groups-rings-fields

[chilly ocean]

you figured it out though!

the hard part is realizing that you need binomial thm

[white oxide]

oh that was instant

[chilly ocean]

then you figure out the exponent that works

[white oxide]

oh lol i should've tried out the exponent

but anyways thanks for the help!

[void cosmos]

both work

[south patrol]

Optimisation being the key word lol

i think someone else mentioned an optimal integer

[white oxide]

wait n being the order of a and m the order of b right

[void cosmos]

why would that matter

[white oxide]

ig doesn't matter

[south patrol]

maybe they didn't

Lol

[white oxide]

but i mean

[void cosmos]

yea its just the matter of existence right?

[void cosmos]

not order

[white oxide]

oh yea my fault

not a group

but like

[void cosmos]

a is nilpotent when there exists an integer positive n such that a^n = 0

now say a and b are nilpotent

show that a+b is

[white oxide]

say a^4 = 0 and b^3 = 0 right, and you claim that (a + b)^(4 + 3 = 7) = 0 right

where say n = 4 and m = 3

but i tried verifying that and it doesn't work

or am i misinterpreting

[void cosmos]

u are miscalculating

[chilly ocean]

that should work?

[white oxide]

oh oops

that works

brain fart

i've had my fair share of embarassing moments on this server but this is up there

[void cosmos]

why would this be embarassing haha

[white oxide]

cuz i couldn't see it right in front of my eyes

brain no work

[void cosmos]

nope u got htis

[barren sierra]

one (hopefully last) hail mary ask if anyone has any idea how to do these

cause while I've verified (a) computationally

I haven't got the slightest idea

btw it turns out for (a) that every element can be written in 60 ways in the manner described which is the size of A5

I'm sure that's part of the key but what part it is and what the other parts are I have no idea

[long nebula]

what does a prime power field look like? like how does addition and multiplication in F8 work for example

is there a straightforward way to write out the multiplication table in general

I know that it's the splitting field of x^8 - x in F2

[agile burrow]

Sure. If you construct your field as F_2[x] modulo some irreducible polynomial of degree 3, then each equivalence class has a unique representative of degree less than 3

Then addition and multiplication are done exactly as you'd expect, modulo that irreducible polynomial

[long nebula]

oh hmmm I see

that makes sense

i was hoping there would be an easier way that didn't involve multiplying polynomials

[agile burrow]

In particular, the additive group is (Z/2)^3. More interesting is that the multiplicative group is cyclic

[long nebula]

right

That makes sense

becase classification of f.g. abelian groups

[agile burrow]

Fair enough, maybe there's another way that I just don't know of. I'm not very good with fields

[long nebula]

haha helpful nevertheless, ty!

[barren sierra]

at this point I'm taking the L

and I'll just ask my prof to meet with me at some point and walk me through this

[long nebula]

Let (q = p^r) where (p) is prime. Recall that (\mathbb F_q^\times) is a cyclic group. Must it be the case that for every (\alpha) such that (\mathbb F_q = \mathbb F_p(\alpha)), (\mathbb F_q^\times = \langle\alpha\rangle)?

So I just wanna check that I've understood this question right; I considered (\mathbb F_9 \cong \mathbb F_3[x]/(x^2+1)) and then let (\alpha) be a root of (x^2+1) so that (\mathbb F_9 = \mathbb F_3(\alpha)), but (\mathbb F_q^\times = \langle\alpha\rangle) because (\alpha) is only of order (4). This feels too simple; am I missing something?

asjdf;lka texit doesn't understand \bF

[cloud walrus]

tietzERIC extension

[south patrol]

That seems good

For q = p^2 it's also clear that there are p^2 - p elements generating F_q as a F_p-algebra [literally anything not in F_p], whilst F_q^x has p^2 - p elements and hence only φ(p^2 - p) generators

[barren sierra]

Dumb question

what is (c) asking me to do

[long nebula]

yay awesome

[long nebula]

ohhh okay I was confused if all of the elements not in the original field generated F_q as a F_p-algebra

[barren sierra]

is it asking me to write the vector of traces from (b) as linear sums of the rows of the given table?

[long nebula]

but it makes sense that they all should

[south patrol]

That isn't what they are asking but that is how you answer the question xd

Like ur rep is a sum of irreps and you need to find what the sum looks ike

Lol interesting they say vector of traces

[barren sierra]

Wait what are they asking then 💀

I'm very bad at representation theory I have no idea what's going on since this is my first time seeing it

Evidenced by this + my prior question

[south patrol]

.

[feral agate]

If you have $a,b \in I$ where I is an ideal of some ring, I don't understand why $a-b \in I$.

[cloud walrus]

FrankF

[south patrol]

I is a group under addition

[feral agate]

oh ok

[feral agate]

What does the star mean?

[chilly ocean]

units in the ring

[feral agate]

units?

[chilly ocean]

elements with multiplicative inverses

think about what they are,|| in this case they are just +1, and -1||

[celest furnace]

You can show the units of R[x] are the units of R for any integral domain R

[south patrol]

• can mean non-zero elements for some authors too oop

[barren sierra]

really?

I've never seen this

[barren sierra]

I got that the a_i's are rational, not integer. Is that to be expected? idk I expected integers for some reason but maybe I'm being dumb

[celest furnace]

Potato makes a good point because I think most people use \cross instead of *

[feral agate]

Shouldn't it be the set ${f \in Z[x] \mid a_0 = +1 \lor a_0 = -1}$?

[barren sierra]

People use \cross?????

[cloud walrus]

FrankF

[celest furnace]

$\bZ^{\cross}$

[cloud walrus]

UwUChad

[celest furnace]

This isn’t normal?

[barren sierra]

so you're saying 1 + x is invertible in Z[x]?

[chilly ocean]

what would be the inverse of x+1?

[feral agate]

Yeah you're right

[chilly ocean]

yeah it is good exercise to prove that $R[x]^\times = R^\times$ in an integral domain

[cloud walrus]

R-module

[celest furnace]

Ah it was \times I was thinking of not \cross lol

[chilly ocean]

yeah i guess the notation is strange because tbh if i saw $\mathbb{Z}[x]^*$ i would probably think the dual of $\mathbb{Z}[x]$ as a $\mathbb{Z}$-module

[cloud walrus]

R-module

[white oxide]

how is i(a)/i(b) an element of D? isn't it by definition an element of F?

my b i didn't include context, but fraleigh defined an isomorphism i: D --> F where D is an integral domain and F is a field constructed from a subset of the cartesian product on D by i(a) = [(a, 1)]

is it because (a, 1) = a?

and similarly (b, 1) = b?

[celest furnace]

Ah this notation sucks

You should just think of this as like actual fractions as they work almost the same way

[white oxide]

ahhh ok

[celest furnace]

Pretty much your question comes down to this theorem:

[white oxide]

that makes sense thank u

[white oxide]

oh ok

is that dummit and foote

[celest furnace]

Yes

[white oxide]

bet

i should read gallian on this chapter as well

[celest furnace]

Hard to self-read 😭 I recommend YT videos

[white oxide]

i think it's a bit weak but that just might be me being stupid and i'm in no position to judge

yea i heard lmao

oop was about to say it reads like a textbook

[celest furnace]

Not as hard as lang though

200 pages of D&F group theory in 60

[white oxide]

lmao that's insane

doesn't he have like

lin alg in 40 pages or smt

😂

[celest furnace]

https://youtube.com/playlist?list=PL8yHsr3EFj51pjBvvCPipgAT3SYpIiIsJ this guy has been super helpful for me (added bonus: he is a fields medalist!)

[white oxide]

ha

i'm taking a class with him next sem

that class in particular

[celest furnace]

YOU WHAT???

[white oxide]

yeee

i go to berkeley

[celest furnace]

Wow 😳 savor every moment

He is a legend

[white oxide]

i didn't realize that faculty is insane

i have to go to office hours but from what i've heard he's basically on the spectrum

which makes sense

fields medalist

[celest furnace]

On the spectrum?

