it into an algebra using the convolution

[untold sapphire]

you can also do it a different way - by first linearly extending the functions on $G$ to $\mathbb{C}[G]$

[cloud walrus]

bdobba

[untold sapphire]

since the set $G$ is a basis

[cloud walrus]

bdobba

[sinful mirage]

what does it mean to linearly extend?

[untold sapphire]

hmm

ok, i'd really recommend studying some linear algebra if you're doing representation theory

[sinful mirage]

I tried putting algebra structure both on CG and functions on G and tried proving they are isomorphic

[untold sapphire]

to linearly extend means to take a function on a linear combination of basis elements , say $f(a x_1, b x_2,\dots c x_n)$ and then simply apply it to each basis element in a linear fashion

[cloud walrus]

bdobba

[untold sapphire]

you're going to have a really tough time in rep theory if you don't brush up your linear algebra

it's like 80% linear algebra

not trying to scare you lol

[sinful mirage]

i'll try to do it the way you showed me

then my attempt doesn't work,right?

putting algebra structure on both and trying to show they're iso

[untold sapphire]

possibly not, sorry I haven't had the time to read it thoroughly - currently trying to make my way through a paper my advisor sent me

[sinful mirage]

okay,sure,no prob

thanks for the help!

[untold sapphire]

no worries

[unique juniper]

whats a general way of showing two fields are not isomorphic?

what im trying to do is

find two pairs of roots of $x^4 - 2x^2 - 2$ that generate non isomorphic extensions

[cloud walrus]

是的

[old lava]

depends on fields, there isn't really 1 algorithm to determine if fields aren't isomorphic

you can prove fields aren't isomorphic in lots of ways depending on which fields they are

in your case, could just try to construct a homomorphism between the two fields

and show no such homomorphism exists

[unique juniper]

$\mathbb{Q}(\sqrt{1+\sqrt{3}}) \cong \mathbb{Q}/\langle x^4 - 2x^2 - 2\rangle \cong \mathbb{Q}(\sqrt{1-\sqrt{3}})$

this is what i am confused by

[daring ibex]

\langle \rangle works better than < >

[unique juniper]

yeah it does, oh well lol

but yeah, this is telling us that they are isomorphic?

[cloud walrus]

是的

[old lava]

yes

actually, no

[unique juniper]

\sqrt{1 + \sqrt{3} } and \sqrt{1 - \sqrt{3} } are both algebraic over Q and x^4 - 2x^2 - 2 are their minimal polys

so are they isomorphic or not?

[chilly ocean]

they are roots of the same irreducible polynomial so they should be isomorpic

[golden pasture]

hopefully this would make it clearer: they are isomorphic as fields, here we explicitly selected a different embedding into some algebraic closure

[unique juniper]

isee

so am i misunderstanding this question?

[golden pasture]

have you touched galois groups yet

[unique juniper]

yes

[golden pasture]

i think what they want you to show

is that say the roots are r_1,r_2,r_3,r_4, then Q(r_1,r_2) not isomorphic to Q(r_3,r_4)

[chilly ocean]

aren't those the same?

[unique juniper]

lol

yeah

i dont get it

[chilly ocean]

I guess find complex root

and maybe try to show Q(real root) isnt iso to Q(complex root)

[unique juniper]

butttt

uh i thought we concluded they were

[chilly ocean]

Where?

[unique juniper]

.

maybe i am not understanding

[chilly ocean]

oh, well there are 2 more roots right?

[unique juniper]

but they are the same?

in the same fields

maybe question error

because the polynomail is irreducible

all extensions should be isomorphic

[golden pasture]

all extensions isomorphic as in

Q(r_1) = Q(r_2) = Q(r_3) = Q(r_4) as fields

[unique juniper]

i mean taking a root, generating Q(a root)

they all should be the same

[golden pasture]

yup

[golden pasture]

there's a difference here

[unique juniper]

what should r1 r2 r3 r4 be then?

[chilly ocean]

wait so why are those isomorphic? Because they are roots of the same irreducible polynomial?

[unique juniper]

yea

[golden pasture]

As a simpler example, consider the polynomial $x^4+1$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[golden pasture]

its roots are $e^{\frac{(2n+1)\pi i}{8}}$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[golden pasture]

but lets say i choose the root $e^{\frac{\pi i}{8}}$ and $e^{\frac{5\pi i}{8}}$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[golden pasture]

oops the field i made is the same as when i only chose the root $e^{\frac{\pi i}{8}}$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[unique juniper]

but even considering pairs

of roots

theres no pair that will be not isomorphic?

