groups-rings-fields

[candid dove]

Huh

[unkempt stream]

What are the orbits of Stab(S) in S?

[candid dove]

Stab(S) actis trivially on S?

Isn't that what stab is?

Like it permutes S

[unkempt stream]

Different action

[candid dove]

But yeah

[unkempt stream]

There we go

[rocky cloak]

Well, I think some of those will end up being equal.

[unkempt stream]

Stab(S) acts on S by sending elements of S to elements of S

It doesn’t fix every element of S

[candid dove]

Yeah I meant that like

[unkempt stream]

So Stab(S) acts on S

[candid dove]

Okay

Yeah

[unkempt stream]

What are the orbits?

[candid dove]

Elements of S of some kind ?

[unkempt stream]

Describe them explicitly

[candid dove]

Uhhh

[unkempt stream]

Don’t think too hard

[candid dove]

Idk

[regal zodiac]

Oh u mean i will be able to create x^n-1(y+1) with x^n(y+1) qnd the other members ?

[unkempt stream]

Stab(S)x lol

What are those sets called

[candid dove]

Aren't they called orbits of x

[unkempt stream]

Cosets

They are cosets of Stab(S)

[candid dove]

Oh okay

[unkempt stream]

What can you conclude about S itself

[rocky cloak]

Wouldn't surprise me at least

[candid dove]

It's a disjoint union of cosets of stab(S)?

[unkempt stream]

Yes!

So what does that say about |S|?

[candid dove]

Oh |stab(s)| | S?

Ohkay

Yeah

[unkempt stream]

And we stated |S| is coprime to |G|

[candid dove]

👍

[regal zodiac]

But even then, (x^3+3x^2 +2xy y^2 -1, ) will Never be simple in ( x^3 +3x^2 +2xy , y^2 -1 , x^n(y+1)) will it ?

[candid dove]

Right cool

[unkempt stream]

But |Stab(S)| divides both so it is necessarily 1

Yep

[candid dove]

Okay it wasn't that hard lol

[regal zodiac]

I'll take a look

[candid dove]

But wasn't like a one liner atleast

Or maybe if your lines are long

[unkempt stream]

Tldr:

if we have a finite group G, every subset S is a disjoint union of cosets of its left-multiplication stabilizer Stab(S), so |Stab(S)| | |S|

That is the fundamental lemma for Sylow’s theorems

[candid dove]

I might have actually forgotten sylow's proof

[unkempt stream]

Which basically partitions the set of subsets of order p^n based off the power of p for the orders of their stabilizers

[candid dove]

Right

[unkempt stream]

I wonder if you can use the same bs to do solvability arguments

[wraith cargo]

nobody remembers that sylow theorems are FAKE i HATE them

[unkempt stream]

They’re not too bad :c

[candid dove]

Group theory goated

[regal zodiac]

[wraith cargo]

Ummm show every group of order 36 is simple 🤓 🤓 🤓 🤓 🤓 🤓 🤓 🤓

(literally poison)

[candid dove]

I might consider going into geometric group theory which is not group theory entirely but yeah

[unkempt stream]

I have been trying to do some silly bs where I identify subsets of cardinalities dividing the order of the group by the prime decomposition of their stabilizer’s order lmao

And doing arguments based off of that to try and find normal subgroups

Like how if there is one Sylow subgroup then it’s normal

[candid dove]

I think group theory atleast for finite groups is essentially combinatorics

[unkempt stream]

Basically yes

[candid dove]

Which is based

[wraith cargo]

be warned if you talk about finite groups too much you'll summon demons who will use words like "fusion systems" to confuse you

[unkempt stream]

@delicate orchid

[candid dove]

Wait are we specifically talking about some members

Ohok

[arctic trail]

goated

[arctic trail]

Poisson?

[arctic trail]

wait what

this isn't evne right

dawg I was so confused

[rocky cloak]

Think they meant no group of order 36 is simple

[wraith cargo]

Yeah I meant that I didn't remember the exact statement anymore lol

[unkempt stream]

Me when cyclic group

[wraith cargo]

I just pulled something out of my ass

[unkempt stream]

36 = 6^2

[arctic trail]

dawg

small prime divisors

how am I even supposed to use Sylow

y'ain't even throw a bone

[unkempt stream]

My favorite way to show wild isomorphisms is to use Sylow group actions

Groups act on their own Sylow groups lmao

[candid dove]

Guys if y'all could check #advanced-algebra message 🙌

[unkempt stream]

You can show there is no simple group of order 24 using a psychotic Sylow action argument

I need to remember it

Oh I know lmao

If |G| = 24 and G is simple, then every group action by G is faithful.

G has either 1 or 3 Sylow 2-groups, and there must be 3 due to simplicity. That means G acts by conjugation on the 3 Sylow subgroups faithfully, meaning G embeds into S_3, a contradiction as |G| > |S_3| = 6 lmao

Speaking of which, is there a name for the kernel of a group action?

For modules it’s the annihilator

[delicate orchid]

What’s up

Fusion systems are easy it’s centric linking systems which are fucked

[unkempt stream]

And this implies that every action from a simple group is faithful implying that it acts faithfully by conjugation on all it’s Sylow group collections

Idk what the implications of that is

[delicate orchid]

every action of a simple group is faithful
The trivial action:

[unkempt stream]

Well beyond the trivial one obv

[delicate orchid]

A less boring answer, action on the cosets of a proper subgroup

[unkempt stream]

Since that’s the only nonfaithful one

But there’s a kernel for the action no? So it has to be faithful as there can’t be a non proper/trivial kernel

[delicate orchid]

Stabilisers need not be normal

[unkempt stream]

Stabilizers sure

But what about the “total” stabilizer

[delicate orchid]

These aren’t representations they’re G-sets

[unkempt stream]

Stabilizing the whole set

[delicate orchid]

Yeah… a subgroup fixes all of its cosets…

[unkempt stream]

I’m super confused

[delicate orchid]

Well it fixes one anyway

The other ones you have to do transfer nonsense to figure out where they go

[unkempt stream]

Eh fuck group theory

[vapid vale]

tru

[south patrol]

Groups are fun

[tardy hedge]

Do you like groups or rings more

[south patrol]

Well

Eh weird to compare them lol

But I don't do group theory and am more interested in AG so in that sense rings

Hbu

[arctic trail]

the heck is a g-set

[dim widget]

set with g action

[arctic trail]

I mean

representations are in bijections with group actions (?)

[dim widget]

no?

[arctic trail]

they're not?

[dim widget]

Every representation consists of a group acting on a set, but not every group action on a set is a representation?

Given a group acting on a set there is an associated permutation representation, but not every representation arises this way

so no

[arctic trail]

Sure

You can always linearize a group action into a linear representation.
And every linear representation has a group action associated with it.
surely this now follows from like Schröder–Bernstein

I wasn't saying it was a Functorial relation

[dim widget]

lol

[arctic trail]

I'm too sleepy for this

But I thought statements in group actions usually had broader representation theoretic generalizations

Like burnside's lemma

[dim widget]

I think Burnside's lemma is just an example of something which can be proved as a statement both about permutation representations and about group actions.

[arctic trail]

fair, burnsides lemma is essentially just interpreting an inner product

[tardy hedge]

Based on my experience so far, I like learning about rings more

Im excited to learn algebraic geometry

[rotund aurora]

Consider a polynomial $x^3+ax^2+bx+c$ with rational coefficients with roots $\alpha_1,\alpha_2,\alpha_3$ and Galois group $\Z/3\Z$. Can you calculate $\alpha_1^2\alpha_2+\alpha_2^2\alpha_3+\alpha_3^2\alpha_1$ and $\alpha_1^2\alpha_3+\alpha_2^2\alpha_1+\alpha_3^2\alpha_2$ in terms of $a,b,c$?

[cloud walrus]

croqueta3385

[rotund aurora]

I mean you can given that you can write the roots in terms of radicals, but I would like a simpler expression

[dim widget]

yes

by Newton's theorem every symmetric polynomial can be expressed in terms of the elementary ones

[rotund aurora]

yes, but these polynomials are not symmetric

[dim widget]

What is Q[x, y, z]^{Z/3}?

[rotund aurora]

A^2+B^2 and A+B are symmetric

[dim widget]

yep

[rotund aurora]

wher A=alpha_1^2 · alpha_2+..., B=alpha_1^2 · alpha_3+...