[next obsidian]

He’s calling him autistic lmfao

[white oxide]

yeah like i heard he has mild aspergers

that's what my friend told me who took a class w him

[celest furnace]

He stutters a little in his vids maybe that’s what he’s taking about

Can’t believe you get to take a class with the GOAT

[white oxide]

do you know david eisenbud

[white oxide]

nvm turns out he's not teaching it next sem

[white oxide]

but this guy is

[celest furnace]

RIP

[celest furnace]

Doesn’t ring a bell

Oh Wikipedia it says “Eisenbud is the son of Eisenbud” lol

[white oxide]

LMAO

might just take a second course in algebra before jumping to grad level algebra then tbh

[celest furnace]

Trying to self teach myself so I can skip to grad

[white oxide]

yea imma grind hard asf this summer

thinking about doing artin

[celest furnace]

Super nice Ph.D. student is helping me I would recommend reaching out to someone like that

We meet on fridays to discuss the section I read (and do some of the exercises) it really helps seeing someone who has like a perfect understanding of the material teach you

Saves a LOT of time

[white oxide]

hm i should definitely do that

[celest furnace]

Check pins it seems someone posted all their homework

Also he was my TA who I went to office hours with if you’ve ever done that target those people first

[white oxide]

oh ok i'll try it out bett

[white oxide]

why exactly does b^p-1 = 1 in Zp? i understand that b^p-1 = 1 in the group consisting of elements 1, 2, 3, ... p - 1, but why does it still hold in Zp?

and why couldn't we have b^p = 1?

[thorn delta]

if G is a finite group and g is in G then g^|G| = 1 always. the group of nonzero elements of Zp has order p - 1

so b^{p-1} = 1 for any b != 0 in Zp implies b^p = b for any b != 0 in Zp

so i mean, b^p does equal 1 when b = 1

[white oxide]

right, but why in particular does that hold in Zp?

shouldn't it only hold in that group?

[white oxide]

oh

my b

you just explained it right there didn't see that

wait what

how is that true

cuz take Z7 and 5 in Z7

nvm

my fault

lol

thanks!

[thorn delta]

npnp

[knotty bane]

Is there a reason why in proofs invoking the finiteness of some set, people seem to generally opt for two indices instead of one in showing that repeated application of some operation to an element must 'loop back around' to that same element?

As a concrete example, say you wish to prove that every nonzero element in a finite ring R (with identity) is either a zero-divisor or a unit. So, let x in R st. x /= 0. If x is a zero-divisor, the result follows. Hence suppose x is not a zero-divisor. By finiteness, you may write that x = x^n for some n > 1. Thus, x^n - x = 0 <=> x(x^(n - 1) - 1) = 0, and since x is not a zero-divisor, we must have x^(n - 1) = 1. From this it follows that x^(n - 2) is the inverse for x, i.e. x is a unit.

Most often, though, I see that people write x^m = x^n with 0 <= m < n, and then go on to derive a similar conclusion. Is there something I'm missing here logically speaking, or it just a matter of taste?

[For that example, shout out to the proof just using the map f(x) = ax, by the way]

[upper pivot]

yes

so basically you cannot say a priori that x=x^n

why? you are just saying that {x, x^2 ...} is finite right

so this doesnt mean that x= x^n for some n, but that some 2 elements are the same

[knotty bane]

riiiiiight

yeah, think I get it now

[woven obsidian]

Is this statement just wrong? Here A is the polynomial ring over a finite field

Eg if the finite field is F2, and $m_1 = x^2+x+1$ and $m_2 = x(x^3+x+1)$

[cloud walrus]

AoiKunie

[wooden ember]

Probably why they said except possibly in the case q=2

[woven obsidian]

@wooden ember They said except possible in the case q=2 and m1 and m2 have relatively prime degrees

I.e. both should hold

[wooden ember]

I think they meant those cases distinctly?

[woven obsidian]

No you need q=2

[wooden ember]

Fair enough

[woven obsidian]

And I can prove that you also need that the degrees of the prime factors are relatively prime

But the counterexample is correct, right?

[wooden ember]

Maybe they meant for m1 and m2 to be irreducible cause yeah your example@looks correct

[woven obsidian]

Yeah I guess that'd make more sense, ty for the input

tried to prove the statement for a long while

[white yoke]

Why is the only division algebra over the complex numbers just itself?

[south patrol]

I assume you mean finite dimensional division algebra?

[white yoke]

Yeah

@south patrol do you know why?

[delicate orchid]

it holds by artin-wedderburn I'll see if I can find the proof

[white yoke]

I am familiar with Artin Wedderburn theorem

[delicate orchid]

ok I think this works but get someone to double check this

Let D be a division algebra over C. The artin-wedderburn decomposition of D can only contain a single matrix algebra as we're decomposing a division algebra, and if this matrix algebra was of dimension bigger than 1 it would have zero divisors - thus we must have D \cong M_1(C) \cong C

[south patrol]

Oh really Artin-Wedderburn huh

Wasn't expecting that for some reason lol

[white yoke]

But for example in the real numbers we can have the quaternions H coming up and they are 4 dimensional. What's special about C?

[south patrol]

C is algebraically closed.

[white yoke]

I thought it would be because of this. But how does it follow

[south patrol]

And in that case, Artin-Wedderburn gives you that all teh matrices have entries in C

Whereas in general you get like finite division algebras over your thing

[white yoke]

I see

[south patrol]

But yeah this feels hm

[white yoke]

Yeah it's cool though, I don't think I need this

[delicate orchid]

there's something missing - I couldn't find the original place I saw the proof

[south patrol]

The way I was thinking lol was like

If D is a finite dim division algebra over C then in particule it is a finite dim division algebra over R, so it's C or H (or O depending on conventions lol)

And then you just rule those out

But that's overkill

[delicate orchid]

ok very idiotic question but I can't be bothered thinking about it, do the division algebras over R have any non-trivial ideals (as algebras)

[south patrol]

Division algebras have no non-trivial proper left (or right) ideals

For the same reason as for fields

Lol this is cursed cause I've been working with normed division algebras where one seems not to evevn demand associativity

[white yoke]

Yeah i consider those too lol

[south patrol]

Oh

[delicate orchid]

I thought not... hmmmmmm

[south patrol]

Hm

Wait how do you know that Artin-Wedderburn appleis wew

Maybe I'm missing smth lol

[delicate orchid]

that's like the only thing I remember from the proof I saw

the rest was me

there could be a much easier way to do this

[south patrol]

Yeah lol fair

Just appeal to Hurwitz and be a meme like me

[delicate orchid]

do I google it chat

[south patrol]

Cause then you can just use topology

jk not quite

[white yoke]

I was just wondering because I'm gonna give a presentation on my masters thesis and this could be a question they ask

[south patrol]

Lol wikipedia also deduces it from Hurwitz like i did

[white yoke]

At the end

[delicate orchid]

I will le google it

[south patrol]

OH

[delicate orchid]

what

[south patrol]

yeah wew your argument is fine lmfao

[delicate orchid]

WHAT

[south patrol]

Because D is a division algebra, it is in particular simple lol

[delicate orchid]

A WORLD FIRST!!!

[south patrol]

And then just apply the rest of ur argument

& ur gucci

[delicate orchid]

ok but how does that change over R?

oh yeah duh

[south patrol]

Artin-Wedderburn specialises for algebraically closed fields to give you matrices over the field

[delicate orchid]

matrices are still in C

yussssss

[white yoke]

I seee

Thank you guys

I have to go now

[south patrol]

Adios

[delicate orchid]

adidas

[south patrol]

Kinda cool how you can see hurwitz theorem topologically ish

like normed division algebras give rise to continuous maps between spheres which one can rule out

[feral agate]

Could someone explain why the elements in the quotient ring are roots to x^2+1?

[south patrol]

The elements in the quotient aren't all roots to that polynomial

This is just saying that if x^2 + 1 were reducible then you could write it as (x-a)(x-b) for some a and b and it'd have a root

[feral agate]

Yep thanks

[feral agate]

is $x^{34}-1$ reducible in $Z$?

I am not sure how to think think about that

[cloud walrus]

FrankF

[formal ermine]

1 is a root

[south patrol]

Also, like

Maybe I'm being silly

[feral agate]

My bad I forgot to mention in this context. I am really confused because Z/101Z is a set

[south patrol]

But I'm pretty sure x^a -1 | x^b - 1 iff a|b

[feral agate]

and shouldn't the reducibility of $x^{34}-1$ depend on Z/101Z?

