groups-rings-fields

[delicate bloom]

ah we don't have to 'cancel' we can just write w^3 = i

we know that it must be that i^4=1 as it's defined this way because otherwise there'd be some number less than 4 such that i^k=1

but that would mean (w^3)^k = 1 for k<4

and so that would make w^(3k) = 1, but since we know 12 is the smallest number that makes w^n=1 it must be that 3k=12

[daring ibex]

why so?

[delicate bloom]

everything after that is my explanation

maybe that's a bit confusing how I've said it

we know that 12 is the smallest number n that makes w^n=1

[daring ibex]

yeah i feel like this step assumes i = w^3

[delicate bloom]

no I am using it as it's defined as a symbol there

[daring ibex]

Ah

[delicate bloom]

I see the confusion though let's just avoid it

w^n=1, n=12 is the smallest number that makes this true

[daring ibex]

yeah

[delicate bloom]

that means the smallest number that makes (w^3)^k =1 is k=4

so we know w^3 can't be something like 1 or -1

[daring ibex]

hm

[delicate bloom]

let's be more convincing

let's call s = w^3

[daring ibex]

Sure

[delicate bloom]

and look at the polynomial x^2+1

[daring ibex]

yeah

[delicate bloom]

can we factor this as (x+s)(x-s)?

if so, since we're in a field that'd mean (x+s)(x-s)=0 has only two solutions, s and -s

and both satisfy x^2+1=0

[daring ibex]

Hm

okay makes sense

[delicate bloom]

well just write it out and see, (x+w^3)(x-w^3) = x^2 - w^6 = x^2 - (-1)

[daring ibex]

hm

[delicate bloom]

whoops we might have to show w^6=-1 since that's the same kind of problem as showing w^3=i

but it's easier this time

we know w^12 = (w^6)^2=1

so that means w^6 must satisfy x^2-1 =0

but we know w^6 != 1 since that would mean the order of w is 6, not 12

so it must be -1

[daring ibex]

ahh yes that makes sense

and then we can identify i with w^3, showing that Q(i) embeds into Q(w_12)

[delicate bloom]

yeah

[daring ibex]

I think this generalizes to Q(i) embedding into Q(w_4n)

The only possibly degenerate case is Q(w_4)

which ill check rn

[delicate bloom]

yeah, that's exactly the 'obvious' criteria I had in mind earlier

but for all we know at this point there could be weirder ways of constructing i within Q(w) so it's not definitive

[daring ibex]

err in the case of fourth root of unity wecould just say i = w

[delicate bloom]

yeah, so in Q(w_4n) what is the k you pick to represent w^k = i?

[daring ibex]

we choose k = n

[delicate bloom]

yeah

now let's back up a sec, I picked w with order 12 earlier but

but nothing, I'm tired, I don't know what I was thinking to say next there lol

[daring ibex]

happens to me too

lol get some sleep

[delicate bloom]

I'm gonna head to sleep, maybe you can think of some ideas for how to continue until tomorrow maybe

[daring ibex]

thanks for the help

yeah sure

[delicate bloom]

yeah you're welcome

[daring ibex]

wait nevermind, you can't do this. I don't think Q(i) embeds into Q(w_4)

tho it should work fine for other n

err no i think it does work

I think we have sth more, we have an iso b/w Q(i) and Q(w_4)

[hidden haven]

Depending on your definition of w_4 you'd have equality or no isomorphism at all

Is w_n the nth complex root of unity, or an abstract nth root of unity?

In the former case, w_4 = i

[rustic crown]

what's an abstract root?

[hidden haven]

Like when you adjoin elements without having an ambient field

To any ring

[rustic crown]

but if we work over Q does that make a difference

[hidden haven]

It does because w_4^2 doesn't have to be -1 then

Oh it's round adjoint actually

[rustic crown]

wait then it no field

[hidden haven]

So it might have to be

Yeah I was thinking Q[w_4]

[rustic crown]

ah yes

minimal polynomial over these need not be irreducible

[hidden haven]

Ye

[daring ibex]

complex root of unity

[hidden haven]

Then w_4 = ±1 or ±i

You need to define it better than just "a complex fourth root of 1"

[daring ibex]

it's primitive root

like the

[hidden haven]

Epic

So it's i

[daring ibex]

Yeah

[hidden haven]

Or -i

And you'll have equality of extensions not just isomorphism

[rustic crown]

nice

[daring ibex]

[hidden haven]

[daring ibex]

Also here, if we consider Q(w_4) then this has 1,w,w^2 as a basis for Q(w_4)

as a v.s. over Q

[hidden haven]

no.

[rustic crown]

no.

[hidden haven]

[daring ibex]

oof

[rustic crown]

1,w is enuf

[upper pivot]

w^2 is -1 oof

[daring ibex]

hrm

Well here's my reasoning

If w = w_4 for notational ease. Then Q(w_4) is iso to Q[x]/<f> where f = 1 + w + w^2 right

[hidden haven]

You can't say that

[rustic crown]

f=w^2+1

[daring ibex]

Though yeah, the w^2 = -1 fucking up the lin independence thing was bothering me

[rustic crown]

Q(w_4) isn't isomorphic to Q(w_3)

[upper pivot]

yeah thats 3rd roots of unity

[daring ibex]

oh word?

[hidden haven]

Where did you get that f

[daring ibex]

well don't we have

[rustic crown]

[daring ibex]

hrm

okay now i'm confused

[hidden haven]

[daring ibex]

hrm

[hidden haven]

So yeah minimal polynomial of i over Q is x²+1

You have then f = x² + 1

[daring ibex]

hrn

okay this is hurting my head

Bcs i was told

two different things now i have to sort them out

[rustic crown]

whut was the first?

[daring ibex]

Probably just me misinterpreting what was said tho

Alright then

1+w+w^2+...+w^{n-1} is only the minimal polynomial when n is prime i think

[rustic crown]

yep

[daring ibex]

Otherwise in our case of n = 4 then there is some stuff like for example x^{n/2} = -1 which can be written in terms of the constant term

i think that's it tho

or

perhaps we have to remove all of the terms with a factor of n

[upper pivot]

yeah thats kind of an idea

[hidden haven]

Even with this 4-1 ≠ 2

[upper pivot]

you want the polynomial whose roots are the primitive roots of unity

[rustic crown]

And in general minimal polynomial of w_n is called the nth cyclotomic polynomial and it has degree phi(n).

