groups-rings-fields

[azure cairn]

you should stop reading this and pick up a copy of Linear Algebra Done Right.

[tulip otter]

the easiest way is to use lang's algebra ofc

[somber goblet]

so you can always refer back to Sn and Cn

[cursive spindle]

ANOTHER BAD BOOK?

[azure cairn]

why's ladr bad

😡

[glass sinew]

man he touches the snake lemma in his rings (+modules) chapter it's so funny

[cursive spindle]

Cuz linear algebra done rong

[tulip otter]

fun fact: linear algebra done right is actually done wrong

[novel star]

tbf lang does that too

[azure cairn]

ladw is pretty bad imo

[glass sinew]

lang is an incel idk

[novel star]

slander

[glass sinew]

me when i open the jar of shit and it has shit in it

literally the title of the book

[cursive spindle]

Lang is just pretentious ass

[tulip otter]

for linear algebra you should use roman

[azure cairn]

well lecture notes imo will usually not beat a textbook

ladr is a lot easier to read and a lot better written (again, in my experience)

[novel star]

lang is ragebait

[tulip otter]

this line is taken from that one picture huh

[azure cairn]

roman looks good but maybe as a Third course in linalg loooool

[glass sinew]

ladr is written with love, ladw with spite

[novel star]

reading should be an exercise in self temperance

as such

[tulip otter]

yea that was meant as a joke

[novel star]

you must read a book that makes you hate life

[azure cairn]

i'd read it if i had time

[tulip otter]

more seriously maybe FIS or greub if you want something harder

[azure cairn]

nope, ladr

if youve already learned about determinants and rref ladr is the Next logical book, fis as well but i think parts of fis are confusing

[cursive spindle]

-10000 social credits

[azure cairn]

[cursive spindle]

Oh that is okay I guess

As a second book

[azure cairn]

ladr in his preface says it's intended as a second course

but i read it as a first course and supplemented with another book

[tulip otter]

its not ok to do a second course in linear algebra

linear algebra is meant to be at most one course

[azure cairn]

[cursive spindle]

Meanwhile here we do 3 courses of linear algebra and we don't even cover quotient spaces

[glass sinew]

linear algebra must be learned linearly, obviously. never double back on it

[azure cairn]

yep abstract algebra is the second course and functional analysis is theThirdCourse

[novel star]

the second course is just homological algebra

[tulip otter]

my condolences

[azure cairn]

[somber goblet]

i like how herstein gives you determinants after putting you through just about every possible theorem you can prove without matrices

[cursive spindle]

It's hilarious how bad are the linear algebra courses

[tulip otter]

just learn linear algebra from an abstract algebra textbook

[azure cairn]

artin ,

[somber goblet]

he assumes everything is finite dimensional too so idk what the point is

[cursive spindle]

A cohomological approach to linear algebra

[novel star]

van der waerden

[glass sinew]

only understood vector spaces after reading about field extensions so this is true

[novel star]

you are strange

[tulip otter]

work with vector spaces as modules over fields

[cursive spindle]

Working with modules with poor linear algebra background is a great way to burnout

[novel star]

how much lin alg do you really need tho

[glass sinew]

lin alg background is propaganda just go straight into aa twin ❤️‍🩹

[somber goblet]

real

[tulip otter]

meanwhile I am starting commutative algebra without actually doing linear algebra properly

[somber goblet]

FR

[novel star]

unironically did this

[azure cairn]

what is commutative algebra btw

[cursive spindle]

It's okay I've been there and it was a bad idea

[novel star]

commutative rings

[tulip otter]

algebra which is commutative

[novel star]

pre-geometry

[azure cairn]

ok idk ring theory so i guess idk if ill like it yet!

[somber goblet]

whats the point in learning 10 axioms for a vector space when you can just make them an abelian group with extra structure

[azure cairn]

i cant list the vsp axioms off the top of my head

i dont think

[cursive spindle]

Ring theory fight of people complaining about whether to add 1 or not

[tulip otter]

thats just stating the axioms in a nicer way lol

[somber goblet]

yeah

[tulip otter]

rng moment

[novel star]

genuinely anyone who doesnt assume unital is a masochsit

[cursive spindle]

Yes so basically the analysis department

[glass sinew]

are textbook writers allergic to the word rng or something

[somber goblet]

idk i think its a lot nicer when you can relate subspaces to subgroups, quotient spaces to quotient groups, linear maps to homomorphism, null spaces to kernel

[tulip otter]

the same department that uses N={1,2,..} and N_0 for {0,1,..}

[azure cairn]

it was very fun when i made those connections 🎉

[somber goblet]

why learn specialized abelian groups before you learn about groups in general

[novel star]

why learn about groups when you can learn about inverse semigroups

[cursive spindle]

The road from "yeah bro rank nullity theorem" to "everything is just first isomorphism theorem wow"

[azure cairn]

instead of N_0 i use boldface W

[tulip otter]

https://media.discordapp.net/attachments/917866853198594131/1471604215905259761/hmmmm.webp?ex=69ac8aba&is=69ab393a&hm=e2edca5fc18a493e47ca89d532c83cd44f0533323d430ef3c4ea2f4359a762ac&=&format=webp&width=72&height=72

[azure cairn]

(ragebait)

[glass sinew]

lie theorists when the truth theorists walk in

[cursive spindle]

Point and laugh

[azure cairn]

hello

[novel star]

truth theorists when tarski theorists walk in

[somber goblet]

true (i think anything weaker than a group is too general for most contexts)

[azure cairn]

who started the self reacting trend

[novel star]

manifolds

[glass sinew]

idk

[cursive spindle]

nGroupoid

[somber goblet]

?

[tulip otter]

:chad:

[novel star]

manifolds are topological spaces compatible with a suitable pseudogroup of transformations

i.e.

[azure cairn]

[novel star]

an inverse semigroup

[tulip otter]

sounds like a lie ngl

[somber goblet]

pseudogroup?

[cursive spindle]

I wish it was

[somber goblet]

algebraists have gone too far

[tulip otter]

too good to be true

[novel star]

*category theorists

[glass sinew]

how much homological algebra did you have to do to reach this point

[tulip otter]

enough to reach this point

[glass sinew]

*asylum residents

[novel star]

mac lane does this notion as an example in his book on sheave

[cursive spindle]

Yes.

[azure cairn]

is this algebra

i remembered this when seeing "cohomology" multiple times

[glass sinew]

LEGENDARY introduction

[novel star]

least funny mathematician

[azure cairn]

how did ts get accepted with these grammatical errors

[glass sinew]

some people use question marks like commas

[novel star]

the aura was blinding

[azure cairn]

https://cdn.discordapp.com/attachments/1282082024106229843/1339706119114723500/togif.gif

[cursive spindle]

Craziest thing I "learned" so far is something called automorphic cohomology

[somber goblet]

fire intro to some lecture slides i saw

[glass sinew]

field theory is an agricultural studies course

[tulip otter]

is there cohomological automorphy

[cursive spindle]

tfw Galois theory has a great lore

[somber goblet]

facts

[azure cairn]

why is algebraic geometry have a channel if it's only 7 pages worth of theory

[novel star]

hermann weyl never read yaoi

[azure cairn]

LMAO?

