groups-rings-fields

[iron vessel]

I'm not sure how to compute the degrees involving Q(b)

[hidden haven]

That's the minimal polynomial of b over ℚ

[iron vessel]

ahh, so it must be of degree 3 then

[hidden haven]

Yeah, the bigger extension will be degree 3

[iron vessel]

so [Q(a):Q(b)] = 3 right?

[hidden haven]

ℚ(a)/ℚ(b)

Yes

[iron vessel]

Ok, and to find the minimum polynomial I must find an irreducible monic polynomial in Q(a) such that a^3 is a root right?

[hidden haven]

No

Polynomial over ℚ(b) such that a is a root

[iron vessel]

wait is it just x^3 - a^3?

[hidden haven]

Yes

[iron vessel]

Isn't it reducible?

[hidden haven]

Better to write b instead of a³ there

[hidden haven]

That's exactly why it's better to write b there lol

a is not in the coefficient field

[iron vessel]

That makes sense

While I have you may i ask you another question?

[hidden haven]

Sure

[iron vessel]

Part a, it might be easy but I am not sure how to answer it rigorously

Should I just find a polynomial that sort of generates i and 3sqrt(2) by plugging in i3sqrt(2)?

and if there is then they are included?

[hidden haven]

So if icbrt2 is there in the field, so is its cube

From there you can deduce that i is in the field

Once i is in there, you can divide by i and get cbrt2

[iron vessel]

that's it? so like "(icbrt2)^3 = (-2i) so Q(i) is contained"?

and similarly, "Since i is contained, then icbrt2/i = cbrt2"?

[hidden haven]

Yes

You go from -2i to i by dividing by -2 ofc

[iron vessel]

Yeah. That makes sense, thank you!

[hidden haven]

[weary terrace]

Let $\rho$ be a representation of a finite group $G$. Say $\rho=\sigma\oplus\tau$ where $\sigma$ and $\tau$ are both irreducible representations of $G$.

If $Z(G)$, the center of $G$, acts non-trivially on $\rho$, does it mean that $Z(G)$ acts non-trivially on both $\sigma$ and $\tau$?

If not, under what conditions would it be valid?

[cloud walrus]

Gromov

[coarse storm]

Is this equivalent to looking for an irreducible representation whose kernel contains the centre?

[weary terrace]

If the center is contained in the kernel then its action is trivial which is the opposite of the conditions I offered.

[pastel cliff]

is there some Z_m that isnt a field?

by that i mean integers mod m

[delicate bloom]

every m that isn't a prime

[pastel cliff]

mildly related but what is the additive inverse of an element in Z_2

[delicate bloom]

also worth mentioning, every finite field has a power of a prime number of elements, but Z/p^n Z is not a field for n>1

[delicate bloom]

take a guess

there are only 2 elements in Z/2Z

[pastel cliff]

so if it's 0 then 0 obv

and for 1 is it also just 1

[delicate bloom]

yup

[pastel cliff]

cuz 1 + 1 = 2 = 0 mod 2

cool

[delicate bloom]

😎

[pastel cliff]

ok Z_4

at the very least i think it's a commutative ring right

well actually are all Z_m commutative rings?

[tribal moss]

Right.

Because their multiplication is derived from the multiplication in Z, which is commutative.

[pastel cliff]

so why isnt it a field

[viscid pewter]

no multiplicative inverses

[delicate bloom]

try writing out a multiplication table

there are some multiplicative inverses

[viscid pewter]

what's the inverse of 2 in Z_4

[viscid pewter]

yeah ok

[delicate bloom]

don't give away answers let them do it lol

[pastel cliff]

that's what im doing

ohhhhh

in a naive sense every column/row of the table would need to have a 1, which Z_4 does not

gotcha

these might be unrelated but can a field have zero divisors

[tribal moss]

No, because a zero divisor cannot have a multiplicative inverse.

[pastel cliff]

oh neat

ig that explains why Z_m is only a field if m is prime

bc if it's not prime, then the primes then the factors of m will be zero divisors

[cold hatch]

hI guys , i need some help in calculating the coordingates and angle

as you can see in the diagramm , i tried to find x2 and y2 . but when i created a code on MATLAB to solve it . it is not giving the desired angle output
the given is L1,l2,l3 ,x3,y3
theta 3 is not given
the main thing is , the input would be the coordinates and output would be the angle of each joint . but to calculate the angle for theta 2 ,i have to know the x2,y2 ,,,,which i do not know ,how to calculate from L1,l2,l3 ,x3,y3

[pastel cliff]

idk if this is the right channel

[frail zealot]

that picture is unreadable cause it's not high enough resolution and this is definitely not the right channel

[cold hatch]

where should i post than?