[cloud walrus]

FrankF

[south patrol]

Wdym "is a set"

[hot lake]

yeah but 1 is still a root even modulo 101

[south patrol]

It's a ring here in particular

[feral agate]

I mean an element is a set in that quotient ring

[south patrol]

Well elements are always sets under normal constructions but sure

Basically the glue you may be missing is like

1 sec i can do a diagram

[feral agate]

okay ser

[formal ermine]

wait is 101 prime

oh wow it is

that feels wrong

[feral agate]

[delicate orchid]

it's my favourite prime

[delicate orchid]

the way to view this is you're taking polynomials with coefficients in F_101 and then setting x^34 = 1

[feral agate]

Yeah, I think I have understood it

Thanks all

[south patrol]

weird

[hollow mica]

101 gives off major prime number vibes, idk what you guys are on

[rotund aurora]

I agree with mrean

[formal ermine]

it feels like it should be divisible by 3, 13, or 17

[elder wave]

the quersumme is 2

[kind jacinth]

what does this even mean

this is how the chapter of lagrange theorem starts

[formal ermine]

what are you confused about

[kind jacinth]

im not sure what the meaning of this is

like what are they doing

[formal ermine]

defining a function

and then showing that it is a bijection

[kind jacinth]

g and a are both elements of G

whats the difference between g and a

[formal ermine]

g is a fixed element

a is the input of the function

[kind jacinth]

so im defining a function for a fixed g

[formal ermine]

yes

left multiplication with g

what it's saying is essentially that for any group G and any g in G we have gG = G

(in fact this is an iff)

[kind jacinth]

ohh so no matter which fixed g i select ill get G again

and i guess its a permutation because a->ga ..

[formal ermine]

permutation here means gG = G lol

you'll get G again but "in a different order"

[kind jacinth]

ah i see

[kind jacinth]

and this is true right? G again in a different order depending on g

alright then this makes sense now thanks!

[kind jacinth]

The Euler φ-function is the map φ : N → N defined by φ(n) = 1 for n = 1,
and, for n > 1, φ(n) is the number of positive integers m with 1 ≤ m < n and gcd(m, n) = 1.
this isnt a bijective function then right?

because for the same n u can have different results where gcd(m,n) = 1

[formal ermine]

no it's not a bijection

it is not even surjective

[rotund aurora]

https://en.wikipedia.org/wiki/Carmichael's_totient_function_conjecture

this may be of interest

[drifting pivot]

for some finite cyclic group G, how can i show that the number of elements g for which g^n = 1 (n is a positive integer) is equal to gcd(n, |G|)?

[void cosmos]

|g^n| = k/gcd(n,k) where k is the order of the group

ops

[drifting pivot]

can you explain a bit more?

[void cosmos]

say g is a generator for a group G

then the order of g^n is as given

for any n

post integer

[drifting pivot]

wait isnt the thing in my question different?

its just gcd(n,|G|)

yours is |G|/gcd(n,|G|)

wait yours is correct, i checked the internet

then how can we show this is equal to gcd(n,|G|)? @void cosmos

[void cosmos]

try to do it yourself

[drifting pivot]

i asked because im stuck

[void cosmos]

use this as ah int

as a hint*

[drifting pivot]

order of the group is |G|

so |g^n| = |G|/gcd(n,|G|)

im stuck here, not sure how to show this is equal to gcd(n,|G|)

[restive birch]

think about the cyclic subgroup generated by g^n

that might help

[delicate orchid]

consider the fact that the map g -> g^n is a homomorphism because G is abelian

[feral agate]

Would it be correct to say that the size of this set is 2^{deg(x^3+x^2+1)}=(2^3)=6?

So in general if you have $(Z/nZ)[x]/(p(x))$ it is of the size $n^{deg(p(x))}$

[cloud walrus]

FrankF

[formal ermine]

the polynomial may have repeated roots

[feral agate]

So this is true assuming that p(x) is irreducible

[formal ermine]

only in char 0

[feral agate]

Is there an easy way to see whether x^3+x^2+1 mod 2 has roots?

[pastel cliff]

this is pretty old but

[formal ermine]

it's a finite problem

sooo lmao

[empty rose]

there are only two numbers in Z/2Z to check

[pastel cliff]

here, repost

that sum is a matrix?

[feral agate]

why though, can't other integers give a root?

[pastel cliff]

that sounds wacky to even say

or is it about how det interacts with other multilinear maps

[empty rose]

...well there aren't any other integers in Z/2Z

[feral agate]

I mean like x^3 mod 2, I don't see why x being 0 or 1 are the only ones giving an unique answer.

[pastel cliff]

they're the only choices

[feral agate]

So we have (x mod 2)^3 mod 2?

[pastel cliff]

there's really no other way to say it

there are only two elements in integers mod 2

0 or 1

any polynomial mod 2 can only have 0 or 1 as coefficients and as solutions

if you're confused you should review what an equivalence relation is and how it works with modulos

[empty rose]

x is in Z/2Z, so x is either 0 or 1

what kind of other integer could give you a different result?

[drifting pivot]

if we have a prime p for which 2 is a primitive root, can we say that x^2-2 = 0 has no solutions in F_p?

[feral agate]

It is not trivial to me why x can only have two classes. I understand that Z/2Z has two elements and out of each element you can pick an element in it to perform calculations with it as a representative. However, x does not carry that restriction as in it can take on any value right?

[delicate orchid]

that looks more like a matrix multiplied by a vector

[delicate orchid]

the quotient map pi : Z -> Z/2Z is a ring homomorphism, thus pi(x^3) = pi(x)^3, and since pi(x) can only take two values we need only check those two values
I think this is what you're after

[empty rose]

...what's "any value"? and how do you cube "a value"?

let x be the operation of negation, as a function from the set of integers to itself (so for instance x(1) = -1)
what's x^3?

[barren sierra]

I think you should do this

[long nebula]

in a normal extension K/F, does every minimal polynomial of an element of K/F split in K?

my feeling is yes but I'm not sure why

[agile burrow]

Lol what's your definition of a normal extension?

[long nebula]

definition of normal here is that it's the splitting field of some collection of polynomials

[agile burrow]

Ah

[long nebula]

I'm like sure this is true but I can't remember why

every element is of the form x^m, so counting the number of elements for which g^n = 1 is equivalent to counting the number of 0≤m<|G| for which mn is a multiple of |G|

hopefully that helps?

make sure you see why

[rustic crown]

yee it says like det is a "universal" multilinear alternating map :3

[summer path]

is universal >.<

[long nebula]

I'm not sure you have to use lagrange's theorem for this, but yes

[feral agate]

I managed to formalize what I am actually confused about, which is the substitution map p: Z/2Z[x] -> Z/2Z defined by p(f(x)) = f(a) for some a in Z and all f(x) in Z/2Z[x]. Everyone says that for some reason f(0) and f(1) must give you the representatives of all classes in Z/2Z. However, I don't see what is underlying that claim. Basically, why f(0) and f(1) cannot give the elements in the same class of Z/2Z

[agile burrow]

so I'm thinking like if F/k is a splitting field, let x be in F and let p(x) be it's minimal polynomial. If y is some other root of p(x), then there's an isomorphism k(x) -> k(y) which extends the identity on k. Viewing F as a splitting field over k(x) and F(y) as a splitting field over k(y), uniqueness of splitting fields let's us extend to an isomorphism F -> F(y), which is in particular an isomorphism of k-vector spaces, so dim F = dim F(y) as k-vector spaces.

But then we obviously have inclusion F -> F(y), and since dimensions are equal we get that F = F(y) so y is also in F

[long nebula]

m = r|G|/n and then count the number of possibilities for r that make this satisfy 0≤m<|G| with m an integer; it might be instructive to try an example, like |G| = 12 and n = 4

[delicate orchid]

oh so you're asking why you can't do, for example, f(2)? Why f(2) has to be equal to f(0)?

[feral agate]

Assuming that f(0) and f(1) give the elements in the same class of Z/2Z, we would need to test other elements in Z to get the representatives of all classes in Z/2Z, so f(2) would be one of the tests.

[long nebula]

hmmm I will take some time to read this

ty!