[glass sinew]

1pt font megamarathon in those 7 pages btw

[azure cairn]

lemme skim it

[novel star]

lang speaks of this

[tulip otter]

I should sleep

[tulip otter]

you probably should too

[somber goblet]

algebra lore

[azure cairn]

[cursive spindle]

Algebraic geometers hype the field a lot but when they get asked what's all the fuss about they draw schizophrenic diagrams

[glass sinew]

half the reason i care about galois theory is because i like galois himself

[cursive spindle]

(I am one of them)

[novel star]

https://tenor.com/view/truth-nuke-truth-true-soyjak-wojak-gif-14828627546768133163

[tulip otter]

I would like to see you saying that in nG's face

[somber goblet]

real

[azure cairn]

so basically his killer robbed us of multiple years of mathematical progression

[tulip otter]

there will probably be something like

https://media.discordapp.net/attachments/917866853198594131/1471604215905259761/hmmmm.webp?ex=69ac8aba&is=69ab393a&hm=e2edca5fc18a493e47ca89d532c83cd44f0533323d430ef3c4ea2f4359a762ac&=&format=webp&width=72&height=72

before admitting

I dont think he will agree tho, he will find a reason which no body in the chat understands except him

[cursive spindle]

tfw Galois was smart enough to cook up some great math but not as strong to win a duel 🥀

[glass sinew]

"secret opencry" closedcry

[novel star]

the bullet induced an automorphism called death

that fixed the body

[glass sinew]

galois should have read his findings to his duellist so then he would have been confused, giving galois an opening to shoot

[azure cairn]

so if the field extension is the body what is the field

[cursive spindle]

See that's why I go to gym cuz if I ever get the opportunity to be in a duel I'd win

[novel star]

the field extension is his general being

the body is the field

[glass sinew]

BANG💥BANG💥BANG💥BANG💥BANG💥BANG💥BANG💥BANG💥 *reload* BANG💥BANG💥BANG💥BANG💥BANG💥

[azure cairn]

mb

[novel star]

how many fields are there?

[cursive spindle]

🌾

[cursive spindle]

As many as you want

[somber goblet]

the bullet was transcendental over the ground field

leading to a countable-degree extension into the afterlife

rip

[tulip otter]

now add the word medalists, the question becomes how many fields medalists are there? Suddenly the answer is : very little

[crystal vale]

How does it imply F_p(\alpha) contains F_p^n?

[cursive spindle]

Just buy one at your local store

[novel star]

just glancing at it, but maybe something with the frobenius endomorphism

[somber goblet]

i think there's something to do with fermat's little theorem? i dont remember

[velvet hull]

(alpha + a) - alpha = a

[crystal vale]

But I have to show alpha +a in F_p(\alpha)

I think F_p(\alpha) already normal extension

So all roots of given polynomial will be there

And i assumed given polynomial irreducible

Right?

[glass sinew]

trying to make a diagram of the 3rd iso thm and idrk how well this gets the point across

[crystal vale]

Then I don't get why I need a second hint? Because from first hint I can conclude n divides p^n

[novel star]

doesnt group have a zero object

[velvet hull]

Yeah but for groups it’s usually denoted 1 instead of 0

[glass sinew]

idk, i've always written 1 := {1}

[novel star]

this vexes me

[velvet hull]

And it’s technically not a zero object

Since Grp is not an abelian category

It’s just the initial and terminal object

[cursive spindle]

Idk how useful is writing out the diagram for 3rd iso

It looks cool I guess

[glass sinew]

just trying to motivate it for myself since the typical statement of it was really dry on my first reading

[cursive spindle]

The way I like to think about it is naively dividing is actually a correct isomorphism

[velvet hull]

Quotient groups are just fractions

[cursive spindle]

yeah

[glass sinew]

(A/B) \times B \cong A 😌

[azure cairn]

truke

[crystal vale]

Oh no, I need more information

[novel star]

when A is a projective module yes

[glass sinew]

sick

[azure cairn]

commutators r so hot

[crystal vale]

Ah, i don't see how to show the second hint?

[somber goblet]

i should try to prove this

[novel star]

not true in general unfortunately

[somber goblet]

oh wait

yeah

[novel star]

only for exact sequences that split

[somber goblet]

$(A/B) \ltimes B \cong A$ ?

[cloud walrus]

lexi

[somber goblet]

imagine the times symbol is pointed the right way idr

[novel star]

Z/4Z is a counterexample

with Z/2Z

we must be able to split

for example

when calculating the automorphism group of the heisenberg group over a finite field

you can get a split sequence from the inner automorphism group

[crystal vale]

I am dumb

[somber goblet]

would the corresponding exact sequence look like this

[novel star]

x->p*x gives the desired automorphism

[novel star]

yes

[somber goblet]

awesome

[novel star]

in abelian categories replace semidirect with direct sum

[crystal vale]

Sorry, is it x -> px?

[somber goblet]

i see

[novel star]

yes

because p^n=1 mod p^n - 1

[crystal vale]

I see

Thank you

[azure cairn]

are there any other definitions of group operations on G/N that would be useful besides ones that make the canonical map a surjective homomorphism with kernel N

[somber goblet]

the quotient group is also nice because the operation is the 'same' as G, it's just being done between cosets of N, where the normality of N ensures this operation is well-defined and results in a group

[azure cairn]

yep

[somber goblet]

G/N is a set so you can put whatever operation you want on it but... why?

[azure cairn]

just wondering if theres any other definition for a group operation on G/N besides the most common one

[azure cairn]

well that's what i'm asking, is there any other common op

[somber goblet]

the structure we give G/N has a bit of category theory behind it as well, being the quotient in the category of groups

[somber goblet]

im sure there's something but especially considering the definition of a normal subgroup is introduced precisely so that quotient groups are well-defined under their canonical operation

i can't see anything else that would be particularly useful

[novel star]

not really

quotient groups are useful precisely because they satisfy a suitable universal property

[prisma ibex]

[prisma ibex]

no what are you smoking

[prisma ibex]

in general you can consider extensions E of A by B, that is short exact sequences of groups 0->B->E->A->0. Equivalence classes of such extensions are classified by the group Ext^1(A,B), and the semidirect product E=B\rtimes A=A\ltimes B is the zero element of this Ext group

not every extension is a semidirect product and not every semidirect product is a product

[fleet cairn]

i kinda dont wanna read another book for LA, is the linear algebra in aluffi's book enough as a "first undergrad level math course of LA"
(id assume this channel is best place to ask)

[azure cairn]

what is this even meant to mean LOL

for more context its part of the quotient groups section and we have defined coset multiplication to be like (aN)(bN) = (ab)N

ty

too tired to understand this rn tbh maybe best if i sleep

finally got to this though :3

[ashen heron]

Probably just showing that the cosets partition G and maybe it wants you to think about how abN isnt just aN or bN, what would happen if it intersected either? cosets that intersect are identical, what does that say

[azure cairn]

i noticed the lines in \bar{a}, \bar{b}, \bar{ab} very recently

but why arent they overlapping in the same space

i know this already

all cosets are disjoint

yes i know

im asking why the coset for ab isnt overlapping with the coset for a and b

wait

ah never mind

the composition makes them equal**

yeah i see

ok yea i get it, i understood when i saw the lines lop

ty both

[glass sinew]

never ask artin to draw anything ever again

[runic lotus]

Does anyone have a playlist or something they recommend for a first course on group theory? The book I’m following is called abstract algebra

[novel star]

i incorrectly applied the abelian category statement of splitting to groups

[runic lotus]

Thanks so much!!