[upper pivot]

i.e the cyclotomic polynomial

[daring ibex]

For this n is prime

[hidden haven]

Isn't it cyclatomic

[rustic crown]

p sure not

[upper pivot]

its actually a bit non-trivial to show that cyclotomic poly are the minimal polies for nth primitive roots of unities

[hidden haven]

Yeah but for w_4 you said minimal polynomial 1+w+w²

[daring ibex]

ya

That's wrong, only works when it is prime

[rustic crown]

yea that irreducible proof was pretty clever

hunting for that prime was hard

[daring ibex]

oki that helps

[hidden haven]

I'm saying even if 4 were prime you were supposed to go up to w³

[rustic crown]

even if 4 were prime

[hidden haven]

[daring ibex]

for large enough 4 and small enough 5, 4=5

[daring ibex]

i cee

[upper pivot]

oh yeah i knew a few proofs but ill be honest i forgot all the proofs lol, i know one used something to show its irreducable lol

[rustic crown]

(4-1)! is not 0 (mod 4) hence 4 is prime

[unique juniper]

$x^p - a$

[cloud walrus]

是的

[unique juniper]

trying to show that this is irreducible over a field of characteristic p, a is an element that isnt a pth power when the field is infinite

over the finite case i can do it

but when infinite

since its not seperable

its not a product distinct irreducibles of course

but

now

how would i show that x^p -a itself is irreducible

not just a product of (atleast 1) repeated irreducible

[chilly ocean]

I think it is separable right? Not that I know how to show irreducible from this

It is not separable, because if xi is some root (in algebraic closure) then xi^p = a, so x^p-a=x^p-xi^p = (x-xi)^p

Ah, okay right

So if it is reducible then (x-xi)^r lies in k[x] for 0 < r < p, we obtain that all coefficients of (x-xi)^r lies in k, in particular -r xi lies in k, and since r ≠ 0, xi lies in k

[sturdy mirage]

can you define an ideal in a polynomial ring to be: generated by at least one of two given polynomials $f(x), g(x)$? so its the set of $<f(x)> \cup <g(x)>$? would this be an ideal? is this a principal ideal?

[cloud walrus]

reking

[chilly ocean]

(f)+(g) is definable

[sturdy mirage]

i read that every ideal in a polynomial ring is a principal ideal, but i don't see how this would be true here

a principal ideal is generated by only one element, right

[chilly ocean]

it is basically (gcd(f,g))

[sturdy mirage]

what is the gcd of two different, irreducible polynomials?

[chilly ocean]

1

[carmine fossil]

Only true if the polynomial ring is F[x] for a field F

[sturdy mirage]

riight..

[carmine fossil]

For example in Z[x] (2,x) is not principal

[chilly ocean]

(f) U (g) is not an ideal in general, but all possible linear combinations p f + q g is an ideal generated by gcd(f,g)

[sturdy mirage]

i suppose i mistakenly thought <f> U <g> would be an ideal, but i see how that fails

[chilly ocean]

ye, (2) U (3) in Z doesn't contain 5, but the ideal should be closed under +

[sturdy mirage]

thanks!

[chilly ocean]

you are welcome :з

[unique juniper]

I don’t understand this

[chilly ocean]

what "reducible polynomial" means?

"q is reducible" means that q = fg in k[x] for some f with 0 < deg f < deg q

and we know that in algebraic closure q = (x-xi)^p

it means that if q in the base field is reducible then (x-xi)^r lies in k[x]

[unique juniper]

What about irreducible

[chilly ocean]

irreducible means that it is not reducible

[unique juniper]

I mean

I’m trying to show

That

X^p - a is irreducible in an infinite field of char p

[chilly ocean]

you trying to show that "a≠xi^p -> x^p - a irreducible"

contrapositive "x^p - a reducible -> a=xi^p"

[unique juniper]

sure

[chilly ocean]

so you have to show that if some factor of x^p-a lies in k[x] then xi lies in k

[unique juniper]

is xi supposed to be a root

[chilly ocean]

yes

xi^p=a

in algebraic closure any polynomial have a root

[unique juniper]

I dont get it

so what i understood was

you are saying that we should show that there are no roots of the polynomial

in the field

?

@chilly ocean

[chilly ocean]

Let xi be a number in k-bar such that xi^p=a. You have to show implication "x^p - a reducible -> xi lies in k"

So, assume x^p -a reducible in k[x]

in k-bar[x] we have x^p-a = (x-xi)^p

so if (x-xi)^p is reducible in k[x] it means that some (x-xi)^r lies in k[x] for some 0<r<p

Is it clear?

[unique juniper]

no

how do you know that its just a single root

[unique juniper]

this

[chilly ocean]

in k-bar[x] we have at least one root of x^p-a (because k-bar is alg.closure) lets name it xi

so we have xi^p=a

so x^p-a=x^p-xi^p = (x-xi)^p

[unique juniper]

hm

ok

so then

since we assume its reducible that means that (x-xi)^p is in the base field??

[chilly ocean]

since we assume it reducible means some factor of (x-xi)^p is in the k[x]

[unique juniper]

sure

[chilly ocean]

all factors of (x-xi)^p looks like (x-xi), (x-xi)^2, ..., (x-xi)^(p-1)

(non-trivial factors)

[unique juniper]

yeah

[chilly ocean]

so we assume that some (x-xi)^r lies in k[x] for 0<r<p

[unique juniper]

ok

[chilly ocean]

and if we expand (x-xi)^r it would be (x-xi)^r = x^r - r xi x^(r-1) + ...

(x-xi)^r lies in k[x] implies that -r xi lies in k

[unique juniper]

ok

[chilly ocean]

but -r xi lies in k implies xi lies in k, because 0<r<p so it is invertible

[unique juniper]

ok

[chilly ocean]

that's the proof, we proved that "x^p - a" reducible imples "xi lies in k"

[unique juniper]

i see

[unique juniper]

how exactly

[chilly ocean]

hm, you can multiply -r xi by (-r)^(-1)

and (-r) is not 0 because 0 < r < p

[unique juniper]

how do you know that r is in the field

[chilly ocean]

it is just natural number, so, it is just 1+1+...+1 r times

[unique juniper]

oh yes right

i see

that works, thanks

[chilly ocean]

you are welcome :з

[unique juniper]

I was expecting something different

i thought maybe .. showing that it just has 1 irreducible factor, itself

Thank you!