[pastel cliff]

#❓how-to-get-help

[hidden haven]

#geometry-and-trigonometry

[turbid pond]

for a metric space, does the symmetric group coincide with the isometry group?

specifically im thinking about group actions on a metric space, and im wondering if "action by isometry" and "action by symmetry" mean the same thing

[vast quiver]

Acting by isometry is a stronger thing. Normally, a group action is an assignment $G\times X \to X$ which satisfies an associativity and identity axiom. Acting by isometry has these same axioms, but also requires that the mapping $x \mapsto g\cdot x$ is an isometry (not just a bijection/permutation, which is the only thing required by default)

[cloud walrus]

Joseph

[vast quiver]

Another way of talking about this is that a group action is equivalently given by a homomorphism $G \to \mathrm{Bij}(X)$, where $\mathrm{Bij}(X)$ is the group of bijection/symmetric group on $X$.

Then, an action by isometry is more strongly a homomorphism $G\to \mathrm{Isom}(X)$, a group homomorphism to the isometries of $X$.

[cloud walrus]

Joseph

[vast quiver]

There are many examples of bijection that are not isometries. For example, $\mathbb R$ is a metric space, and you can have the bijection $f$ that just swaps 0 and 1, nothing else. This is a bijection, but it is not an isometry because $d(-1,0) =1$ but $d(f(-1),f(0))=d(-1,1)=2$.

[cloud walrus]

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

[turbid pond]

I see. I think I may be confusing the symmetry group and the symmetric group

[vast quiver]

Hmm usually they refer to the same thing? Sometimes people use some ambiguous notation. Like when people say “symmetries”, sometimes they mean automorphisms in the appropriate context. Like if we’re talking about metric spaces, we’ll have it be isometries, or if we’re talking about graphs, it might be graph automorphisms. Idk

[turbid pond]

hmm if we have a square, wouldn't the symmetry group be the dihedral group with 8 elements, but the symmetric group Bij(X) be something that doesn't need to preserve edges?

[vast quiver]

Yeah, you’re right. It’s just that not everyone uses the term “symmetry group” in a totally consistent way.

[chilly ocean]

p^n actually

[vast quiver]

I think people do take symmetries of a metric space to be isometries, so maybe in your terminology, action by symmetry is the same as action by isometry? What sort of reference are you using?

I think this is a terminology issue

[tribal moss]

No, Z_m is not a field if m is a nontrivial prime power. There is a field with p^n elements, but it is not the integers modulo p^n when n>1.

[chilly ocean]

Ye mb

[turbid pond]

yeah its probably a terminology thing first and foremost, ill ask my instructor. thanks tho :)

[pastel cliff]

wdym

[iron vessel]

Can anyone recommend a nice representation of a module over a principal ideal domain? Or any idea on how I can make such a representation

[languid moss]

I got an F in my graduate abstract algebra.

[chilly ocean]

Pog

[tribal moss]

https://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain#Non-finitely_generated_modules suggests there are strong limits to how "nice" you can get unless your module is known to be finitely generated.

[pastel cliff]

just double checking my understanding here: Z is a commutative ring, throwing in mult. inverses make it a field..?

[iron vessel]

I sort of meant like a visual representation

If there exist

[delicate mauve]

{n, 1/n | n in Z} is not a field. You have to take the closure under rings operators

And that only makes sense when you have some ambient field in which it embeds to begin with

[tribal moss]

If you throw in enough stuff to give everything multiplicative inverses and stay a ring without collapsing any elements, then you end up with Q.

[delicate mauve]

But in general for integral domains you can form fields of fractions

That is the smallest field up to isomorphism that contains the ring as a subring

But if you don't have an integral domain you can't form a field for obvious reasons

[barren sierra]

Any tips? I'm taking that next sem lol

[languid moss]

Since my class was very fast pace. I would suggest hitting the books the first week of classes. There are some nice youtube videos of algebra. I never took undergrad abstract, but from someone who had a friend take the undergrad it seems they go slower. By the end of the semester my friend was learning about orbit and stabilizer. While I learned it like the in six or seven week of classes. But don't worry about it. Yo will do great

[barren sierra]

ah ok

I've taken undergraduate so that should help

I plan to actually stick with the textbook

my undergrad class jumped around the text a TON

which made it hard to follow

[languid moss]

Oh you're taking the grad version?