[ashen heron]

https://youtube.com/playlist?list=PL8yHsr3EFj51pjBvvCPipgAT3SYpIiIsJ&si=uQBJ6ie3yZExbFmZ

[prisma ibex]

this isn't true in Abelian categories either

[runic lotus]

Amazinggg

[ashen heron]

lots of books are titled abstract algebra better tell the authors name

[novel star]

doesnt the splitting lemma give an exact sequence backwards when any of the morphisms split

[prisma ibex]

not the way you drew it no not at all

but yes you get a splitting

[novel star]

i thought they just redrew the arrows backward

[prisma ibex]

yes that's what people always do

like this

[novel star]

but not parallel for whatever reason

[prisma ibex]

$\begin{tikzcd} 0 \arrow[r] & A \arrow[r] & E \arrow[r] & B \arrow[r] \arrow[l,shift right] & 0 \end{tikzcd}$

[cloud walrus]

nGroupoid

[novel star]

yes

[frigid epoch]

fiber bundle with a global section 🤯

[prisma ibex]

me when I mix analogies out of confusion

[frigid epoch]

ikkkkk lol

wasn't meant to be serious

[novel star]

is there such a thing as a "projective group"?

[prisma ibex]

what do you mean by projective here

[novel star]

like projective module

every short exact sequence with the group in the middle splits

[prisma ibex]

sure this definition makes sense

Grp isn't Abelian but you don't quite need all this structure to talk about projective objects

[novel star]

Grp feels almost abelian for some reason

[prisma ibex]

yeah they very nearly are

[knotty badger]

Grp is an exact category

and an abelian category is precisely an exact additive category

[prisma ibex]

it's more than just exact

it's regular exact protomodular

this is enough to get most of the usual homological algebra proofs to make sense

[novel star]

thats quite nice

[prisma ibex]

there are still some painful issues though

[novel star]

i wonder how much stuff one can squeeze out

is there an analog of freyd-mitchell or something for such nice categories

[prisma ibex]

there is a book by Borceux and Bourn (Mal'cev, protomodular, homological and semi-abelian categories) that squeezes everything out that you can

[prisma ibex]

no not really

[maiden crater]

I don't get why N(r) can't just be N(d) here, why must it be 0

[prisma ibex]

one really painful issue with groups compared to Abelian groups is that non-Abelian group cohomology is a complete mess

like it's fine for H^1 and barely okay for H^2

[limber tapir]

The definition of a euclidean valuation is that N(r) < N(d) or r = 0

[maiden crater]

oh right

so N(r) is forced to be 0

[limber tapir]

N

No

[novel star]

i lowk dont know anything about group cohomology

is it at all related to galois cohomology

[prisma ibex]

yes Galois cohomology is a special case of this in a certain sense

[limber tapir]

r must be zero by minimality of N(d)

[wise pasture]

:trollface:

[prisma ibex]

Galois cohomology is group cohomology for absolute Galois groups

[limber tapir]

Not just N(r), r itself must be zero

[prisma ibex]

in general it's a bit subtle to get continuous group cohomology to work nicely but for profinite groups it's relatively straightforward

in general topological spaces are just the wrong approach for trying to define algebraic constructions like this, thankfully we know better these days

either way it's nice to know some low degree group (co)homology these things tend to come up a lot

like Ext^1 classifying extensions is a hugely useful construction

also H^2(G,A) classifies central extensions of G by A

[frigid epoch]

dude profinite spaces covering regular old spaces is my favourite thing you've shown me

[prisma ibex]

[novel star]

i suppose this is the universe telling me i should learn group cohomology/whatever context it appears in

[prisma ibex]

especially if you know what the words Galois cohomology mean

if you like number theory then one huge payoff is beng able to learn class field theory

those proofs are mostly cohomological in nature

[novel star]

i remember hearing that kummer theory is motivated by number theory too

i should probably learn some basic number theory lol

[prisma ibex]

yeah Kummer theory is another great example of how Galois cohomology controls things in number theory

another fun example is how H^2(Gal(\bar{F}/F),Z/2Z) classifies quaternion algebras over F

[novel star]

number theory sounds like the trenches

[prisma ibex]

Br(F)=H^2(Gal(\bar{F}/F),G_m) classifies central simple algebras over F more generally

it's really nice once you learn enough machinery

[novel star]

i hope that i can learn the machinery before the brain melts

[prisma ibex]

especially once you learn enough geometric techniques and analogies

loads of tight analogies between number fields and 3-manifolds once you have the cohomological machinery

[novel star]

why does cohomology appear everywhere

[prisma ibex]

same reason why linear algebra appears everywhere

[novel star]

ah

[prisma ibex]

to understand very nonlinear geometric things we often have to linearize them and study invariants which capture the geometry without being too hard to compute

[novel star]

well ig its time to read a course in arithmetic

[frigid epoch]

there are lots of local-global phenomena in maths and cohomology "measures" how nice local things may fail to fit together globally

that's the way I see it, roughly

[prisma ibex]

yes that's another nice justification for cohomological and sheaf theoretic methods

like part of why these things appear in number theory so often is precisely because of these local to global problems in arithmetic

[frigid epoch]

have you seen de rham or singular cohomology @novel star?

[novel star]

yes

[frigid epoch]

okay good

both or just one or...?

[novel star]

both

[prisma ibex]

trying to solve something all at once about global arithmetic over Q or Z is often too hard, usually you need to study what happens locally over Q_p and R first, and then try to understand how the local informaton glues back to global information

[knotty badger]

someday i want to understand this

[prisma ibex]

the local to global step is where all the actual deep number theory is contained, in the form of reciprocity laws

[frigid epoch]

I mean

hairy ball theorem

[supple ice]

have you seen habiro cohomology? lmao

[prisma ibex]

most unhinged example to name

[frigid epoch]

lol

[novel star]

that kinda reminds me of a problem i saw in lang

and the problem was to prove this result in the rationals

[prisma ibex]

really funny how you can quickly get used to insane stuff like continuous THH over KU around this Habiro story, but the fundamental challenge of just like, solving q-difference equations, never really goes away

[novel star]

but it was known already by the 30s that it holds for finite fields

this one

[crystal vale]

Lang

[novel star]

yes

[maiden crater]

I'm a bit unsure of how to do this

every element of D^{-1}R can be generated by the genertor of R and the some, is it not

[rocky cloak]

The elements of D^-1R can be seen as fractions a/b. For a given ideal of D^-1R what can you say about the set of numerators of the elements in that ideal

[maiden crater]

they come from all of R?