[dim widget]

Anyway the way I would go about this is that you want to understand some polynomials which are in K = Q[a_1, a_2, a_3]^{Z/3Z}

in terms of the symmetric polynomials L = Q[e_1, e_2, e_3]

K/L is a quadratic extension so there is a quadratic formula for the elements you want in terms of the coefficients

[rotund aurora]

right

[dim widget]

here it's easy to work out by hand: Tr(A) = A + B and Nm(A) = AB I assume

[rotund aurora]

yeah

[dim widget]

so that's your expression then

[rotund aurora]

right. (T-A)(T-B) is a polynomial in T with symmetric coefficients. One can then use the quadratic formula

[dim widget]

A^2 - Tr(A)A + AB = 0 then you are left with expressing the coefficients in terms of the coefficients of the polynomial you started with

[rotund aurora]

More generally, I am interested in calculating norm forms given roots alpha_1,...,alpha_n of some polynomial of degree n. Like prod_sigma ((sigma alpha_1 )x_1+...+(sigma alpha_n) x_n) where sigma runs through all elements of the Galois group. Are there reasonable algorithms?

[dim widget]

I don't know of any reasonable algorithms for this which are particularly fast

[rotund aurora]

Are there formulas in general?

like you are only allowed to take radicals and stuff

the corresponding extension K/L need not be solvable

[dim widget]

for the norm? not just in terms of the roots, you have to know how galois acts on them.

[rotund aurora]

what if I tell you it's cyclic and the degree is prime?

[dim widget]

i guess it depends on what exactly you want to know

if you only care about some power of the norm you can pretend its S_n

[dim widget]

hmm interesting

[rotund aurora]

yeah, I did realize that

but then, I suppose you can have different forms f,g with f(x)^n=g(x)^n, right?

[dim widget]

yes definitely

well actually

not over Q

like if you are looking at norm forms over Q then they have unique Q-valued nth roots

if n is odd

[rotund aurora]

I don't think this is true actually

unless I'm being stupid

[dim widget]

why's that?

[rotund aurora]

Say for the case cubic cyclic. You are taking a product of $\prod_{cyc}(\alpha_1 x+\alpha_2 y+\alpha_3 z)$ (cyc is meant to permute cyclically the $\alpha_i$). This is not symmetric in $x,y,z$, and there is no reason why powers of this should be symmetric

[cloud walrus]

croqueta3385

[dim widget]

the product will be symmetric though because it will lie in Q

[rotund aurora]

yes, but you are not allowed to permute the alpha_i freely

In particular, if A!=B (A, B as above) then it's not symmetric

[dim widget]

uhhh

[rotund aurora]

like prod_sym ... is not a power of that

or not necessarily

[dim widget]

So my logic was

we want to compute Nm^{H}_{S_n}

ah I see

yeah I was being foolish

[rotund aurora]

btw, if f(x)^n is symmetric, then f(x) is symmetric, right?

[dim widget]

I don't exactly understand the question

like f(x) is a standin for a polynomial in many variables

[rotund aurora]

oops meant to write f(x_1,...,x_n)

mb, sorry

[dim widget]

okay

I would guess that is true for n \geq 5 anyway if youre looking at f(x)^m and m > 2

otherwise x_1 - x_2 isn't symmetric but its square is

in general this would work for \prod_{i < j} (x_i - x_j)

the point is by galois theory if f(x)^n is symmetric but f(x) is not then that tells you that there is a Kummer subextension of Q(x_1, \dots, x_n)/Q(e_1, \dots, e_n)

but then there would be a normal subgroup of S_n, yada yada

[rotund aurora]

yeah I was writing that. But like, you might as well just prove it for C(x_1,...,x_n), no? This way when you adjoin f to the symmetric polynomials, your extension is Galois (because you have roots of unity)

and so it's a quotient of S_n

[dim widget]

it doesn't matter either way because Q(x_1, \dots, x_n) is normal

so allegedly if one root of a polynomial is in it, so are all of them

[rotund aurora]

yes but you want Q^sym(x_1,...,x_n) < Q^sym(x_1,...,x_n)(f) to be normal

[dim widget]

no I just want to know that SymPoly(f) generates a subextension of Poly_n which is normal and whose galois group is solvable

like the point is the normal closure of X^n - p(x) is always solvable

anyway

[rotund aurora]

yeah but can't you just use that there are no nth roots of unity in Q then?

[dim widget]

I suppose

[rotund aurora]

I was wondering if you can classify the cubic forms in three variables that occur as norm forms of cubic cyclic fields

[dim widget]

up to what?

[rotund aurora]

idk what's the standard thing. I suppose one should ignore change of basis for example. I think that amounts to acting by GL_3 or something?

[dim widget]

yeah

[rotund aurora]

such fields are given by cubic (irreducible) polynomials with square discriminant if I'm not mistaken

[dim widget]

Okay i think for the case you gave of cyclic degree p there is some hope for this problem

[dim widget]

yeah

[serene dune]

$\raisebox{.2em}{SO(n+1)} \big/ \raisebox{-.2em}{SO(n)} \cong S^n$

[cloud walrus]

yeshua

[serene dune]

any hint of the proof ?

action is matrix multiplication right ?

alr, imma try that

on a short note

$\mathbb{R} / \mathbb{Z} \cong S^1$

[cloud walrus]

yeshua

[serene dune]

what about this one ?

[rocky cloak]

See if you can find a homomorphism from R to S1.

It might help to think of S1 as a subset of the complex numbers

[serene dune]

yeah i have a intuitive geometry going on my head for this one

$\varphi : \mathbb{R} \to \mathbb{C}, \ \
x \mapsto e^{2\pi ix}), \
kernel(\varphi) = \mathbb{Z}$

[cloud walrus]

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

[void cosmos]

guys

im having a stupid moment here

K is the spltiting field of an irred poly f of deg n

irred seperable polyu

or whatever

now for my argument i have that Gal(K/F) is abelian

so by the fundamental theorem i now have that all interemediate extensions are normal

why is it that these extensions are F(a_i) where a_i is a root of f

this is probably easy but im just being tupid

[south patrol]

You should need more hypotheses on your extension I think, since all the extensions F(a_i)/F are of degree n

whilst K/F could be of much higher degree

I mean for example consider the splitting field of the cyclotomic polynomial phi_n over Q - this is always going to be an abelian Galois extension of Q and not every subfield is generated by a root of phi_n (e.g. almost all roots of unity are not real but there are non-trivial real subextensions)

[void cosmos]

... yeah

ur right

but wait

can i say that K is F(a_i) for some root ?

K is the splitting field itself

and now by the fundamental theorem ik it's normal so can i just say it's simple ?

adjoining any root of f

[rocky cloak]

Yeah, since the Galois group is abelian, every intermediate extension is normal, so F(ai) is normal

[void cosmos]

thank you king

[chilly radish]

Being normal is equivalent to being closed under the galois action, thus F(a_i) being normal implies that g(a_i)\in F(a_i) for every g\in G, but since f is irred G acts transitively on the roots, so F(a_i)=K

[void cosmos]

oh wow yeah

ur smart

ty

Being normal is equivalent to being closed under the galois action

so cool

[chilly radish]

(In general you can phrase it as every map K->\bar F actually maps into K)

If you don't wanna do it relative to some galois extension

[thorn ermine]

How do I grasp the concepts of fields, Integral domains and division rings more intuitively? It’s a bit hard to keep track of the definitions and I’m unable to devise proofs for why a certain thing like C (complex numbers) would be a field or ID or division ring

[cloud walrus]

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

[rocky cloak]

phi(r) = (r, x) seems like a good idea, so just continue with that

[pastel swift]

I just checked it now, I get 12 unique elements with that, and they obey homomorphism properties and is bijective so I constructed an isomorphism

Maybe the stack exchange post mistyped something

[tardy hedge]

I think you do just need to memorize the definitions and their implications at the start, and then have a few good concrete examples of each at hand

An integral domain that is not a field is Z for example

[thorn ermine]

Also like, how can I prove that an n x n matrix over real numbers is a field or integral domain only if n = 1? Like I can provide a counterexample that for n = 2 there are matrices that don’t allow it to be a field/integral domain but isn’t there a formal proof that everything n > 1 is not a field or integral domain when it comes to n x n matrices?

[delicate bloom]

I'd suggest playing with some matrices that have 0s on the diagonal

or mostly 0 entries for that matter

[thorn ermine]

True that’s a way of showing a counterexample that I was thinking of, but I thought formally since I can’t write down counterexample for n = 2, n = 3, n = 4 etc etc I should use something like induction?

[delicate bloom]

well I think we can make a simple example of some various form and formalize that

what examples have you found let's see if we can shape them up a bit?