If you're taking the grad version, then hit the books. Don't stop studying

[barren sierra]

yea lol

[languid moss]

Well you will be somewhat okay. As someone who never took it as undergrad. Damn it was hard but at the end I actually liked it.

Sadly never put a lot of effort to it. That was my mistake

but even if I did put effort I doubt i would have passed

[brisk jasper]

Suppose we were given an incomplete Cayley table. How much of the table has to be filled out for one to be able to complete it (unambiguously) using just the group axioms? Or rather, is there a condition which is equivalent to the problem being ambiguous. (Out of my depth here, this might have implications for some experiments I'm doing in another field. If the question itself is not well defined please tell so I can arrive at a better one)

[covert vector]

I have no idea; however I remember @fading wagon came up with a challenge problem for this server once upon a time which claims that; theres an algorithm to completely determine the cayley table of a group of size n; from just 2n entries

actually i guess the result is a bit stronger than that; since you apparently dont even need to know which element is the identity for it to work . was always curious about that problem

it is problem 145 in the document found here: #changelog message

[brisk jasper]

Ah I wasn't specific but I didn't assume the identity was known (though it might be easy to recognize).

[covert vector]

perfect then it matches

[brisk jasper]

Oh yeah this is super relevant. But we don't have the solution for this challenge, lol?

[covert vector]

Lol correct; but pretty sure ppl who made challenge problems were supposed to know the solution to their own problem

so i pinged them; maybe they could clue in as a local expert of sorts

[brisk jasper]

Haha cool, thanks a lot 🙂

[next obsidian]

The problem there is better than being handed a random Cayley table

Eg, you can fix an element then query its multiplication with every other element, this guarantees you can pin down the identity in n-1 queries

[covert vector]

Why is that a sad cat react

n-1 sounds like a lot of queries

[hidden haven]

Idk what chmonkey said sounded like challenge problem easier

[hidden haven]

You can do n/2 by multiplying different pairs each time

rather than keeping 1 element fixed

[covert vector]

I was trying to think of a way to slim it down

Oh. Smart

[hidden haven]

[covert vector]

Eventually you'll find a*b=a

Which means b is the identity

🧠

[next obsidian]

My point is just that you have control over what entires of the table you can see

If you get given a random partial Cayley table of size 2n you don’t have the same info as when you can pick the 2n entries

[covert vector]

O yea True

For example being given 2n-1 things just being products with identity. Lol.

[next obsidian]

Yeah

That last entry is gonna have to put in some serious work

Lol

[covert vector]

Hahahaha

Group ends up being Z/2Z

[brisk jasper]

Lol in the context I'm using this you don't have the privilege of asking the oracle.

So it's a more broad problem I guess

[hidden haven]

You need at least half in general if I am not wrong because for any abelian group N, you have N x Z/2Z and N semidirect product Z/2Z where Z/2Z → Aut(N) is the inversion homomorphism

and I think half of the entries in their Cayley tables match

[coarse storm]

My eyes keep reading it as Z/ZZ.

[hidden haven]

[coarse storm]

Mm. I do want to sleep.

[next obsidian]

I’m a chmonkey

[brisk jasper]

Surely you don't need half of the table for an abelian group, wouldn't you know the whole table by symmetry.

[hidden haven]

The semidirect product won't be abelian

[brisk jasper]

Oh ok that went kinda past my head then.

[next obsidian]

Also you don’t know a priori the group is abelian

If you somehow did then you can use symmetry

[hidden haven]

The reason I want N to be abelian is to ensure that there is a non trivial homomorphism Z/2Z → Aut(N)

Which I can take as 1 maps to inversion automorphism

in the abelian case

[brisk jasper]

I'm here because I got a neural network to fill in a whole table by looking at ~30% of elements. But it only does this if you perturb it a lot in a specific manner. If you train it naively it just does perfectly on the 30% visible ones but fails miserably at the unseen pairs (~0% accuracy pretty much).

Makes me think that the network configuration which encodes the group structure is somehow more stable than the one which just memorizes the seen data, so I'm trying to pin down what exactly happens there.