[chilly ocean]

hm?

[unique juniper]

wait no

[chilly ocean]

I think that's right

[unique juniper]

sorry i understtand XD

yes you are!!

[chilly ocean]

🙂

[unique juniper]

i just errr got confused :)

thank you!!!!

[chilly ocean]

it's ok to be confused, no problems : )

[unique juniper]

Sorry I’m confused again

We need to the assumption that xi a root isn’t in the field

We don’t have that?

[chilly ocean]

Hm, I don't see how we need that, we need to prove that xi is in the field assuming that x^p-a is reducible

Compare: "we need to prove that a^2+b^2=c^2 assuming a,b,c are the sides of rectangle triangle"

We don't need to assume that a^2+b^2≠c^2 separatly

[unique juniper]

Hm

If rainy implies cloudy, not being cloudy implies not being rainy

Right?

[chilly ocean]

right

[unique juniper]

So we need to assume that it’s not cloudy for it to not be rainy?

[chilly ocean]

sure

[unique juniper]

So x^p -a is reducible if xi is in k

[chilly ocean]

the initial statement was: "if xi not in k then x^p-a irreducible"

oh sorry

[nova plank]

No. It could be cloudy and not be rainy.

[unique juniper]

I mean not cloudy implies not rainy sorry

[nova plank]

👍

[chilly ocean]

the contrapositive to "if xi not in k then x^p-a irreducible" is "if x^p-a reducible then xi \in k"

[unique juniper]

Omg

Yes

My bad

[chilly ocean]

🙂

[unique juniper]

I’m so sorry

And thankful :D

[chilly ocean]

don't worry like that, I like to explain :з

[chilly ocean]

How can I find the degree of $Q(i) \subset Q(i, 3^{\frac14}, 2^{\frac13})$?

[cloud walrus]

Ledog

det

[chilly ocean]

okay yeah I c

after drawing a diagram

for some reason I was doing some wierd stuff first

tyy

[rustic crown]

[chilly ocean]

oh and is this obvious that this extension is a splitting field for (x^4-3)(x^3-2)?

[rustic crown]

"obvious" is a weird word. but yea that is true.

[old lava]

a factor of (x^2+1) as well

is needed

no?

considering the i and all

[chilly ocean]

over Q(i)

[old lava]

ah

[rustic crown]

you'll have omega and you have and 4th root of 3 so you can also recover i from there

wait i said it backward

[chilly ocean]

yeah that's what I was thinking

[rustic crown]

.<

[chilly ocean]

[rustic crown]

you have i and 4th root of 3 so you can get (-1+sqrt(3)i)/2 and i which are 3rd and 4th primitive roots of 1

[chilly ocean]

yep

[rustic crown]

[pine patio]

how do i count the number of homomorphisms from S_3 to D_8?

i found that S_3 has 3 normal subgroups: A_3, {e} and S_3

and this $S_3/{e}\cong S_3, S_3/S_3\cong {e}, S_3/A_3\cong C_2$

[cloud walrus]

AKS

[pine patio]

so i need to use the first isomorphic thm

[next obsidian]

Is D_8 size 8 or 16

under your conventions

[pine patio]

size 8

[next obsidian]

okay, well you can tell ther'es no injections first

based on size considerations

do you see why?

[pine patio]

6 doesnt divide 8

[next obsidian]

right

so now how many maps are there which have kernel = S_3?

[pine patio]

1

[next obsidian]

right

so now we can focus on maps with kernel A_3

but these are exactly maps from C_2 -> D_8

and this should be something you can figure out yourself hopefully :)

You might want to figure out in general how many maps there are from a cyclic group into any group

it's pretty useful to figure that out

[pine patio]

yes so {e, r^2}, {e, rs} etc

[next obsidian]

yeh

[pine patio]

thanks!

[next obsidian]

Just remember that you've already counted the zero map

so don't accidentally count the map C_2 -> D_8 which is the zero map

since that actually corresponds to the zero map S_3 -> D_8

[untold sapphire]

V nice

[next obsidian]

bdobba

Does a dim 1, Noetherian local ring necessarily have to be regular?

I think the answer is no, but idk how to show it

[green stone]

anybody here studied wreath products?

[old lava]

I did a singular exercise on wreath products in d&f

so you could say I'm an expert

[green stone]

i trust your expertise. im having a bit of trouble visualizing a wreath product of infinite groups

Z wr Z is going to be some subgroup of Sym(Z x Z) i know.

[old lava]

rip, my expertise fails me

I wish I could've helped

[green stone]

dont sweat it. i bet i'll get it eventually. im starting to think i can kinda visualize it.

[next obsidian]

TFW Dummit and Foote didn't make you a wreath product expert

[green stone]

does dummit and foote cover wreath products at all? I have a copy coming in the mail, but I'm using a different textbook at the moment

[carmine fossil]

There is one question

and exactly one question on wreath products

[old lava]

yep

there's one question on wreath products in total

[green stone]

im using a book called A Course in the Theory of Groups by Derek Robinson. One of my professors recommended it to me, and it has a decent handful of problems on wreath products. it covers a lot of ground on group theory in general too. im enjoying it

my professor had the author as one of his professors which is sort of cool

[old lava]

that's probably a more thorough exploration of groups than d&f

d&f groups is really just basically all the group theory an undergrad should know realistically

the book you're using is a grad group theory book

[green stone]

it is a grad level book. i'm certain it does cover more group theory too, but i bet the d&f book will be pretty useful still.

[old lava]

d&f is useful for a first course

but not if you wanna learn wreath products

[next obsidian]

https://simple.m.wikipedia.org/wiki/Cohomology

[old lava]

there's a simple wikipedia page on cohomology?

damn lmao

oh nevermind

it's one line

[proud bear]

[short iris]

what is an algebra

[sick valve]

Just to be sure, is this right?
$${1,2} \times \mathbb{Z}_{+}={{1,1},{1,2},...,{2,1},{2,2},...}$$

[cloud walrus]

𝔙eryhappyperson

[chilly ocean]

the elements should be ordered pairs

not sets containing two elements

[sick valve]

Ohh, sorry.