[thorn ermine]

Like say “I claim n = 1 holds for all real numbers” then separately say like “for every n >= 2 it does not hold, then for the k + 1 part show that n = 3 doesn’t work”

[delicate bloom]

that's not what I think of as an example, I'm thinking of writing down a specific 2x2 matrix

[thorn ermine]

Like a two by two matrix that doesn’t work?

[delicate bloom]

right, has some property that violates being in a field

[thorn ermine]

I think [1 , 0 // 0 , 1] And [0 , 1 // 1 , 0] would violate its properties

Because this would give zero divisors

And we’re not allowed zero divisors in fields

[delicate bloom]

yeah sounds good, so specifically how do you make zero divisors from this and then next can we generalize this to nxn matrices?

if that feels too rough we can always go back to the drawing board and find some other 2x2 matrices that are easier to work with too

[thorn ermine]

Hmm well if the first matrix is a and the second is b then this makes zero divisors because every non zero element in the matrix is multiplying by a zero and cancelling out to 0

Not sure how we could generalize it

[delicate bloom]

this is not true

the first matrix is the identity matrix so when you multiply it with the second, you get the second matrix as the result

[thorn ermine]

Don’t you just get the zero matrix if you multiply them?

[delicate bloom]

no because matrix multiplication is typically not defined to be entrywise like that

what you're doing is something called the Hadamard product, not really what people mean when they talk about matrix multiplication

I'd say go back and review linear algebra, for instance, how do you multiply a 2x2 matrix by a 2x1 vector?

clearly you can't just pair up entries right

[thorn ermine]

Ohhh damn yeah I’m stupid

[delicate bloom]

all good haha

[thorn ermine]

I was doing it in my head completely forgot

Alright yeah so I see how anything multiplied by the identity matrix gives the second matrix

So like how can we generalize that into something we need for all matrices

[delicate bloom]

well there are a bunch of ways we could do this I think, but I guess since your original 2x2 example is sorta invalid let's go back to the drawing board and find an example again

I think the strategy of finding zero divisors is a good one

[thorn ermine]

And wait in the first place, the fact that it doesn’t give us the zero matrix kind of makes the example useless to us right? Cuz we’re trying to prove that fields can’t have zero divisors

[thorn ermine]

True

[delicate bloom]

yeah we're on the same page now I think, so see what you can cook up, brb making some tea

[thorn ermine]

So something, that when multiplied by the identity matrix gives us a 0 for all n > 1

[delicate bloom]

well the identity matrix might be best to avoid in this regard

since identity matrix will just multiply by anything to return that anything, so it'll only output 0 if you input 0

[thorn ermine]

Ah yea that’s true

Would that be invalid to use tho 🤔 because the fact that it only gives you zero if one of the matrices you input is 0, aka if b = 0 for example, then it proves the fact true that fields have no zero divisors (because no zero divisors means that a = 0 or b = 0 always)

I feel like I’m creeping up on a proof by contradiction solution 🤔

[delicate bloom]

well there are zero divisors for sure

to expand a bit on my hint earlier, try to find a 2x2 matrix with mostly 0s in it that squares to the zero matrix maybe

[thorn ermine]

I mean I guess you could do [0,0 // 1,0]

And then square it to get 0

[delicate bloom]

yeah that looks good!

can we make a similar 3x3 example

[thorn ermine]

Yup! Same thing with a 1 on the bottom left of the matrix always gives a 0 matrix

No matter the dimensions

[delicate bloom]

cool, ok so now we just need to make that a little bit more rigorous I think, and we're home free

[thorn ermine]

I just have no clue how I could show that for every n x n without brute forcing and writing it for everything tho lol

[delicate bloom]

do you know the summation definition for multiplying matrices?

[thorn ermine]

I don’t think so

[delicate bloom]

so like AB=C you have $$\sum_{j=1}^n a_{ij}b_{jk} = c_{ik}$$ for the entries

[cloud walrus]

Merosity

[thorn ermine]

Yeah no no idea what that does

[delicate bloom]

that's literally the same thing as matrix multiplication for how you get the entries, like $$A=\begin{pmatrix}a_{11}&a_{12}\a_{21}&a_{22}\end{pmatrix}$$

[cloud walrus]

Merosity

[delicate bloom]

I admit it's a bit complicated looking but your example is perfect for this

since it means most of those entries are 0 except for 1

[thorn ermine]

I see

So what would this do for us in our proof

[delicate bloom]

So in that 2x2 A matrix I wrote there, what would be i and j for a_{ij} = 1 for your matrix with the 1 in the lower left corner?

[thorn ermine]

Not sure 🤔

[delicate bloom]

a_{ij} is the entry in the ith row and jth column, so what row and column is your 1 in here?

[thorn ermine]

Ohh 21?

[delicate bloom]

exactly

[delicate bloom]

so our goal is to show that this formula here always gives 0 when you use it to compute A^2

[thorn ermine]

Uh huh

[delicate bloom]

precisely, we're looking at showing this is 0 $$\sum_{j=1}^n a_{ij}a_{jk}$$

[cloud walrus]

Merosity

[thorn ermine]

Right

[delicate bloom]

we know the only term that we can possibly have in there that would be nonzero is if we have a_{21}a_{21} otherwise you're multiplying by 0 right

does this term ever occur?

[thorn ermine]

The term 21 = 1?

Like you mean it occurs 2 times because of our matrices?

[delicate bloom]

You may have to spend some time working through this on your own to see that this definition lines up with how you compute matrix multiplication.

[thorn ermine]

Yeah

[delicate bloom]

for instance what will be the entry in the first row and column of the product? It'll be this:
$$a_{11}a_{11} + a_{12}a_{21}$$
You can see the "inner" indices matching as j=1 and j=2 there.

[cloud walrus]

Merosity

[delicate bloom]

in our case, a_{11}=0 and a_{12}=0 so the whole sum is 0 for that case, if that makes sense

Maybe this is a bit much so I'll cut to the chase and not drag it out, and then you can work through the details and ask more later.

In general your example becomes a_{n1}=1 with all other entries being 0. This means in the matrix product you have a sum of terms of the form a_{ij}a_{jk} and the only way to make this nonzero is if both are a_{n1}a_{n1}=1*1=1 which is impossible unless n=1 since the "inner" j index has to be equal.

so it gives us a legitimate counterexample for n>1 of a matrix that squares to 0

[thorn ermine]

I see

Let me try and work that formula into my proof and try and make sense of it

Thanks for the help

[delicate bloom]

yeah you're welcome

[acoustic vortex]

How do I find the solutions to roots of unity. By this I mean, what is the solution to (x^n-1)=0 in radicals of degree less then n (or "principle radical values"). (This means e^2pi*i/n or cos(x)+isin(x) wouldn't be answers, or nth root of 1).
https://math.stackexchange.com/questions/92587/cyclotomic-polynomials-explicitly-solvable
https://math.stackexchange.com/questions/92870/non-trivial-solutions-for-cyclotomic-polynomials
According to those posts should be solvable "in radicals" because of galois theory and that doesn't just mean 1^(1/n) and they can be solved using langrange stuff.

I was reading through them I didn't quite get it.
Using some method how would I get the 1000nd root of unity (or something) with my previous conditions?

This is one of the solution to 7th roots and judging by it's complexity and length, it probably means that for larger n, I will need a program to solve it.

Wolfram alpha gives up on solving them when n=11

I originally posted this to a help channel, but was told to post it here instead

[rotund aurora]

You are meant to find a normal series for (Z/nZ)^x with cyclic factors

Do you know Galois theory? If not, you should study it, this is not trivial

For example, if p is of the form 2^n+1, like when p=17, then it is possible to choose the "cyclic factors" to be Z/2Z. This means that you can write a pth primitive root by only using square roots and the other usual arithmetic operations

[acoustic vortex]

ok

I don't know any Galois theory

That being said, one of those answers apparently outlines a method to achieve it

https://math.stackexchange.com/questions/4083331/exact-expressions-of-sine-and-cosine-of-frac-pin-with-n-11-13-19-23-an

I found an answer that does what I want I think

[acoustic vortex]

nvm

[errant wedge]

All maximal left ideals are left primitive right?

[errant wedge]

nvm sorry im sleep deprived, u can just take R/M and that's a simple left mod with annihilator M, whoops

[rocky cloak]

Like TTeppa implies, left primitive ideals are always two-sided ideals, where's maximal left ideals need not be.

[errant wedge]

Yes my bad, thank you

[rocky cloak]

But a maximal left ideal that is two-sided would be

[errant wedge]

Mhm

LOL

Nah I got an extension from him, I just submitted it like 30 mins ago

Q3 was cool asf

Fwiw Bergman could sue me for my solution

I got so many hints from friends, this assgn was brutal

So true

I did everything except 1b

My 2b is super scuffed tho

Mhm one of my friends did stuff with ore extensions but Ik nothing abt them

Yes

[languid trellis]

I'm struggling to get the details clear on this example, what is phi and how is it surjective?

