groups-rings-fields

[worthy solar]

Unless this involves some (dumb) shifting of the argument

But I don't know it involves 2 complex pairs. I just am at the situation that the product of the 2 linear factors forming a quadratic is not reducible over Q

[chilly ocean]

so end_R(R)= R^op

[rocky cloak]

As a left module, yes

[chilly ocean]

What about division algorithm in C[x,y

[glad osprey]

yeah, the standard argument to show this is irreducible is to shift x |-> x + 1 then use Eistenstein

[rocky cloak]

What about it?

[chilly ocean]

Does it work

[rocky cloak]

Kind of, you can always divide by monic polynomials

But I guess, works to do what exactly, is the question

[elfin wraith]

Holy Groebner Basis

[chilly ocean]

How it works

[chilly ocean]

How to do the division

[chilly ocean]

the ring $C[X,Y,Z,W]/<XY-ZW>$ is integral domain?

[cloud walrus]

Curvature

[delicate orchid]

we may never know

my hint is to have a think about if <XY-ZW> is a prime ideal or not

[chilly ocean]

i doesnt look like prime ideal but , not able to argue

[thorn jay]

Why not?

[delicate orchid]

I don't think I've seen a more prime looking ideal in my life

[thorn jay]

The humble (x)

[delicate orchid]

it is in the prime of it's life... the humble ideal.....

[rocky cloak]

I guess change of variables Y -> Y+X, then Eisenstein at the ideal (Y, Z) would do the trick

Or just Eisenstein directly at the ideal (Z) should work

[delicate orchid]

that's clean yeah do that

[chilly ocean]

you havent seen a life of 7th year phd student

[worthy solar]

PhDs last that long

[chilly ocean]

i can show you some profiles

7 years in combinotorial topology

[worthy solar]

this all after bachelors? That is a lot of school

[chilly ocean]

no after msc

[worthy solar]

💀

what are the prospects after all that effort

[chilly ocean]

nothing just to realizer this is all shit

[chilly ocean]

I AM doinG it at prime ideal $<Y,Z,W>$ and getting $X^2+YX-ZW \in C[Y,Z,W,][X]$

[cloud walrus]

Curvature

[rocky cloak]

ZW is in the square of that ideal though

[chilly ocean]

I was trying to do for polynomial in one variable

[rocky cloak]

I don't know what you mean by that

[chilly ocean]

i was trying to get a prime ideal P and a polynomial R[X]

[rocky cloak]

Yeah, so thinking of it as C[Y, Z, W][X] is a good idea, then you just need to pick a prime P

[tardy hedge]

You’re a 7th year phd student?

[chilly ocean]

NO

[tardy hedge]

OKAY

[chilly ocean]

so <Y,Z> is working now

[rocky cloak]

That would do

[south patrol]

I am a 17 year phd student

[chilly ocean]

in india its impossible

[chilly ocean]

whats ur area

[rocky cloak]

Anything is possible if you set your mind to it

[chilly ocean]

no the system will not let you do it

[rocky cloak]

Just gotta work around the system, change your name a few times

[chilly ocean]

well we also need some good mentors who tells us a right path

here people will be preparing for entrance exams only at every stage

but that 7th year phd guy now knows almost every subject in pure maths

[warm badge]

let p be a prime, and Zp denote the group of integers mod p. show that for each positive integer n there are at most n solutions of x^n = 1 (mod p) in Zp. don't know any field theory to work with, this was an exercise in the automorphisms section

that = is congruence

[rocky cloak]

If this was in the automorphism section you might know that the automorphism group of Zp is cyclic?

[warm badge]

yess i do

[rocky cloak]

And you know that the automorphisms are given by multiplication by an element x.

So x^n = 1 is just asking how many automorphisms have order dividing n

[chilly ocean]

how to show prime ideals are gen by at most 2 elements in C[X,Y]

using only rings

this question was in my algebra ppr sem 1

[tardy hedge]

NO

[quiet tundra]

Hii can anyone help me with question no 19

[tardy hedge]

wow interesting question

i havent seen somethign like that before

[chilly ocean]

I just know I am 23 now

And sem 1

[thorn jay]

Can confirm im the phd

[quiet tundra]

Our Algebra teacher has given us questions like these which will be given on our Finals. He is super strict and hard to approach. And the worst part is that we can nowhere find the solutions over the internet

[chilly ocean]

What is the only zero divisior is 0 then result holds in that case

[thorn jay]

You need a non-absorbing zero divisor

[tardy hedge]

Yea im definitely interested to see the solution to that one. I also want to think on it more myself

[chilly ocean]

You have n+1 elements in that case , so we can assume there is atleast one non zero zero divisior

[thorn jay]

Because you'll do "something something for zero divisor z, zR is a set of zero divisors"

And then its a combinatorial argument, probably

I suck at combinatorics

[chilly ocean]

Adding a zero divisior to non zero divisior just remains inside the set of non zero divisior

[thorn jay]

Noncommutative ring

[quiet tundra]

We haven't studied combinatorics

[chilly ocean]

I am using excercise 18

[thorn jay]

Well it is just set theory

[thorn jay]

That requires the other to be nilpotent

And the ring to be commutative

[chilly ocean]

Let's do it for commutative first

[thorn jay]

No?

Why

[chilly ocean]

That's how we solve problems first looking at special cases

[thorn jay]

I guess, but explicitly using exercise 18, something you cant use in general seems counterproductive

Id say

[chilly ocean]

Nothing is counterproductive ,

[elfin wraith]

By combinatorics they just mean basic counting, you dont really need to study that per se (I do though, im so bad at counting arguments)

[tardy hedge]

one two three four five six

i am doing it i am doing it

[thorn jay]

"Basic" i fucking suck at any kind of it

[tardy hedge]

no u can do it enpeace

it first goes one, then it goes two

[thorn jay]

EIGHT!!! FUCK

[elfin wraith]

I might give that problem a try though, ive not see that and it seems nice, and if I need to look at anymore "alggeo" today ill go insane

[tardy hedge]

FUCK

im sorry dude 😦

[thorn jay]

Solve it with alggeo

[quiet tundra]

What is alggeo

[thorn jay]

Algebraic geometry

I was jesting, though

Algebraic geometry usually deals with commutative rings anyways

[quiet tundra]

😭🙏

[quiet tundra]

Got it

[thorn jay]

(Unless you're a chad doing noncommutative geometry)

[elfin wraith]

I also say "alggeo" because the course is ostensibly algebraic geometry, but I dont think it deserves to be called so

[thorn jay]

[elfin wraith]

I pray to be, just need to actually get into a masters at some point

[thorn jay]

How so?

[elfin wraith]

We dont even do topology

[quiet tundra]

Our course has both involved

The teacher never differentiated while teaching

[elfin wraith]

Well, we often do, but "secretly" which just makes things 100x more confusing. Its not terrible for a first course, but eh, its not great either

[thorn jay]

"Nonzero zero divisor" both left/right? Either left or right?

[thorn jay]

That sucks

[quiet tundra]

Either left or right

[thorn jay]

😭 why

Thats ass

[thorn jay]

Oki

[elfin wraith]

I think im instantly less interested if it can be a one sided zero divisor that seems more annoying than I have time for right now

maybe not, lemme see

[thorn jay]

I knoww

[quiet tundra]

He is an ass, comes to class just to do revision for himself and doesnt care of we understood or not. He is the reason I failed my Group Theory paper last semester and now barely surviving this Ring theory

[thorn jay]

Ah wait, if x is a not right left zero divisor of R, then the right monoid action of R on R
x -> xr
Must be injective, and xr is a zero divisor for all r, so
|R| = |xR| <= n
But of course n < |R|, so |R| = n, which obviously cant be true

Hence every zero divisor must be left and right

Wait thats actually a really cool fact

[chilly ocean]

We had this fact in class

And the proof was same using a map

[thorn jay]

Yeah i say "monoid action" but thats math jargon for "this mapping which happens to be a monoid action" lmao

[chilly ocean]

Wait lemme see the notes

[elfin wraith]

Sorry I think I must be missing an assumption youre making somewhere because this isnt true in general

something something matrices

[thorn jay]

It uses the fact there are a finite amount of zero divisors

[elfin wraith]

Yep yep now I agree

[thorn jay]

:33

[elfin wraith]

Ok, im going to give myself until half past to try this problem then I need to do actual work

[chilly ocean]

Why xr is zero divisor there

[chilly ocean]

, u assumeed

[thorn jay]

Consider the group action by R^+ on R, so on a zero divisor z:
z -> r • z
By orbit-stabiliser:
|Rz| = |R| / |r in R where r•z = 0|
=> |R| = |Rz| • |r in R where r•z = 0|
But both sets on the RHS are sets of zero divisors + 0, so
|R| <= (n+1)^2
As desired

[thorn jay]

Let y be nonzero such that yx = 0, then yxr = 0. Hence xr is a zero divisor

[elfin wraith]

Thats incredibly nice

very slick

[quiet tundra]

Thanks alot

[thorn jay]

Didnt even need the fact that every zero divisor must be both left and right

[elfin wraith]

I need to think of orbit stabiliser more often, it shows up in nice places so randomly

Fair play, I did not see that approach at all and im not sure I would have

[chilly ocean]

Group approach came in mind, but I m in bed

[thorn jay]

I do too!!

I was thinking about the fact that
r • x = r' • x <=> (r - r') • x = 0
And that gave the enlightenment to use orbit-stabiliser

[chilly ocean]

Can anyone come to my question

[rocky cloak]

So given a prime P you can take the intersection with C[X], then there's two cases
The intersection of (X - a) or 0.