[chilly ocean]

e.g., {1, 1} is just {1}

[sick valve]

Okay.

Just btw, typing lots of regular sets in LaTeX is annoying.

Thank you.

[next obsidian]

Module but it’s also a ring

[paper flint]

[golden pasture]

a R-algebra is a object in the category Ring^R

unironically tom dieck defined an algebra as that

[hidden haven]

Any non pth power in K is a root of an inseparable irreducible deg p polynomial over K^p

(K^p is a field so you can draw a field tower diagram involving that, see which of the extensions are separable)

[chilly ocean]

I wondered that there is some argument possible based on "frobenius map is probably almost surjective, in some sense", this sounds sort of like that, although my field theory is not good enough to understand

[hidden haven]

Yeah seems similar to that result for perfect fields but idk

Didn't get this

Any field homomorphism is injective

If y is a non pth power in K, then y^p = a for some a in K^p

x^p - a is the minimal polynomial of y then

I used x for both the variable and the element

y^p is in K^p by definition of K^p

You don't need y to be a non pth power to say this

Yeah

Yeah, minimal polynomial over K^p

The part that's relevant here is that that polynomial is inseparable

Also it's the minimal polynomial only if y is not in K^p, because else it reduces

F is not relevant yet

My claim is y is inseparable over K^p

You know its minimal polynomial

Can you figure out if that has repeated roots?

How does minimality imply inseparability?

Yes

x²+1 is irreducible over Q but is separable

It's a polynomial in K^p, but in a splitting field, it will have repeated roots

You know one root is y

Yep exactly

y is the repeated root

So whenever K ≠ K^p in characteristic p, K/K^p becomes an inseparable extension (in fact purely inseparable)

Now K/F is separable and K/K^p is purely inseparable. See if you can conclude anything about separability of K/F(K^p)

Yeah, why is that?

Purely inseparable means that every element has an inseparable minimal polynomial

Whenever you have K/E/F, and y in K, the minimal polynomial of y over E must divide the one over F

K/F(K^p)/F

or K/F(K^p)/K^p

The tower has 3 levels, the lowest level has 2 fields which are both subfields of the middle level

[cloud walrus]

Moldilocks

[hidden haven]

used online app to draw with mouse and generate the code

yes

oh I labelled the wrong edge as sep in diagram

but K/F(K^p) being separable is important

can you use this to argue that it must be the trivial extension?

something like that

you can reason through it, pick a y in K, you know its minimal polynomial over F(K^p) is separable but over K^p is not, and you know one of them divides the other

not each other

the one over F(K^p) divides the other one

since its the intermediate extension between K and K^p

[stone fulcrum]

Where are you getting that?

[hidden haven]

yeah

[hidden haven]

q.uiver.app

[stone fulcrum]

Bless

[hidden haven]

we know the inseparable minimal polynomial's exact form

the separable one is a factor of that

what are the possibilities?

yepp

so minimal polynomial of any element in K over F(K^p) is degree 1

yep

happens

welcome

[hidden haven]

oof

yeah lol if anything I would have confused irreducible and inseparable

because in english those have similar meanings lmao

oh right makes more sense

why are things in math called separable when nothing is being separated

goddamn baguette people

[rustic crown]

separability of polynomial makes a lot more sense than separability of metric spaces

[hidden haven]

true

still very

[rustic crown]

lol

[daring ibex]

Question how am i supposed to do this. I've been trying different approaches for about a day now but i haven't come up to a complete solution

a hint would be appreciated

[rustic crown]

if alpha is irreducible then there is nothing to prove!

[daring ibex]

Well yeah that is trivial

[rustic crown]

else can you think why would there be an irreducible dividing alpha?

[daring ibex]

I was thinking of inducting on the number of prime factors

[golden pasture]

prove that Z[i] is a euclidean domain

[daring ibex]

Of N(\alpha)

[daring ibex]

hmm

alright then

[golden pasture]

you are given the norm, jus need to verify it works

[rustic crown]

once you get an irreducible factor, look at alpha = pi * beta, and then think about N(beta)?

[hidden haven]

||Or you can also say both factors have smaller norm rather than specialising to the case of one being irreducible||

[rustic crown]

oh lol

[daring ibex]

Hm so you're saying if \alpha is irreducible then this is trivial, so consider \alpha not irreducible. I.e. writing for a,b not units, we have \alpha = ab? Which implies N(\alpha) = N(ab) = N(a)N(b)

And then consider that?

[golden pasture]

easiest just prove euclidean tbh

[daring ibex]

What does that show tho

I think euclidean follows fairly easily geometrically

[daring ibex]

I mean if you mean euclidean => pid => UFD that's not really needed

I've shown Z[i] is a UFD already thru a geometric argument in another exericse, i'm just unsure how to do this using induction

on N(\alpha)

[daring ibex]

never mind i think this is it, N(a), N(b) are necessarily smaller than N(\alpha), and a,b are not units so they are not 1. Then we just say that by induction we are done

[rustic crown]

Haven't seen the notation before, but if I understand it right then isn't p=3 with F = Q and K = Q(i) a counterexample?

Then what about F = K = Fp(x)

Ye

If F has char p and K/F is a field extension then K has to have char p

[rustic crown]

this is a counterexample

[golden pasture]

recall euclidean implies pid directly

waitwat

oh as in like

how can characteristic be a prime power lol

p

Recall

how do we define char of a field

1+1+...+1=0

so it really doesnt matter what extensions we throw at it

(x^p=0 -> x=0 cuz you know, field has no zero divisors)

it's a counterexample at φ(α)->α^p is an isomorphism

[rustic crown]

The image of Frobenius would be F_p(x^p)

[golden pasture]

yup

wait

isnt that an embedding of F_p(x) into F_p(x) tho

ig it isnt bijective as a map from F_p(x) to itself

it is only injective

but not surjective

as x has no preimage

injective means like

f(a)=f(b) -> a=b

[scarlet estuary]

left

[golden pasture]

i can never rmb what is left and right

directionally impaired moments

[scarlet estuary]

f(a) = f(b) implies a=b whenever there exists a g such that g(f(a)) = a, g(f(b)) = b

[golden pasture]

wait but g doesnt necessarily exist tho

[scarlet estuary]

?