$\pm \sqrt{1 \pm \sqrt{3}}$

[cloud walrus]

是的

[unique juniper]

choosing pairs we get the exact same field or isomorphic??

[golden pasture]

nope

ok maybe in more detail

the 2 pairs would have 1 element in common

not 2 disjoint pairs

[unique juniper]

oh

[golden pasture]

maybe that was where the confusion was haha

[unique juniper]

lol

maybe lol

thanks

i see

it

[golden pasture]

haha yea wording for the qn could have been better

[chilly ocean]

wait so what did they mean I dont get it

[unique juniper]

So there’s 4 roots

Show that taking 2 pairs of roots we can generate 2 non isomorphic fields

Pairs can intersect :D

[chilly ocean]

So what are those pairs

[unique juniper]

Sqrt (1 + sqrt 3), sqrt (1 - sqrt 3) and sqrt(1 + sqrt 3), -sqrt( 1 + sqrt3)

Is one

[chilly ocean]

but isnt Q (Sqrt (1 + sqrt 3), sqrt (1 - sqrt 3)) = Q(sqrt(1 + sqrt 3), -sqrt( 1 + sqrt3)
)?

[unique juniper]

I don’t think so …

Hope I’m not wrong here lol

I haven’t checked

[chilly ocean]

I dont see why Q(root) =/= Q(other root), I thought they are all isomorphic if they are all roots of the same irr polynomial

[unique juniper]

They are all isomorphic

But we are taking pairs here

[chilly ocean]

okay yeah so any pair will be of the form Q(r,r1), Q(r,r2)

[unique juniper]

Yeah

[chilly ocean]

But I don't see how they are non iso

[unique juniper]

$\mathbb{Q}(\sqrt{1 + \sqrt{3}}, \sqrt{1 - \sqrt{3}}) \ncong \mathbb{Q}(\sqrt{1 + \sqrt{3}})$

i wanna say this is true

[cloud walrus]

是的

[unique juniper]

hmm

IDK

[golden pasture]

all you need to prove

is that $\sqrt{1-\sqrt3}\notin\mathbb Q\left(\sqrt{1+\sqrt3}\right)$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[unique juniper]

yes but idk lol

i feel like

ive seen before

that its in it

[golden pasture]

i feel like theres a very straightforward proof

but the only one i can think of

Let M be the splitting field of the poly

it is the field extension genned by adding all the roots

[unique juniper]

can we actually do it by considering degrees

[golden pasture]

now lets look at Gal(M/K), then the gal group must have some element that does $\sqrt3\to-\sqrt3$, $\sqrt{1+\sqrt3}\to-\sqrt{1+\sqrt3}$, $\sqrt2\to-\sqrt2$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[golden pasture]

hence [M:K] is at least 8

99% sure this is too convoluted

[unique juniper]

wait

oh

that makes sense

[golden pasture]

i feel like a more obvious proof should exist

but cant seem to find

[unique juniper]

yes me too

i think ive seen this before!!

[sinful mirage]

is there a way to show that if the convolution commutes,then f is a class function without choosing a basis? i.e. $f*f'=f'f \implies f$ is a class function ( here the convolution is defined as: $(ff')(g):=\sum_{h \in G} \limits f(h) f(h^{-1}g)$

I know how to do the proof if I choose a basis,but i'm curious if it can be done without choosing one

[cloud walrus]

ProphetX

[sinful mirage]

managed to do it

[pine patio]

how do i show the isomorphism?

[rustic crown]

try to show that S_k * S_{n-k} stablizes the subset S = {1, 2, 3, ..., k}. Can the stablizer be any larger? Now what can you say about stablizer of a subset in the orbit of S?

[pine patio]

so if alpha is in stab(S) then alpha = (sigma)(tau) where sigma takes S to S, and tau takes S^c to S^c

so we can identify sigma to be something in S_k, and tau to be in S_n-k, but i'm not sure how

[pine patio]

i'm still confused

[rustic crown]

yep that the idea... Take A to be the subgroup of permutation of elements of S and B to be subgroup of permutation of elements of S^c. A is isomorphic to S_k and B is isomorphic to S_n-k.