In the first case you have C[X, Y]/(X-a) = C[Y] is a PID, so you just need one more element in addition to (X-a).

In the other case I'm not sure if you're allowed to use localization, but localizing at C[X]\{0} means it corresponds to an ideal in C(X)[Y] which is a PID

[next obsidian]

There’s an easier solution jagr

[rocky cloak]

I'm sure there is

[next obsidian]

If you are a height 0 prime, then you are just (0)

[chilly ocean]

Very convoluted

[next obsidian]

If you’re height 1, because you’re a UFD you’re generated by 1 element

Proof: Let f be in P, and then consider the irreducible factorization f = Prod f_i^n_i

By primality some f_i is in P, but (f_i) is a prime ideal contained in P

So if P is height 1, then P = (f_i) is principal

Now if you’re height 2, take a height 1 prime contained in P, this is principal

So look at (f) < P, then mod out by (f)

oh, ChmonkaS

I wanted to immediately say this is a PID and so you are generated by 1 more element

Anyway, I say this was gonna be simpler only because you avoid localization

[chilly ocean]

Why stabilizer stabilizer was those r where r.z=0

[next obsidian]

I mean maybe you just say Nullstellensatz

For the height 2 primes

[rocky cloak]

I'd argue it's very straight forward.

You have something in C[X, Y] and you reduce it first to C[X] then add in Y.

[next obsidian]

But this also uses dim C[x,y] = 2

I think your proof is simpler, GG

[chilly ocean]

What do you even mean by adding here

[rocky cloak]

You first consider P\cap C[X] then P

Anyway, to avoid localization:

Consider a polynomial in P of smallest Y-degree. Then since polynomials in X is assumed not in P you can freely multiply by powers of them. Then you can do polynomial division as normal (treating X-polynomials basically like a unit)

[tough raven]

In a UFD, any prime ideal is generated by irreducibles, say by k of them, and killing one irreducible at a time, we get a chain of prime ideals with k inclusions, so k ≤ Krulddim = 2?

Hmm, this assumes if R is a UFD and p, q, r are distinct irreducibles, then R/(p) is a UFD and the images of q, r are distinct irreducibles.

... which is false by algebraic geometry examples (such as ℂ[X, Y]/(Y^2 - X^2(X+1)) not being integrally closed).

[thorn jay]

thats not the stabiliser, but is the same size as the stabiliser

Again by orbit-stabiliser

[next obsidian]

Yeah exactly

I think doing induction based on the # of variables as Jagr suggested is the correct approach

[warm badge]

but how do i count that?? by isomorphism to Up, multiplicative group of integers modulo p, then this must be asking me how many elements in Up have order dividing n.

[rocky cloak]

Well, it's a more general thing.

You know that Up is cyclic, so it's isomorphic to Z/(p-1).

But in general for a cyclic group, how many elements are there of various orders

Maybe think first about the case where n divides p-1.

How many elements in Z/(p-1) satisfies nd = 0?

[astral galleon]

What does the double arrow mean ? Is it the action of the function pointwise ?

[rocky cloak]

Double headed arrow usually means surjective function (or maybe just epimorphism).

I'm not sure what you're asking in your second question

[astral galleon]

Oh I was asking if it denoted what has happening to each of the elements individually. I've seen the arrow notation but not it written like that

But my misunderstanding is cleared up thanks !

[hazy cloak]

you're thinking of $\mapsto$, as in:

$f: \mathbb{R} \to \mathbb{R}$

$x \mapsto x^2$

[cloud walrus]

that_one_gal9

[tough raven]

If K is a field of characteristic other than 2 and h ∈ K[T] is square-free and non-constant i.e. not in K, is K[T][sqrt(h)] necssarily an integrally closed domain?

t^(1/4) is not in the field of fractions of ℚ[t^1/2], though.

[south patrol]

Oh lol I am dumb

I misunderstood/misread this question

[rocky cloak]

So K(T, sqrth) = K(T)[sqrt h].

a + bsqrth is integral over K[T, sqrth] then so is a - bsqrt h, just by applying the Galois automorphism.

Therefore 2a is integral, and since 1/2 is in K a is integral. And since a is invariant under the Galois action is integral over K[T], hence in K[T].

This leaves the case of bsqrth. You would need to somehow argue that b is in K[T]...

b^2 h is in K(T) and integral, so is in K[T]

Then I guess use that it's a ufd and h square free

[tough raven]

Ah, I see. It just comes down to 2 being invertible in K[T].

Since I tried my hand at being general anyway:

Let A be an integrally closed domain with field of fractions K. Let K(alpha) be a quadratic extension with alpha^2 = a alpha + b. Let z = x + y alpha ∈ K(alpha). Then z^2 - (2x+ay) z + (x^2 + axy - by^2) = 0. So if y ≠ 0 so that z ∉ K, z is integral iff 2x + ay, x^2 + axy - by^2 ∈ A. (In fact if y = 0 the condition implies x^2 ∈ A ⇒ x ∈ A, so the equivalence is true even if y = 0.)

Now suppose a = 0 i.e. we adjoined a square root (we can do this WLOG in characteristic not 2). Then the condition simplifies to 2x, x^2 - by^2 ∈ A. (This should look familiar: indeed, the polynomial we found above is the characteristic polynomial of multiplication by z, so in general (for a quadratic extension) we got z is integral ⇔ tr(z), N(z) are integral). This implies that 4by^2 ∈ A.

If additionally we assume that A is a UFD and make b square-free, then we can conclude that 2y ∈ A. So (in characteristic not 2) equivalently, 2x, 2y ∈ A and (2x)^2 - b (2y)^2 ∈ 4A which is some homogeneous quadratic equation in A/4A. So the integral closure is
{(x+y sqrt(b))/2 : x, y ∈ A, (x, y) mod 4 is a solution to x^2 - b y^2 = 0}.

It seems difficult to say more for a general A, unless we know A/4A well. (For example, if 2 is invertible, we just get A + A sqrt(b). If A = ℤ, we can brute-force solutions for each b.)

Should see what happens in characteristic 2 some time.

Also, is true for any finite extension of the field of fractions that z is integral ⇔ its characteristic polynomial has integral coefficients?

[chilly ocean]

A very intresting question is asked in my class we have to use there exist irreducible polynomial of degree 3 over Z_P and we cant use field theory

we can use group theory

[astral galleon]

Ahh that makes sense i get it now 👍

[acoustic igloo]

this seems related to a problem i did before

Define $\phi:\mathbb{Z}_p\to\mathbb{Z}_p$ by $\phi(n)=n^3-n+1$\
$\phi(1)=\phi(-1)$ so $\phi$ is not one-to-one and hence not onto\
Let $a\in\mathbb{Z}_p$ not in the image of $\phi$\
Then $x^3-x+1-a$ does not have a zero, otherwise $a\in\phi[\mathbb{Z}_p]$

[cloud walrus]

Axe

[acoustic igloo]

this fails if p=2

[chilly ocean]

same poly works for p=2

[acoustic igloo]

cool

oh yeah because phi(0)=phi(1) in that case

[cloud walrus]

🇵🇱 uuybnuuy

[acoustic igloo]

is it really closed?

is G abelian?

[topaz solar]

So, what are you asking

It’s true if one of A or B is normal

So, find a counterexample with 2 subgroups that aren’t normal

[chilly ocean]

it holds when AB=BA

[ivory ore]

||try to look in Q8 ||

[ivory ore]

quaternions

[vagrant zinc]

you have already found the inverse just say this

[ivory ore]

and then try to look in S3

you would be hopeless in Q8, ||AB=BA for all subgroups A,B in Q8||, that would lead you to a nice result i hope

this is almost tangent to your query tho

[acoustic igloo]

there is a counterexample in S3

[ivory ore]

ofc, the involutions

i did the finite version in previous problem, however no idea how to pull this out

[frigid shard]

Let $G$ be a finite group with representations $\rho, \sigma$. Let $\varphi : \rho \to \sigma$ be a linear (not necessarily equivariant) map. Then it is well-known that the map $\tilde \varphi = \sum_{g \in G} g \varphi g^{-1}$ produced by averaging is equivariant. My question is whether this would ever be $0$ for non-trivial $\varphi$, and if so please kindly provide an example

[cloud walrus]

n1lp0tence

[warm badge]

if n divides p-1, there is an element of order n, let it be a. another element having order n, generates the same cyclic subgroup as a, meaning it's the number of integers less than n and prime to n. φ(n) elements have order n.

any element having order dividing n, must also be a divisor of p-1. so it's the sum of all φ(m) where m divides n. does that sound right? what about n not dividing p-1? then there is no element of order n, so no element of order dividing n and hence there are 0 elements satisfying nd = 0. how do i show that the sum of all φ(m) where m divides n is atmost n?

ok lol just found out that the sum is n

[chilly ocean]

Define group action of group G, on cosets of H by left multiplication and the kernel because index is finite it has proper non trivial proper kernel , this kernel contains H and and

$\bigcup_{g\in G}gHg^{-1}\subset K$

And K is proper in G

[cloud walrus]

Curvature

[rocky cloak]

The kernel of the group action on G/H is

$\bigcap_g gHg^{-1}$

It doesn't usually contain H

[cloud walrus]

jagr2808

[chilly ocean]

Ohk i think i missed something

[rocky cloak]

The kernel is also called the core of H

[chilly ocean]

Yes it's inside H

[elfin wraith]

Im pretty sure $f = 4(x^3+y^3+z^3)$ works for this, but im unsure how to justify the irreducibility of $f$, it certainly feels irreducible, but im unsure how to actually prove that. I know that we also also write every irreducible singular cubic in one of two forms, so maybe thats the better approach but im not sure. This also might be more appropriate for #algebraic-geometry but I thought id ask here since its really the irredcuibility im getting hung up on

[cloud walrus]

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

[elfin wraith]

In the exam for this question they did say you dont need to justify it (and thank god its a 15 part problem, with another 2 questions after it) but I still think I should know how to do this, beyond just it feeling irreducible. I guess since F_7 is small enough I could feasibly brute force it but that sounds hellish

[proper tide]

How the set of all functions from R to R (set of real numbers) with compact support is a ring without identity??