[rocky cloak]

It's a subset of R yes

Anything else we can say?

[maiden crater]

they all are generated by one element

[rocky cloak]

Well, what I was looking for was that it was an ideal

[maiden crater]

sure, it is

[rocky cloak]

But that means it's generated by a single element r yes.

So then the question becomes what's the ideal in D^-1R generated by r?

[maiden crater]

I'm unsure

[rocky cloak]

Well in general if you have a ring S and an element r in S, what is the ideal generated by r?

[maiden crater]

S?

[rocky cloak]

So you're saying the ideal generated by 2 in Z is all of Z for example?

[maiden crater]

oops no

I wanted to say (r) but that seemed stupid

[rocky cloak]

Well that's the notation for it

But which elements does it contain?

[maiden crater]

${ \sum s_irh_i \mid s_i,h_i \in R}$

[cloud walrus]

wai

[rocky cloak]

And if we're in a commutative ring we might simplify this a bit

[maiden crater]

${\sum s_i r; s_i \in R}$

[rocky cloak]

Can even simplify it a little more

[cloud walrus]

wai

[maiden crater]

huh?

[rocky cloak]

Multiplication is distributive

[maiden crater]

so ${ar \mid a \in R}$

[cloud walrus]

wai

[rocky cloak]

So then the ideal generated by r in D^-R would be things off the form ar/b

How would this ideal be related to the original ideal?

[maiden crater]

would be a principal ideal of it

oh 🤦 I thought I had to show D^{-1} R is principal ( that is generated by one elment)

which made no sense

TYSM!

this shows it's a PID yeah

so sorry 😭

[crystal vale]

if F is a field contained in the ring of n \times n matrices over Q then i don't get how F can be vector space over Q?

[delicate orchid]

This field is characteristic 0. The result follows

[crystal vale]

so it contains isomorphic copy of Q

so scalar multiplication defined as qA as qI \times A

[quiet pelican]

Not necessarily
Only if I is the multiplicative identity of the field

[regal zodiac]

[quiet pelican]

For example you can have Q embed in 2x2 matrices via (q, 0; 0, 0)
Then while multiplication by qI and q(1, 0; 0, 0) look the same, technically the latter is how multiplication by q “should” be defined

[rocky cloak]

I guess contains a field should mean that it has the field as a subring. Otherwise I guess there's no point for it to have the same multiplication either

[crystal vale]

then how do I show [F: Q] <= n, any hint?

maybe characteristic polynomial helps here

so degree of char polynomial is n

so degree of minimal polynomial of A <= n

right?

[crystal vale]

For part a, I am trying to use the given hypothesis that K/F is Galois extension that's means the corresponding subgroup H in Gal(closure L/ F) is normal .

And if G is the Galois group of closure L/F, let's say q is the distinct prime divisor of |G|.
Now I am trying to make subgroup H' such that H \subset H' and [G:H'] has prime divisor q.

[rocky cloak]

Yeah, so with the primitive element theorem you can show F/Q is a simple extension and then this argument works.

[crystal vale]

What is the problem if I don't use the primitive element theorem?

[rocky cloak]

Well, what even is the argument without F being simple?

Like what is A?

[crystal vale]

So we know F is finite dimensional over Q

I see your point

My bad

[crystal vale]

Any hint ?

[rocky cloak]

Something useful is that L'/K will also be Galois where L' is a conjugate of L inside the Galois closure

[crystal vale]

Conjugate of L means?

[rocky cloak]

g(L) where g is a Galois automorphism

[copper kestrel]

did i overcomplicate this problem? should i have just found a counter example

[next obsidian]

Your proof’s logic isn’t really sound

[copper kestrel]

:(

rats

[next obsidian]

Your proof technically goes backwards, you need to start from the last line and then conclude that k(m/n) + l(p/q) = 0, you’ve essentially worked the other way around. Every step is reversible so it’s okay, but you either need to state that, or even more preferable once you’ve brought m,n,p,q into existence you start off by saying you pick l and k such that this holds and then derive that the thin is 0

[copper kestrel]

oh interesting

[next obsidian]

Like just read the line Then:
k(m/n) + l(p/q) = 0.

What? k and l don’t exist yet, what does this mean?

You need to either bring them into existence first or say that you’re searching for such a k and l

[copper kestrel]

i thought i did bring them into existenexe

in the previous sentence

[next obsidian]

Okay yes sorry but

If you say let k,l then this is a universal quantification, a for all

[copper kestrel]

true

[next obsidian]

So what you’re stating there is that this holds for arbitrary k and l

[copper kestrel]

that makes sense

so should i start off with k = -l then?

then reverse the steps?

[crystal vale]

so that automorphism fixed K?

[copper kestrel]

i probably should just do a counterexample instead of that proof

[rocky cloak]

Fixes F, but it will satisfy g(K) = K

(since K is normal)

[crystal vale]

so i have to show L' is splitting field of some polynomial over K

so L is already splitting field of f over K

[next obsidian]

This won’t work

[copper kestrel]

oki

[next obsidian]

I mean it literally won’t make the equation true

[copper kestrel]

yea

[crystal vale]

then for automorphism g L' will be splitting field of g(f) over K

is it correct jagr?

[next obsidian]

I don’t think a counterexample proof will work though

At least if you try to do it in the way I think

The quantification of property 2 you tried to disprove doesn’t allow it

[copper kestrel]

[next obsidian]

Also to disprove this, after you contradict property 2 you still have to argue that Q isn’t a free group on a single generator

Because your disproof will require the existence of more than a single generator

[copper kestrel]

i shall come back to this problem later then

[crystal vale]

I don't get how this helps me here

[ripe harbor]

If $f: R \to S$ is a surjective ring homomorphism, then $$I \mapsto f^{-1}(I)$$ defines a bijection $${\text{ideals of $S$}} \to {\text{ideals $J$ of $R$ with $\ker(f) \subseteq J$}}.$$ Now apparently from this follows that with the canonical projection $p_I : R \to R/I$ we get $${\text{ideals $J$ of $R$ with $I \subseteq J$}} \to {\text{ideals of $R/I$}}.$$ But why? $p_I$ is the surjective homomorphism, so shouldnt it be in the other direction?

[cloud walrus]

ILikeMathematics

[crystal vale]

Sorry i don't get your question

[rocky cloak]

Next you may want to compare Gal(L/K) to Gal(L'/K)

[rocky cloak]

Bijections go in both directions

[crystal vale]

I think both have the same cardinality

[rocky cloak]

That's right. So they are both p-groups.

[crystal vale]

Yes

[rocky cloak]

Lastly you might want to think about the map
Gal(closure L / K) to product of all the Gal(L' / K)

[crystal vale]

I think there is a relation between Gal(LL'/L) and and Gal(L'/K)

[crystal vale]

I see

[crystal vale]

So it is injective mapping

Because one product is Gal(L/K)

Wait

[crystal vale]

I got the map but I am trying to find the kernel

So g is in the kernel iff g is identity on each L'

And where I will use K/F is Galois extension?