[rustic crown]

As a set map it does

[golden pasture]

ah

yea

[scarlet estuary]

oh yeah it might not be a homomorphism or whatever

[golden pasture]

petition: can we also change function composition order it trips me up so much

[rustic crown]

Yeaaaa

Or if not make a petition to start drawing maps going left

[golden pasture]

i dont mind if it means to use x.f.g for g(f(x)) tbh

[rustic crown]

A --> B shouldn't be used

B <-- A would fix it

[golden pasture]

ehh

not rlly

like when diagram chasibg

you go from your start to end

but you end up having to flip your order

[rustic crown]

Just make your start on right

[golden pasture]

it annoys me everytime without fail

[rustic crown]

((x)f)g looks scary

[golden pasture]

x.f.g is nice tbh

[tough raven]

I like to think of it as having a 'post' inverse i.e. an inverse when applied after it.
Then I remember that the function applied after goes on the left.

(as a set map)

[golden pasture]

like say this right, if i want to go A->B->B'->C', i trace f b g' but when writing it down i need to write g' b f

[scarlet estuary]

just be a chad and define $fg = g \circ f$

[cloud walrus]

Namington

[golden pasture]

actually is that like

a thing you can do

[scarlet estuary]

this technically forms an algebra of functions

a very stupid one

[golden pasture]

if so im going to do it daily for my sanity

[scarlet estuary]

i would not advise it

but the FBI probably isnt gonna prioritize you

[golden pasture]

sad

maybe we should start a petition like the petition to remove ξ

[tough raven]

I saw a textbook mention the possibility of writing arrows from right to left in diagrams

But then it said that would also go against convention too much

[hidden haven]

Some texts do that

[tough raven]

And that it would live with the discrepancy

[golden pasture]

i only know of herstein which does x(f)(g) which tbh is superior

[hidden haven]

It takes 0 time to get used to

Only issue is ambiguity if you see excerpt from some text but that doesn't seem a huge issue

[hidden haven]

Best example I've seen is a pdf on elementary theory of category of sets

It was really nice to read

[golden pasture]

oo

[hidden haven]

And notation was genuinely a large part of it

Also no need to put brackets in ETCS when applying a function to an element because application becomes an instance of composition

So it just becomes xfg for g(f(x))

Which is another extremely nice thing

[golden pasture]

honestly

the more i see RPN

The more i like it

unfortunately it gets ambiguous for like

Z(f) vs Z(f,g)

maybe a mix of RPN and () is ideal

[hidden haven]

As in?

The no brackets thing is somewhat specific to ETCS

[golden pasture]

so like sometimes we use the same symbol for 2 different operations, one unary one binary

so like Z here can either refer to the function with one input Z(-) or the function with two inputs Z(-,-)

[hidden haven]

What's the issue if we write (-)Z and (-,-)Z?

[golden pasture]

oh

that works

haha

[tough raven]

It should work if you use RPN but bracket expressions ending in operators with multiple arities

I think

[hidden haven]

[tough raven]

[golden pasture]

i think that would be neatest

so like

(5,3)*3+=18

or just 5 3 * 3 + = 18

[tough raven]

18=

[hidden haven]

Infix notation is still good for numbers tho lol

[golden pasture]

lmao true

but once you write and use enuf rpn in code you kinda learn to read it mentally

[tough raven]

(5 3 *) 3 + 18 =

[tough raven]

In code? What language?

[golden pasture]

so like

infix sucks for computers

RPN is way easier to parse

so typically you convert everything to rpn

[hidden haven]

Because no bracket ambiguity?

[golden pasture]

play in rpn/expr tree

then convert back

not rlly cuz infix has order of operations stuff

[tough raven]

True
But most programming languages just make the parser right?
Which one actually asks you to use RPN?

[golden pasture]

oh like internally

a sane programmer would parse infix to rpn then play with it

(to be fair after rpn the next step is typically expression tree)

[sick valve]

Is $$\mathscr{D}={B \times C \ | \ B \in \mathscr{B} \ and \ C \in \mathscr{C}}$$ the same as saying $$\mathscr{D}={\mathscr{B} \times \mathscr{C}}?$$

[cloud walrus]

𝔙eryhappyperson

[tough raven]

No

The former has a lot of elements in most cases.
The latter has only one element.

The latter is more like

[cloud walrus]

Raghuram

[sick valve]

Thank you

I am still not 100 percent sure if I get this. What if I had something like $${{1,2},{3,4}} \times {{5,6},{7,8}}?$$

[cloud walrus]

𝔙eryhappyperson

Raghuram

Raghuram

[sick valve]

Thank you so much

[hidden haven]

High effort

Raghuram

[rustic crown]

Hey Moldi

[hasty sorrel]

Does anyone here have any interest in cyclotomic polynomials over finite fields? I have many curiosities about how they factor in prime fields of different sizes.

[rustic crown]

I'm sure a lot of them do

[hasty sorrel]

them?

[rustic crown]

"people here"

[hasty sorrel]

Oh. If anyone has any resource to study specifically this, I would like that. I found some patterns that I want to know if they are true or not.

Also, nice profile picture.

[rustic crown]

[slender sable]

Find all linear transformations T : F → F where F =R or C .can anyone tell how to approach this?

[rustic crown]

I have only seen simple stuff that follow from gcd(x^n - 1, x^m - 1) = x^gcd(m, n) - 1 and how x^q^n - x splits over F_q^m

[rustic crown]

linear wrt which ring?

Z-linear, Q-linear, R-linear, F-linear?

[hasty sorrel]

Yes, and that all leads into Galois Theory I believe. I am currently studying that to see if it can help answer my questions.

[rustic crown]

oh okie

[hasty sorrel]

I am kind of self-taught though... I don't actually study math in school, though I am a graduate student.

[rustic crown]

how is it being a graduate student?

i mean how does it feel like

how is it different from undergrad?