It says here what it is but I'm still struggling to "map it onto" the example

Say H = Z(D_8), then phi(r) = rH. How do i relate this to the klein 4-group?

[true island]

In a general case, I think an extension is:
$1 \rightarrow H \rightarrow G \rightarrow G/H \rightarrow 1$, so when they present something like (in your first picture) it is implied that the quotient group $D_8/Z_2$ is isomorphic to $Z_2 \times Z_2$

[cloud walrus]

Ran

[hidden wind]

ok so if i wanted to be very formal about it i would have to consider their images under the evaluation hom ?

images of elements of the ideals that is

[quiet pelican]

yes

[true island]

ok so to see what phi means find a subgroup of D_8 isomorphic to the klein 4group, once you will find it you'll likely realise what's the projection from D8 from that subgroup and and it will be Phi
Usual candidates for such a subgroup appearing in an exact sequences like this is G/H, so in this case $D_8 / Z_2$, it might help to view $D_8$ as $Z_4 \rtimes Z_2$
(sry I myself have not been able to find such a subgroup, nor to show that $D_8/Z_2$ is isomorphic to the klein group)

[languid trellis]

Yeah I agree. I'm pretty sure that the subgroup is generated by r^2, sr, sr^2 and sr^3. Need to work on the details still, but thanks (;

[true island]

I need help with this one:
Let $S_7$ be a sylow 7 subgroup of $\mathfrak{A}_8$, Show that its normalizer is isomorphic to $C_7 \rtimes C_3$

[cloud walrus]

Ran

[true island]

I've observed that there are in total 8 sylow 7 subgroups in Alt(8)
but I can't understand the result because the orders doesn't seem to be coherent (the order of the normalizer would be that of Alt(8) divided by the cardinal of the orbit of S7 (which is 8) [orbit-stabilizer theorem] and the order of C_7 (semi-direct-product) C3 is 21 ???

[true island]

also $C_n$ refers to the cyclic group of order $n$ ($\cong Z/nZ$)

[cloud walrus]

Ran

[quiet pelican]

I don’t think there are just 8 sylow 7 groups
A 7 cycle is uniquely determined by a choice of number to omit (8 choices), and an ordering of the 6 numbers in the cycle that aren’t the smallest (6!)
Each sylow 7 subgroup has 6 7-cycles in it, and they intersect trivially
So there are 8*5! sylow 7 subgroups
Notably, this gives the right order for the normaliser

[true island]

Thank you!! I found where I did a mistake: I thought A8 acted faithfully on the set of sylow 7 groups meaning A8 is "injected" in Sym(number of sylow 7 groups)

[tardy hedge]

if a is not contained in pi for 1<= i <= n, just pick something in a and that necessarily will not be in any of the pi’s so then x^n will be in a but not be in any of the pi’s , so a cant be contained in the union of pis

Doesnt that do it?

Ohh wait

I see where i messed up

If a is not a subset of b it does not mean that every element of a is not in b

Just that at least one element of a is not in b …

[ashen heron]

the latter

[tardy hedge]

Ya

Quantifier moment

[tardy hedge]

What the hell even is this sum supposed to mean

Ive been stuck on this proof for days. So confusing

Ok ...

At this point I just need to go line by line here and catch where I am getting confused

The result is true for n-1. Then for each i there exists xi in a such that xi not in pj whenever j not i. Is this using induction hypothesis ?

"for each i" meaning i from 1 to n-1?

[charred iris]

The result is true for n-1 ideals. We apply it to the n-1 ideals given by forgetting about p_i, yielding an element xi

so e.g. if n=4, we find x1 that isn't in p2, p3, p4 (n-1 = 3 ideals), and also an element x2 not in p1, p3, p4, and an element x3 not in p1, p2, p4, and x4 not in p1, p2, p3

[tardy hedge]

Oh. So x2x3x4+x1x3x4+x1x2x4+x1x2x3 is in a but not in p1 or p2 or p3 or p4.

and for example since x2x3x4 is not in p1, x2x3x4+x1x3x4+x1x2x4+x1x2x3 cant be in p1 cause then that would imply x2x3x4 is in p1

so in this proof we are not only using the fact that its a prime ideal, the fact that its an ideal in general is also relevant ?

and in the case where we have like x1 not in p2,p3,p4 but it also happens to not be in p1 then we are already done in that case

If everything up there in my understanding is correct^ then i think i understand this now. And thank you so much edward

[glad osprey]

If A and B are groups, for Hom(A, B) to be a group under pointwise addition, we only need B to be abelian, right?

[coral spindle]

That's sufficient, yeah

[slim kayak]

If the right component is a group, you get a group. It also inherits being abelian

[coral spindle]

I frankly don't know if there is a subtler necessary-and-sufficient condition for this

[slim kayak]

(actually, cant we just repeat the formal mumbo jumbo thats used for group schemes as representable functors to conclude that B must be a group object?)

[tardy hedge]

For proof of 2), why are we supposing p not a subset of ai for all i?

Shouldnt it be ai is not a subset of p for all i?

[hidden wind]

the claim we want to show is p is a superset

[tardy hedge]

Negation of that is p not a subset of ai for all i?

Ive been studying math all day i need a break

[hidden wind]

get some rest

[slim kayak]

I am confused now too. Note that the inclusion is both negated and flipped as written here

Like, that is both not a proof by contradiction and the bit right after the first sentence doesnt follow from it

[hidden wind]

oh yeah you’re right and the next sentence seems to be describing the negation of the opposite inclusion

[slim kayak]

Wake up!

[hidden wind]

pff no don’t

[slim kayak]

Okay now that you are informed that you were right, go rest

[tardy hedge]

Yea it was homecoming for my school on saturday so i was out all day

Had to grind like hell today and yesterday loo

Lol

[unkempt stream]

damn

[thorn ermine]

Do yk any prof Alex brandt

[rocky cloak]

If you want Hom(A, B) to be a group for all A, then that would work.

Otherwise you could pick A such that the only maps have image in the center / some fixed abelian subgroup.

Like A = C3, B = S3

[somber sleet]

Really easy questio: how does one prove that Z is commtuative?

[inner steppe]

What is the difference between a function f(x) and f(x) ∈ F [x] (polynomial ring?) I understand the definition of rings and sort f understand why a polynomial can be a ring but what I don't understand is what difference does it make, so far in my book everything that holds for f(x) holds for when f (x) ∈ F [x] and vice versa??? I don't get the difference at all.

[mighty kiln]

How do you define ×

[mighty kiln]

Different polynomials can be the same function

[somber sleet]

I do not know, I was asked this question in the lesson, but they didn't introduce Z as a ring (first course linear algebra)

That's why I am asking

[mighty kiln]

I think it should be obvious once you define ×

[inner steppe]

What do you mean? I am not following?

Are you asking me the properties of it?

[mighty kiln]

Sorry that was replying to the other person

[inner steppe]

What do you mean by this, anyways?

[somber sleet]

Is x defined as the sum? Inductively?

[mighty kiln]

One can define × on the naturals via cardinality of cartesian product

And then extend to Z

[grave sedge]

(if your definition of naturals is "something that satisfies the peano axioms" then you define it recursively)

[somber sleet]

I don't have a definition of Z, I only have this

[mighty kiln]

Say x² and x⁴, as polynomials in F3[Z] vs as function F3 → F3

They're different elements of the polynomial ring but the same function

[somber sleet]

Does it make sense to first inductively show that the addition in N is commutative, then that addition in -N is commutative and the use above definition

[grave sedge]

If you already have N then you can construct Z as the set of pairs (x,y) of naturals quotiented out by the equivalence relation given by (x1,y1)~(x2,y2) iff x1+y2=x2+y1 (where the pair (x,y) basically represents the integer x-y, except you don't really have subtraction because you are using naturals)

[grave sedge]

Yes that's how it is done

[somber sleet]

How do I explain this to first semester studis

they're gonna punch me

[mighty kiln]

Usually an algebra book would just assume everything learnt in hs

[grave sedge]

You probably don't

Like

It's more or less a foundational thing

[somber sleet]

I know, but I was asked this question during the lesson and I was like: jeez did this like 4 years ago and never seen it again

[grave sedge]

Idk how you can give an answer which is both an actual explanation and avoids actually doing all the constructions

[mighty kiln]

"See page randint(30,100) of [proofs and logic book]."

leave before they open the book

[grave sedge]