[ripe harbor]

Oh, I was taking the actual map instead of f for the kernel accidentally. Then yes it works out, thanks

[rocky cloak]

I mean we've already used that for L'/K being galois

[crystal vale]

Oh yes K is normal

[crystal vale]

So what I think is, closure L is a closure of L over F, so if L = F(a_1,..,a_n) then closure L' = F(b1,...,b_m) b_i's are the conjugates of a_j's.

So for each b_i, there exists automorphism such that a_j -> b_i, for appropriate j.

Hence if g is identity on each L' then it implies g is identity on closure L, because if g is identity on each b_i then g is identity on closure L.

Is it correct?

[somber goblet]

i see

[crystal vale]

b), if F = Q and K = Q(√3), L = K and closure L = Q(√3, i), so K/F has degree 2, L/K has degree 1 but closure L /F has degree 6.

Is it correct?

Thank you jagr

[karmic dove]

I know an alternate method: this is not a simple extension , but a finite extension, so we will get infinitely many intermediate fields. So I know the existence of infinitely many intermediate fields. But the question demands for an explicit representation of the intermediate subfields. Hint?

[next obsidian]

Well you could pore into the proof of why a field extension with finitely many intermediary subfields is simple and from there realize an algorithm to do such a thing

I think doing this will give you enough information to figure this one out

[somber goblet]

can you characterize a ring via its ideals

[wraith cargo]

Wdym?

[somber goblet]

it seems like you can tell a lot about a ring via which sets are ideals and which aren’t

you have a certain lattice of ideals as well

[supple ice]

just not always enough to stop different rings from pretending to be the same

[somber goblet]

ah

[wraith cargo]

Tho I mean also

You always have R as a trivial ideal

So I'm not sure what kind of answer you want a priori

[chilly radish]

I think they mean given the lattice of ideals can you reconstruct the ring

This is already a non-starter because of fields

[wraith cargo]

Yeah I mean

Take M_n(F) as well lol

So it's not even a thing about fields

[chilly radish]

Well sure any simple ring

But fields are an example where even if you look at left and right ideals you don't get new information

Unlike M_n(F)

[crystal vale]

Since the Galois group of cyclotomic field over Q is abelian group therefore every subextension should be normal but this extension is not normal therefore it can't be subfield of cyclotomic field.

Right?

[quiet pelican]

That works

[crystal vale]

Any hint?

[uneven sun]

You know what the Galois group.

$$\text{Tr}(\xi_{n})=\sum_{g\in G} g\xi_{n}$$

[cloud walrus]

Calisto

[uneven sun]

if n=1 then $\xi_{1}=1$

[cloud walrus]

Calisto

[uneven sun]

The galois group is trivial and so you have it by default

[azure cairn]

is this because the shift map is injective but not surjective

like you can just choose $(l \circ s)(n) = n$ by

$$l(k) = \begin{cases} k - 1 & k \ge 2 \\ \text{anything} & k = 1 \end{cases}$$

because of the fact that 1 has no preimage in s

[cloud walrus]

Altanis

[uneven sun]

The only possible surjection that would make this works leaves the codomain

f(n)=n-1 would be the natural choice

[cursive spindle]

What book is this?

[azure cairn]

artin still

[cursive spindle]

...

[azure cairn]

why?

[cursive spindle]

It's so over

[azure cairn]

idgi

[cursive spindle]

The way N is defined

[uneven sun]

actually yeah this is sufficient. Every surjection has a right inverse

I believe that is a necessary and sufficient condition

[cursive spindle]

but michael artin is based so i'll let it slide

[azure cairn]

ah

nope that's the correct definition

[cursive spindle]

hmm

[azure cairn]

in any case it's just defined like that here

i think

i dont think he uses \bN elsewhere unless he specifically makes it clears it's {1, 2, ...}

[azure cairn]

yeah i had an answer to the question, was just making sure it was due to its lack of surjectivity that it had no right inverse

[cursive spindle]

yeah anyways what you did is good enough i think

[azure cairn]

cuz its similar to the shift operator on ell^2 or even just matrices/operators that arent injective/dont have full col rank or arent surjective/dont have full rank rank

is there some "algebraic" way im meant to do this or am i supposed to just shove this into the euclidean algorithm

[cursive spindle]

euclidean algorithm

but there is an algebraic way to make sense of GCDs

[uneven sun]

This is a very important thing to be comfy with

[cursive spindle]

If you look at Z as a ring then you can take the ideal (a,b) and if you find an ideal generated by one element say d such that (a,b) = (d) then d is your gcd

[somber goblet]

thats interesting

[cursive spindle]

this (a,b) is precisely the linear combination ra+sb

[somber goblet]

its cool that you can phrase so much elementary number theory in terms of ideals

[uneven sun]

Number theory…..

[cursive spindle]

Yeah number theory shines when you use tools like this

Otherwise it's just blindly walking at midnight

[somber goblet]

it kinda feels like ideals are the 'right' way to look at these things

instead of the numbers they can represent

[azure cairn]

is it..?

[cursive spindle]

yes it is

[azure cairn]

🤢

[uneven sun]

It comes up a lot in Galois theory proofs for example

[somber goblet]

what does

[cursive spindle]

Yes they are

[azure cairn]

i just know the number d s.t. Zd = Za + Zb is the gcd of a and b

and m = lcm(a, b) Zm = Za AND Zb

[uneven sun]

Like the properties of a field extension depending on whether or not a number divides another number or like even some definitions use gcd

The Euler phi function ends up relating to cylcotomic polys

[somber goblet]

you can rephrase factorization into subideals

[azure cairn]

i dont think this book even taught the euclidean algoirhtm yet actually gimme a sec

[cursive spindle]

Historically, ideals were invented to solve a problem that occured when you looked at arbitrary ring of integers

[azure cairn]

ok it was mentioned but wasnt like given much time**

[somber goblet]

you can draw a subset lattice and it will be the same as the lattice of factors which is really cool

[cursive spindle]

The issue is that UFD is not always guaranteed so the way to go about this is to introduce ideals

[azure cairn]

i have no idea what any of this is So i feel like if the euclidean algorithm is important fo rfuture topics it'll just get re-emphasized

[cursive spindle]

maybe computational wise it's kinda pointless sure

but know how to do an example

[azure cairn]

isnt it just dividing the larger number by the smaller number a bunch of times till u get no remainder

[cursive spindle]

pretty much yeah

[azure cairn]

321 / 123 = 2 r 75
123/75 = 1 r 48
75/48 = ...
...

then you choose the remainder before 0 and thats ur gcd i think

[cursive spindle]

yeah then you terminate and do backwards substitution to get your linear combination

[azure cairn]

cuz gcd(321, 123) = gcd(75, 48) = ... = gcd(some number, 1)?

or not 1 uhh

just some simple gcd expression

[somber goblet]

is there an idea of linear independence in modules

[azure cairn]

oh idk how to do that

i should prob check that out lol

ah ok i see

[somber goblet]

is there anything you learn in elementary number theory that you don't in group/ring theory

[cursive spindle]

some niche things

[somber goblet]

also i realized you can phrase $\operatorname{lcm}(a, b)$ as $(a) \cap (b)$

[cloud walrus]

lexi

[cursive spindle]

for the most part you can learn about 80% and that's more than enough

[somber goblet]

like what?