[coarse storm]

nods. Sure, sure.

pretends to understand.

[brisk jasper]

Ofc stochastic optimization doesn't know what group axioms are, but there should be a lower bound on the amount of seen samples which implicitly convey the group structure.

Lol I'm posting this because some people might find it interesting. Machine learning is nothing but maths in the end

[hidden haven]

lower bound is n^2/2 + 1 (or 0.5) but you could probably do better if you specialize to certain kinds of groups

[brisk jasper]

Maybe some caveats on that last sentence lol

[brisk jasper]

Lol yeah, I guess if you construct a purposefully hard sample. But a random 30% apparently seems to be enough >99% of the time

[upper pivot]

Oh this is interesting, id be interested to see a distribution of this based on some database

[brisk jasper]

https://mathai-iclr.github.io/papers/papers/MATHAI_29_paper.pdf this is the paper which first figured this phenomena out

It's on the shorter/simpler side

Managed to reproduce it this morning for S5

[upper pivot]

Neat, time to run this over the entire database of finite groups

[brisk jasper]

Lol they did quite a few binary ops actually, don't think they're all groups

[chilly ocean]

merry christmas

[fading wagon]

actually I submitted the solution too, I'll dm

[covert vector]

[hidden haven]

Why not post here what are you hiding

[lethal dune]

answer of Riemann hypothesis

[fading wagon]

the answer of the previous challenge problem wait is it still running I forgot

[hidden haven]

Challenge problems are all archived

But I was joking

[fading wagon]

so can I post the solution here lol?

[hidden haven]

Yeah probably

[trim grove]

i know that prime ideals of Zn are of type <p> where p is prime , but what they want to say in that underlined line, can someone please explain?

[chilly ocean]

it's a bit of an abuse of language

[delicate mauve]

When you have a ring R and an ideal I, with a map f: R -> R/I, then the maximal ideals of R/I are exactly the images of the maximal ideals of R under f

That is, if J' is a maximal ideal of R/I, then f^-1(J') is a maximal ideal of R containing I, and conversely if J is a maximal ideal of R containing I, then f(J) is a maximal ideal of R/I

This follows between the more general order-preserving (under inclusion) bijection between ideals of R containing I, and ideals of R/I, where the map is given by the image under f

So if you have a quotient ring, to study its ideals it's enough to study the ideals of the parent ring satisfying a certain property

[trim grove]

can you please tell how this is abuse of language?

[delicate mauve]

It's an abuse of language because maximal ideals of Z containing p^2qZ aren't actually maximal ideals of Z/p^2qZ

[chilly ocean]

an ideal in Z/whatever cannot be an ideal in Z because they literally live in different sets

[delicate mauve]

But their images are

[trim grove]

okkk i got it

btw thankyou for help

[trim grove]

so it should be "the maximal ideal of Z/p^q Z are those pre-images of maximal ideals of Z containing p^qZ"?

[chilly ocean]

no, images

[weary terrace]

Trying my luck again, this time I have a solution that I'd appreciate if you validate. The attached image is a reminder to my question. my solution is in the next message.

By Schur's lemma, since $Z(G)$ is abelian, the restriction of $\rho$ to $Z(G)$ is one dimensional. Hence, $Z(G)$ acts through $\rho$ by some scalar $\lambda$. So if $\rho = \sigma\oplus\tau$, then $Z(G)$ acts through both $\sigma$ and $\tau$ by $\lambda$, i.e. nontrivially.

[cloud walrus]

Gromov

[wind parrot]

Its not true. For instance, if G is abelian, take sigma to be the trivial rep and tau some non-trivial 1-dim rep

Your solution doesn't work because schur only says that irreducible reps of abelian groups are one-dimensional, not every rep

[chilly ocean]

Merry christmas

[weary terrace]

Thanks, I knew I missed something.. Do you happen to have any other idea?