[south patrol]

Just add and multiply pointwise

[proper tide]

Can you please explain it..

[rocky cloak]

(f+g)(x) = f(x) + g(x) defines an addition of functions

(f*g)(x) = f(x)*g(x) defines multiplication

[glad osprey]

isn't it also a ring under composition?

[rocky cloak]

I guess assume f can be factored, then evacuate at y=z=1. Then you get a factorization of x^3 + 2 over F7, but it's irreducible (it has no roots)

[rocky cloak]

Composition doesn't distribute over addition

[elfin wraith]

Is there a general way forward as far as showing irreducibility over multivariate polynomial rings, or do we just, in general, need to work a little more directly like that? This question kinda made me realise ive never really looked at irreducibility outside of "obvious" cases

[rocky cloak]

I don't know of any general strategy, but I'm not sure I would know

So idk

[glad osprey]

ah, I guess that's only true for endomorphisms, not for functions in general?

[elfin wraith]

Thats fair, a little too commutative

[rocky cloak]

But I guess, use Eisenstein / reduce to fewer variables would be good tricks

[elfin wraith]

Yeah that seems like a very sensible way forward. Most of the cases ive seen have had a single term depend on a varible so you assume you can write your function f(x,y,z,w) as (wg(x,y,z)+h(x,y,z))p(x,y,z) and get a contradiction, so i suppose assume it is and reduce varibles until you can use a nice single varible test or hit a contradiction before then

[glad osprey]

Is R^+ the additive group of R? Then Rz = { r + z | r in R } right? But why is this a set of zero divisors? How do you conclude the last line?

[thorn jay]

Yeah i need to properly formulate it gahshsgs

Ill get back to you on it

[glad osprey]

Nice, as long as there's group actions involved, I'm interested

[thorn jay]

Yippwe

[rocky cloak]

I guess it's not really a group action. It's just the homomorphism of R-modules

[tough raven]

There is probably an algorithrm factorisation into irreducibles in multivariate polynomial rings over ℤ. The techniques used in that would occasionally be helpful for manual calculations as well.

[elfin wraith]

I had a quick look to see if sage could do it and apparently it cannot, obviously doesnt mean no algorithm exists (and not even that sage cant do it, the docs are awful) but it does make me slightly less hopeful

Well I say hopeful, its not particularly important to me that one does exist, but I guess it would be nice

[tough raven]

Someone else who finds Sage discoverability terrible 🍻

[thorn jay]

Ah yeah

Left R-modules

But only the abelian groups are needed, really

I.e. the counting argument comes from Langrange

(Which is an instance of orbit-stabiliser ig lol)

[rocky cloak]

I guess it's first isomorphism theorem for G-sets

[thorn jay]

:0 of course

[glad osprey]

hmm, I'm still not following you consider phi(r) = rz as a homomorphism of R-modules? How do you proceed?

[elfin wraith]

Yeah I shouldn't say that they're terrible, the information is there somewhere but best of luck finding it lol

[thorn jay]

By langrange
|zR| = |R| / |ker phi|
=> |R| = |ker phi| • |zR|
But both ker phi and zR consist of nonzero zero divisors and 0, so
|ker phi|, |zR| <= n+1
By the hypothesis, and the result follows

[glad osprey]

I see, nice

[proper tide]

But how I will find that there is no identity of this ring?

[thorn jay]

Because f(x) = 1 does not have compact support

[rocky cloak]

Say f is an identity. Then for each x there exists a compactly supported function with g(x) nonzero.

Then f(x)g(x) = g(x) implies f(x) = 1. But if f(x) = 1 for all x it's not compactly supported

[chilly ocean]

a very nice dinner for all of you

Believe me its worth to spend some time

[south patrol]

Lol death by notation

|G|||Xn| must be a troll

[rocky cloak]

If you just mod out by the core of H it reduces to the finite case.

[chilly ocean]

DAMN FELT THE SAME

[chilly ocean]

i can send more nightmare dreams

[south patrol]

But presumably like you consider how ||G acts on X_n||

[thorn jay]

Oh yeah thats cool

[chilly ocean]

deadly

[chilly ocean]

yes

[chilly ocean]

the solution is just to mimick the burnside theorem proof

It is asked to use galois theory here but coudnt get it (how?)

and yes n is not 4

[south patrol]

Sure

[chilly ocean]

why dont you post it for others

[south patrol]

Oh I mean I haven't written down anything tbh, only on mobile lol

May think about it later heh

[chilly ocean]

sorry i thought you solved it

[south patrol]

I thought I had but was mistaken

Lol

[tough raven]

By?

[thorn jay]

||conjugation I presume||

[chilly ocean]

how to make spoiler

||ghanta||

[elfin wraith]

Just spent close to an hour working out why this question asked me to work out what group this one element set was, only to realise I copied the question down wrong

[chilly ocean]

which que

[elfin wraith]

Computing the group of torsion points of some elliptic curve over Q, but I missed a constant in the equation of the curve and just ended up getting no points that could even potentially be torsion points and kinda trivialised the problem, turns out theres an additional 3 points it could be if you do it correctly

[chilly ocean]

forgot group law for elliptic because it just came up from no where

[barren sierra]

Sage docs are utter ass

I should put my money where my mouth is

and start contributing to improve them

[tough raven]

That assumes you can learn enough to know what's there so you can write about it in the docs. Kudos if you do.

[barren sierra]

I mean mainly what I find lacking is examples

the examples they give of some usage of commands is very esoteric

[elfin wraith]

and its worse the more "niche" the function is, you spend an afternoon looking for the page then you get one or two examples which are so dense and poorly explained you spend the rest of your day working out how it works. I think most of my issues with sage are skill issues on my part, but im glad to know its not entirely that

[lusty marlin]

Thank god someone said this, I had sort of assumed that I was just incapable of making good use of them

[glad osprey]

Also the search function is completely useless, and it's hard to understand how things are structured. Say if I'm reading about polynomials (https://doc.sagemath.org/html/en/tutorial/tour_polynomial.html) and I want to learn more about the PolynomialRing constructor. There are no links to that function from that page, and if I try to search for it: https://doc.sagemath.org/html/en/reference/search.html?q=polynomialring A long list of completely useless results...

maybe the easiest way to find stuff is to scroll through this list: https://doc.sagemath.org/html/en/reference/py-modindex.html

[coral shale]

maybe AI search time

[elfin wraith]

Thank god its written in plain english python otherwise it would be actually unusable

[acoustic cargo]

protip: use find and grep in the html/en doc tree to narrow your search

[earnest pecan]

I think this works:

Note that $f_n(x) = 1$ precisely whenever $x \in S := {1, 2, \dots, n}$. Suppose for contradiction that $f_n(x) = g_n(x) \cdot h_n(x)$. For any $k \in S$ we must have $g_n(k) = h_n(k) \in {\pm 1}$ in order to satisfy $f_n(k) = 1$. Define $S_1 := {k \in S : g_n(k) = 1}$ and $S_{-1} := {k \in S : g_n(k) = -1}$.

Suppose $n$ is odd, so that either $S_1$ or $S_{-1}$ must contain more than $\frac{n}{2}$ elements. Without loss of generality, we assume that $#S_1 > \frac{n}{2}$. It follows that $g_n(x) - 1$ and $h_n(x) - 1$ both have more than $\frac{n}{2}$ roots, so that $\deg(f_n) = \deg(g_n) + \deg(h_n) \geq \frac{n}{2} + \frac{n}{2} = n$. This is a contradiction, so we conclude that $f_n$ is irreducible for any odd $n$.

Am not sure how to deal with the case where $n$ is even rn.

[cloud walrus]

Catch23

[earnest pecan]

Hate it though, very inelegant. Did you think of anything nicer?

[chilly ocean]

If n is even then observe it is square of some poly

[amber maple]

what does sigma B actually mean here it kinda just shows up in the books im looking at and it doesnt seem to be defined that well

is it just sigma acting on B only

[next obsidian]

That’s the restriction of a map lol

B is a subset of E so you can take an automorphism of E and restrict its domain to B

@amber maple

The only content of this is that the image of B is equal to B, this says that sigma restricted to B actually also gives you an automorphism of B

[amber maple]

checks out

the proofs in this book are a little... lazy, so its annoying to follow

less rigorous than i like :(

[next obsidian]

I mean I can prove this to you lol

B is generated by the roots alpha_i of f(x), and because sigma fixes E this means f(sigma(alpha_i)) = 0 so all sigma(alpha_i) = alpha_j(i)

This means generators of B get sent into B, so all of B maps into B via sigma

But B is a finite extension of E so if you have E < sigma(B) < B and both B and sigma(B) have the same degree over E the two are equal

Tadah

[crystal vale]

let G be a finite group with a proper normal subgroup N that is not contained in the centre of G. Prove that G has a proper subgroup H with | H | > \sqrt | G |.

take n in N which is not in Z(G), then by orbit Stabilizer theorem, we have | G | = | cl(n)| | C(n) |, since N is normal subgroup therefore | cl(n) | < | N | , and n not in Z(G) implies C(n) is proper subgroup of G, so we have | G | < |N| |C(n) |, now take subgroup which has maximum order among them, so we get H, such that | G | < |H|^2.

is it correct?