[delicate orchid]

ah I see right, it's becuase when you do f(a) = n_0+n_1a+n_2a^2+...+n_ka^k with a in Z, if you think about the n_is as actual cosets in Z/2Z, n_i*a^i will give the same coset as n_i*(a^i+2Z) - so if a, a' are in the same coset in Z/2Z then n_ia^i = n_ia'^i for all i, and thus f(a) = f(a'), so we just need to check a = 0, 1

[feral agate]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[delicate orchid]

you can see they'll give the same coset by being really explicit and writing out n_i as n_i+2Z and then (a^i+2Z)(n_i+2Z) = a^in_i+2Z, and also a^i(n_i+2Z) = a^in_i+2Z by the Z-module action or whatever

[feral agate]

Ah okay, I will do some more research but I am not confused anymore. Thanks a lot!

[delicate orchid]

I'm sure this can be generalised to any quotient ring btw

[kind jacinth]

i just started learning about orbits

to me orbits and cosets seem similar

is there a relation between them?

[south patrol]

Yes

If G is a group and H a subgroup, then H acts on (the set) G by h.g = hg

The orbit of an element g under this action is precisely Hg

[kind jacinth]

then what is the difference between the two

[south patrol]

This is a special case of an action

[kind jacinth]

oh so only for this action they are the same

[south patrol]

Yeah

But anyway so like

I'm not sure if you've seen it yet, but I imagine soon you'll do the orbit-stabiliser theorem and you can view that as a generalisation of Lagrange's theorem

Since if you take the action to be the action I just described then you get Lagrange's theorem from that

So like, this is quite an important special case

[kind jacinth]

it is coming next! thanks for the help

[south patrol]

np

[glossy crag]

Are there infinite fields such that every element is a root of unity?

[south patrol]

lol

https://math.stackexchange.com/questions/4193256/field-whose-every-non-zero-element-is-a-root-of-unity for more generally

But for example the algebraic closure of F_p should work

Indeed every non-zero element of F_{p^n} is a root of unity and the alg closure of F_p is a colimit of those

@glossy crag

[atomic mesa]

How do we know G/K circled is a group?

I know G/H is a quotient group if H is normal in G

[chilly ocean]

because the kernel is always a normal subgroup

let h be in the kernel of phi, and g be in G, is ghg^-1 in the kernel of phi?

[atomic mesa]

Always? I thought kernel of only canonical homomorphisms are subgroups

[chilly ocean]

the kernel of any homomorphism is a normal subgroup, it is a one line check

in fact the other direction is also true: every normal subgroup can be exhibited as the kernel of a group homomorphism

[white yoke]

φ(ghg^-1)=φ(g)φ(h)φ(g)^-1=0 because φ(h) =0

[atomic mesa]

Ah it is…

[white yoke]

This algebra 4?

[atomic mesa]

Ah ok I was thinking this as well

[atomic mesa]

Yeah 😂

[chilly ocean]

close.. correct is : $\phi(ghg^{-1}) = \phi(g)\phi(h)\phi(g^-1) = \phi(g) * e* \phi(g)^{-1} = e$

[cloud walrus]

R-module

[white yoke]

Thought I recognised ur username 😂

[white yoke]

Ah yeah I was thinking of ring homomorphism I guess

But e is group identity

[atomic mesa]

We did this, it seems like a weaker statement than yours…

[white yoke]

Is Talbot teaching this year?

[atomic mesa]

No, we have Yusra. Thankfully she’s very very good

[chilly ocean]

it is the same, to every group homomorphism you can produce a normal subgroup (its kernel) and to every normal subgroup you can construct a group homomorphism exhibiting that subgroup as its kernel (the canonical quotient map)

[white yoke]

Ah nice. Talbot's notes were quite good so I can send if you don't have them

[atomic mesa]

If u could that would be awesome :D

[white yoke]

I'll DM

[atomic mesa]

Oh I see.. Actually you’ve answered everything here

Ijust checked and we didn’t even prove the proposition ._.

[chilly ocean]

yes the connection between normal subgroups and kernel (that they are the same thing in different guises) is perhaps the most important part of group theory 🙂

[atomic mesa]

Noted, thank you

[chilly ocean]

does anyone know a good intuitive way to think about the adjoint representation map of a lie algebra? im trying to understand the inner automorphisms of a lie algebra but I already dont feel like im grasping the geometry of the map ad : L -> gl(L)

[next obsidian]

Isn’t it literally just conjugation?

[chilly ocean]

what im taking away is that the image of an element x, ad(x), measures how much it does not commute in L?

[next obsidian]

Or well I mean I guess like

[chilly ocean]

i mean the pairing with the bracket map

[next obsidian]

Yeah that’s basically conjugation

[chilly ocean]

x -> [x , *]. how is that conjugation?

[next obsidian]

Like in a group this does exactly the same thing, measure how much it doesn’t commute

Sorry I’m being opaque. Conjugation in a group = how much does it not commute

Similarly for the adjoint rep

[chilly ocean]

ok im following, this was my original feel, but then i couldnt understand why in the defn of an inner automorphism why to include composition with the exponential map

if the adjoint pairing is already measuring what you want

like the inner automorphisms of a group are the automorphisms which arise by conjugation, and hence the maps which measure the failure of commutativity, but in the lie algebra case the inner automorphisms are defined as the group generated by exponentials of the adjoint map of nilpotent elements

[next obsidian]

I’m kinda noob with Lie stuff so

I’ll tap out and hope someone better comes by

[chilly ocean]

haha np thank you sm though, much appreciated

[next obsidian]

Ask about commutative rings next time so I can be useful 😠

[rustic crown]

so me also noob with lie stuff

but lie algebras are exactly the stuff you get when you try to study lie groups using some linear algebra

you look at the tangent space at the identity (or equivalently left-invariant vector fields) which naturally becomes a lie algebra

and conjugation in the lie groups corresponds with some adjoint rep stuff at the algebra

and exponential map lets you go from lie algebra land to lie group land

rest idk >.<

[rustic crown]

me too, but when chmonkey is sleeping, so i can be useful

[rotund aurora]

chmonkey never sleeps

[next obsidian]

I never sleep

[rotund aurora]

[summer path]

,ti chmonkey

[cloud walrus]

The current time for Chmonkey is 09:00 PM (EDT) on Fri, 24/03/2023.
Tubular Cat is 1 hour behind, at 08:00 PM (CDT) on Fri, 24/03/2023.

[rustic crown]

so det never useful?

[next obsidian]

☠️

[frigid lark]

Maybe when Chmonkey is busy doing math

[summer path]

det is cute, so det is always useful

[chilly ocean]

so this is why it is called an "inner automorphism" because on the lie group level it looks like conjugation?

and the appearance of the exponential map is because it takes you from the lie group to the lie algebra

that intuition helps!

[tough raven]

(This is just my guess, this being a “why” question and my not being very experienced with Lie algebras:) so that it's an automorphism.
[x,-] is a derivation, ie d[y,z] = [dy,z] + [y,dz] where d = [x,-], but if you want a homomorphism of Lie algebras you need σ[x,y] = [σx,σy]. If σ = exp(d) for d a derivation (nilpotent so the series makes sense without doing topology), σ will be a homomorphism, and as a bonus invertible since exp(-d) is an inverse for exp(d).

(To very partially motivate why you'd try exponentiating, in a sense Taylor series expansions say that f(a + h) = ( exp(h d/dx)f )(a), so the exponential of “derivative” is “shift the argument by 1”, which is an automorphism.)

[tough raven]

Oh, that might provide better motivation for taking the exponential.

[chilly ocean]

how come the adjoint [x,-] not already invertible? i am struggling to come up with lie algebra where [x,-] = 0 does not imply x = 0

[long nebula]

The Galois closure of a finite separable extension is always a finite extension, right? Because you just take the splitting field of finitely many minimal polynomials

If so I think I finished going through the proof of the primitive element theorem!

[next obsidian]

Let [x,y] = 0

For any pair x,y

Here’s a less stupid example

[chilly ocean]

i meant in a nonabelian lie alg lol

[next obsidian]

Given a ring R (not commutative 😢) you can let [x,y] = xy - yx

Then [x,-] = 0 iff x is in the center of R

[chilly ocean]

ah and if R contains a copy of a field in its center then this is a counter example

[next obsidian]

I mean

The center is always non-trivial

1 exists

Or just take square matrices

For a specific ring

[chilly ocean]

but to make R into a vector space youd need a field in its center i guess right?