[cursive spindle]

continued fractions

[somber goblet]

oh

i never understood the reasoning behind those

[cursive spindle]

Also

Quadratic reciprocity

[somber goblet]

ah

[cursive spindle]

This one is niche in the sense that you don't really see it in group/ring books

But it's very important

[somber goblet]

yeah it sounds important

[velvet hull]

Yes, it is the exact same definition

It just no longer enjoys some of the nice properties you see in linear algebra

[somber goblet]

can you talk about a module having a "basis"

[velvet hull]

No, not in general

Because maximally linearly independent and minimally spanning are no longer equivalent conditions

[azure cairn]

i think theres a term called free module for modules that have bases?

idk if that's relevant tho

[velvet hull]

It works in that case, yes

[somber goblet]

oh i see

[cursive spindle]

I'd say learning it from an elementary number theory book is not the best way to get a good insight on it

it'll just feel like random result

[velvet hull]

e.g. {(0,2), (2,0)} in Z^2 is a maximally Z-linearly independent set but does not span

[somber goblet]

im trying to understand fundamental thm on finite abelian groups

i want to know the proof because its quite important

[cursive spindle]

You're learning Galois theory correct?

[somber goblet]

yes

[cursive spindle]

I'd postpone quadratic reciprocity

till you are comfy with it

The natural way to see it is through a bit of galois theory

[somber goblet]

ik anything in Q[x] with roots in Q has a corresponding polynomial in Z[x] with roots in Q

[cursive spindle]

hmm I can talk about it if you're curious

[somber goblet]

ok...

[cursive spindle]

This will also be your first example of a representation

[somber goblet]

ok i see

[cursive spindle]

Okay consider Q(zeta_n)/Q

[somber goblet]

im up to constructible numbers with galois theory

[somber goblet]

cyclotomic extension?

[cursive spindle]

yup

What's the Galois group isomorphic to?

[somber goblet]

i know this

its $(\mathbb Z / n \mathbb Z)^\times$

[cloud walrus]

lexi

[cursive spindle]

right

now we are interested in group homomorphisms from this group to C^x

[somber goblet]

complex numbers?

[azure cairn]

3 = 321(-18) + 123(47) :D

[cursive spindle]

yes

[somber goblet]

you can make that isomorphic to $\mathbb T \times \mathbb R$ cant you

[cloud walrus]

lexi

[somber goblet]

using exponential map

[cursive spindle]

?

[somber goblet]

$\mathbb C^\times \cong \mathbb T \times \mathbb R$

[cloud walrus]

lexi

[somber goblet]

where T is circle group

[cursive spindle]

sure but this won't really help here

[somber goblet]

ok

[fading acorn]

[somber goblet]

we know this group is abelian

[cursive spindle]

maybe as an example take n=4

give me all such homomorphisms

[somber goblet]

ok the group is {1, 3} so its isomorphic to C2

[cursive spindle]

yes and where can you send 1?

[somber goblet]

1 is the identity of C2 so it has to go to 1 in C

[cursive spindle]

right and how about 3?

[somber goblet]

and 3 is of order 2 so we need to find an order 2 element of C

which must be -1

so its {1, -1}

[cursive spindle]

okay you have the interesting homomorphism

[somber goblet]

theres also {1}

[cursive spindle]

you can also take everything to 1 right

[somber goblet]

you can always do that

[cursive spindle]

yes so in total we have 2

[somber goblet]

ok...

[cursive spindle]

how would you extend this to the integers?

What's the obvious thing to do here?

[somber goblet]

extend what to the integers? the starting group?

[cursive spindle]

yes

[somber goblet]

well 0 has to map to 1

and then i pick wherever 1 goes to and that determines how the rest of the group maps

[cursive spindle]

hmm what i'm trying to consider is the following: chi(a+m) = chi(a) where m is the modulus here

[somber goblet]

but the members of any group are in bijection with maps in $\operatorname{Hom}(\mathbb Z, G)$

oh

[cloud walrus]

lexi

[somber goblet]

im not sure what you mean

what's the modulus

[cursive spindle]

it's 4 in this case

[somber goblet]

oh wait like in Z/nZ

got it

[cursive spindle]

yes

[somber goblet]

so this map has to have a "period" of n?

[cursive spindle]

so what we did so far is we defined a function chi: Z->C which is multiplicative and periodic with period m

yes

exactly

also by how we started chi(a)=0 iff gcd(a,m)>1 right

[somber goblet]

multiplicative? like $\chi(ab) = \chi(a) \chi(b)$

[cloud walrus]

lexi

[cursive spindle]

yes

[somber goblet]

woah

[somber goblet]

if gcd(a, m) > 1 then a is not in (Z/nZ)^x

[cursive spindle]

yup

right so anyways the takeaway here you will have phi(n) such functions

[somber goblet]

do we?

[cursive spindle]

yes we do

in fact these form a group

[somber goblet]

Hom((Z/nZ)^x, C)

under pointwise multiplication?

[cursive spindle]

yes exactly

[somber goblet]

how do we know we have phi(n) of these

[cursive spindle]

I'll let you think about this

[somber goblet]

the group doesnt necessarily have a single generator to use

but maybe the periodicity puts a restriction?

[cursive spindle]

oh that's the confusion i see

[maiden crater]

how would I do this?( without assumung its ab/(a,b) and proving that is it)

[cursive spindle]

no just think about them as Hom((Z/nZ)^x,C^x)

I said the latter just fyi

[somber goblet]

ok

[maiden crater]

I suppose contradiiction would work here?

Suppose $a \mid m; b \mid m$ and $l \nmid m$ So $m= lp+r$ , not sure what to do beyond this though

[cloud walrus]

wai

[somber goblet]

i have no idea how to show this

[maiden crater]

I suspect minimality somwhere

[somber goblet]

i feel dumb

[cursive spindle]

are we okay with the fact that any homo from Z/nZ -> C^x is determined by where the generator goes?

[somber goblet]

yes

[cursive spindle]

okay

[somber goblet]

for the additive group

[cursive spindle]

so you decompose G into cyclic factors

[somber goblet]

we are assuming fundamental thm on finite abelian groups?

or wait a minute

do we need that

[cursive spindle]

you technically can do it without it

[somber goblet]

if $a \in G$ has $a^n = 1$, then so does $\chi(a)$

[cloud walrus]

lexi

[somber goblet]

so everything has to map to some root of unity

[cursive spindle]

yes

[somber goblet]

$\mathbb C^\times$ doesn't have any finite subgroups with elements outside the unit circle does it

[cloud walrus]

lexi

[cursive spindle]

by this I assume you mean chi(a^n)

a typo

[somber goblet]

yes

$\chi(a)^n$

[cloud walrus]

lexi

[somber goblet]

as well

[cursive spindle]

yeah

[somber goblet]

ok...

that has to put a restriction on things

[cursive spindle]

yeah you're just mapping things to S^1

what I wanted to touch upon is quadratic residues

so a such that x^2 = a mod p

[somber goblet]

ok

like square roots in prime fields

[cursive spindle]

yeah "square roots"

[somber goblet]

yeah

[cursive spindle]

right so the idea is to use the characters somehow

[somber goblet]

the homomorphisms to C

[cursive spindle]

yeah but which one exactly

[somber goblet]

we want one s.t. phi(x)^2 = phi(a)?