[waxen hedge]

You only know that Z(G) act not trivially on the direct sum, so there must be at least one component where the action of Z(G) is not trivial, but in general there is no reason such that it's true on every component

I don't know about conditions for this specific question

[chilly ocean]

if M is a finitely generated torsion free module over a PID why is it free? no jordan normal form and no structure theorem allowed

[cloud walrus]

wren

[next obsidian]

By induction you can show a projective module is free

If a ring satisfies the property that every finitely generated module is principal, a torsion free module is flat

By using that a finitely presented flat module is projective the conclusion follows

Tl;dr: I think it’s really hard if you’re trying to avoid the structure theorem

If you’re local then you’re fine but then you’re just over a DVR, and to get it in general I think you need to know facts about projectivity being equivalent to locally free

[chilly ocean]

then why did wikipedia say that the fact that finitely generated torsion free modules over a PID are free is a step in proving the structure theorem 🙁

[sturdy marsh]

Prove that sub things of R^n are free, and stick the torsion-free module inside R^n

the first thing can be done by induction

to prove the second thing, look at a maximal linearly independent subset of a generating set

and find an injection from M to the submodule generated by the maximal linearly independent subset

[lethal dune]

Q17, is there some trivial way I'm not seeing?

GL2(Zp) has order (p²-1)(p²-p) so testing the divisors is the best option here?

finding the common pattern of the power and etc..

[oblique river]

can you put it into jordan normal form?

[rustic crown]

yee jordan normal form is like the proper way to do it, but you could also use frobenius. so by cayley hamilton,
(A - 1)^2 = 0
(A - 1)^p = 0
A^p = 1^p = 1

[lethal dune]

nice

lol I calculated A^2 and A^3 and tested for Z2, Z3 and concluded p

[rustic crown]

proof by example(s) >.<

[lethal dune]

works for correct option type questions

[rustic crown]

reminds me of jee lol

[lethal dune]

lol old tricks

the cayley hamilton approach is actually nice

[rustic crown]

yea but we kinda got lucky >.<

the jordan form is

[1 1]
[0 1]

and it's pretty simple to show that n^th power of this is

[1 n]
[0 1]

[lethal dune]

true?

ok I'm not sure

[rustic crown]

i'm scared when i see test questions >.<

[lethal dune]

yeah I forgot most of the field part, don't even remember the structure of F×

[rustic crown]

if |G| = n, then we need to find q such that n divides q - 1

so pick a prime p not dividing n, and then q = p^phi(n)

[rustic crown]

knowing it's cyclic is all you need here >.<

[lethal dune]

yes

ig I didn't think the question through

exam hall is exhausting

[rustic crown]

i'm giving online exams ever since March 2020

[lethal dune]

yeah same, 2yrs of online class,, my brain is empty now

[woven delta]

Is n divides q-1 enough here?

[lethal dune]

filled with "yes sir" "ok sir" "am i audible" shits

[woven delta]

Are groups of units of finite fields cyclic?

[rustic crown]

yee

[woven delta]

Oh okay, then it is enough

[rustic crown]

any finite subgroup of multiplicative group of any field is cyclic

[woven delta]

I vaguely remember that fact but wasn't sure

[woven delta]

Was interpreting this to be about G not F_p^x lol

[rustic crown]

oh oops

[rustic crown]

even (Z/p^nZ)* is cyclic

[woven delta]

[rustic crown]

[upbeat juniper]

hold up

is that supposed to be an easy fact

[rustic crown]

nu

[upbeat juniper]

[rustic crown]

i was checking if i was not confusing F_q with Z/qZ

but it's not like super hard as well, just a weird induction argument

maybe there are cuter proofs

[upbeat juniper]

I see

[lethal dune]

lol that's exactly where messed it up

[woven delta]

😢

[regal arrow]

Given an Abelian group G and an endomorphism d of G such that d^2 sends everything to the identity, the homology is given by ker d/im d. I'm trying to see examples of this using Z or (Z/2Z)^2 as prototypical examples but I can't find anything other than the isomorphism sending everything to the identity. :/ Does anyone know more interesting examples off the top of their head?

[hidden haven]

For abelian groups, g ↦ g^n is an automorphism for all n ∈ ℤ

n = -1

[rustic crown]

i feel "sends everything to the identity" means it's the zero map

[hidden haven]

Oh right

[regal arrow]

Sure, never heard the term "zero map" used before.