Now like you said the stablizer is the subgroup AB. Since both A and B are normal in AB and A and B intersect trivially AB is isomorphic to A x B which is isomorphic to S_k * S_n-k

[pine patio]

i've managed to show part i) and ii). for iii) i think i need to use the first isomorphism theorem but i'm not sure which homomorphism to pick

[old lava]

there's only two groups of order 6 up to isomorphism

Z/6Z and S_3

prove that S_4/V_4 isn't abelian, and you're done

[chilly ocean]

quick question - can galois group of polynomial of degree 3 be Z/6Z? I'd say no, cause it would seem like you need degree of the polynomial to be 6 or sth (permuting roots of poly deg 2 and poly deg 3), but not sure

[old lava]

no it couldn't, the galois group would be S_3

if it were order 6

[chilly ocean]

Yeah, but not quite sure why, any proper argumentation for that (or maybe my reasoning is enough)?

hmm maybe I could consider factorization of poly deg 3?

[old lava]

because any galois group of a degree 3 poly must be a quotient group (I think it's a quotient) of S_3

[chilly ocean]

[old lava]

I think in general it's galois group of a degree n poly is a quotient (or is it a subgroup) of S_n

[chilly ocean]

Hmm actually yeah its a subgroup

It embedds in S_n

[old lava]

ya, I got mixed up in the lattice flipping

and shit

[chilly ocean]

[old lava]

but ya, whatever it is Z/6Z isn't a subgroup or a quotient group of S_3

considering they're the same order and obviously not isomorphic

[chilly ocean]

Galois stuff is pretty neat so far.

[old lava]

it is pretty cool

[oblique river]

independent of lattice stuff

S_n only has three quotients (for n > 4)

itself, the trivial group, and Z/2Z

so if you ever find yourself saying the phrase "quotient of S_n" make sure that's really what you mean haha

[old lava]

I mean the quotients appear at the "top" of the lattice, while subgroups at the bottom

and that's how I normally think of it

so I got a bit mixed up since the lattice of subgroups and the lattice of field extensions are like flipped and I'm not too familiar with it yet

[oblique river]

yes but I'm not talking about lattices

i'm talking about groups

and S_n doesn't have many quotients as a group

which could have been a hint that "quotients of S_n" is not what you were looking for

[lunar coyote]

for this question what was the point in doing division with remainders

[old lava]

true

[hidden haven]

That is the whole proof, how would you do this without division with remainders?

[lunar coyote]

i was just wondering whats the intuition behind this like how did they know to use division with remainders

[hidden haven]

You want to show that the lowest degree polynomial in the ideal generates the ideal, ie all other polynomials in the ideal are multiples of the lowest degree polynomial

Then it's natural to divide by the lowest degree polynomial and prove that the remainder is zero (if it weren't, it would have lower degree than the lowest degree polynomial in the ideal and will be in the ideal, contradicting "lowest degree" of f)

[lunar coyote]

so this whole proof was just to show it's in some form f(X), there is no need to show that it's an ideal

[hidden haven]

I didn't get that

It's given to be an ideal

You have to show every ideal is (f(X)) for some f

[lunar coyote]

hmm ok

also is the 2nd line incorrect

shouldn't it be r = g-qf

not r = f -qg

[hidden haven]

Yeah

[green stone]

the mathieu groups

[acoustic pine]

are all local rings localizations of some ring?

[latent anvil]

yes, every ring is a localization of itself with respect to the multiplicative set {1}

[acoustic pine]

Sorry, I should say "ignoring the trivial case"

[latent anvil]

I'm not sure what counts as trivial. if R is some ring and S any other thing then R is the localization of R x S wrt the multiplicative set {(1,0)}

there's lots of ways to make a ring a localization of something

if you want to localize at a prime, any local ring is the localization of itself at its maximal ideal

but the multiplicative set here is just the set of units of R, so is that trivial?

[acoustic pine]

I guess in my mind, trivial is just taking S={1}

[hidden haven]

perhaps non trivial = "natural map into the localisation is not an isomorphism"

[latent anvil]

well then the R x S example works

[thin geyser]

stuck on finding the inverse in 2b

[delicate bloom]

what'd you say for 2a?

when you quotient out by m, you're basically saying x^2-4=0

so you can say x^2 = 4 and then raise both sides to the 5th power to get x^10 = 4^5 and then you have x*x^9 = 4^5, divide by 4^5 and you have (x/4^5)x^9 = 1

[thin geyser]

i argue the gcd of m(x) and f(x) ~ 1 for 2a

that was so helpful for 2b though thank you, i was pulling my hair out trying to run the extended euclidean algo