*chi

[cursive spindle]

yeah the point is that we take the nontrivial chi such that chi(x)^2 = 1

if g is a generator of (Z/pZ)^x then we can write a = g^k then a is a quadratic residue iff this k is even

and odd otherwise

[somber goblet]

yeah

[cursive spindle]

right so if our chi outputs 1

we know we landed on a quadratic residue

[somber goblet]

oh wait so chi(x) = 1 or chi(x) = -1

[cursive spindle]

both

remember each homomorphism does something different right

[somber goblet]

yeah

[cursive spindle]

in (Z/4Z)^x we have
the trivial one:
chi(1)=1, chi(3)=1
non trivial one:
chi(1)=1, chi(3)=-1

the second one is called the quadratic character

[somber goblet]

i see

[cursive spindle]

right that's like half of the story

im surprised you stayed this long : )

[somber goblet]

im trying...

my head hurts

[cursive spindle]

yeah the first time you see it it's kinda weird and daunting

but let's rewind back to what we did so far

We started with Q(zeta_n)/Q

then considered the galois group

there is an isomorphism from that group to (Z/nZ)^x

then from there we took another map to C^x

so really what you should be thinking is characters of (Z/nZ)^x should be thought as characters of this special galois group

[somber goblet]

ohh

i hear the term 'character' a lot

what does it mean

[cursive spindle]

it just means homomorphisms from G -> C^x

[somber goblet]

oh ok

[cursive spindle]

okay here's the second part of the story

I think it's best explained with a familiar example

[somber goblet]

i hear it as "trace of matrix" in rep theory

[cursive spindle]

yeah different sources do different things

[somber goblet]

oh ok

[cursive spindle]

so the point is that this galois group is actually encoding deep arithmetic info for certain ring of integers

now you might ask how so?

for this let's consider Q(i)/Q. The corresponding ring of integers is Z[i] correct?

[somber goblet]

yes i think so

[cursive spindle]

yeah do you remember my ramblings about primes not being primes?

for example 5 = (2-i)(2+i)

but some stay primes like 3

[somber goblet]

yeah

Z[i] isnt a UFD

or smth

[cursive spindle]

yup

okay so Q(i)/Q can be thought of as a cyclotomic extension right

[somber goblet]

umm

maybe?

[cursive spindle]

yeah which one?

what zeta_n

[somber goblet]

wait we're adjoining i

which is just zeta_4

[cursive spindle]

yes good job

now we consider characters of Gal(Q(zeta_4)/Q)

which is the same as characters of (Z/4Z)^x

but we already talked about these!

okay take the quadratic character which gives chi(1)=1 and chi(3)=-1 as usual

now

5 = (2-i)(2+i) correct?

[somber goblet]

yes

[cursive spindle]

what happens when we do chi(5)?

(remember we are taking things modulo 4 here)

[somber goblet]

chi(5) = chi(1) = 1

[cursive spindle]

yup

let's try 13 = (3-2i)(3+2i)

what is chi(13)?

[somber goblet]

chi(13) = 1 as well

[cursive spindle]

nice

okay let's check the ones that stay prime and see what we get

for example 3 and 7

what is chi(3) and chi(7)

[somber goblet]

-1 for both

[cursive spindle]

yup

do you see what's going on here?

[somber goblet]

chi tells us if something is prime or not?

[cursive spindle]

YES

[somber goblet]

or has unique factorization

[cursive spindle]

no it just tells us whether p is splitting or not

[somber goblet]

what the hell

how

[cursive spindle]

yeah that's the real reason why you should care about quadratic reciprocity

[somber goblet]

and chi(2) is set to 0?

[cursive spindle]

yup

this tells you that 2 is ramified

which we already know

[somber goblet]

ok

this is weird

[cursive spindle]

yeah but do you find it kinda interesting

[somber goblet]

there is something deep and weird going on here

yes

[cursive spindle]

yup

So I guess another way to look at these things is taking the quotient Z[i]/(p)

[somber goblet]

ok

maybe

[cursive spindle]

yeah and you can check that this is isomorphic to F_p[x]/(x^2+1)

[somber goblet]

because it doesnt matter what order you take quotients?

is there a theorem for that

[cursive spindle]

yeah it's exactly what you think

it works nicely

[somber goblet]

i should try proving that :3

[cursive spindle]

rewrite Z[i] as Z[x]/(x^2+1) and you go from there

but the thing to note here is that this boils down to understanding what happens to F_p[x]/(x^2+1) for every prime p

this is basically the question x^2 = -1 mod p

so when is -1 a quadratic residue mod p!

[somber goblet]

oh right x^2 + 1 often splits over prime fields

hmm

so we're just asking if $x^2 - a$ splits over $\mathbb F_p$

[cloud walrus]

lexi

[cursive spindle]

yes and that is what you might do in elementary number theory

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

this legendre symbol is the quadratic character

so in that notation the legendre symbol (-1/p) is the main culprit of our story

[somber goblet]

were we talking about the dirchlet character

[cursive spindle]

yes

the quadratic character mod p is the legendre symbol

[fossil beacon]

Can I say every infinite subgroup if SL_2(R) has an element of infinite order.

[fossil beacon]

Feels like "yes"

Not a very easy question

[somber goblet]

i feel like there's something topological you can do here

[cursive spindle]

yeah anyways I just wanted to shed some light into this topic cuz it's very poorly done in elementary number theory

[somber goblet]

yeah i see

its quite fascinating yes

number theory is cool and i feel like most elementary introductions don't do it justice

from what i can see of what "real number theory" looks like

[spark veldt]

This is called a universal property, so I imagine it must be somewhat important. Though I don't immediately see how. Where would this be used, normally? Sorry I've only had very minimal exposure to category theory, so idk exactly what is the significance of this universal property (of polynomial rings)

[cursive spindle]

in general one might except similar rules like this for general ring of integers

[somber goblet]

universal property means it uniquely defines what the polynomial ring looks like

[lusty marlin]

Can't we embed the group of roots of unity inside SL₂(ℝ) as rotation matrices?

[fossil beacon]

How does that help

[lusty marlin]

This is an infinite group and all of its elements are of finite order.