[hasty sorrel]

I am really enjoying it. I am actually a computer science major, but I prefer studying things that are as closely related to abstract algebra as possible.

I think it's better than undergrad, because you have a bit more freedom about what you will study, especially if you have a fellowship.

The classes feel very similar though, I don't really enjoy class so much.

[slender sable]

tell about R and C

[rustic crown]

okie so lets see R-linear maps from C --> C
any element of C looks like a + bi, so an R-linear map T is completely determined from where it sends 1 and i as T(a+bi) = aT(1) + bT(i)

And we can choose T(1) and T(i) any way we want, there are no constraints, this will give an R-linear map

and in general F-linear maps between F^n --> F^m look like m x n matrices with entries in F

[hasty sorrel]

Forgive me if I am mistaken, but isn't the constraint that T is linear? As in it follows the homomorphic properties on addition and scalar multiplication?

Oh, you said that in the previous message.

Sorry about that...

[slender sable]

cool cool

[rustic crown]

oh i mean the choice of T(1) doesn't affect the choice for T(i). They both can be arbitrary complex numbers.

So what I just said is, if V is an F-vector space with basis B, and W is another F-vector space. Then you can extend any set function B --> W to an F-linear map V --> W

So this gives a one-to-one correspondence between set-functions {B --> W} and F-linear maps {V --> W}

[hasty sorrel]

Oh, that's true. They are independently mapped (linearly independent)

[rustic crown]

@hidden haven I wanna tell something... you up?

[rustic crown]

https://tenor.com/view/eevee-pokemon-cute-anime-sad-gif-17272241

[golden pasture]

it's basically the same as studying cyclotomic polys over p adics

we do know precisely how Q_p(\zeta_n) behaves as well!

(if you're interested in poking, recall that all finite fields of order p^n are isomorphic and you have the frobenius morphism x->x^p in these fields, in particular, F_q is precisely the field fixed under the morphism x->x^q over \overline{F_p})

[chilly ocean]

@rustic crown I wanna tell something... you up?

[hasty sorrel]

I have not read much about p-adics, but I will look into it. Thanks!

[golden pasture]

:D

neukirch chapter 2 would be your friend here haha

possible "gateway drug" into class field theory with lubin tate as well heh

http://www.math.uni-bonn.de/people/ja/lubintate/lecture_notes_lubin_tate.pdf
have been following these notes mainly, pretty nice exposition

[rustic crown]

yep

[chilly ocean]

ok watch this https://www.youtube.com/watch?v=iwO0crHrFoc&ab_channel=HoussemKouki

[rustic crown]

[limpid edge]

If you have the Q (the rationals) adjoined sqrt(2) and sqrt(3)

what is the unique alpha s.t. Q(alpha) = this same extension field?

[carmine fossil]

det,do you have a fav eeveelution?

[rustic crown]

I don't wanna evolve my eevee.

[chilly ocean]

sqrt(2)+sqrt(3)

[rustic crown]

(unique alpha? if alpha works, then any non-zero rational times alpha works)

[chilly ocean]

oh yea its not unique

[old lava]

it's not too bad an exercise to prove that Q(sqrt(2), sqrt(3)) = Q(sqrt(2) + sqrt(3)), but doubtful that the alpha is unique

[limpid edge]

oh it's not unique then

[chilly ocean]

and any rational plus alpha

[limpid edge]

but yea that works thanks

so generally

if you have Q adjoined a bunch of irrationals

it's equal to Q adjoined the sum of them

[old lava]

bunch of algebraic numbers, firstly

the thing you posted said the degree of the extension is < infty

[chilly ocean]

and if there's infinitely many there's no sum

[old lava]

and ya, that's the other thing

gotta be more precise than "a bunch of irrationals"

[limpid edge]

yeah oops lol

finite bunch of algebraic numbers

[chilly ocean]

@chilly ocean Can I dm you?

[limpid edge]

oh yeah

[chilly ocean]

yea lol

[limpid edge]

i haven't formally learned any of this stuff from a textbook or class

and i know that fields and polynomials are super connected

like "algebraic numbers" and stuff

but i don't actually know where it comes from

does anyone know a good introduction to this stuff?

[maiden ocean]

any text on galois theory

[chilly ocean]

I don't really know what stuff you mean, but I like dummit foote and it might be what you are looking for

[limpid edge]

like algebraic numbers are (?) numbers which are roots to polynomials with coefficients in some specific field

probably a bit wrong in this definition

galois theory ok

that's what it's called

[limpid edge]

https://nms.kcl.ac.uk/ashwin.iyengar/df.pdf this?

[chilly ocean]

yeah, maybe you would enjoy this book called Introductory Algebraic Number Theory by Kenneth S. Williams.

[limpid edge]

ok this is good stuff

[chilly ocean]

But probably better to first get a bit comfy with rings/fields

[limpid edge]

from dummit

[chilly ocean]

Yeah, currently reading it as well.

[limpid edge]

you think it's better to go through this first before ^^?

[chilly ocean]

I'd recommend reading the ring theory part first.

yep lol

[limpid edge]

there's also this in between the ring section and fields section

[chilly ocean]

you can skip part 3

[limpid edge]

but like are modules relevant to the level of study I'm doing

ok

[chilly ocean]

Unless you never learned linear algebra before I guess

[limpid edge]

I uh, went through a third of axler

[maiden ocean]

yes but theyre not particularly relevant to introductory galois theory so

[old lava]

yes you should do rings before you do field or galois

and module theory knowledge is good to know

[maiden ocean]

oh by "they" i meant modules here

not rings

definitely know rings

[old lava]

but module theory isn't necessary

rings is like 110% necessary

lol

[limpid edge]

are modules just another algebraic structure

they sound like category theory

[chilly ocean]

there is 0 knowledge of chapter 3 assumed in chapter 4 of DF

[old lava]

the secret is, it's all category theory

[maiden ocean]

Modules are to rings as vector spaces are to fields

[chilly ocean]

only basic linear algebra I guess

[old lava]

untrue statement, you need to know some lin alg

ya

[limpid edge]

well what a vector space is and that's pretty much it I think

[chilly ocean]

that's enough to know

[old lava]

you basically only need to know

1. what a vector space is
2. what dimension is
3. what a basis is

and that's about it

[limpid edge]

yeah that's about as far I've gotten in axler so I'm fine i guess

does DF go through number fields at all?