"commutativity in Z follows from commutativity in N which in turn follows from a bunch of induction arguments"

[kindred vector]

I mean there's also the option of not explaining by instead teaching them that sometimes you can (and should) black box things

[grave sedge]

Why do it here when you can do it with something like jordan's curve theorem

[kindred vector]

They'll learn these constructions later but for now its a waste of time

[somber sleet]

HAHAAHAHAHAHAHAH

[somber sleet]

yes you are right, I will mention that next week

[somber sleet]

It really disturbed me yesterday because they kept bombing me with questions and didn't let me really do what I planned to

It wasn't even the first time of me teaching

[kindred vector]

Learning that lesson earlier rather than later is better imo. I've seen people not learn it at all during undergrad which is kinda scary - once you reach a certain level you inevitably have to do it

[grave sedge]

Yeah it was mostly a joke

[kindred vector]

Haha there's such a thing as too eager I suppose

[kindred vector]

Ah my bad

[grave sedge]

except black boxing jordan's is done in every complex analysis course ever

[somber sleet]

I completely agree hahaha

[kindred vector]

Yeah JCT in complex analysis is a good example lol

[grave sedge]

Other than that i'm totally in favour of avoiding "foundational" issues (in the wider sense) when they are not really issues

[somber sleet]

I will mention this next time

You knwo they didn't even introduce rings and groups (which we did in our first course)

How are they suppose to work with matrix groups and stuff like that

I actually have an other question regarding $\mathbb{Q}(i)$

[cloud walrus]

damn_guuurl

[somber sleet]

I know that it is a field since it is a field extension and blabla

but if onw want's to prove it with axioms, how does one show that the addition is commutative? More specifically, why does this hold? $(a + ib) + (c + id) = (a+c) + (ib+id)$

[cloud walrus]

damn_guuurl

[grave sedge]

You define addition this way

"complex numbers are pairs of reals where you define addition as (a,b)+(c,d)=(a+c,b+d) and multiplication as (a,b)*(c,d)=(ac-bd,ad+bc), have fun proving they are a field"

(replace complex with Q(i) and reals with rationals if you wish)

[somber sleet]

This is much easier, why did I complicate my life

do you guys maybe know a book where all this stuff is listed?

I would like to send it to my students so that they don't ask questions 😄

[grave sedge]

Iirc the first chapter of baby rudin constructs R and C but starting from Q

And it might not be that easy of a read

[wraith nexus]

tao's analysis i & ii cover the constructions quite thoroughly

(complex # in tao ii section 4.6, and N/Z/Q/R in book i)

but its quite long

however should be easier read

[night tartan]

in a polynomial ring, we consider f(x) as just a string of meaningless symbols

i.e. X, 2X^2 + 3X^3 + 1, X^p + 2X etc.

obviously we can easily induce a function based on these symbols

but a big difference to look out for is that 2 polynomials could represent the same fnction

i.e. take F = Z/pZ

then X^p - X and 0 are 2 different polynomials

but by FLT, X^p - X = 0 for all n in Z/pZ

so X^p - X and 0 are equal for all elements of Z/pZ even though they represent 2 different polynomials

[south patrol]

By flt rather than FLT

jk

[rocky cloak]

FlT perhaps

[unkempt stream]

FLT?

[wraith cargo]

fermat's little theorem

[tardy hedge]

Damn bc it was wrong?

I searched up errata for atiyah macdonald and this didnt even come up

[south patrol]

Yeah this seems to be a mistake lol but it is rectified by just swapping the not subset for a not supset

[tardy hedge]

If ring R is commutative can a free module over R have bases of different size?

[void cosmos]

no

need R to be commutative tho

and unital

[tardy hedge]

Ok. But, in this case you may have a maximally linearly independent set that is NOT a basis

For maximally lin ind sets to always be basis, u need R to be a division ring

[void cosmos]

im not following

what case

and why

[tardy hedge]

If R is commutative a free module can have a set that is maximally linearly independent but cannot be extended to a basis

This is not true for vector spaces

Maximal lin ind sets in vector spaces are bases

[chilly ocean]

Does there exist a non-abelian group $G$ such that we have $a^3 = e$ for all $a \in G$ ?

[cloud walrus]

Saïd

[chilly ocean]

Can you please give me a hint or an exemple to direct my intuition, I personally looked at the structure of the permutation groups and my claim was that (after testing with some constructions with 3-cycles in S_7) either the cycles were disjoint and thus giving a group with all its elements having order 3 but it is abelian due to the "disjointness" or the cycles are not disjoint but then their composition is decomposable into a product of transpositions thus making that element of order 2

and thus my intuition is somewhat telling me that the assertions (being non abelian and having all elements of order 3) are not compatible yet I do not know how to prove it and while trying my hardest I kind of have a change of heart and think that it maybe possible for it to exist xD and so I am stuck and in need of either a hint of a class of exemples (like the group of matrices or else) which could give me more intuition on the internal structure

thank you for your time and attention

[void cosmos]

idk all i can say is

if |G| is finite then ig u would have |G| to be some power of 3

lmlao

[chilly ocean]

but then are theses groups intrinsicaly non abelian ?

I do not know much about finite groups and this exercise was layed down right after the most basic definitions of groups

and so I guess that it should be solvable with some basic notions ?

[void cosmos]

i can probalby

think of a counterexample

to that

[south patrol]

wdym

you can just do Z/3

[void cosmos]

probably Z/3

lmfao yeah

[south patrol]

Every abelian example is (Z/3)^X for a set X

[void cosmos]

yeah exactly

[topaz solar]

They were looking for non-abelian

[south patrol]

Yes

[void cosmos]

yeah i was just answering this

[topaz solar]

With choice

[south patrol]

Pretty sure it must be abelian but the proof is annoying

[void cosmos]

why can't u like

take some matrix group over Z/3

[chilly ocean]

do you mean that the group must be the set of the morphisms from Z3 to X ?

[topaz solar]

Isn’t there an annoying way thing where a ring with x^3 = x being commutative

[south patrol]

Because the Z/3 structure shows everything has 3x = 0

but the noncommutative multiplication is, well, multiplication

However yes I think this does give an example

1 sec

[void cosmos]

im sorry i dont follow

i meant like

[south patrol]

maybe this is what you meant actually

lol

[void cosmos]

GL(Z_3) or something

or like

something upper trinagular

or something of that sorts

but i can't like think of it straight

[chilly ocean]

in Z_3 isn't 2^3 = 2 mod 3 ?

and so it isn't really under the scope right ?

[void cosmos]

XD did u delete my msg hahaha

[south patrol]

no

unless it was by accident

lol

[void cosmos]

hahah yeha it was so funny

whatever

but yeah ig in my mind i would try to like

see how the matirx looks like so that if the entries has order 3 the matrix has order 3

with multtiplication

like it wouldn't be any general matrix it would have to be like

some upper triangular or like

[chilly ocean]

I seeeeeeeeeeeeeeee

[void cosmos]

diagonals dont work right

[chilly ocean]

yes maybe could work

[void cosmos]

ig diagonals do work ?

[chilly ocean]

i'll try with 2 dimentional matrices over Z_3

[void cosmos]

diagonals over Z/3?

[chilly ocean]

that sounds like a good lead in fact

[void cosmos]

yeah im going to grab something to drink

[void cosmos]

yeah try it out king

[chilly ocean]

the non comutativeness of matrices and mix that with some mod 3 arithmetics (which I do not master well xD)

but yea that is exactly the type of hints i looked for

thank you so much

[south patrol]

actually

oh nvm

oh yeah lol

||Heisenberg group||

[chilly ocean]

BRUH

what's that some kind of entangled group which we can't deduce the elements bafore making their experience ?

[south patrol]

Lol

[chilly ocean]

and you take the elements over Z_3 ?

aaah so it is just as @void cosmos first proposed right ?

since taking H_3(F3) would give a finite group of p^3 elements right ?

[void cosmos]

lmfao im just so stupid

diagonals are abelian

like im literally insane

[south patrol]

Yes but 3x3

[chilly ocean]

yes yes

[south patrol]

3^3 yes

[void cosmos]

yeah i was like in the kitchen and then i thought about diagonals over 2x2

[chilly ocean]

i am now trying to find some subgroup which will satisfy both

[south patrol]

Wdym satisfy both

[chilly ocean]

or else imma have to try with all 27 matrices xP

[south patrol]

H_3(F_3) works

[chilly ocean]

this has the misfortune of including a lot of order 2 elements

[void cosmos]

lmfao yeah heisneberg group

[chilly ocean]

non abelian and all elements of order 3

[south patrol]

Yes the Heisenberg group works

Well order dividing 3 ofc

[chilly ocean]

well it's not so obvious that it ain't abelian or am I lacking some godly intuitioin here ?

well I understand that thanks to Lagrange the order of the elements must divide 27 and so we'd have order 3 and 9 right ?

and so shouldn't we only take the elements of order 3 ?

and see if they constitute a subgroup in the first place

[topaz solar]

9 divides 27 so it’s possible, but do any elements of that order actually occur in this group

[chilly ocean]

i don't know xP

i am trying with order 3 for now and they are commuting god damn it xD

[prime quail]

https://www.weddslist.com/groups/misc/pcubed.html

what is the up arrow mean on the website?