[hidden haven]

F

[rustic crown]

so a nice example is endomorphism of A x A sending (a, b) --> (b, 0)

[hidden haven]

On Z/p²Z you could have multiplication by p

[rustic crown]

(for abelian groups people prefer additive notation)

[regal arrow]

Aight, gracias gamers

[wooden ember]

You can also take g^2 on (Z/4Z)^n

And for that matter you can pepper in any extra combination of Z/2Z ‘s in that product

[chilly ocean]

M has a set of generators and you're taking a maximal linearly independent set from those generators. are you taking this to be maximal across all possible generating sets for M. also unless we already assume that M is free, we don't know that this is all of M, so how ok earth does M inject into the submodule that they generate

[trim grove]

i'm not getting what that coset of x means here?,i also tried to make that Q[x]/<x^2+1> structure is it will be Q X Q , form , because elements of
Q[x]/<x^2+1> will look like remainder when elemnet of Q[x] divided by x^2+1
this is confusing me , please help.

[hidden haven]

Q[x]/(x^2+1) is not isomorphic to Q x Q

These are only isomorphic as vector spaces over Q

not as rings

the first is a domain, the second is not

coset of x means its equivalent class

in the quotient

[trim grove]

can you please tell me the structure of Q[x]/<x^2+1>?, i am facing difficulty while driving the structures these quotient rings.

[hidden haven]

It is isomorphic to Q[i]

You can prove this by the first isomorphism theorem

[trim grove]

by taking isomorphism whose kernal is <x^2+1>?

[hidden haven]

yes

x maps to i

[trim grove]

is there any other way , because every time thinking isomorphism will be dificult?

[hidden haven]

Adjoining a variable and then quotienting by a polynomial can be thought of as adding a root of that polynomial to your ring

but other than that

you should just think of it as a quotient

you will have to get used to treating quotients as their own thing

There is a way to think of quotients as just a universal object but if you haven't seen universal properties then you could have a hard time doing that

If this kinda thing scares you then just stick to thinking in terms of cosets

you will see universal stuff eventually

[trim grove]

hmmm,looks like this is the main problem

[hidden haven]

Ye so maybe just stare at the construction of the quotient and solve some problems until you feel comfortable

You can also try proving properties like R/I is a domain iff I is prime

that can get you some initial familiarity

[trim grove]

i am familer with this and R/I is field iff I is maximal ideal

but dont know how they will hlep

[hidden haven]

nice

I mean looking at the proofs you can learn how to work with quotients

see if you can recreate the proofs of these facts on your own

[trim grove]

okk so here <x^2+1> is irredcuible so its quotient should be a field , but here Q X Q is not even a integral domain , so my suposition is wrong?

[hidden haven]

That is correct

[trim grove]

Thank you @hidden haven , so in my question they just asking for irreducibilty , here R=Q[i], which is a field, so for order 2 and 3 degree polynomails i have to just check their roots are in Q[i] or Not ?

[hidden haven]

I think the problem should be talking about irreducibility in R rather than over R

As in, all the given elements are elements of R

And if they are non zero, then they are units, since R is a field

And units are not considered irreducible I think? I'm not sure

But yeah everything is either zero or a unit, if you see this then it's just a matter of playing with the definitions

[trim grove]

these are two different things?

[hidden haven]

Irreducibility over R usually means irreducibility in R[t] where t is a variable

Over R means you're talking about polynomials over it

Rather than within R itself

Idk if this is always true but I've never seen someone saying over R when talking about elements of R itself

[median pawn]

This questions looks more algebraic than analytic so I've posted it here

[cloud walrus]

Hausdorff

[median pawn]

How do I proceed?

Here's a relevant definition

[coarse storm]

My first thought is that it should follow from showing that the sequence is recurring in a specific way.

[median pawn]

That is my thought too

I saw this, but didn't really understand why p is a generator for Z/qZ

[coarse storm]

You can probably adopt a proof of Fermat-Euler to see this.

[median pawn]

Okay, proved it, nice

[hybrid lichen]

can someone give me a nilpotent group example rather than Q8 and Heisenberg group?

[wooden ember]

I want to say $E_{p^2} \rtimes \mathbb{Z}/p\mathbb{Z}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

Where E_p^2 is elementary abelian

I’m just having a bit of a brain fart doubt on whether there’s a non trivial homomorphism here

$GL(2,\mathbb{F}_p)$ has order divisible by $p$ so yeah it’s aight

[cloud walrus]

𝓛ittle ℕarwhal ✓

[sturdy marsh]

Pick any generating set {x_1, ..., x_k, y_1, ..., y_r}. WLOG assume {x_1, ..., x_k} is a maximal linearly independent subset of {x_1, ..., x_k, y_1, ..., y_r}. For any y \in {y_1, ..., y_r}, by maximality, we have a nontrivial linear combination a y + a_1 x_1 + ... + a_k x_k = 0. a must be nonzero as otherwise it would contradict the fact that {x_1, ..., x_k} is linearly independent. Therefore for all y_i, we have a nonzero a_i such that a_iy_i is in the submodule generated by {x_1, ..., x_k}. Set a = a_1a_2...a_r. Then we have a map from M to <x_1, ..., x_k> given by m \mapsto am. The map is injective as M is torsion-free.