or is that just galois theory

[golden pasture]

just galois iirc

any algebraic number theory text should intro number fields

i.e. marcus/neukirch

i suggest reading jacobson for intro to algebra it's nice

[limpid edge]

ok yeah I have seen the first chapter of marcus

and they spend the whole chapter talking about x^2 + y^2 = (x + yi)(x - yi)

[golden pasture]

tbf the first chapter is only 4 pages

it is more of like

motivational speech

[limpid edge]

ok i didn't actually read it so i'll take your word for it lol

I need to get a good background (DF) before I start that

[chilly ocean]

I'd recommend intro alg nt by Williams it's an easy read with a lot of problems with partial soluttions on the author's site

[maiden ocean]

i still havent read ireland rosen sorry zoph

maybe this summer

[chilly ocean]

but it will be a better read after you know rings a bit I suppose

[golden pasture]

but anyways should learn like groups/fields/galois grousp from some intro alg book anyways

[golden pasture]

just call it trivial and move on

[carmine fossil]

What if Ari is zoph's alt

[golden pasture]

thonk

wait zoph is chinese right

[limpid edge]

if you're going into algebra

[golden pasture]

his name sounds chinese

[limpid edge]

what other stuff do you learn after basic abstract alg, galois theory, and (intro) alg nt

[maiden ocean]

i mean there are many options

[maiden ocean]

ya

[golden pasture]

id assume most algebra courses would look something like

[maiden ocean]

you could do comm alg, you could do AG, you could AT

uh

[golden pasture]

groups
rings
modules
fields+galois stuff
maybe some rep, maybe some com alg, may some cute topics

[maiden ocean]

im sure other things are good

oh yes rep theory

lie stuff

[golden pasture]

jacobson does the maybe some cute topics route

[maiden ocean]

homological algebra

[golden pasture]

eh most really just do like finite group rep right

[limpid edge]

my plan is like

to get a foundation in abstract alg and alg nt

then go back to linear alg

[golden pasture]

uh

[limpid edge]

and then finally analysis

because i know zero analysis

[golden pasture]

do you know like

what is a vector space

[limpid edge]

yeah

[golden pasture]

ok nice

[limpid edge]

i've went through a bit of axler

[golden pasture]

ahhh

by lin alg do you mean modules over a pid then

[maiden ocean]

no wonder you're delaying learning lin alg then

[carmine fossil]

axler believes in choice but not determinants

smh

[golden pasture]

honestly rep theory is like

lin alg

but on steroids

and drugs

[maiden ocean]

steroids are drugs

redundant!

[limpid edge]

yea rn

[golden pasture]

i never thought lin alg could be so hard

[limpid edge]

i know that matrices

[golden pasture]

until i started trying to learn finite group rep theory

[limpid edge]

are for representing linear transformations

and that only

[golden pasture]

i made it as far as figuring like maschke theorem and lost interest

[chilly ocean]

matrices can represent other stuff

[limpid edge]

lmao the first text I find on finite group rep theory

F

[golden pasture]

HAHA

i use like

jacobson

for my finite group rep as well

lol

jacobson book 2 is hugeeee

[limpid edge]

i see

[limpid edge]

like what

ik ppl pack (random?) info into a matrix

like the jacobian or something

[maiden ocean]

volume 2 seems omegachad

[golden pasture]

lin alg is basically embedding groups into matrix groups

[golden pasture]

ikr?

literally in the preface jacobson is like

oh you have a few options

can spend like a semester on finite group rep

or one on universal algs

or some on hom alg or com alg

actually wild

[limpid edge]

have you guys seen

evan chen's napkin (compilation of notes taken by him)

https://venhance.github.io/napkin/Napkin.pdf

[golden pasture]

it feels more like

these are my notes

than ike

an actual reference

[limpid edge]

well yea

by no means is it written like a textbook

[golden pasture]

it's cute

but whenever people use it as a reference im like

wat

[chilly ocean]

i peeked at napkin but idk seemed too superficial

[golden pasture]

^^^

[limpid edge]

i use it as a reference on

[golden pasture]

it's more of a peak into like

what he is doing

[limpid edge]

what higher math looks like

oh yeah also

[golden pasture]

hmm

i guess thats possible

for me i literally

open some book i hope to read

figure what prereq im missing

fill in the prereq recursively

lol moth the only reason i started tom dieck was cuz the proof of something decently big in 3-manifolds required funny homotopy notions

[limpid edge]

i should learn multivar calc sometime soon

i kind of skipped it because it looked so boring

and now linalg is terrible

[golden pasture]

eh

[limpid edge]

im hoping on like

[golden pasture]

you may want to learn multivar before like analysis i think?

honestly it isnt too bad

just like work through some examples

[limpid edge]

i hate like

doing stuff without knowing what's going on under the hood

[maiden ocean]

napkin

[golden pasture]

cursed topology

[golden pasture]

i think it is good to get an intuition of like how the integrals and derivatives interact with each other before like

proper proper formalizing it

[maiden ocean]

what is this for ari

rep memes?

[golden pasture]

https://arxiv.org/pdf/math/9807001.pdf

moduli space memes

[maiden ocean]

i see

i dont actually i didnt click on the link

but i will pretend to see

[golden pasture]

[limpid edge]

quick question

how mathematically mature do you think this paper is

http://math.iupui.edu/~roederr/correlations_v3.pdf

[golden pasture]

honestly its actually quite a cool topic

wdym

"mathematically mature"

[limpid edge]

like

rn some of the stuff i have no idea what they mean

"complex analytic manifold"

when do you learn this stuff

complex analysis?

or like topology

[golden pasture]

you'll see this when you start studying like

smooth manifolds and riemann surface stuff

[limpid edge]

and when is that

in the standard analysis progression

[chilly ocean]

What the name of the group generated by 1^(1/2^∞), 1^(1/3^∞), 1^(1/5^∞), ... ?
(all possible roots of unity)

[golden pasture]

but yea looks very much like something in say smooth manifold territory

[golden pasture]

$\mu_{\infty}$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[limpid edge]

I see

[golden pasture]

is there really a standard analysis progression?

i feel analysis diverges into many paths quite early doesnt it?