[regal zodiac]

For the problem 5, does sigma really has to be of infinite order or different from identity is enough ?

[rocky cloak]

I think if sigma has finite order it's much easier to prove.

... Actually proof is kinda the same I guess

[regal zodiac]

Could i show you what i did?

[rocky cloak]

Sure

Same should also be true if you replace sigma with any subset of Aut(L)

[regal zodiac]

Thanks !
Take c in L\K. Then c is algebraic over K, then let f be the minimal polynomial of c over K. We remark that sigma(c) is also a root of f. thus let's consider the set { c , sigma(c), sigma^2(c) ........ sigma^i(c) } where i is the Last index such that all these elements are different. Let g = (x-c).....(x-sigma^i(c)). Then if E denotes the map induced by sigma : L[x] -> L[x], we can check that E(g) = g and thus g is in K[x]. Which concludes because g = f and thus f splits over L.

[regal zodiac]

Yes i also think so, that's why the infinite order thing was triggering me

[rocky cloak]

Yep, perfect.

Notice also that the same argument can be used to show that L/K is seperable.

Hence the characterization of Galois extensions as normal, seperable, (algebraic)

[regal zodiac]

Yes that's true, it also Proves L/K is separable, thank u so much !

[tardy hedge]

1.8 was stating that the nilradical of A is the intersection of all prime ideals of A. A comment before this said that the radical of a is the inverse image of the nilradical of A/a.

So, nilradical of A/a = intersection of prime ideals of A/a

so phi^-1(intersection of prime ideals of A/a) = intersection of phi^-1(pi) (by just properties of inverse image?)

The inverse image of a prime ideal is prime and the lattice theorem says this is an ideal containing a

So thats how we get that result ?

[wraith cargo]

yeah basically

[tardy hedge]

Does this sound good? Sorry if messy

Thanks

I need to review how ideals get transferred by homomorphisms

For an ideal (m) in Z and a prime divisor p of m, (p) is a prime ideal containing (m) right?

[rocky cloak]

Yup

[elfin wraith]

Could anyone walk me through whats going on in the proof here? I really cant make sense of it

[cloud solar]

I have a linear algebra problem but no one knew to solve it or give an idea on #linear-algebra so i hope no one gets mad if i leave it here hoping for a hint. Find all matrices A with integer entries such that rank(A^k)>=Tr(A^k) for every k>=1.

[vapid vale]

well if you know trace is the sum of eigenvalues, and you know how eigenvalues scale with taking powers, what can you say about the eigenvalues

[cloud solar]

All real eigen values have abs<1

[vapid vale]

right

or >= but fine

[cloud solar]

Yeah right

But then?

Here i am stuck

[unkempt stream]

Pretty sure you can do some sort of bullshit here with a semidirect here of Z/n with Z/n^2 for any n > 1 to have a nonabelian G satisfying x^n = e for each n

I am wrongo

[coral steeple]

I am stuck trying to prove that if H has finite index in a f.g. group G, then H is f.g. How do you relate the cosets of H with a generating set for G?

[warm ember]

any subgroup of a finitely generated group is finitely generated?

or

only for abelian

[coral steeple]

Not my G sadly

[errant wedge]

Does the proof of 2=>3 not rely on 1 when we take theta inverse?

[errant wedge]

nvm a friend pointed out it should be g and im just blind

[heavy minnow]

no, look at this example: https://en.wikipedia.org/wiki/Generating_set_of_a_group#Finitely_generated_group

[grave sedge]

This is probably nontrivial to prove algebraically, but there is a nice topological proof

First, notice that your G is a quotient of a f.g. free group F, and H corresponds to a finite index subgroup of F if you do the inverse projection

So it suffices to show that a finite index subgroup of a fg free group is fg

https://en.m.wikipedia.org/wiki/Nielsen–Schreier_theorem the arguments here should prove that

(it's a corollary of the second part of the theorem)

[tardy hedge]

I forget why every non-unit is contained in a maximal ideal?

Can you just create the ideal generated by this non unit and then use result that every ideal is contained in a maximal ideal

Is there another way

[mighty kiln]

Yes

[hidden wind]

borcherds mentioned something along the liens of "k[x]-modules are just linear transformations" as an "obvious example" and frankly i've no idea what he means

what does this mean

[rocky cloak]

A k[x]-module M is an abelian group with an action of k[x].

Since k is a subring of k[x] this means in particular that M is a k-vector space.

Now x commutes with k, so the action of x on M is a map that commutes with the action of k. Aka a linear transformation.

[hidden wind]

whoa

[rocky cloak]

And that completely describes the action, so a k[x] module is exactly the information of a k-vector space and a linear transformation on that space

[hidden wind]

thank you i will have to spell out the details of this for myself

[tardy hedge]

Dummit and foote talks about this in section 10.1

In great detail

As usual for dummit and foote lol

I always liked this example

[hidden wind]

tfw my uni library doesn't have dummit & foote ????

[tardy hedge]

Thats a bruh moment

[next obsidian]

There’s a different way to think of this

Start with an abelian group M, and then End(M) is a ring, the product is composition and the addition is just adding functions

An R-module structure on M is just a ring map R -> End(M), the way it works is that given such a map f, you define the action
a•m as f(a)(m) [recall that f(a) is in End(M), so it’s a map M -> M]

The fact that distributivity, associativity, etc holds is the fact that f is a ring homomorphism

Conversely given an R-module structure on M, you can define a map f:R -> End(M) as follows

For ever r in R, let f(r) be the map that takes m to r•m

The fact that M is an R-module implies that f will be a ring homomorphism

[hidden wind]

[next obsidian]

So now assume that you have an R-module M, this is equivalent to a ring map R -> End(M) by the above

To extend this to an R[x]-module on M (which just means that r•m is the same in R[x] as it is in R) is to ask that you have a map R[x]-> End(M) such that the triangle formed with R -> R[x] and R -> End(M) commutes

The map R[x] -> End(M) is the R[x]-module structure

And to ask the triangle commutes is just saying that r•m is the same when considered over R[x] or over R

Now finally, we come to the universal property of R[x]. Such a commutative diagram is a map of R-algebras, and R[x] is the free R-algebra on one generator

Said another way, to make the triangle I described, all you have to do is say what x goes to, and you can pick literally anything

Because the fact that the diagram commutes means that for f:R[x] -> End(M), you already know what f(r) is

And now f(rnx^n + … + r1x + r0) = f(rn)f(x)^n + … + f(r1)f(x) + f(r0)

You know what f(rn) is because it’s forced, and once you know f(x) you know f(x)^n

So putting this all together, an R[x]-module structure extending the R-module structure is just a ring map R[x]-> End(M) making a triangle commute

But that’s the same thing as picking where x goes to in End(M), aka picking an endomorphism of M

[stone sky]

I’m looking here and i’m studying cosets and subgroups. If H is a subgroup of G and aH=bH, i know the orders of the two cosets are the same, but the orders of a and b don’t necessarily have to be the same order do they

[long geyser]

consider the example H = {e}

probably a more illustrative example is A_n in S_n, two permutations can have wildly different orders while still having the same parity

I just realized H = {e} is a horrible example

I meant the other trivial case, H = G

[stone sky]

So in group theory is my best bet finding counterexamples?

i just can’t get a good grip on this area of math to save my life.

[night tartan]

well only if ur dealing with a statement that u don't think is true lol

[stone sky]

i just can’t seem to prove anything without an example or counterexample

[night tartan]

like for this particular one u gave, if u've done some group theory questions u might lean towards this fact being false

but even if u weren't too sure u'd probably try it out for certain groups

[stone sky]

right.

[night tartan]

like obviously the trivial group won't help here

what about the next simplest group, C_2?

it has 2 subgroups {e} and C_2

[night tartan]

clearly, rG = eG here but r and e have different orders

then we'd be done

u might then wonder 'are there any non-stupid examples'

[stone sky]

probably

but don’t we want the easiest one

[night tartan]

and S_n is another good place where u can find counterexamples

[night tartan]

also what exactly did u mean by the order of the coset?

are u taking H to be a normal subgroup and considering the cosets as elements of G/H?