[subtle ivy]

any p-group

[chilly ocean]

this seems so like cheating but I get it at the same time

[sturdy marsh]

lol

[chilly ocean]

I can see how it works with examples too though

[cloud walrus]

mns

mns

mns

mns

[polar copper]

Can you show that $(hk)(kh)^{-1} \in H \cap K$ thus it must be $e_G$?

[cloud walrus]

Bradyfish

[polar copper]

Think about what you can learn about $(hk)(kh)^{-1}$ based on the fact that it is equal to $e(k_1k^{-1})$

[cloud walrus]

Bradyfish

[simple mulch]

interesting thanks

[polar copper]

npnp feel free to @ me if that's not enough

[upbeat juniper]

@rustic crown this is the most annoying problem I've done in a while

[regal arrow]

Are finite groups of integers considered subgroups of Z?

[hidden haven]

What do you mean by finite groups of integers

[regal arrow]

Z/nZ

[hidden haven]

They are not subgroups of ℤ

[regal arrow]

Cool

Thanks a ton

[covert vector]

Any p grouo

Finite p Group

Any abelian group

Groups of upper triangular matrices with 1s along the diagonal

I guess that's heisenberg

It works for nxn in general tho

Not just 3x3

Any group for which each sylow subgroup is unique (normal)

[trim grove]

can someone please explain that underline, words?

[obsidian sleet]

is it possible to just count the number of elements dividing p^2 and that is enough?

maybe im just dumb lol

but once you have p^2 q-sylows then there are p^2(q^2-1) elements of order dividing q^2

so then there are p^2 elements of order dividing p^2 (including the identity)

so necessarily there is one p-sylow

this is probably not correct

need to get good

oh wait

i would have to show that the q-sylows intersect trivially

ok

[weary terrace]

Lemma: Let $G$ be a finite group and $\rho$ a representation of $G$. If $Z(G)$, the center of $G$, acts non-trivially on $\rho$, then $\dim \text{Hom}_G(\rho\otimes\tau,\tau)=0$ for all irreducible representations $\tau$ of $G$.

A solution attached:

[cloud walrus]

Gromov

[weary terrace]

My question: In the solution, why may we assume that $\rho$ is irreducible?

[cloud walrus]

Gromov

[hidden haven]

Write ρ as a direct sum of irreducibles, distribute the tensor, distribute the hom. The original set is 0 iff each hom(summand ⊗ τ, τ) is 0

[weary terrace]

Omg you're right, I forgot that the Hom is distributable.
Tnx, sry for the dumb question 🤦‍♂️

[weary terrace]

Wait, I might be missing something again:
If $Z(G)$ acts nontrivially on $\rho$ it doesn't mean it necessarily acts nontrivially on each of its summands.

[cloud walrus]

Gromov

[hidden haven]

It acts non trivially on one of them, and I think from there they deduce that it acts non trivially on τ* ⊗ τ?

[weary terrace]

But Z(G) does act trivially on τ* ⊗ τ.

[hidden haven]

Oh wait right

Hmm I don't see it

[weary terrace]

Thanks anyway 🙂

I think the lemma is wrongly stated. It must require Z(G) to act nontrivially on each summand.

[wooden ember]

If there are p^2 q-sylow subgroups then the index of each of their normalizers is p^2 which means their normalizers have order q^2 and hence they are their own normalizers

So they are self normalizing

[untold cloud]

Hi, guys, is there any good reference for the proof of PBW theorem?

[weary terrace]

Is this the place to ask about Lie groups?

If so, I know that SU(2) is a double covering of SO(3). Now, I'm looking for the lifting map from SO(3) to SU(2). So, given A in SO(3), what is the corresponding A' in SU(2)?

[tribal moss]

By the nature of things there are two corresponding elements in SU(2).