[next obsidian]

I mean I guess

I didn’t really concern myself with that part haha

The word for this is an algebra over a field btw

Specifically here we need a noncommutative algebra

So like, the quaternions over R also works

[chilly ocean]

i see .. [1,-] = [0,-] = 0 and 1 is not 0 so ad cannot be an automorphism

[next obsidian]

Yeah

[chilly ocean]

i didnt get what you were saying for a sec now i get it

[chilly ocean]

new word every day

[next obsidian]

Or well like

ad an auto morphism doesn’t make sense at all

ad:g -> gl(g)

So the codomain and domain aren’t the same

But it’s not injevtive is what you’re observing

[chilly ocean]

what i mean is ad(x) is not an automorphism

for a fixed x

[next obsidian]

No ad(x) is always an automorphism I think

[chilly ocean]

because the inner automorphisms are defined by exp(ad(x))

and i was wondering the point of the exp map, and my guess was to make things invertible

[next obsidian]

Wait no

You’re right

but not because of what you said

[chilly ocean]

so i was curious why ad is not already invertible over a non abelian lie alg

but you gave the reason

[next obsidian]

This is saying ad(0) = ad(1)

[chilly ocean]

yes

[next obsidian]

This is not saying that ad(x) isn’t an automorphism

This is saying ad(-) isn’t injective

[chilly ocean]

oh i want ad(x)(-) to not be injective , not ad(-)

[next obsidian]

Yes

But this is still possible

[chilly ocean]

very tricky

[next obsidian]

I mean okay you did get a very good example

ad(1) is 0

So it’s not an automorphism

And it’s possible to have ad(x) ≠ 0 and not injevtive

I believe one way to do this is to do something like

Take a product R x S of algebras

And look at ad((x,0)) where x is not in the center

If you apply this to something of the form (0,s) it will always spit out 0

Because (x,0)(0,s) - (0,s)(x,0) = 0

Because the stuff live in different components

But if you apply ad((x,0)) to something of the form (y,0) where xy ≠ yx

You end up with (xy - yx,0) ≠ 0

Does that make sense?

[chilly ocean]

i dont think i understand the reason why [x,-] can be noninjective yet.. like what would be an example of [x,a]=[x,b] but a different from b and x nonzero

[next obsidian]

Here

[chilly ocean]

i think that is my main confusion rn lol

[next obsidian]

[x,y] = xy - yx here

In a ring R x S

If x = (r,0) and y = (0,s)

You end up with [x,y] = (r,0)(0,s) - (0,s)(r,0) = (r•0 - 0•r, 0•s-s•0) = (0,0)

[chilly ocean]

that much i follow

[next obsidian]

So if you want [x,y] = [x,y’]

Just take x = (r,0) and y,y’ = (0,s) and (0,s’)

[chilly ocean]

ah and both brackets equal zero but s != s'

[next obsidian]

Yes

[chilly ocean]

wow how did u come up w that

[next obsidian]

I mean I just

Idk, I knew the example of [,] given by multiplication and wanted to come up with the dumbest way to make [x,y] = 0 and neither 0

And I went “oh wait it’s multiplication so I can just put them in different components!”

[chilly ocean]

yeah i see i was trying to focus on matrix groups or explicit examples of lie algebras that i had seen but they all had the injectivity thing happening

[next obsidian]

Yeah

¯_(ツ)_/¯

[chilly ocean]

ty !

so in the end the ad(x) is NOT an automorphism of L

[next obsidian]

No

[chilly ocean]

but exp(ad(x)) is

[next obsidian]

That’s correct

I believe so

[chilly ocean]

and i guess somehow the feel for the inner automorphisms (the ones generated by exp(ad(x)) for x ad-nilpotent) is that they are the effect of the conjugation action on the lie group but seen as acting on the tangent space at e i.e. acting on the lie algebra

its a nice link between conjugation and commutators

[next obsidian]

So to expand on this a bit

The “canonical” Lie algebra is this xy - yx thing

As det said Lie algebras come from Lie groups, and the canonical example of Lie groups are matrix groups

GL_n, SL_n, SO_n etc etc

What ends up happening is the Lie algebra for these end up being either M_nxn, or sub algebras thereof, and the lie bracket is given by literally doing xy - yx

Now think about what happens if you apply the exponential to this

Oky wait lmfao

I thought after applying exp you get something of the form ghg^-1h^-1

But that’s not true xD

But okay whatever the first part is still good

[chilly ocean]

i believe every lie algebra is a subalgebra of some M_nxn in fact

[next obsidian]

I think this is true as well

Maybe you need some hypothesis

Definitely finite dimensional

[chilly ocean]

in humphrey's , lie algebras are finite dimensional and he just says something like "its proved in bourbaki in general, but will be obvious soon for all examples we care about, so we dont prove it"

[next obsidian]

Lol

Sounds about right

[tough raven]

In fact, that's never injective for non-zero x because [x,x] = 0.
But there's also the fact that it's not a Lie algebra homomorphism, so it can't be an automorphism even if it's bijective.

Hint: ||consider the zeroes of both sides||.

[tough raven]

The values of t for which each side evaluates to 0.

The roots?

[regal willow]

Hey, any idea how to prove that this set is a cyclic group?

I'm kinda stuck on how to find that matrix ^n = En, thought of using cos and sin, but dunno

[formal ermine]

either list all elements (don't do this) or find an isomorphism to a cyclic group

[rustic crown]

Use that ||group of units of a finite field is cyclic||

[frigid lark]

This server has , thought it was just a nitro emote

[rustic crown]

my second message in the server was

[summer path]

chill

[next obsidian]

Which follows from the general fact that finite subgroups of mult groups of fields are cyclic

[south patrol]

Which follows from the general fact that any finite group of order n with <= d elements of order dividing d for each d|n is cyclic

[lethal dune]

which is a consequence of group theory

[formal ermine]

which is a consequence of zf(c)

[rustic crown]

oh how do you see this? i can only see it quickly for finite abelian ones

or if you change it to "number of elements of order dividing d is at most d" then the same proof works

[south patrol]

Oops yeah lol

I think the way I learnt to do this proof originally was to use properties of the phi function

Oh yeah I realise this is actually stronger since we haven't assume the group is abelian

Lol why that reaction

[rustic crown]

yea

[coral shale]

the phi function shall remain locked up in #elementary-number-theory

[south patrol]

One can just show that sum over d|n of φ(d) is n by partitioning a cyclic group of order n into elements of order d, of which there are φ(d)

[rustic crown]

if N(d) is the number of elements of order d, then you're only telling me that N(d) <= d instead of N(d) <= phi(d)
so don't immediately see how sum of N(d) = n and sum of phi(d) = n would tell something

[south patrol]

Yeah I made another typo lol

/ mistake

Anyway

[rustic crown]

i was worried if i forgot group theory >.<

[south patrol]

Lol sorry

But yeah the point is that with these constraints you have either 0 or φ(d) elements of order d I think for each d | n

[rustic crown]

yee that true

[south patrol]

And then by bounding you see you must have φ(d)

In particular φ(n) elements of order n

hehe

[rustic crown]

[south patrol]

But yeah i've never seen that theorem applied other than 1) as a first year group theory problem basically (but in another way) or 2) a less quick proof of the result Chmonkey quoted lol

i guess it's the most elementary way to show that finite subgroup of F^x is cyclic but less quick once you have a classification of fg ab grps

hm anyone got any cool Galois theory questions

[rustic crown]

there's that another elementary way where you use prime factorization to show given elements of order a and b there is an element of order lcm(a,b)

but if find this argument with phi cuter

[rotund aurora]

Hi

[empty rose]

i might be able to come up with a bad galois theory question if i can remember enough words to construct a grammatically valid question that's also about galois theory
other than that i don't have anything

[coral shale]

trisect my tetrahedron

[rotund aurora]

So I think linear groups and lie things play a prominent role in quantum mechanics and physics stuff. Would it be nice to read some of this to get a better grasp on the linear groups?