[limpid edge]

I wouldn't know

[chilly ocean]

I remember that it have some name like "plucker group" or something like that

[limpid edge]

I just know there's like real and complex analysis

[chilly ocean]

But the symbol is good enough, thx

[limpid edge]

and apparently riemannian geometry now

[golden pasture]

all i rmb is seeing it in context of like Pontryagin dual stuff lul

[chilly ocean]

Anyone knows what the notation K^p means in "Let K a field of char p such that K is not K^p"? i don't think it's Fp cuz book uses different notation for that

[golden pasture]

very much in manifold-land

you'll see it in diff geo

[limpid edge]

I see

[golden pasture]

$$F^p=\left{x^p|x\in F\right}$$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[golden pasture]

isit in context of like

$F$ of char $p\neq0$ is perfect iff $F=F^p$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[limpid edge]

so there's

real + complex analysis, then diff. geo, and manifold theory is tied into diff.geo

[golden pasture]

youll see diff geo appearing when studying smooth manifolds defo

or just learn GR

[chilly ocean]

it's in context of x^p-a irreducible if a not in K^p

[limpid edge]

isn't it the other way around lol, diff geo then GR

[chilly ocean]

but why would you say K != K^p? K^p not a field in general i think,sounds kinda weird thing to compare

[golden pasture]

eh most GR courses introduce the diff geo you need, with varying degrees of "rigor"

[golden pasture]

rmb, char p

x->x^p is an endomorphism of fields

so as an example, $$\mathbb F_{p^n}=\mathbb F_{p^n}^p$$ but $$\mathbb F_p[x]\neq\left(\mathbb F_p[x]\right)^p$$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[golden pasture]

and indeed, $\mathbb F_{p^n}$ are perfect but we have stuff like $\mathbb F_p\left[x^{\frac1p}\right]$ as a inseparable extension of $\mathbb F_p[x]$

[cloud walrus]

煮豆燃豆萁，豆在釜中泣。本是同根生，相煎何太急？

[chilly ocean]

@golden pasture Wrote it like this, hope everyone will understand

[golden pasture]

the yellow color why T_T

[maiden ocean]

noooooo the Q

_<

[golden pasture]

but looks ok

[chilly ocean]

I always imagined rational numbers in my head as yellow dots, don't know why

[golden pasture]

oo

[limpid edge]

what are you talking abou

rational numbers are green

irrationals are red

[chilly ocean]

Irrationals doesn't have color because this is not an algebraic system

[limpid edge]

wait color actually has mathematical significance here lol

[chilly ocean]

Yes, but only for me :з

[past temple]

is it true that if i have an extension K/F

such that [K:F] = 2

then K = F(a), where

min(F,a) = x^2 - a

?

[golden pasture]

yes

[chilly ocean]

wait what but a isn't in F

so more like x^2 - a^2?

[old lava]

it's definitely true in any field of characteristic not 2

it may still be true in a field of characteristic 2, but it's just not as immediately obvious

but, it is obviously true in any field not of characteristic 2 by virtue of quadratic formula

[oblique river]

what do you mean by this?

there are degree-2 extensions of fields of characteristic 2 of the given form, and there are degree-2 extensions of fields of characteristic 2 which are not of that form

[old lava]

I don't know if it's true in a field of characteristic 2

[oblique river]

oh, it's definitely not true

take any finite field of characteristic 2

[old lava]

ya, I just didn't wanna say it's true or not

I'm pretty sure it's false

[oblique river]

it has a unique quadratic extensions and it's not of that form

[old lava]

oh ya, it's kinda obviously false in char 2 case

only polynomial of that form in F_2 is x^2 + 1, which is reducible

I just said "it may be true" because I didn't wanna make a claim either way

[oblique river]

an extension of the form F(sqrt(a)) for some a in F will never be galois over F if F has characteristic 2

[old lava]

but ya, @past temple, that's your answer, it's true if the field is not characteristic 2, it's false if the field is characteristic 2

to see why it's true in field not characteristic 2, just use quadratic formula

[cloud walrus]

squirtlespoof

[chilly ocean]

Doesn't it follow from the fact that w_i's are the only roots of that polynomial?

And automorphisms permute roots

[hidden haven]

oh god what have I done

I'm up now

[past temple]

got it

[limpid edge]

ok I will come back here

after I finish basic abstract algebra

aka sections 1,2, and 4 of DF

and then I will bother everyone with my alg nt questions

or maybe I don't have any (😮 ) because of my new AA foundation

[cloud walrus]

squirtlespoof

[hidden haven]

It is pretty tricky

Try to count the number of homomorphisms from F(alpha) to algebraic closure of F

(That fix F)

What can alpha map to?

Yeah, and is there anything else?

Why can't alpha map to something else?

Why?

that is not true

alpha is allowed to map to one then youll lose injectivity

yep

notice that if f(alpha) = 0 for a polynomial f with coefficients in F, then applying the homomorphism h on both sides you get f(h(alpha)) = 0

since h fixes all the coefficients

so this gives you a necessary condition, that alpha must map to a root of its minimal polynomial

but can it map to any root of its minimal polynomial?

yep

can you see why it can map to any other root?

yes

cool

so you get that F(alpha) is separable implies there are deg(min alpha) many homomorphisms from F(alpha) to F closure fixing F

The converse also turns out to be true

That is the trickier part, but see if you can think of any ways to prove it

no, algebraic closure of F

thats why you can say that all the other roots of min alpha exist in it

otherwise you may not have anything to map alpha to

[hidden haven]

hint: ||start by picking an element beta in F(alpha). Given that the number of homomorphisms is equal to the degree of the extension F(alpha)/F, you want to show that beta is separable over F. It suffices to show that F(beta)/F is a separable extension. Draw the tower diagram with homomorphisms to F closure and see if you can infer anything||

[rustic crown]

lol i slept by then

did you sleep again ._.

[hidden haven]

I'm up

[rustic crown]

yea so i was saying this...

you know these right

[hidden haven]

Yeah?

Like the indexing?

Or the problems