[delicate bloom]

you're welcome, yeah it can be easier to just mess with the algebra with these kinds of things

might even be tempting to say x=+-2 if that's useful, but probably not since you can't fix a sign on it

[hidden haven]

That's not the reason it's not helpful

When you say x² = 4 you really mean [x²] = [4]

But with ±2 you can't make any such statement, because [x] is a completely new element independent from both [±2]

By independent I just mean different equivalence class

[daring ibex]

a question

isn't this supposed to go up to p-1

[chilly ocean]

omega is a pth root of unity? w^p=1, and (w-1)(1+w+w^2+...+w^{p-1})=0, so w^{p-1} = -1-w-w^2-..., ie w^{p-1} can be written in terms of the lower powers

[daring ibex]

oh word

hm

this is funky bcs the book says this

Doesn't this imply that 1, w, w^2,...,w^{p-1} should be linearly independent over Q

[delicate bloom]

the minimal polynomial is not p, it's p-1

n=p-1 so n-1=p-2

[daring ibex]

wdym? Are you saying that the minimal polynomial is x^{p-1} = 1?

why is that

[delicate bloom]

no look at this factorization

we are factoring out (w-1)

since 1^p=1 is not a root we're adding

[daring ibex]

hmm

so the minimal polynomial is 1+w+...+w^{p-1}

[delicate bloom]

to put it another way, the minimal polynomial is $\frac{x^p-1}{x-1}$ which is degree p-1

[cloud walrus]

Merosity

[delicate bloom]

yeah

[daring ibex]

i see, since we already have the root of x- 1 in our field

[delicate bloom]

yup

[daring ibex]

And we can repeat the same exact thing if p is not a prime as well

okay this makes sense

thank u

[delicate bloom]

sort of

the cyclotomic polynomials are simple if it's a power of a prime

have you heard of mobius inversion?

[daring ibex]

not really

ive heard the name

nothing more tho

[delicate bloom]

ah, well I'll give a quick sketch then to give you an idea of stuff that's involved if you want to derive a cute formula

[daring ibex]

hmm

thanks

[delicate bloom]

if you multiply all the cyclotomic polynomials that are at divisors of n, you end up with all the roots of $x^n - 1$ $$x^n-1 = \prod_{d|n} \Phi_d(x)$$

[cloud walrus]

Merosity

[delicate bloom]

this product can be inverted if you take its logarithm, you need something called the mobius function which encodes inclusion-exclusion over divisors

$$\log(x^n-1) = \sum_{d|n} \log \Phi_d(x)$$
$$\sum_{d|n} \mu(n/d) \log(x^d-1) = \log \Phi_n(x)$$
$$\prod_{d|n} (x^d-1)^{\mu(n/d)} = \Phi_n(x)$$

[cloud walrus]

Merosity

[delicate bloom]

I guess just to explain enough to confirm it works, $\mu(n)$ is -1 raised to the number of distinct primes dividing n, if it has repeated prime factors, it's 0

[cloud walrus]

Merosity

[delicate bloom]

so if we try n=p you can confirm $$\Phi_p(x) = (x^1-1)^{\mu(p)}(x^p-1)^{\mu(1)} = (x-1)^{-1} (x^p-1)$$

[cloud walrus]

Merosity

[daring ibex]

hmmm

i'll look into it

this looks interesting

one more thing, the fourth root of unity adjoined to Q should be iso to Q(i) right?

however

we have that 1,i,i^2 should be linearly independent over Q, but i^2 = -1 so 1,i,-1 is independent over Q but you can write -1 as a linear combination of 1 and i

err no

what am i doing

[delicate bloom]

write what you think the minimal polynomial is for i

[daring ibex]

x^2 + 1

i see the problem i think

argh

[delicate bloom]

maybe try explaining to me why if f is irreducible and deg(f)=n that we have 1, a, a^2, ... a^{n-1} and we exclude a^n

[daring ibex]

Hm

this is just

the same trick as before i think

[delicate bloom]

yeah, what's the trick for reducing a^n into a linear combination of 1, a, a^2, ..., a^{n-1}?