[delicate orchid]

wtf is ramification theory

[delicate orchid]

you read more about linear groups to get a grasp on those things not the other way around

[south patrol]

Lying over property for integral ring extensions

[rotund aurora]

but applied things give a lot of insight

sometimes

[south patrol]

Oh how does this work

Oh okay lol

Trying to remember the easiest way to show existence of primitive roots mod p

[rotund aurora]

primitive roots are actually crazy

[empty rose]

...or i could ask grammatically invalid questions i guess
are my antennae field extensions?

[south patrol]

Lol

That is grammatically valid though

[delicate orchid]

who up validating they grammar

[rustic crown]

is there a nice way to go from p to p^n?

[south patrol]

It just makes no sense

[south patrol]

Eh I think Hensel's lemma

Or similar

Like

[rotund aurora]

ye

[south patrol]

So you can show that any prim root mod p^2 is a prim root mod p^n for any n

And then u only need to lift once

[rustic crown]

ah right

[south patrol]

Which isn't hard

But I think there is a better way in general

[rustic crown]

i had this more constructing thing in mind, which was if g is a generator then g or g+p is a generator for all p^n

[south patrol]

But idk I speedread elementary number theory

Yeah ithjnk that's what I learnt

[empty rose]

well yes but also no, i'm counting types as part of the grammar

[south patrol]

Wdym by types lol

I don't see how this is ungrammatical in any way

[rotund aurora]

if g is a primitive root mod p^2 then it is a primitive root modulo p^n for all n>2

[empty rose]

the same kind of issue as like
"is the number 4 a subgroup of the number 5"?
4 and 5 are not groups, so asking if they're subgroups of things is nonsense

[south patrol]

Yes but it's grammatical lol

[empty rose]

well it's still clearly an invalid question in some way

there's no way to answer it, and if you did have an answer you wouldn't be able to do anything with it

[carmine fossil]

4 is not a subgroup of 5

Because <4> is not a subgroup of <5>

[empty rose]

well i don't think 4 and <4> are the same thing
or 5 and <5>

[empty rose]

...why is this true? i can't think of a particular reason it wouldn't be true but it feels surprising

[carmine fossil]

I wonder if that is due to Hensel lifting

[rustic crown]

it jsut induction ig?
if g is a primitive root mod p^2, then g^phi(p) is not 1 mod p^2.
from this you get g^phi(p^(n-1)) is not 1 mod p^n.

so order of g mod p^n will divide phi(p^n) and by induction is divisible by phi(p^(n-1)), it can't equal phi(p^(n-1)), so done

[empty rose]

...i feel like something is missing there and i can't tell if there actually is or if my brain is just not working
i should probably sleep

[rustic crown]

[south patrol]

there r some calculations needed tbf lol

[rustic crown]

yea, i skipped the induction

[south patrol]

Okay ye lemme do that quickly if one is inch rested uh

[rustic crown]

g^phi(p^(n-1)) is not 1 mod p^n.
this is not immediate lol

[south patrol]

basically sps $g^{p-1} \ne 1 \mod p^2$, then write $g^{p-1} = 1 + ap$ for some $a \not = 0 \mod p$, then you can just show that $(1+ap)^{p^{k}} \equiv 1 + ap^{k+1} \mod p^{k+2}$ I think, with $k=0$ trivial and inductive step from binomial expansion lol

[cloud walrus]

potato

[formal ermine]

what does sps mean

[south patrol]

suppose

lol sorry

What I do in exams lol

[oblique river]

it's what you say to get a cat's attention

spspspspspspspsps

[rustic crown]

[pastel cliff]

what do we mean when a say a module is torsion

[rustic crown]

there is a non-zero divisor in your ring which kills the whole module

wait maybe i'm wrong :p

it could also mean for each element in your module, there is a non-zero divisor in your ring which kills that element

like every element in the abelian group Q/Z is a torsion element, but idk if the whole thing should be called torsion module

@pastel cliff >.<

[rustic crown]

okie wikipedia says it's this

see this quickly >.< i gave you wrong info >.<

[pastel cliff]

i see it det

thank u

[coral shale]

meow !

[formal ermine]

when is $H \oplus G/H = G$?

[formal ermine]

G is a group

H is a normal subgroup

yeah not in general

the problem I originally intend on solving is concerned with sess of sheaves that induce sess of abelian groups, and it boils down to this

[agile burrow]

For abelian groups, splitting lemma tells you that this is equivalent to having a section/homomorphism G/H -> G such that the composition G/H -> G -> G/H is the identity. Equivalently, you have a homomorphism G -> H such that H -> G -> H is the identity.

For groups in general, you'd get that G has the structure of a semi direct product (probably not relevant, but I'm throwing it out because it's interesting)

[formal ermine]

ah ok thanks I didn't know about the splitting lemma

[agile burrow]

Mm yeah, it's worth working through if this is your first time encountering it. It's also useful if you want to think about projective and injective modules

[formal ermine]

mhm

[oblique matrix]

Hey all!

I can't find anything about "positive sets" online and I don't really know where to look. I suspect the following to be true but I can't really prove it since I don't know how to incorporate axiom 2) and 3)

Let R be a commutative ring. Let P be a subset of R satisfying the following properties:

1. for every r \in R, exactly one of the following 3 statements hold: i) r \in P, 2) r = 0, or 3) -r \in P.
2. if a \in P and b \in P, then a+b \in P
3. if a \in P and b \in P, then a*b \in P.

Is it true that if a positive set exists in a commutative ring it must be unique?

[rustic crown]

nope

if you have a non-trivial automorphism in your ring, then you can use it to translate one positive set to another

simplest example is that of Q(sqrt2) with the usual order induced as a subset of R

but you can define an element a+b*sqrt2 to be in P if a-b*sqrt2>0

[formal ermine]

do these axioms even work? if your ring contains a nilpotent element then 1) breaks

[rustic crown]

yea so it means that (non-zero) ordered rings are domains :p

(that's how you prove Z is an integral domain btw :0)

a nice and cute axiomatization of Z is that it's an ordered commutative ring where the positive elements are (non-empty and) well ordered.

[rustic crown]

(to see that, let R be a ring with those properties. notice positive elements non-empty means 1 is not 0, and since 1*1=1, it can't be negative. so 1 is positive. now if m is the smallest positive element, then m = 1 otherwise m < 1 implies m^2 < m. since your ring is ordered, it can't have positive characteristic which means we have an injection Z --> R. to show this is surjective, consider the smallest positive r which is not in the image. since r-1 > 0, r-1 is in the image, but then so is r, so r can't exist. Z --> R surjects onto positve elements, so must surject on the negatives as well :3)

[delicate orchid]

le 2nd cohomology has arrived

[formal ermine]

heh?

[delicate orchid]

the 2nd cohomology group of a group classifies the group extensions, of which the semidirect products are a subset

[formal ermine]

what's the rigorous definition

[delicate orchid]

of which one

[formal ermine]

2nd cohomology group

[delicate orchid]

it's the quotient of the 2-cocycles by the 2-coboundaries, which I will now also explain

[elder wave]

I don’t think you need to go there

[delicate orchid]

take it away timo I'm gonna go play video games

[elder wave]

But feel free to do a writedown if you feel like it

It’s the dual to homology

You’re welcome

👍

[delicate orchid]

👍

hope this helps 👍

[elder wave]

👍

[formal ermine]

ah that makes it clear

could've just said that at the beginning

[agile burrow]

Group cohomology

[oblique matrix]

Thank you!

[agile burrow]

If you're willing to take on faith what 2-cocycles and 2-coboundaries are, then the connection to group extensions isn't too hard to realize

[rustic crown]

oh so group cohomology is deriving a hom functor?

[agile burrow]

Yes, but I don't think that's the natural way to see the connection to group extensions

[rustic crown]

(that wasn't your original question tho >.<)

[agile burrow]

Given a group G and a G-module A, you can define a normalized 2-cocycle to be a function f: G x G -> A such that gf(h, k)- f(gh, k) + f(g, hk) - f(g, h) = 0.

Now given an extension 0 -i-> A -> E -> G -> 1, there's an action of G on A induced by conjugation inside E. Explicitly, given g in G, let e be some element in the preimage in E. Define the action of g on a by e a e^-1 (it's a quick exercise to show that this is well-defined). This turns A into a G-module.