[velvet hull]

everything looks right except I'm not sure how you get the inequality to be strict

[crystal vale]

because | cl(n) | < |N|

[velvet hull]

I don't immediately see why equality isn't possible

[crystal vale]

n not in Z(G) and e not in cl(n)

[ivory ore]

alr

[crystal vale]

8 a part, we can deduce that f(p/q) = f(1)^{p\q}.

So f is uniquely determined by f(1), so we have infinitely many homomorphisms.

Is it correct?

[kind temple]

yes, that’s correct

[crystal vale]

Okay thank you

[crystal vale]

8, b part

Let f:Q -> Z be a group homomorphism, then f(x+y) = f(x) + f(y).

Now it implies f(1) = qf(1/q) for all q in Z{0}.

Since the image is in Z, so it implies q divides f(1) for all q in Z{0}, which is not possible if f(1) ≠ 0.

So f(1) must be 0.

And we can conclude that qf(p/q) = pf(1), so qf(p/q) = 0 implies f(p/q) = 0 ( q≠0 ).

Hence there is only one group homomorphism Q to Z.

[kind temple]

that is also correct

nice

[crystal vale]

Okay Thank you

[crystal vale]

There is a mistake f is not uniquely determined by f(1)

[chilly ocean]

If f(1)=1 then what is f(1/3)

[crystal vale]

The point is f(1) cannot be any non-zero value

[chilly ocean]

if it is some non zero then there is a loop

[crystal vale]

I proved this

[chilly ocean]

thats why i asked it

what is f(1/3)

[crystal vale]

See this

[crystal vale]

It is 1/3 but it is not possible because f(1/3) must be in Z

[chilly ocean]

thats correct infact it was one of my interview question,

[chilly ocean]

i was talkin about part (a)

[crystal vale]

I see

[chilly ocean]

not b

b is right

[velvet hull]

the idea you have here can be summarised by the following observation -
as rings, Q is torsion, but Z is free. Hence there can only be the trivial ring homomorphism
(note. Q is not a torsion group. see below, but the argument can be modified to work in group theory context)
and indeed, that is what you essentially show in your argument, that the image of a torsion group must always be torsion

[crystal vale]

Then f(1/3) = f(1)^1/3, so now you have to choice to take which root

Therefore f is not uniquely determined by f(1)

[chilly ocean]

infact every subgroup of Z is cyclic but Q isnt

only possibility for the image to be cyclic is trivial

[earnest pecan]

How so?

[crystal vale]

But the finitely generated subgroup of Q is cylic

I am making a mistake here

[chilly ocean]

same oberservation, you method does proove f_n(x)=(g_n(x))^2 as g_n(x)-h_n(x) has n roots so use identity theorem and get g_n=h_n over reals

[chilly ocean]

yes then what

you have to map whole Q

[earnest pecan]

Ah thats pretty neat

[chilly ocean]

still irreducibility is not proven

[hasty aurora]

so advanced huh

[thorn jay]

[jade mason]

I don't think torsion group is the word you're looking for

[velvet hull]

oh yeah you're right, poor choice of wording

[thorn jay]

Divisible group, right?

[chilly ocean]

I heared torsion part, torsion module , what is torsion group

[velvet hull]

exactly what you expect it to mean

[chilly ocean]

it seems it means every element of that group is of finite order

[jade mason]

Yeah

[chilly ocean]

now $g_n(x)$ has half the degree as $f_n$ and implies $g_n(i)=+1$ or$-1$ for $1\leq i\leq n$ let $n=2k$ if any of +1 or -1 cant be attained more that k times so you can assume both are attained k times exactly

[cloud walrus]

Curvature

[earnest pecan]

Yeah you’re right. Am not sure how to derive contradiction from f_n = g_n^2.

[chilly ocean]

WE get f(x)-f(y)=2 for some x and y integers with choices 1 to n

and we will get f(x)+f(y)=0

[earnest pecan]

For any x,y integers 1 to n, we have f(x) = f(y) = 1, so that f(x) - f(y) = 0 and f(x) + f(y) = 2. Moreover, with g(x), g(y) = +/-1, we get the same result: f(x) = f(y) = g(x)^2 = g(y)^2 = 1.

So I think you’ve flipped the two, but even still I’m not sure how this leads to contradiction?

[cloud walrus]

🇵🇱 uuybnuuy

[rich granite]

Are you sure the question doesn't specify k to be an integer?

ϕ: Z[X] → Z[X] that maps x to 1 is a ring homomorphism but ϕ(x) =1 ≠ x × ϕ (1)=x

So it has to be k ∈ Z

[rich granite]

And then R is to be interpreted as a Z-algebra for this to make sense

But at least then 1. is true

[chilly ocean]

Can you write down the full exapisn of polynomials f(x)-f(y)=2, contradiction follows immediately

[thorn jay]

Even from a UA standpoint those are the same

[rich granite]

Wouldn't they only be polynomially equivalent

[thorn jay]

The isomorphism of equational theories is literally just an extension of the nullary operations by k := 1 + ... + 1 (k times, with a minus if k < 0)

I.e. the most "the same" two equational theories can be without being exactly the same

[rich granite]

Yeye

[thorn jay]

Or unary ig who cares, just add a •x

[rich granite]

Me when the category is isomorphic to the undercategory under the initial object

[earnest pecan]

But where do you get f(x)-f(y)=2 from?

[chilly ocean]

there are some integer x and y such that f(x)=1 and f(y)=-1

[earnest pecan]

How do you know that f(y) = -1 has integer solutions?

[chilly ocean]

sorry for mixing up the notations

[chilly ocean]

i should say we have some i and j such that g_n(i)-g_n(j)=2

[earnest pecan]

Okay okay I agree w/ that. For some integers i,j we have g_n(i) - g_n(j) = 2 and g_n(i) + g_n(j) = 0. Now how do you say the contradiction follows?

[chilly ocean]

$g_n(i)-g_n(j)=\sum a_k(i^k-j^k)$

[cloud walrus]

Curvature

[chilly ocean]

you just have to write it and see the right side, i-j divides it

i will be intrested if this can be solved using eisestine critrion or some other criteria

[vagrant zinc]

I put + because I don't know what type of operation it is, but the ideal one is that one

[earnest pecan]

Err I don’t follow this

How?

[chilly ocean]

$g_n(i)-g_n(j)=\sum a_k(i^k-j^k)=2$

[cloud walrus]

Curvature

[chilly ocean]

i-j divides 2

[earnest pecan]

Yes I see the contradiction but how do you get $g_n(i) - g_n(j) = \sum a_k(i^k - j^k)$?

[cloud walrus]

Catch23

[chilly ocean]

$g_n(x) =\sum a_k x^k$

[cloud walrus]

Curvature

[earnest pecan]

Ah of course lol

Well thank you

[chilly ocean]

i assume there exist a solution using some other trick

maybe eisestine

we can use the tht there exist a prime between n and 2n always

[earnest pecan]

Does that hold for any n? Didn’t know tjat

[thorn jay]

Ye thats true lmao

[sturdy spear]

moment when two people talk about cat theory

[chilly ocean]

yes

[south patrol]

"Bertrand's Postulate", which now has various proofs

(I think the first proof used a good amount of analytic number theory but there's a much more elementary one due to Erdős for example)

[earnest pecan]

I see i see

[vestal stream]

is there an explicit construction for the following : given a complex number z such that Re(z)>0, and Im(z) != 0, can i find P in R[X] with positive or zero coefficients such that (X^2-2Re(z)X+|z|^2) P(X) has positive or zero coefficients ?

[violet spade]

Maybe try $X^2-2Re(z)X+|z|^2=(X-z)(X-\overline{z})$

[cloud walrus]

joel

[violet spade]

If you mean multiplication, just take P(X) = 0. Then the product is clearly zero haha

I'm not sure i understand....

@vestal stream

[acoustic rose]

How do I show that the set of algebraic numbers is a field?

[tardy hedge]

I think you can use tower law for fields and properties of field extensions

to show a,b in F makes a+b in F etc

[acoustic rose]

sorry could you explain this a little more

[tardy hedge]

Oh I just mean to show its a field, you need to show its closed under addition, multiplication, division etc. In this case that means you need to show a+b, a/b etc is algebraic. You can do this by looking at field extensions

Because we know finite field extensions are algebraic

[lime junco]

yall, i just wanna make sure im not tripping

is Ext^1(P), where P is projective

always trivial?

by Yonedas

since if we characterize it via extensions, theres only one extension up to yoneda equivalence, because any extension splits by projectivity

[median ether]

when a \in xH and a \in yH (left cosets in G, a \in G), how to explain correctly xH = yH? i see x and y has dependency between them (x = y*(composition of elements of H)). y*h'*h generates every element in yH. by dependency with x, it generates xH.

[trail cave]

How would I show that pi + sqrt2 is transcendental over Q?