[daring ibex]

(a-1)(1+...+a^{n-1}) = -a^n - 1

[delicate bloom]

nope that isn't the trick

that's just because x^p -1 is not irreducible, it factors into (x-1)(1+x+...+x^{p-1})

[daring ibex]

oh

hm

[delicate bloom]

we are starting knowing that f is irreducible, so that won't work

[daring ibex]

idk then

hrm

[delicate bloom]

ah ok good I'll show you

like for instance x^2+x+1 we know is irreducible and has the roots w and w^2 right

we have that 1, w, w^2 doesn't give us a basis because we can always reduce with the polynomial like so:

w^2+w+1=0

w^2 = -1-w

[daring ibex]

yeah

[delicate bloom]

so w^2 is a linear combination of -1 and -w

and this will always work for a degree n polynomial if we have its root adjoined

maybe a more concrete example, let's say you have a degree 4 irreducible polynomial, might look something like $$x^4+Ax^3+Bx^2+Cx+D$$ with root $a$

[cloud walrus]

Merosity

[delicate bloom]

then $a^4+Aa^3+Ba^2+Ca+D=0$ so we can just rewrite $a^4 = -(Aa^3+Ba^2+Ca+D)$

[daring ibex]

oh wait, we just factor out x-a

[cloud walrus]

Merosity

[delicate bloom]

no never

[daring ibex]

oof

[delicate bloom]

it is irreducible from the start

[daring ibex]

Ah 🤦‍♂️ true

[delicate bloom]

the root plugged into the polynomial gives us a relation to the lower powers

I think you're still stuck looking at the wrong thing

[daring ibex]

yeah, makes sense

yeah sorry im just restarting looking at this stuff so this is very fuzzy for me

[delicate bloom]

you have nothing to apologize for

this is just how learning goes, trying to clarify what's in your mind like this, you're good lol

[daring ibex]

hm

thanks

[delicate bloom]

alright quick test

let's pretend x^7 + bX^3 + 1 is irreducible with root a

what is a^7 in the basis {1, a, a^2, ..., a^6}

[daring ibex]

well then $a^7 + ba^3 + 1 = 0$ and so $a^7 = -1-ba^3$

[cloud walrus]

uoft

[daring ibex]

right

[delicate bloom]

yup, you got it

[daring ibex]

but what if this is reducible

does that cause any issues?

i wouldn't think it does right

[delicate bloom]

totally changes the problem

[daring ibex]

bcs then we can just factor into irreducibles

and then do the same thing

[delicate bloom]

yeah

[daring ibex]

So we only consider irreducibles bcs every case just boils down to this

[delicate bloom]

yeah I guess you could say that

might not look so clean as this

adjoining a root of an irreducible polynomial doesn't necessarily get you the splitting field of that polynomial

simple example, if we take x^3-2 and think of Q(cbrt(2)), we are missing out on complex roots

even though x^3-2 is irreducible over Q

but Q(cbrt(2)) is a field with basis {1, cbrt(2), cbrt(4)}

[daring ibex]

hmm

i see

like we'll have to adjoin all the roots then?

directly or indirectly

well i guess "directly or indirectly"

doesn't make sense

[delicate bloom]

if we want x^3-2 to split completely into linear factors, yeah

you can factor x^3-2 over Q(cbrt(2)) and you'll get a new irreducible quadratic times a linear factor

and then adjoin a root of that to Q(cbrt(2))

maybe try finding what that quadratic is

[daring ibex]

well you just divide x^3 - 2 by x- cbrt(2)

Actually the whole reason im thinking about this was i was under the impression that Q(i) could be written as Q(w) for w some n-th root of unity

[delicate bloom]

that's a fun question, it won't be possible in general

[daring ibex]

hrm, i'm not sure how i'd go about it other than saying for the fourth root of unity $\mathbb{Q}(w) = {a_0 + a_1w + a_2w^2 \mid a_i \in \mathbb{Q}}$

[cloud walrus]

uoft

[delicate bloom]

there are a lot of "obvious" cases where you can definitely do it

[daring ibex]

Hrm

wdym

[delicate bloom]

let's say w is a 12th root of unity, can we make i?

[daring ibex]

well yeah

since i^12 = 1 we should be able to write it in terms of w

[delicate bloom]

there's a super explicit way to do it

[daring ibex]

how so?

[delicate bloom]

with just one term

[daring ibex]

w^4 mebe

[delicate bloom]

getting warmer

w^4 is actually a cube root of 1

since (w^4)^3 = w^12 = 1

[daring ibex]

Hrm

(i^4)^3 = 1 too

that doesn't help tho

i^4 = 1 and (w^3)^4 = 1 so (w^3)^4 = i^4 i guess, but not sure what to do from here

can't just straight up cancel the fours