Fixing a set-theoretic section s: G -> E such that s(1) = 1, we can ask for how far s is from a homomorphism. Explicitly, s(gh) and s(g) s(h) both map to gh in G, so they differ by an element of A in E. We consider the factor set associated to the extension, which is the map f: G x G -> A defined by f(g, h) = s(gh) s(g)^-1 s(h)^-1. Note that f is trivial iff s is a homomorphism, in which case the extension is split and E is a semidirect product of A and G.

Now we have a set-theoretic bijection A x G -> E by (a, g) -> a s(g). To pull back the group structure of E to A x G, the action of G on A forces the multiplication to satisfy i(a) s(g) i(b) s(h) = i(a) i(gb) s(g) s(h) = i(a + gb) i(f(g, h)) s(gh) = i(a + gb + f(g, h)) s(gh). This determines the group law on A x G. In particular, associativity of the multiplication forces f to be a 2-cocycle.

Conversely, if you start with some function f: G x G -> A and define a multiplication on A x G by (a, g) (b, h) = (a + gb + f(g, h), gh) then A x G is a group iff f is a normalized 2-cocycle (this is a 2-cocycle such that f(g, 1) = f(1, g) = 0). This yields a bijection between normalized 2-cocycles and group extensions with a normalized section.

The last piece of the puzzle is that if you choose two different sections s and s': G -> E then the corresponding factor sets differ by a 2-coboundary, which isn't hard to just write out explicitly. That is, equivalence classes of extensions of G by A are in bijection with H^2(G, A)

So it's kinda strange because Ext^n corresponds to module extensions of length n, but since groups are weird, H^2 gives you group extensions of length 1

I will say that while I've written this up with an abelian kernel, a lot of this extends to the case where you want to study extensions 1 -> N -> E -> G -> 1 where both N and G are nonabelian. It turns out that if the center of N is trivial, then there's a unique extension associated to a map G -> Out(N) (this is the data of the action of G on N, except now N isn't a G-module since it's nonabelian). More generally, you'll end up studying the second cohomology of G with coefficients in the center of N.

Probably the most interesting thing though is that you can broadly ask about the existence of certain extensions. So like, given groups G and N, and a homomorphism G -> Out(N), you might wonder if there even exists an extension 1 -> N -> E -> G -> 1 such that the induced action of G on N is described by the homomorphism G -> Out(N). There turns out to be a cohomology class in H^3 that you can associate with this data, and such an extension exists iff this cohomology class is trivial

Super explicitly, given a group G and a G-module C, each element in H^3(G, C) can be used to construct a group N such that Z(N) = C and a homomorphism G -> Out(N) which induces the action of G on C when restricted to Z(N) such that the corresponding cohomology class associated to this data is the one you started with

It's really kind of miraculous that it works so well

[agile burrow]

If you want to see group cohomology as a derived functor, you can form the group ring ZG. Then group cohomology H^n(G, M) = Ext_{ZG}^n(Z, M), so it's the derived functors of Hom_{ZG}(Z, -)

[rustic crown]

(will read this message tomorrow >.< a lil tired rn)

[rotund aurora]

Just find the primitive 2m-th roots of unity in terms of the promitive m-th roots of unity

[primal tusk]

can somebody help explain what this is asking

like what is O(h)

[agile burrow]

It probably refers to the orbit of h under the action of D_6

[primal tusk]

is an orbit basically just a coset

[tender wharf]

its a subgroup actually

[agile burrow]

Oh, the orbit of a point is the image of the point under the action. So like, D_6 acts transitively on the vertices of the hexagon, so the orbit of one of the vertices is the set of 6 vertices

Another example, Z/2Z acts on Z by sending an integer n to -n. The orbit of n is then {n, -n} if n is non-zero. If n = 0 then the orbit is just {0}

A basic fact is that the orbits partition the set which your group is acting on

[primal tusk]

im trying really hard to comprehend this with my one remaining brain cell

im not following the whole Z/2Z sending n to -n thing though

like does that just send things to 0 or 1 ?

[agile burrow]

So Z/2Z has two elements. The identity element has to act trivially by definition of a group action. We're defining the non-identity element to act on Z by sending n to -n

You can verify that this is a group action

[primal tusk]

ah im realizing i dont even know what a group action is i need to go read up on that first lol

[white oxide]

how important is fermat's little theorem in a introductory course to abstract algebra lol

[tender wharf]

you actually can prove it using lagranges theorem

[white oxide]

oh interesting

[frigid lark]

When your lecturer uses it to prove CRT

cause the introductory algebra course doesn't have rings in it

[summer path]

an introductory abstract algebra class will likely have rings in it, no?

or is that not what an intro course to abstract algebra class is

[frigid lark]

Not at my uni

[summer path]

what did your intro aa class cover >.<

[frigid lark]

Groups, up to Sylow/finite abelian groups and wallpaper groups of all things

[summer path]

i see

i think it's pretty reasonable for an intro to aa course to cover some subset of groups, rings, modules, field theory, galois theory though

[frigid lark]

In one semester?

Oh, didn't read the subset part

[summer path]

yeah lol

also it's possible for an intro to aa sequence to take 2 semesters as well

[frigid lark]

That sounds good

Yes

[white oxide]

ye

that's my first semester algebra course

[solar glacier]

Question

In proving Q(sqrtd1) iso to Q(sqrtd2) iff d1=d2 where they’re both square free integers

Can I say the first fuel extension is iso to Q[x]/(p(x)) for some irreducible polynomial p and similarly the latter is iso to Q[x]/(q(x)) for some irreducible a

Which are iso by assumption and isomorphism is transitive

So the polynomials must have the same root

Ie., d1=d2

Supposing d1=d2 is the trivial direction

Does this argument work

Anyone ?

If I can sssum those are iso

[empty rose]

i might just not know something, but, where did this come from?

[delicate bloom]

I don't follow, you can have different field extensions that are iso coming from polynomials with different roots

[solar glacier]

Hmm

Was so at least on the right track somewhat

Was I *

[delicate bloom]

idk, hard to tell, it feels like you're dodging too much of the question though

[solar glacier]

Cause that’s what I answered on an exam lol

Wondering what kind of partial points I’ll get then

[empty rose]

...you didn't use the fact that d1 and d2 are square-free

[delicate bloom]

I don't see any reason to give partial credit

[empty rose]

also wait hang on

Q[x]/(x-1) and Q[x]/(x-2) are isomorphic right? because they're both basically just Q

so you're claiming that if Q is isomorphic to Q, then 1 = 2

[solar glacier]

Oh I see my error in logic now

[delicate bloom]

yeah that's what I'm saying here basically

[solar glacier]

😢

[delicate bloom]

but I see why that's opaque lol

[solar glacier]

But

[delicate bloom]

another similar example is look at R adjoin the roots of any irreducible quadratic, you get the same field C

[solar glacier]

Sqrtd1,sqrtd2 not in Q

[delicate bloom]

x^2+1 and x^2+x+1 have different roots in particular

[empty rose]

sqrt2 and sqrt2+1

[solar glacier]

I’m talking@about the d’s in my problem

[empty rose]

(i mean sqrt2+1 isn't the square root of an integer but it isn't particularly obvious that there isn't any counterexample that looks like this)

also sqrt2 and sqrt8 aren't the same number but will give you isomorphic field extensions

that's why the requirement that they're square-free is there, because otherwise the thing they're asking you to prove would be false

[solar glacier]

Was I headed the right direction with saying they’re iso to Q[x] modded out by an irreducible poly

[empty rose]

i have no idea, i don't actually know how to prove that statement

[solar glacier]

Then I’d have to some how ustilize the fact that their swore free

Square

Rip 🪦 my midterm

I got a few proboems somewhat correct

[summer path]

i mean you might get partial credit just by the fact that it's a midterm and i think people tend to be more generous than they probably should be with partial credit

but that depends on who's grading

[solar glacier]

Yeah this guys huge on partial credit as am I as a grader for classes I TA for

[thorn delta]

i think i have an idea for this one if you want to hear it

[formal ermine]

https://www.calstatela.edu/sites/default/files/users/u1061/cyclotomic.pdf this has a proof

theorem 15

oke

have you proven that x^n - 1 = prod dth cyclotomic polynomial for d divisor of n in class

so x^(2p) - 1 = phi_2p phi_p phi_2 phi_1

yee

typo >.<

yee that's just rewriting the product

ye

[formal ermine]

now we need to find out what phi_p phi_2 phi_1 is

I assume you've talked about pth cyclotomic polynomials in class?

yee so phi_1 phi_2 = x^2 - 1

for phi_p you can use the same theorem again

x^p - 1 = phi_p phi_1

ye

which is x^(p - 1) + x^(p - 2) + ... + 1

now multiply in x^2 - 1

[atomic mesa]

Sorry to interrupt, just want to ask if these sets are the same?