[chilly ocean]

You have to use pi is trancesdental

[trail cave]

Yes but how exactly

I showed using pi is transcendental that pi is also transcendental over Q(sqrt(2)) but I don't see how for this one

[chilly ocean]

What is the degree of Q(π+√2): Q(√2)

[mighty kiln]

Sums, products, and reciprocals of algebraic numbers are algebraic

[trail cave]

Sorry I'm not sure, would it be the degree of the minimal polynomial of pi+sqrt(2)?

[trail cave]

That's what I'm trying to prove though in this case otherwise the problem is trivial right

or at least I don't think I can assume that here

[chilly ocean]

If the extension is algebraic so every element is algebraic π+√2 is also algebraic as Arki said π+√2 - √2 is also algebraic

[trail cave]

ah ok I see

thank you

[chilly ocean]

is the reduction map $SL_2(\mathbb{Z})\to SL_2(\mathbb{Z}/N\mathbb{Z})$ is surjective ? for $N\geq 2$

[cloud walrus]

Curvature

[thorn jay]

Is this well-defined?

As a homomorphism I mean

Is it a homomorphism

[mighty kiln]

Yes

[thorn jay]

Ye i see that now, is it also a homomorphism? I suppose so because x -> x/N is a ring hom too?

[mighty kiln]

Yes

[thorn jay]

Cool :0

Thats a general fact for a ring hom h : R -> S I can imagine

[ivory ore]

$\text{Aut}(Q_8) \cong \text{PGL}(2,3) = \text{GL}(2,3)/{\pm I} \cong S_4$

[cloud walrus]

longboard kayak

[ivory ore]

got this on wiki page

how to argue for the isomorphism for the expreme left and right

i can see i,j,k are inter permutable upto sign

ahhh

$\text{Inn}(Q_8) \cong Q_8/Z(Q_8) \cong V_4$

[cloud walrus]

longboard kayak

[ivory ore]

how to show that the Out is S3

also can we argue from the "geometry" ?

Q8 acting on the 3dimensional real space {\pmi, \pmj, \pmk}?

[chilly ocean]

yes it is well defined

[chilly ocean]

i have takken this question from the book neil koblitz modular forms, there are other similar type of questions in that book

[thorn jay]

Ah thats why I'm ending up with an integer equation
Lol

My idea was to look at the morphism of exact sequences

0 -> SL_2(Z) -> GL_2(Z) -> C_2 -> 0

0 -> SL_2(R) -> GL_2(R) -> R* -> 0

And by assuming the desired surjection to force an injection of C_2 into R*, and if R = Z/NZ with phi(N) is odd that would be a contradiction

The way to force an injection is for the kernel of the map between the GL_2's to contain only special linear matrices

This is probably not the right way to tackle it though lol

You'd need to show the impossibility of the integer equation

n^2(ad - bc) + n(a + d) + 2 = 0

(If phi(n) is odd)

[chilly ocean]

you are claimin map is not surjective?

[thorn jay]

If the map is surjective and this has no solutions and phi(n) is odd, then we get a contradiction

[rocky cloak]

I guess one way would be to show that an outer automorphism is determined by how it permutes the three subgroups of order 4

[chilly ocean]

actual answer is yes but

[thorn jay]

Oh thats cool

[thorn jay]

Nvm then, cool proof that this then must always have a solution if phi(n) is odd

[chilly ocean]

since the actual ans is yes, contradiction is not right approach we have to construct some matrix with required image

[thorn jay]

I know

But because we know the answer is true, this gives a proof that if phi(n) is odd, the above equation must have a solution, as if it didnt, the map couldnt be surjective

[ivory ore]

{\pm1, \pmi(j,k)} right?

honestly forgot the definition of Out to make the connection

[rocky cloak]

Out is Aut/Inn

[ivory ore]

one thing is poking me

if we let k be unchanged and interchange i with j

does it not affect the algebra

[rocky cloak]

That will not be an automorphism

[ivory ore]

k has to be replaced by -k

[rocky cloak]

Yup, then you're all good

[ivory ore]

ohh ig this thing gets under the carpet if we quotient out by the Inn

and we only get to see permutation of 3 objects which is not fully automorphism

$1 \to V_4 \to \text{Aut}(Q_8) \to S_3 \to 1$

[rocky cloak]

Yeah, the automorphisms are like permuting i, j and k, except they kinda change signs randomly

[cloud walrus]

longboard kayak

[ivory ore]

is it spliting?

[rocky cloak]

And the action of the inner automorphisms is exactly changing those signs

[rocky cloak]

I would very much doubt it, but I haven't checked

[ivory ore]

i mean if you try to reverse construct the Aut is there one option left for us?

ahh ok let me see how S3 can act on aut of V4

i think AutV4 too is S3

without any knowledge whatsoever i feel like these objects would nicer to study if i know representation

[thorn jay]

Any objects are nicer study using representations

[ivory ore]

idk, maybe you guys are genius, but my stupid brain always wants nice things, speaking of which, i always used to think that B in a subgroup of G in the case of

$1 \to A \to G \to B \to 1$

Q8 broke my dream

[thorn jay]

I love nonsplit extensions

[rocky cloak]

Wait till you hear about Z/4

[ivory ore]

crazy thing is i knew the concept of semi direct product/ nontrivial action, still my brain made me convinced that B has to be indeed in G

i mean c'mon

[thorn jay]

Semidirect product is precisely when B is naturally (via a section of the projection) in G

[ivory ore]

$1 \to C_2 \hookrightarrow \ Q_8 \twoheadrightarrow V_4 \to 1$

[cloud walrus]

longboard kayak

[ivory ore]

so this is NOT a semidirect product?

[rocky cloak]

That's correct. There aren't many maps V4 -> Q8

[ivory ore]

oh at the definition of semidrect product it is mentioned both has to be subgroup

hmm, i see

so not all the extentions have to be semidirect product?

[rocky cloak]

Also C2 is in the center, so the only way for it to be semidirect is if it is direct

[ivory ore]

ohyeah

[rocky cloak]

I think you can probably argue that most of them are not

Especially if you're dealing with p-groups

[ivory ore]

so the core misconception for me was that i used to think all the extensions are some semidirect product

i tried classifying order 8 groups with that approach

now i can see the bug

it's so rewarding to do mistakes in algebra

[rocky cloak]

My Google-fu tells me that Aut(Q8) is indeed the semidirect product of its inner and outer groups and that it's isomorphic to S4

[rotund aurora]

The first proof was given by Chebyshev and it's completely elementary (and short). See Theorem 2.4 in Montgomery's multiplicative number theory book and the discussion that follows

[ivory ore]

got lucky, btw what is google-fu ?

[chilly ocean]

google f*** u

[rocky cloak]

Google-fu is one of the ancient Chinese martial Arts, first developed for use in combat, but later finetuned for the purpose of finding information online.

[ivory ore]

interesting

i used to use approach0 at times

[rocky cloak]

https://en.m.wiktionary.org/wiki/Google-fu

[crystal vale]

I did this question in this way, any better approach?

[south patrol]

Oh sure lol nice

I will look thanks

[chilly ocean]

assume m=n then |G|=1+2m shows m divides 1 hence m=1 this gives G is Z_3 now assume (WLOG)m>n so |G|-m=1+n but m divides left side so is right side and m divides 1+n this forces 1+n=m so |G|=2+2n similarly |G|-2n=2 gives n divides 2 hence n is either 1 and 2 (NOW YOU CAN COMPLETE IT)

[crystal vale]

m divides 1 + n how it implies 1 + n = m?

[chilly ocean]

if n=1 (NOT possible)(AS GROUP OF ORDER 4 IS ABELIAN) and n=2 gives |G|=1+3+2

[chilly ocean]

m>n and divides 1+n implies m<= 1+n

[crystal vale]

i see

thanks @chilly ocean , this one is better approach

[chilly ocean]

order of conjugacy class divides order of group is main thing to use

[crystal vale]

yes Orbit Stabilizer theorem

[chilly ocean]

please share this pdf

[ivory ore]

so essentially you got two constraints right?

all of them adds to the group order

[chilly ocean]

why dont you do that

[ivory ore]

and if you multiply three of them individually with respective index of centralisers you get the group order

wdym?

[crystal vale]

yes

[ivory ore]

as the number here is small im thinking of manually working all those out

[crystal vale]

if i take Z/2Z \times Z/2Z\times..., then this is not PID but every prime ideal is maximal in this ring, correct?

[rocky cloak]

Same holds true for any commutative artinian ring (so in particular finite rings)

[crystal vale]

i will prove it for Artinian ring, yes in finite ring every prime ideal is maximal ideal but i didn't work much in finite ring so now i want non PID finite ring, maybe we can take non-integral ring Z/6Z, but finite ring which is integral but not PID

so we can say quotient of PID is not PID because since it must be integral domain but Z/6Z is not

[crystal vale]

I have an interview in the upcoming month for int phD in mathematics. And I am very serious about this interview so if anyone wants to take my mock interview, he/she will be welcome. Let me know if you are interested.

If this message shouldn't belong here I will delete it.

[stuck egret]

yeah

ig

[vestal stream]

yep i meant multiplication, P has to be different than 0 though

[crystal vale]

You interested?

[stuck egret]

Yep

[crystal vale]

Thank you ❤️

[tardy hedge]

True that! He is working hard here all the time

[chilly ocean]

I wish I could work hard during my days

[chilly ocean]

What is the meaning of this, how to relate it

[chilly ocean]

Couldn't understand it

Your delightness is too much for me

[chilly ocean]

I thought you troll me

[tardy hedge]

Haha

No James is a nice fella

[chilly ocean]

everyone is nice , the influence of worst situations can make them bad(for instance)

[soft tiger]

I’m trying to solve exercise E5 here and I’m confused as to why they tell us to use part 4 since m=ord(a) and n=ord(b) are not necessarily coprime in this case.