[somber goblet]

not necessarily

rotation by irrational multiple of pi

[lusty marlin]

I meant embedding the group generated by e^(2πi/k) for integral k

[somber goblet]

thats a countably infinite group

oh wait i see

[lusty marlin]

Yes, they were asking for an infinite group

They didn't talk about countability

[lusty marlin]

From whatever I've seen, irrational rotations are not called "roots of unity"

I may be wrong about that though

[somber goblet]

right yeah

i see now

yes that should work

[cursive spindle]

tfw SL_2(R)/SO(2) = H

[fossil beacon]

Iwasawa decomposition

[cursive spindle]

deleting means you used noggin

good job wai

[maiden crater]

I got confused 😔 , I foror that it's closed under multiplication from teh ring, not addition

[cursive spindle]

yeah no worries

can you pinpoint exactly the ideal?

write it explicitly

in terms of generators

[azure cairn]

waow i just realized the nth roots of unity are cyclic subgroups of (C, x) and angles form cosets mod 2pi

[somber goblet]

H as in half plane or quaternions

[cursive spindle]

(upper) half plane

[azure cairn]

this exercise is so cool

(b) is $\exp(i \pi (n - 1))$ right

[cloud walrus]

Altanis

[azure cairn]

$\inner{z} = \left{\exp(i\frac{2\pi k}{n})\right}{k = 0}^{n - 1}$, so then the product of each root of unity would be
$$\prod{k = 0}^{n - 1} \exp(i\frac{2\pi k}{n}) = \exp(i \frac{2\pi}{n} (0 + 1 + 2 + \dots + (n - 1))) = \exp(i \frac{2\pi}{n} \frac{n(n - 1)}{2})$$
$$= \boxed{\exp(i \pi (n - 1))}.$$

[somber goblet]

rip

[azure cairn]

this is ragebait

[cloud walrus]

Altanis

[azure cairn]

finally

[warm badge]

and so all finite subgroups of the multiplicative group C is cyclic, which is cool when i first thought bout it

[azure cairn]

woah that's cool

[fossil beacon]

?

[somber goblet]

note that this is either 1 or -1

i believe this is $(-1)^{n+1}$

[cloud walrus]

lexi

[glass sinew]

oh wait, can we say R[x] is a vector space over R with an infinite basis

wait that's just a module isn't it

nvm

[somber goblet]

do we have a concept of a module that's a ring

like a R-algebra

[azure cairn]

LMAO i just realized

[somber goblet]

oh thats literally what its called

[cursive spindle]

you can do better than this lol

oh lexi already said lmao

[azure cairn]

$e^{i \pi (n - 1)} = (e^{i \pi})^{n - 1} = (-1)^{n - 1}$

[cloud walrus]

Altanis

[somber goblet]

yep

[glass sinew]

there is a short group-theoretic proof. let $Z_n=\langle a\rangle$ be the cyclic group of order $n$. the product $1\cdot a^1\cdots a^{n-1}$ reduces to $a^{n/2}$ if $n$ is even; otherwise, it reduces to $1$ if $n$ is odd. in your subgroup of $\mbb{C}^\times$ these are -1 and 1

[somber goblet]

yeah i was thinking you could pass this through the canonical isomorphism

[glass sinew]

yep

[somber goblet]

i mean they're implicitly doing that since the isomorphism is exp...

[glass sinew]

yep!

[somber goblet]

but you can totally do all your work in Cn

and its nicer

[glass sinew]

inclined to believe that was the idea of the exercise

[somber goblet]

are you sure you have it the right way around between even/odd

[glass sinew]

mb it is the other way around lol

[cloud walrus]

bsharp

[azure cairn]

you get a^(n(n-1)/2) if you multiply all the group elements yes

so then you split into cases dpeending on if n is even or odd

uh leme use paper rq

[azure cairn]

um

ok

so n even

$a^{n(n-1)/2} = a^{n/2} a^{n-1} = a^{n/2} a^{-1}$

[cloud walrus]

Altanis

[azure cairn]

?

[azure cairn]

where did the a^{-1} term go for n even

[glass sinew]

you're looking for $a^{(n-1)(n-2)/2}$

[cloud walrus]

bsharp

[azure cairn]

what

[glass sinew]

$Z_n=\langle a\rangle={1,a,\cdots,a^{n-1}}$

[cloud walrus]

bsharp

[azure cairn]

omfg

lol

yes ok i see

[somber goblet]

$a^{kn} = 1$

[cloud walrus]

lexi

[somber goblet]

for any $k \in \mathbb N$

[cloud walrus]

lexi

[azure cairn]

$a^{(n - 1)(n - 2)/2} = a^{(n-2)/2}a^{n - 1} = a^{n/2 - 1}a^{-1} = a^{n/2}$

[cloud walrus]

Altanis

[azure cairn]

for n even

[azure cairn]

wait what no

1 + ... + (n - 1) is n(n-1)/2

,w sum from k = 1 to n - 1 of k

[cloud walrus]

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Requested by altanis_

[azure cairn]

i see

ok for n odd $a^{n(n-1)/2}$, you have integer $n(n-1)/2 \equiv n \pmod n$ so its $1$

[cloud walrus]

Altanis

[azure cairn]

then for n even we have $a^{n(n-1)/2} = a^{n/2} a^{n-1} = a^{n/2} a^{-1}$

[cloud walrus]

Altanis

[azure cairn]

?

omg im an idiot

a^(xy) = (a^x)^y not a^x a^y 😭

[cursive spindle]

many such cases

[azure cairn]

ok i see

then for n even we have $a^{n(n-1)/2} = a^{(n^2 - n)/2} = a^{n^2/2} a^{-n/2} = (a^{n})^{n/2} a^{-n/2} = a^{-n/2} = a^{n/2}$

[cloud walrus]

Altanis

[azure cairn]

:D

then a^{n/2} for roots of unity would be the 180deg rotation so -1

very cool ty both

    \begin{problem}{Let $a, b \in G$ for some group $G$. Show $|ab| = |ba|$.}
        First, suppose $ab$ generates a finite cyclic group, so $|ab| = n$ for some $n \in \bZ^+$ (equivalently, $(ab)^n = 1$). Then note
            $$(ba)^n = b(ab)(ab)\cdots(ab)a = b(ab)^{n - 1}a = b(ab)^{-1}a = 1,$$
        meaning $|ba| \le n$. Suppose $|ba| = k < n$. Then
            $$(ba)^k = b(ab)^{k - 1}a = 1 \implies (ab)^k = 1,$$
        but $n$ is the smallest positive integer for which $(ab)^n$, and as such, no such $k$ exists. Thus $|ba| = n$.
        
        Now suppose $ab$ generates an infinite cyclic group, so there is no $k \in \bZ^+$ for which $(ab)^k = 1$. For the sake of contradiction, suppose $ba$ generates a finite cyclic group, and so there is some $\eta \in \bZ^+$ for which $(ba)^\eta = 1$. Then
            $$(ba)^\eta = b(ab)(ab)\cdots(ab)a = b(ab)^{\eta - 1}a = 1.$$
        Rearranging yields $(ab)^{\eta - 1} = b^{-1}a^{-1} = (ab)^{-1}$, which implies $(ab)^\eta = 1$, which is contradictory. Thus $ba$ generates an infinite cyclic group. In any case, $ab$ and $ba$ generate groups of the same order.
    \end{problem}

is this proof OK?

[cloud walrus]

Altanis

[azure cairn]

i would imagine i dont need to show |ab| = |ba| have the same cardinal assigned to them

cuz artin just said a group is finite order of n or "infinite order"

[cursive spindle]

time to steal your latex setup

[azure cairn]

do u want it?

its pretty horrendous tho

there are Comments. let me fix that