[formal ermine]

oo that's smart

I multiplied out the whole thing lol

but using x^p - 1/x - 1 is way better

ye

.<

I didn't know that you had this fact available

oke so

pth cyclotomic polynomial is x^(p - 1) + ... + 1

plug in -t for x

yee and like you said p-1 is even

so we get t^(p - 1) - t^(p - 2) + ... + 1

heh?

if p - 1 is even then

(-t)^(p - 1) = (-1)^(p - 1) t^(p - 1) = t^(p - 1)

[elder wave]

bro thinks he's det

[formal ermine]

ye and now we're done

[formal ermine]

ruhe auf den billigen plätzen

[formal ermine]

for this we could've just done the polynomial division fwiw

for what it's worth

ye

[pliant raptor]

What is an example of a ring $R$ with the property that $$\forall x \in R, \exists n>1: x^n=x$$ that isn't a field?

[cloud walrus]

ImHackingXD

[rustic crown]

F_p x F_p?

[south patrol]

The 0 ring

and Boolean rings give a nice family of examples (ones in which x^2 = x for all x)

For example, let S be a set and put operations on the power set P(S) by A + B := symmetric difference of A and B and A.B the intersection, then A^2 = A for all A in P(S)

[rustic crown]

enslaved (F_2)^S

:p

[south patrol]

Lol

[pliant raptor]

Thank you both

[solar glacier]

yes plz

[sage lodge]

what's the difference between a finite extension and an algebraic extension of a field? is one always the other?

[cloud walrus]

tushar

[formal ermine]

yee finite => alg

[sage lodge]

but are there algebraic extensions that are infinite dimensional?

[solar glacier]

yes i wanna say

R:Q

[formal ermine]

take any finite field and its alg closure

[formal ermine]

but only fg and alg => finite

[empty rose]

...R is not algebraic over Q

assuming you mean the real numbers

that includes stuff like pi

[elder wave]

adjoin the n-th root of 2 for all n to Q for example

[empty rose]

but yeah there are definitely infinite algebraic extensions where you just add "infinitely many" algebraic elements

[sage lodge]

ohh i see

and how can we show that's an infinite extension by the way?

[elder wave]

you can show that the minimal polynomial is X^n-2

[K:Q] is then greater or equal than n for all n

[sage lodge]

over the previous field right, since you adjoin them one at a time

[empty rose]

...if you adjoin them one at a time you're never going to finish?
unless you then do some kind of union thing

[sage lodge]

hmm yeah

[empty rose]

but i interpreted the construction as adding them all at once

[elder wave]

yes

this is over Q

[pastel cliff]

https://tenor.com/view/timi-gif-20112884

[empty rose]

it shouldn't be that hard thinking about it
add xn for each n, then quotient to make them all roots of 2

we're doing pretty infinite things but they don't have to be done sequentially

[rotund aurora]

was there any fast procedure for calculating inverses mod n

or you just use Euler's formula

and repeatedly square to obtain the value of a^(varphi(n)-1)

[pastel cliff]

what is V_T referring to here

(the homework 4 it mentions is the one i decided to drop this semester)

[agile burrow]

V_T is just referring to the vector space V as a k[t] module where the action of t is given by T

So the underlying abelian group of V_T is just V, but to make it a k[t] module, we have to specify how t acts. We define tv := Tv where T is the given linear map

[tawny cloak]

this sounds like it could be a good way. another way is to use Euclid's algorithm to find x, y such that xa + yn = 1 and then x is the multiplicative inverse of a mod n.

i'm not sure which is faster but I think they both have logarithmic complexity

[tawny cloak]

I think computing phi(n) is hard in general though

[pastel cliff]

what is the relationship between characteristic and minimal polynomials?

and in general what is the motivation for considering minimal polynomials in the first place

[fleet pelican]

The minimal polynomial factors the characteristic polynomial. If I remember correctly.

[rotund aurora]

Right, requires prime factorization

Probably euclid division way faster

[tawny cloak]

yes and in fact if N = pq and you know phi(N) then it's pretty easy to find p, q

[fleet pelican]

It encodes certain information about a matrix. There are some theorems and proofs which use it. It's kind of a way to understand how the matrix behaves as a function in the space of endomorphisms.

[tawny cloak]

I haven't really learned this yet but I'm just thinking now and for a vector space V over k and transformation T : V -> V, there's a unique algebra homomorphism k[t] -> Endo(V) where t -> T, so this means p(t) in k[t] acts by substituting T for the indeterminate t. Then the zero set of this map is an ideal of k[t], and k[t] is a principal ideal domain. So this means that for any polynomial p(t) in k[t], p(T) = 0 if and only if the minimal polynomial divides p(t).

[solar glacier]

Question is the ideal of functions whose sum of coefficients is zero a principle ideal in the polynomial ring

Principal**

[fleet pelican]

yes

[solar glacier]

My gut tells me yes because you can generate it with 1,-1s?

[tawny cloak]

every ideal of k[x] is a principal ideal (when k is a field!)

[fleet pelican]

closed under addition and multiplication

[tawny cloak]

in fact x - 1 generates it, I believe

[solar glacier]

Yes a 1 and a -1

[pastel cliff]

interesting! i'll keep reading, mucho thanks :)

[rotund aurora]

are you asking about linear algebra?

cuz like, looking at this things is how you obtain the Jordan canonical form

and that is super useful

[pastel cliff]

it's with regards to JCF for general modules over a PID

[tawny cloak]

yeah this might be an overly abstract answer if you're just learning linear algebra now

I learned linear algebra on my own a long time ago and I think I didn't study it deeply enough. really I should go over it with my newfound knowledge again.

[rotund aurora]

isn't JCF enough motivation lol.

[fleet pelican]

Linear Algebra is god's gift to mankind.

[tawny cloak]

that's why I said "I haven't really learned this yet but" then produced to describe something pretty elaborate

[solar glacier]

I could never learn a subject on my@own lol

[pastel cliff]

yes you can!

[solar glacier]

It’s a huge toolkit course

[tawny cloak]

and the natural numbers are god's challenge to mankind

[tawny cloak]

100% of my math knowledge is self taught lol

[solar glacier]

🤯

I’m not that disciplined smh 🤦‍♂️

[fleet pelican]

Damn

[solar glacier]

I can barely learn it while taking the class

I’m getting DESTROYED in alg top rn

[rotund aurora]

anyway, for matrices of degree 2 and 3, the characteristic polynomial is one of the first things that should come to mind if you are dealing with powers of matrices. If there is some special relation with the trace and the determinant, then you can get really nice expressions

[tawny cloak]

I'm 18 now and I wanna get into a good university but I don't have any formal certifications like really good grades or anything like that cause I had a weird complicated path until now

and all of my knowledge is self taught so

I'm pretty fucked

[rotund aurora]

lol I relate

[solar glacier]

Well I’m 32 getting a phd got my masters and bachelors both in math and I rely heavily@on instruction to aid my learning

[fleet pelican]

That shit is wild. The ppl who invented that went crazy and created weird category theory stuff.

[solar glacier]

Is really is

Like u can picture everything

But putting it into proof format ….

Like writing down certain deform retracts gets tricky

[fleet pelican]

Write it in your application. They usually like ppl with atypical backgrounds. (at least in the US)

[solar glacier]

That part ^^

[tawny cloak]

I'm hoping to succeed in doing something like that. I'm not in the US but maybe it'll work for me here too.

[solar glacier]

It also wouldn’t hurt mentioning some actual explicit mathematics in your letter of purpose

Like what draws you in