I decided to go this direction, but in this work, I’m confused as to how to derive a contradiction since there’s no guarantee that xd is a multiple of m and yd a multiple of n

[trail cave]

How would I do this problem? I saw solutions to similar problems online where they square then cube sqrt(3)-sqrt(5) but I don't see how this helps

[chilly ocean]

$\sqrt(3)+\sqrt(5)\in Q(\sqrt(3),\sqrt(5))$ so $\sqrt(3)+\sqrt(5)+\sqrt(3)-\sqrt(5)=2\sqrt(3)\in Q(\sqrt(3),\sqrt(5))$

[cloud walrus]

Curvature

[chilly ocean]

similarly root(5) is inside Q(\sqrt(3),\sqrt(5))

[rocky cloak]

The trick is to find i and j such that the orders of a^i and b^j become coprime, while retaining the same lcm.

It might be easiest if you think about how you can determine the lcm from the prime factorizations

[rocky cloak]

.

[trail cave]

So just to check are we essentially showing that we can manipulate sqrt(3)-sqrt(5) into sqrt(3) and sqrt(5) respectively to show inclusion?

and the other way is trivial

[alpine plank]

yes

[trail cave]

ok thank you

[chilly ocean]

using fields proprty only

[alpine plank]

an alternate method is to observe that its a subfield of the 2nd field but its minimal polynomial can't have degree 2

[violet spade]

Show sqrt(3)+sqrt(5) in there by calculation of 1/(sqrt(3)-sqrt(5))

[violet spade]

true

[chilly ocean]

yeah that must be 4 nice

[stuck egret]

goop theowe

[tardy hedge]

hi jagr how are u doing today

[rocky cloak]

Okay-ish. But of a cold

Wbu?

[tardy hedge]

Aw that sucks

ive been kind of sick recently too

[rocky cloak]

Such is the season of the rainy spring

[tardy hedge]

I've also been so distracted by this girl I've been seeing. I've only dated her for a month but its been so much so far

Felt kind of guilty i couldnt do much on my thesis stuff this past week i have a meeting with supervisor today lol

[chilly ocean]

same but in 2020-2021

[tardy hedge]

4 year dry spell? get back on the scene!

[chilly ocean]

no i am mature now

imp phase of my life

[tardy hedge]

mature means no female companionship?

[delicate orchid]

Yeah pretty much

[chilly ocean]

yeah,

[tardy hedge]

hahahaha

[thorn jay]

EVERYONE HAS

I SWEAR

[tardy hedge]

wow

[delicate orchid]

I am also ill

[rocky cloak]

Or done with dating, cause you got married and stuff....(?)

[chilly ocean]

that time i could solved group theory but didnt

[thorn jay]

Its spreading through mathcord

[delicate orchid]

I always knew this place had a negative impact on my health

[chilly ocean]

i have end sems of sem 1

[thorn jay]

The good ending

[chilly ocean]

i feel like gohan here

[delicate orchid]

Ain’t nobody seen yo shows!!! We’re fifty seven y/o here

[thorn jay]

Im just uncultured tbh

[soft tiger]

I still find this a bit confusing

[chilly ocean]

watchin ramayan

[rocky cloak]

Dragon ball Z, more like dra-Gen X, immaright?

[thorn jay]

Dragon de-

[delicate orchid]

ts ain’t it unc 💔

[alpine plank]

The hours before meeting your supervisor is the time you learn the most

[tardy hedge]

Hahaha

[thorn jay]

Your thesis was about smt combinatorics right

[tardy hedge]

yh

[thorn jay]

Thats cool, hats off, i probably could never lol

[rocky cloak]

It might be easier to look at examples with few primes.

Like say a has order 12 and b has order 18. For which i and j do a^i and b^j have coprime orders? Can any of these combine to make an element of order 36?

[trail cave]

This is the last question. I've shown I is a prime ideal, and I believe I is a maximal ideal. To show this I need to show there's a homomorphism between Z[x]/I and (I think) Z_3, with kernel I to use the first iso. theorem and show Z[x]/I is a field. I just don't know how to choose the homomorphism

[rocky cloak]

Homomorphism out of Z[x] are particularly nice. They can be fully described by where x is mapped, and you can map x anywhere you like.

[trail cave]

oh and I also need to show that it's not principal (I already showed I = <3,x>)

[alpine plank]

The ideal?

[trail cave]

yes

[rocky cloak]

I guess think about when you can write 3 as a multiple of some other polynomial, and same for x

[trail cave]

can I map it literally to like, [x] mod 3?

[alpine plank]

yea but you have to fix a value for it

[rocky cloak]

Well you have to map it to something that is an element of Z_3 (or just whatever codomain you have)

[chilly ocean]

p(x) goes to p(0) mod 3 the kerner is clearly now (3,x)=I

this ideal is maximal

third isomrphism theorem

[trail cave]

third isomorphism theorem?

I can use the first right?

[chilly ocean]

Z[X]/<X,3>=( Z[X]/<X>)/(<X,3>/<X>)=Z/3Z

[trail cave]

I see, thank you

[chilly ocean]

yes

[trail cave]

I've stayed up all night doing this he sent out an email last night at 5pm saying it was due today instead of Thursday

[chilly ocean]

horrible prof

cant feel the pain

[chilly ocean]

p(x) goes to p(0) mod 3 and first iso is sufficient

[ripe crest]

Why do we say vector space over F, or a k-vector space. Or an algebra over k, or a k-algebra, or an algebra with a base field k.

In my view, if I have a group, ie a groupoid with a single object, and I have it act on, let's say a set, then the action is a functor from Grp to Set.

So I think of it as the group acting on the set.

If I have a k-vector space, then I have a k-action on an abelian group. So I think of k acting over the group.

Why not say it's a vector space under (the) k(-action)

[trail cave]

If I'm being more accurate he's always allowed one rewrite of a homework after grading if turned in on an earlier deadline, but he rescinded that the night before despite implying we would have a chance to turn it in later if we didn't want to do the rewrite

which is unfortunate since this was the one time I was actually going to give up the rewrite and spend more time on it before handing it in

and he implied in class we would be able to do that last week lol

[chilly ocean]

this one can be used if you proove that the ideal is <3,X> which you said that you prooved

[chilly ocean]

I have one example X=(Z_p)^n and G=Z_n act on X pushing the indices forward by j wjere j is in Z_n

[crystal vale]

Active tag

[chilly ocean]

what do u mean by that

[elfin wraith]

I dont know exactly what hes doing, but based on the books and stuff ive seen him ask about its minimally combinatorial (in the sense that you can get by on just algebra) its all pretty nice

[crystal vale]

You got an active tag

[elfin wraith]

Herzog and Hibis Monomial ideals or Miller and Strumfels combinatorial comalg are both nice

[chilly ocean]

oh i m new so i dont know about these things

[crystal vale]

In c part, I think we can use that R is Bezout so p/q = ad/bd, where d = gcd(p,q), right?

[ripe crest]

What are you trying to highlight with this example?

[chilly ocean]

this is an example of action

[ripe crest]

That's not what I was asking

My question was one of terminology

Saying something is a vector space over a field vs a vector space under a field

[chilly ocean]

take <a,b>=d so then ax+by=d so we must have a=dp and b=dq so px+qy=1 and a/b=p/q

[rocky cloak]

I guess the true answer is probably that notation and language isn't really planned out ahead of time.

But a difference between the group case and the vector space care might be in how the subject is used and thought about.

All though it's an equivalent formulation, a vector space isn't often thought about as an abelian group that k acts on. One is not interested in different ways k can act on the group or how different ks acts on it.

Instead one fixes a field k, and then works over that specific context.

[crystal vale]

Yes

[manic cairn]

hey just a small question about names of theorems
We covered a simple corollary of the Orbit-Stabilizer Theorem in our lecture. If i translate the name, we called it the "class equation". Its the fact that:
$$|G|=|Z(G)|+ \sum_{r \in R} \frac{|G|}{|Z_G(r)|}$$
For some Group $G$, its center $Z(G)$ and the centralisers $Z_G(r)$ of the elements of a set $R$ containing one element of each conjugacy class.

I dont have a problem with this statement, i just wonder what its called? I feel like it doesnt have alot of applications, given that all it really states is that all the elements whose conjugacy classes have 1 element are found in the center, which is pretty obvious. Like i said, we called it the "class equation", but i have not found any english name online

[cloud walrus]

Super Matroid

[ripe crest]

Is that really how it's generally done? I've mainly only experienced talking about specific vector spaces, so fixing both k and G.

So are you saying, for vector spaces, k is fixed but the group it acts upon usually isn't?

[ripe crest]

I think it's used in burnside's lemma? Tho seems like you can just prove it via orbit-stabilizer

[manic cairn]

yea i felt like that would be the case

[chilly ocean]

there are two groups of order 8 with same class equation

[rocky cloak]

Well, consider for example linear algebra.

It's a subject all about vector spaces, and linear maps between them and so on.

Exactly what the field is usually plays little role, you just have some fixed field and you're comparing vector spaces over that field.

Similarly for studying algebras. You talk about quotients and homomorphisms of k-algebras.

Contrast with for example the representation theory of a group. Here the representation theory is saying things about the group, and you compare the representation theory of different groups.