[stone sky]

no

maybe the cardinality?

[night tartan]

ah k

well you can prove that all cosets actually have the same cardinality

[night tartan]

but yeah if u do a lot of groups questions, eventually u'll have a grasp of what u think should be true and what u think shouldn't be true

[stone sky]

i have ideas i guess formalizing it is tough

envisioning it is also hard

[tardy hedge]

It is hard

[stone sky]

this unironically made me feel better about my struggles

[tardy hedge]

Im glad

Cause it is hard but over time things make more sense

[south patrol]

Idk all maths is kinda easy

[tardy hedge]

Lol

[stone sky]

true

[tardy hedge]

For Potatoes math is easy

[gentle crow]

if a set equipped with some binary operation is isomorphic to a known group, does this automatically make the set a group?

[coral spindle]

Yes

[stone sky]

yes because the structure is the same

[gentle crow]

thanks

[white oxide]

are algebras just vector spaces with multiplication

i'm not tryna think about algebra rn since i'm doing analysis so a yes or no would suffice

thank u

[glad osprey]

yes, but the multiplication must be bilinear

[white oxide]

ok so just basically distributive property w vectors and associative with scalars

[coral spindle]

People use it to mean different things. One meaning which is relatively common is indeed a v.s. with a bilinear operator on it -- no more assumptions. Another one is a ring with a central embedding of a field, which in particular makes it an algebra in the previous sense, just with a bit more structure.

[white oxide]

i see interesting

i think that's all i need

thanks

[coral spindle]

Generally I would say that, at least in what I tend to see, the second meaning is more common.

[glad osprey]

I think it depends on exactly what you mean by isomorphism. If you mean just a bijection phi such that phi(a + b) = phi(a) + phi(b) then that's not enough to conclude that the set has an identity. I think you also need phi(e) = e and phi(a^{-1}) = phi(a)^{-1}

[coral spindle]

As a rep theorist ig.

[south patrol]

Rep theory

[coral spindle]

Rep theory

[south patrol]

As someone more in geometry/homotopy theory like commutative algebras over R are usually just rings S with a map R -> S

[coral spindle]

Yeah

[south patrol]

Kinda funny how the perspective differs

[coral spindle]

Well ig it's the same?

Like you just need to add the centrality condition for the noncommutative case

[south patrol]

Oh yeah sorry I didn't even read to the end of yours lol

yes a map A -> B landing in Z(B)

[coral spindle]

I did specify field tbh

[white oxide]

in checking if something is a subalgebra does it only suffice to check 0, linear combinations and multiplication

ie 0 in A, af + bg in A for all a, b in the field k it's over and vectors f and g

and fg in A for all vectors f and g in A

[coral spindle]

Yes

[white oxide]

good man

thanks

[gentle crow]

would it be good enough to show associativity at least?

[coral spindle]

N.b. af+bg encompasses 0 in A

[gentle crow]

i'm just working with cayley tables here

[coral spindle]

I don't believe that's correct. Let's say (X, .) is some magma isomorphic to a group G via f : G -> X. Then let b in X have preimage a, so f(a) = b. Then f(e)b = f(e)f(a) = f(ea) = f(a) = b, and evidently the other side holds. So I'm not sure what you mean by this.

Clearly f(e) is an identity in X.

Apologies for the bad typo

[glad osprey]

Aha, you're right. I remember looking at group actions of G on A a while back, and how you can either define them as homomorphisms from G to Aut(A) (with only the homomorphism property phi(a + b) = phi(a) + phi(b)), or as a map from G to A -> A, but then you need the additional axiom phi(e_G) = Id_A. So that's the source of my confusion, but I see that this case is a bit different, since we're talking about a bijection

I think the axiom phi(e_G) = Id_A for group actions just asserts that the image is in Aut(A), it doesn't ensure any additional isomorphism between G and A -> A

"additional isomorphism" doesn't make sense, but I mean, any shared structure that is not ensured by the homomorphism property

[coral spindle]

Well this is still not right, because phi(e) will be idempotent, and it's plain to see that any idempotent is the identity in any group.

Oh no I was misreading

yeah

Indeed there are other subgroups (in the semigroup theory sense) in End(A) which don't share the identity

[weary ember]

Over what type of sets would convolution be a group?

[delicate bloom]

I suppose it depends on what you want to mean when you say convolution. I'd suggest working out as many of the axioms of a group as possible to as far as you can before getting stuck - then seeing what restrictions that gets you.

[topaz solar]

Subsemigroups that happen to be groups :3

[unkempt stream]

semigroup mfers frothing at the MOUTH for a chance to matter at all

[south sinew]

hello, i wanna ask something i have not met before
Can we define a polynomial function over a cyclic group?

This definition is weird.

I mean we can only define one operation in a group, multiplication or addition, right? But polynomials seem to involve both addition and multiplication

So what does it mean to be a polynomial over a cyclic group

[void cosmos]

they would be the same

like x^2+y for example would probably be x+x+y

if ur group is additive

or x^2+y would be x^2y

that's the only way i can think about this

imo

[south sinew]

Thanks，i will buy it

[south sinew]

Wait. Confused
The how do example 2 be explained?

[unkempt stream]

In a sense... Sorta??

Look up group rings

[chilly radish]

I think they're just considering polynomial expressions in the elements of G, so formal sums of products

[unkempt stream]

ah so maps from the free group ring

[chilly radish]

That would be Noncommutative polynomials

It's not clear to me if that's what's being considered

[south sinew]

https://eprint.iacr.org/2023/346.pdf

[chilly radish]

oh ok there's no extra context

[south sinew]

This paper is from an ePrint of Cryptograhpy if you're interested
But, i think the definition is confused

[chilly radish]

this is demented

do they refer to other papers on the subject that maybe have better overviews/direct to literature?

I think that's your best bet

[south sinew]

So, i think it is not good

[chilly radish]

They don't even mention exponents in the description ye

[south sinew]

Yeah

[chilly radish]

This has no references?????

Am I tripping

in that case you might have to reach otu to the authors or do some independent googling about the subject of these polynomial functions

[south sinew]

The only infor they give is:

Seems like they wanna introduce a new definition
But the writing of definition is horrible

[south sinew]

I originally planned to incorporate some tools from this article into my scheme design to address some issues I'm currently facing. However, I eventually found that the quality of this article may not be as high as I thought. Therefore, I decided to abandon this article and look for new solutions.

Thank you for you help and suggestion.

[trail cave]

Any help with this?

[south sinew]

g^5, g^20 are elements of order 6

g^6, g^12, g^18, g^24 are elements of order 5

1. order of the cyclic group is 30, means g^30 = 1
2. To find elements of order 6, you need to find elements in the form of " a^6 =1"
   So, it is easy to find a = g^5 is a solution, a^2 = g^10, however, (g^10)^3 = 1, which means the order of g^10 is 3..
   a^3 = g^15, the order is 2...
   a^4 = g^20, (g^20)^6 = g^120 = (g^30)^4 = 1, is a solution
   a^5 = g^30 =1, so, we can stop retrieving.
3. The same method for the elemets of order 6....

[trail cave]

thank you

this makes sense

[wild jasper]

Would anyone take a look at this solution? It would be most appreciated! https://cloud.eccentric.dk/index.php/s/NArD5tqf86j4zpF

[long obsidian]

If A a ring then A_red=A/nil(A) where nil(A) is the ideal of nilpotent rings. Suppose A is a local ring so it has a maximal ideal m_A. Does it follow that A_red is also local?

[slim kayak]

The ideals of A/I are in 1-1 correspondence to ideals of R containing I.

[hidden wind]

what does this question MEAN is not that first part literally the definition of “Zn is a principal ideal domain”

ohhh right of course

you just saved me (probably) from great embarassment, thanks

[lethal cipher]

Okay, so I am trying to prove this claim.
A couple things: since f o s= id_Q, this tells us f is onto. Using the first isomorphism theorem, we have that Q is iso. to G/N.

So I think that G is going to be isomorphic to the direct product N x G/N (I don't know how to do semiproduct notation, but I am not entirely sure of what the action is

I guess the action would be conjugation of q. Namely, (n_1, q_1)(n_2, q_2)=(n_1 s(q_1)n_1s(q_1)^-1,q_1q_1)

Ah wait, the coset thing is a bit unnecessary

[serene dune]

although there is no information about Q to be normal in G

$1 \to N \hookrightarrow G \twoheadrightarrow Q \to 1$

[cloud walrus]

yeshua

[topaz solar]

Yeah, up to minor typos and maybe the direction of conjugation since I always mix up the left/right

[lethal cipher]

Thanks. I was able to get it. It wasn't too bad

[inner steppe]

Can someone explain this to me like I'm a 5 years old?