[ripe crest]

Exactly what the field is plays little role, [...]
That's true, given that we can only have morphisms between vector spaces with the same underlying field. So that does need to be fixed.

Good point

[ripe crest]

As for the representation theory stuff, that's also a good point. We use the group action to reveal more information about the underlying group itself

[glad osprey]

It's called the class equation in english too, and it seems Aluffi disagrees with you wrt. its applications

[ripe crest]

Whereas for vector spaces, we don't do that

[manic cairn]

well we did finish the lecture with proving it, so we might do some applications tomorrow

looking in the script, yea those are some nice results. Group of order p^2 for p prime is abelian. Thats a cool one

[crystal vale]

Yes

[chilly ocean]

1 application there are p^2+p-1 conjugacy class of non abelian group of order p^3

[chilly ocean]

.\

[storm kiln]

Isn't the class equation also used in proving the existence of Sylow p-subgroups?

Either p | #Z(G) or there exists some g_i such that p doesn't divide |G : C_G(g_i)| i.e. C_G(g_i) is a strictly smaller subgroup of order a multiple of p^alpha due to the class equation, in either case you can proceed with induction

[chilly ocean]

yes you can say

[languid trellis]

newton died a virgin. choose now; be a great mathematician and maidenless, or legacyless with maidens..

[swift prawn]

as far as I know it's just called the class equation

that's what it's been called in every algebra book i've read

[glad osprey]

What does Burnside mean by "simple group of odd order" here? Isn't Z/pZ for p prime simple? I'm assuming he uses a different definition?

[next obsidian]

What

It means a simple group, which has odd order

Oh I see

Yeah non-cyclic prime

It’s kinda buried in there I guess, Burnside’s theorem is that all odd order groups are solvable, so as long as it isn’t cyclic prime it can’t be simple

Wait no that’s Feit-Thompson sorry

[glad osprey]

lol, I don't think Burnside was so stupid that he would conjecture that there are no simple groups of odd order, but I think he's talking about non-abelian groups specifically

yeah, it's probably related to Feit-Thompson, that was where I found it

[next obsidian]

Yeah I mean to go from non-abelian simple groups of odd order don’t exist to the only odd order simple groups are Z/pZ is easy.

Like you just have to show among abelian groups only Z/pZ is simple of odd order, and then you can apply the classification

[glad osprey]

yep, that makes sense

Here is his book btw: https://babel.hathitrust.org/cgi/pt?id=uc1.b4062919&seq=535 in case anyone else wants to read a 114 year old book on group theory

[sturdy spear]

Damn 114 years old

[velvet hull]

recall Lagrange's theorem

[hard hearth]

Depending on the book / teacher, they might not actually write the little line underneath. So it could mean a subset (proper or otherwise)

[velvet hull]

is this the same question as before? because Z/nZ is the additive group of integers

[hard hearth]

If 0 were in the group, would all the group axioms hold? Think about the inverse under multiplication

(Assuming we are talking about groups and not rings)

[toxic zephyr]

why does a stable subfield implies galois? that is, if we have F-L-K and L is stable then L/K is galois.

our definition of galois we've been using is

1. there is some K automorphism that moves any given element outside of K
2. the fixed field of AutK(F) is precisely K (i.e. K''=K)

i got that L is stable

but like... i dont see how that does it

[chilly ocean]

this is equivalent to subgroup that fixes L is normal

[chilly ocean]

well the subgroup that fixes $\sigma(L)$ is $\sigma H \sigma^{-1}$ and we want $\sigma(L)=L$ so we need $\sigma H \sigma^{-1}=H$ as H fixes L

[cloud walrus]

Curvature

[toxic zephyr]

oh shit i missed that. yeah i see that now it's direct
i came up with a whole explanation

F/K galois means that F-E-K has F/E galois but E/K galois iff Fix(E) is normal...

[chilly ocean]

yes

[toxic zephyr]

thanks

[crystal vale]

i don't get this part

okay say a1f1 +...+ a_nf_n = 1, then a_1c_1 +...+ a_nc_n = 1 so how do i show atleast one of them is unit, i know x is not unit then 1 + x is unit in A

iff

i got it

[stuck egret]

Hmmm

Oh it's u hi @crystal vale

[proper jolt]

hey does anyone know if the correspondence theorem for ring homomorphism still holds for non commutative rings?

possibly without identity

[thorn jay]

You mean the correspondence between ideals containing I and ideals in R/I?

Then yes; it is a theorem in universal algebra and thus holds in any algebraic structure (even for structures which do not have ideal-like subobjects)

So for example, for sets, ideals are replaced by equivalence relations.
If you have an equivalence relation R on X, then the equivalence relations containing R correspond exactly to the equivalence relations on the quotient set X/R

For an arbitrary algebraic structure A, the role of ideals is played by special equivalence relations called congruences. We go from ideals to congruences in the following way:
given a congruence C on a ring R, the corresponding ideal is the equivalence class 0/C
conversely, if I is an ideal then we construct the equivalence relation
x ~_I y <=> x-y in I

[proper jolt]

yeh that's what i meant, and thankyou i didn't know that the correspondence theorem was so general

[thorn jay]

Its really cool :)

[thorn jay]

Other generalised things are for example the first isomorphism theorem and the third isomorphism theorem (R/I)/(J/I) ≈ R/J

[proper jolt]

is that like from the universal mapping property?

[thorn jay]

No, thats something else

That has to do with free algebras, just like how the polynomial ring has the universal mapping property for commutative rings, you can have free algebras consisting of more general polynomial-like objects which have the universal mapping property for other class of algebras.

These, categorically, are the objects given by the free functor with an adjunction with Set, but logically tell you exactly what axioms some class of algebras satisfies

[proper jolt]

i haven't even learned free algebras yet, but is it something like free groups, where the free algebra gives you some generators for a ring and then you impose some relation

but anyways thks this is ensightful

[thorn jay]

Oh i mean algebra as in general algebraic structure

The curse of being a universal algebraist lmao

[thorn jay]

Thats basically it yes

[proper jolt]

oh ok i guess i understand a little bit then lol

[thorn jay]

[chilly ocean]

i hate algebra( SAYING THIS )

[coral spindle]

I think there is a typo, but it’s unclear whether they mean Z_12 or (mod 11). In the former case, this is actually a function, but in the latter case this is not well-defined. So I suppose they meant to write Z_12

[amber maple]

doesnt f(x) = 3x (mod 12) make sense? its image is (0,3,6,9) which are in Z_11 right?

unless it isnt well-defined

[thorn jay]

Thats what Boytjie said
3 * 8 = 24 = 2 mod 11 ≠ 0 mod 11

[amber maple]

also anyone know what this notation means when it says HA, is it a left coset? hA with h in H? idk where this came from

[thorn jay]

HA = { ha | h in H, a in A }

If H is normal and A a subgroup then HA < G too

[amber maple]

i thought it would be like 3 * 8 = 24 = 0 (mod 12) so it just maps to 0 in Z_11

[thorn jay]

Z_11 is mod 11

Not mod 12

Z_11 contains only the elements 0, ..., 10

[amber maple]

i mean like you take (f(x) (mod 12)) (mod 11)

[thorn jay]

Right, like that

If thats the case then they could've explained it much better >~>

[amber maple]

i mean if thats the case i think it is well defined

so it might not be a typo

[thorn jay]

Mhm, it would be the composition
Z_8 -> 3Z_12 -> Z_11
Which is well-defined as 3Z_12 = { 0, 3, 6, 9 }

[amber maple]

idek how to interpret this

GN/N really looks like it should just be G

[thorn jay]

What it should be is what this tells you

[thorn jay]

In what context is this

[amber maple]

thats about all the preamble before it just starts doing stuff

[storm kiln]

soluble?? Shouldn't that be solvable?

[amber maple]

thats the full proof

[amber maple]

england vs us

[thorn jay]

Its basically the same

[amber maple]

we say soluble here apparently

[storm kiln]

Wait, in Dutch both translate to the same thing lol

[thorn jay]

Oh hai Jelle

[storm kiln]

Hi

[amber maple]

i get the proof i just am having a hard time interpreting what G_{r}N/N actually is and i dont know what to look up to find it

[storm kiln]

Oh, btw if A and B are subgroups of G then AB is the set containing all elements of the form ab with a in A and b in B

[thorn jay]

Do you know lattice theory?

[storm kiln]

if A or B is normal in G, then this is a subgroup

[amber maple]

no

[thorn jay]

Right, okay

What it basically does is use three facts:

1. If N is a normal subgroup, and H < G a subgroup, then HN = H•N = { hn | h in H, n in N} is a subgoup of G
2. If H1 < H2 are subgroups of G, then H1•N < H2•N, and if H1 is normal in H2 then H1•N is normal in H2•N
3. If H1 < H2 are subgroups containing N, then H1/N < H2/N

< here is the inclusion relation (subgroup ordering)

Then take a subnormal series with abelian quotients of G:
1 = G_0 < G_1 < ... < G_r = G (1)
And "multiply by N" to get another series
N = 1•N < G1•N < G2•N < ... < G_r•N = G•N = G (2)
And then quotient by N to get a final subnormal series in G/N, with notably the same factor groups as (2):
N/N = 1 < G1•N/N < G2•N/N < ... < G•N / N

You can see this as the "most obvious" way to construct a subnormal series in G/N

[amber maple]

i get GN/N being G/N but how is something like G_1N/N not just G_1/N

or is it

[thorn jay]

It isnt

Well "it is" in some sense, but a quotient is only properly defined when the subgroup contains N

[amber maple]

can g_1n form elements in some g_i or something

[thorn jay]

So you take Gi•N, which is the smallest subgroup containing both Gi and N

[thorn jay]

You can see here that only if N is contained in G_i we get a factor group naturally isomorphic to G_i+1/G_i

So it does matter

The theorem
(H/N)/(M/N) ≈ H/M
only works if N < M < H

[amber maple]

this math shit gets serious 🙏

[delicate orchid]

this is essentialy the second isomorphism theorem. G_1N/N = G_1/(N intersect G_1). The intersection is required exactly because N might not be contained in G_1

[thorn jay]

[delicate orchid]

society if I didn't have to care about ts bullshit and they were just equal instead of iso 🌈🌈🌈🌈🌈🌈🌈

[thorn jay]

Just work in the skeleton of Grp

Smh

[thorn jay]

SOCIETY WOULD BE GAY?????

[amber maple]

this kinda makes sense

[delicate orchid]

precisely

[thorn jay]

Omg my typa society

[amber maple]

i dont get how G_1/N would be outside of G_1 to begin with

[delicate orchid]

no, N is

[amber maple]

to require the intersection

[delicate orchid]

how can you quotient by something which isn't a subgroup

[amber maple]

oh is it G_1/(N intersect G_1)

i thought u meant the brackets in a diff place

[delicate orchid]

yeah I went and added parens lol

[amber maple]

lmao

[delicate orchid]

G_1/N intersect G_1 is always empty

[thorn jay]

Dedekind group implied

[amber maple]

{} type shit

[delicate orchid]

what is brother talkin bout 😂 get yo Q_8 lookin ahhhh outta town 😂

[thorn jay]

Classification of dedekind groups is absolutely insane

[delicate orchid]

it's just Q_8 direct a bunch of abelian bulllshit

[thorn jay]

I know

[amber maple]

group theory is just like if doing algebra with a,b in R was slightly wrong

[delicate orchid]

that ain't that bad!

[thorn jay]

It aint that bad but its just so fucking arbitrary lmao

[thorn jay]

Its doing algebra in the group of units

[amber maple]

its not analysis so i take it

[delicate orchid]

well it is slightly strange that in essence the only non-abelian dedekind group is Q_8 but direct producting would be the only way to combine them

[chilly ocean]

Algebra hater

[delicate orchid]

any semidirect or heaven forbid non-trivial cocycle would introduce a shit ton of non-normal stuff

[thorn jay]

Mm

Skill issue

[amber maple]

what is it called when you take GH

the name of the operation

[thorn jay]

I'd call it the join

[delicate orchid]

subgroup product

[thorn jay]

But thats not really the context of it, might be too lattice theory-pilled

[thorn jay]

I guess

One has to be normal though else i dont accept it

I reject its existence

[tardy hedge]

Hi Mathemagicians 💫 🫶

[thorn jay]

Helloo

How are you

[tardy hedge]

I’m good. Ate breakfast 🍳

[thorn jay]

Other timezone.. thats good

[tardy hedge]

Yuhhhhh

1130am

Hello Lovely Mathletes 🫶😼

[rocky cloak]

🇧🇷?

[tardy hedge]

I’m 🇨🇦

I live in Nova Scotia

[rocky cloak]

Shoot and miss. I'll have to practice my psychic powers

[tardy hedge]

Lol

[chilly ocean]

found it on mse

[thorn jay]

Lmao

[chilly ocean]

enough to send someone icu

[glad osprey]

So it's brr, not br

Crossed homomorphisms, is this homological algebra? I'm doing Galois theory, I didn't ask for this

[alpine plank]

Galois cohomology sneak peek

what book is this from?

[glad osprey]

Bhattacharya, Basic Abstract Algebra

Not sure if I like it, it has loads of examples and exercises with solutions, but it also has weird conventions and lots of crappy proofs

[chilly ocean]

as i asked this question((3) of the attached) page before and there are two methods(one answered by some genius here and one using field theory, there is one more answer using cauchy theorem) to proove (3) but we need to use 6.1 to proove (3)

[mental plaza]

Hey can you tell me which book this is from ?

[chilly ocean]

its not a book

some assignment

(1) and (2) jare just linear algebra

[thorn jay]

I cannot for the life of me parse this

Sorry

[chilly ocean]

why man

[thorn jay]

"We have the now that we need to use 6.1 to proove (3)"

What

[chilly ocean]

he sometimes uses extra word in hurry

[thorn jay]

Why do you refer to yourself on third person?

Plural system?

[chilly ocean]

i dont know

its bihari

maybe i have grammer problems but i believe i know enough english to deliever my thoughts

[amber maple]

i fw it

ur pronouns are he/he/he/he/he/he

[thorn jay]

call me sans undertale the way my pronouns are er/er/er/er/er/er

[amber maple]

wouldnt that be freddy fast bear

[thorn jay]

that's hor

[amber maple]

many such cases

[next obsidian]

https://youtu.be/kh5qfSwD5oQ?si=phCz9qsdpaW6m7Z_

[chilly ocean]

abstract nonsense

[white oxide]

Is k[x, y]/(x - 1) not isomorphic to k[x, y]/(x - y)? k[x, y]/(x - y) is isomorphic to k[x], I know that

Wait no... k[x, y] is isomorphic to k[y] which is isomorphic to k[x]

No to what

[void cosmos]

nvm

[white oxide]

So they're not isomorphic or they are

It would be great if they weren't isomorphic if that influences your decision 😊

[south patrol]

This is all just first iso theorem

[white oxide]

Is my reasoning correct

[elfin wraith]

A polynomial ring in 2 variables is not isomorphic to a polynomial ring in 1 variable, so im not sure where youre going there

[white oxide]

Oops I meant the quotient is

k[x, y]/(x - 1) is isomorphic to k[y] right?

Via evaluation at 1

[elfin wraith]

.

A good excercise in first iso to check

[white oxide]

Yeah I know it is by first iso

So in other words my reasoning is correct 😂

[velvet hull]

x -> x-1 is a ring automorphism, so k[x,y]/<x-1> is isomorphic to k[x,y]/<x>
or the evaluation homo + 1st iso also works

[elfin wraith]

evaluation is how i seen it but yeah, that works too

[tough raven]

Both are isomorphic to k[y] (as k[y]-algebras, i.e., the isomorphism is the identity on k[y]), the first by the subsitution x = 1 and the second by the substitution x = y. The second one is also isomorphic to k[x] as a k[x]-algebra by the substitution y = x.

[chilly ocean]

Both r same

[thorn jay]

Universal property says its all the same idea

[arctic trail]

this question is weird

[chilly ocean]

Why , it is weird

It's amazing question

An algebra hater will love this

[arctic trail]

using 6.1 to answer (3)

is weird

[chilly ocean]

This is the purpose , that's why (3) is there because it can be done using 6.1

[arctic trail]

I'm saying it's weird

also is 6.1 true?

[glad osprey]

but will an algebra lover hate it?

[thorn jay]

The question seems like pure algebra

Depends on what algebra you like

[chilly ocean]

no

[chilly ocean]

i am algebra hater so sharing my feelings

LOVES TO HATE AND HATES TO LOVE(CONTRADICTION)

[glad osprey]

I'm currently undecided, if I can't solve it then I'm an algebra hater

but I'm struggling to see how you can use 6.1. Have you had any progress on it?

[chilly ocean]

yes you have to see every cubic polynomial is characteristic polynomial of some 3 order matrix so we have to look at GL(3.Z_p)

we have to choose a subgroup now(a clever choice) and then use 6.1

holy crush

[arctic trail]

how do you relate irreducibility to the concept?

[chilly ocean]

thats why its clever choice

[arctic trail]

I'm not really interested in solving it this way, but I'm intrigued in the solution

can you just give it, spoiled?

[chilly ocean]

i dont have solution yet

i have these ideas but it will take much time to solve nowdays quite busy

[arctic trail]

I suppose that the irreducibility is just so

minimal polynomial = characteristic?

[chilly ocean]

no see jorder matrix

[arctic trail]

?

[chilly ocean]

a matrix with 3 order nilpotent block

[arctic trail]

yes but the polynomial being irreducible translates to the minimal = characteristic, no?

a Jordan matrix is such example

[chilly ocean]

yes opposite is not true

[arctic trail]

of course not

diagonal matrices already give an example

00
01

[chilly ocean]

yes

[rocky cloak]

Thinking about the matrix in rational canonical form, either its minimal polynomial is irreducible of degree 3, or it leaves some 2 dimensional subspace invariant.

Then up to conjugation it leaves the span of the two first basis vectors invariant.

Then apply 6.1 to the subgroup of matrices that leave that space invariant

[chilly ocean]

yes we have to show that set which leave two dimensional subspace invariant form subgroup to complete solution

[rocky cloak]

Well that part is clear
If f(V) = V and g(V) = V, then f(g(V)) = V

[chilly ocean]

i did as above the only part remaining was to proove that it is subgroup

[rocky cloak]

Then I guess you're done

[chilly ocean]

in that part i was confused i was think what if there is some other matrix stabilizing some other 2 dim invariant subspace

[chilly ocean]

i thought same but some extra thought came in my mind

[rocky cloak]

I mean, what then. That doesn't affect anything...

[chilly ocean]

if two matrixes stabilizing two different subspaces of dim 2, how composition will behave ,

this was coming in mind,

[rocky cloak]

Okay, but like, it doesn't have anything to do with the problem right?

The subgroup of matrices that fix the span of the two first basis vectors is a subgroup, so then we can apply the result