How many different generators does the multiplicative group in the field Z2[x] / (x^3 + x + 1) have?
A: 1 B: 4 C: 6
?

[elfin wraith]

Still kinda getting nowhere with this problem,

Let $k$ be a field and $q\in k^\times$. Let $\tau_q$ be the unique automorphism of $k[y]$ such that $\tau_q=qy$. Show that $k[y][x;\tau_q]\cong k\langle x,y\rangle/(xy-qxy)$

[cloud walrus]

Nope

[elfin wraith]

We have a canonical map K[y]<x> ->K[y][x;tau] and under tau this has the same kernel as K_q<x,y> so it’s going to be some universal property type argument, but I’m really not seeing how to go about it

This is essentially all I’ve done (which is nothing) im just really unsure of how to proceed

[tardy hedge]

What country do u live ?

Me too 😄

[elfin wraith]

This was a rough idea I had, just sorta extending the diagram, I’ll give it a try and see if I get anywhere! Thanks

[elfin wraith]

Yeah I mean in some sense it is obvious, we’re quotienting out the same ideal and as you said we can really just map x to x and y to y but that just doesn’t feel right to do

[lime lagoon]

Hello! When doing a subring test, do I need to show that the zero identity of the ring is also the zero identity of the subring? Also, in verifying if the subtraction is closed in a subring, are the elements must be nonzero?

[elfin wraith]

a subring is a subset of your ring, so 0 is still just 0. If 0 isnt in your subset, it couldnt possibly be a subring

As for the subtraction, no you dont need that, the set {0} is a subring of every ring (its the trivial (sub)ring)

[elfin wraith]

$R[x;\tau] = \frac{R\langle x\rangle}{(xr-\tau(r)x |r \in R)}$ (parenthesis for ideals and angled for free algebra)

[cloud walrus]

Nope

[elfin wraith]

Yeah im currently trying to see if I can do anything with the fact we have a canonical map R[x] -> R[x;tau] and the evaluation homomorphism R[x] to R, difficulty is showing that (xr-tau(r)x) is in the kernal of the evaluation

So maybe I need to find a map from R[x] to the quantum plane instead as thse certainly have the same kernel

but actually this is just what I already had and couldnt work out lol

[elfin wraith]

Just going to continue to think out loud here I think, I should probably be using the fact that tau is an automorphism somewhere. In $K_q[x,y]$ we have that $x\left(\sum q_i y^i\right) = \tau_q(\sum q_i y^i) x$ so does this give us an injective homomorphism $K[y]\langle x\rangle \to K_q[x,y]$? I think it possibly does, which means the problem reduces to finding a homomorphsim in the other direction and showing theyre inverse

[cloud walrus]

Nope

[elfin wraith]

If this does work, I think i can get the reverse homomorphism via tau in the same way, so I guess the difficulty is just in showing theyre inverse? (and in working out the exact details of what im kinda hand waving right now I guess)

[winged void]

Hey guys

I have small question

i proved them all

hoevere i need to know the relation between this part and the rest of the exercises

but i do not see to be fair

[rocky cloak]

You may want to think about exercise 3 as comparing
Z[X]/((X+1)(X-1))
and
ZxZ = Z[X]/(X+1) x Z[X]/(X-1)

Then ask yourself what the image of f looks like in the latter

[winged void]

let me think about it

[winged void]

well i tried to see what the difference will be

the only difference that i can see that stll Z * Z is not isomorph and that

i guess because the idem potent elementen are (1,0) and (0,1) and (1,1) not for example

[void cosmos]

what about zero divisors

[winged void]

what do you mean

can you please explain more

[rocky cloak]

The difference between what? I thought you had solved the exercises?

[winged void]

I mean the relation with the rest of the excersice

over the last question

i see no relation

except that the idempotent will be diffrent

[rocky cloak]

Well, what is the image of f in Z[X]/(X-1)?

[winged void]

well this is a bit difficult

i mean

because the only thing that i know that if Z[X] / (x-1) = (x-1)g(x) and that g(x) is a polynoom and thus the zero points is 1 where it equals to zero but the whole image is difficult

right

[winged void]

or did i say something wrong

[rocky cloak]

Yes, but there is a somewhat canonical isomorphism between Z and Z[X]/(X-1)

Do you see what that is?

[winged void]

oh that is Z

but i do not know why

There is something like that in the book

[rocky cloak]

Well, do you know the first isomorphism theorem?

[winged void]

yes offcourse

[rocky cloak]

... And you see how it applies

[winged void]

it says that f / ker (f) isomorph and im(f)

but how this can be applied to prove that it will take long

because first need to find a function

and then the ker of function equals to (x-1 )

and then its surjuctive

to apply isomorph theorem

[rocky cloak]

Yeah, so do you know any maps from Z[X] to Z?

[winged void]

not really to be fair

i cannot think of one

maybe derivative

[rocky cloak]

Okay, well let's just take it one step at a time.

Let's call the map p. What do we think p(1) should be?

[winged void]

p(1) well be sends to 0

right

since 1 is a root of the polynoom

[rocky cloak]

Well, for homomorphism of rings, where do we usually map 1?

[rocky cloak]

Which polynomial is that?

[winged void]

that is this polynomial p(x) (x-1)

[winged void]

1

right

iddentiy to identity

[winged void]

sorry for late reaction i was preparing something to eat

[rocky cloak]

Indeed, so then
p(x - 1) = p(x) - p(1)

So what we thinking for p(x)?

[winged void]

p(1) here i guess its a number

right

[rocky cloak]

p(1) = 1 like we said

[winged void]

true

thus p(x) must be then 2

because 2 - 1 = 1

[rocky cloak]

So what do you want p(x-1) to equal?

[winged void]

p(x-1) should equal to 2 right

[rocky cloak]

Why do you say that?

[winged void]

no i mean that is not true

wait

[rocky cloak]

We wanted a map with kernel (x-1)

[winged void]

that is true

[sweet pendant]

I have a question. Would it be sufficient enough to just show that the function that I constructed from Z -> R preserves addition and multiplication or do I also need to explicitly check that this function does send 0 to 0 and 1 to 1? I am a bit confused. My function is $\gamma(n) -> (n \cdot 1_{R})$. I have checked all the properties and it satisfies.

[cloud walrus]

mss

[winged void]

so we can choose actually the following map f: p(x) \maps to

[winged void]

to be fair i do not see it

[rocky cloak]

So if x-1 is to be in the kernel that means
p(x-1) = 0

Makes sense?

So p(x) = p(1) = 1

[winged void]

make sense that is true

but hwo can we use that

[rocky cloak]

Then similarly p(x^n) = 1, and in general you'll get p(f) = f(1)

Okay, so then we can get back to the problem.

You have this map Z[X] -> Z,
f |-> f(1)

And similarly you have f |-> f(-1)

So together you can make a map Z[X] -> ZxZ
f |-> (f(1), f(-1))

What is the kernel and image of this map?

Hint: ||it's highly related to your exercise||

[elfin wraith]

Ok so ive typed up what im thinking so far so that its hopefully more clear, I think im on the right lines but im kinda stumped as far as showing these homomorphisms are inverse

I could very well be way off as well though, still not convinced that im constructing varphi correctly, but I think its at least something similar to that

[winged void]

let me think about and write it correctly

and i will send it

thanks alot

[chilly ocean]

yo guys having trouble finding the non abelian groups of order 28

so on wikipedia reads em as

but im not sure how to interepet Z7 demidirect Z4

like take (a,b) (a1,b1)
then that would be
a + phi_b ( a1 ), b + b1
right? but what automorphism do is being implied here that gets me that structure?

[chilly ocean]

alright but im looking how Z4 acts on elements of Z7

oh its gone now

[serene dune]

nah i did a mistake

i was doing dihedral

[chilly ocean]

yeah

anyways im thinking that any non trivial automorphism works acc

[serene dune]

well Z4 to Z6

two posssible homo ?

[chilly ocean]

so like (a,b)(a1,b1) = (a + b + a1 mod 7 , b + b1 mod 4 ) would work?

[serene dune]

for nontrivial case

Z4 maps to Z2 \subseteq Z6

[chilly ocean]

can you write some example computations?

[serene dune]

for some $h \in \mathbb{Z}_4, x \in \mathbb{Z}_7
\
h \cdot x = -x \mod 7$

[cloud walrus]

yeshua