[weary terrace]

Yeah, I mean up to these two elements.
I suspect it is rather simple: We can use the isomorphism of SU(2) to the unit quaternions and the mapping from the unit quaternions to SO(3)

Would it suffice?

[tribal moss]

That looks right.

[weary terrace]

So given this A in SO(3), it has two possible corresponding quaternions by this map, which are mapped by this isomorphism to SU(2)

Giving A' in SU(2)

[tribal moss]

I'm not sure there is a simple computational way to go in the other direction. It will need to have an ugly discontinuity somewhere to deal with the two-valuedness. What I can imagine right away would be to go via the Lie algebra -- you can find the rotation axis of the SO(3) matrix as an eigenvector with eigenvalue 1, and the rotation angle as the argument of the two remaining eigenvalues. Selecting the right rotation direction could be tricky, but you might build a right-hand system with your rotation axis as the new z-axis e.g. by taking cross products, and see which quadrant of the new xy-plane the new x vector maps to under your original matrix. This should give you enough ingredients to build a purely imaginary quaternion whose exponential will be one of the SU(2) elements you want.

[rigid cave]

can anyone explain the part that is highlighted in red? I'm pretty sure that the \sigma's are the elementary symmetric polynomials because that's the notation the author used before. But why can a monomial be written as that product?

[tribal moss]

I think in the context of the symmetric polynomial ring we're redefining "monomial" to mean such a product. The only thing he's doing is to prescribe an order of the factors.

[rigid cave]

oh bruh lol

but this seems kind of weird because multiplicaion is commutative right?

[tribal moss]

Yes, which is why you need to select an ordering arbitrarily before we can compare different monomials lexicographically.

[rigid cave]

Oh lmao I’m dumb

[boreal lion]

There's a mistake in the second transformation in the diagrams, right?

like, it did m_3 instead of r_1?

[hidden haven]

yes

[chilly ocean]

geometry

[boreal lion]

Thank you, good sir

[wise igloo]

strad

more like

chad

strad = strong chad

[boreal lion]

Can't talk about chad without bringing up gmod, the chad overseer 😎

[wise igloo]

LOL

[simple mulch]

if A is a normal subgroup of G we can assume A is a subgroup of G 😐

[chilly ocean]

[hidden haven]

is that a question?

[simple mulch]

No

[tribal moss]

Is this a statement.

[simple mulch]

xD

[tawny pine]

like saying a red cat is a cat

[hidden haven]

Why is this purple

Are there purple cats

[chilly ocean]

what if you are colorblind?

[hidden haven]

[wise igloo]

how can you show that a group homomorphism $\varphi:(G, +) \to (H, *)$ satisfies the following: $\varphi(e_G)=e_H$?

[cloud walrus]

gmod

[next obsidian]

Just use the fact that it’s multiplicative

If something satisfies gh = h for any h, then g is the identity

[wise igloo]

oh alright

ty

[tawny pine]

alternatively phi(e)=phi(ee)=phi(e)phi(e)

[wise igloo]

oh okay thats helpful :)

[torn citrus]

Can you guys give me an example of a polynomial g such that when we divide $x^2y + 1$ by that polynomial it will give us a rational function where x and y have the same powers? Such that $g \neq x^2y + 1$ or in the ideal determined by it. Does such a thing exist ?

[cloud walrus]

Naruto_101

[regal arrow]

Difference between R[x] and R(x) for a ring R?

Wiki uses Q(i) for {a+bi|a, b rational} but I've only seen R[x] before.

"But Q is also a field." Yeah okay I know. I just dunno if there's a corresponding thingy for rings, too. Point is, what the heck is the difference between Q[i] and Q(i).

Oh crap there was a question asked here

sorry man..

[thorn delta]

so, lets say you have fields K \subset F, and x is in F. Then K[x] is the subring of F generated by K and x and K(x) is the subfield of F generated by K and x. When x is algebraic over K, K[x] = K(x)

[regal arrow]

Appreciated

[thorn delta]

npnp

[pastel cliff]

ok i understand why Z_m is not a field for composite values of m, it's bc the factors of m wont have an inverse, but why is that

if that makes sense

[thorn delta]

the factors of m are zero divisors, and zero divisors are not units

not sure if it gets any deeper than that

maybe go back to the proof that a finite field has prime characteristic (and therefore has p^n elements for some prime p):
you can add 1 to itself n times, so that 1n = 0, then factor n into primes, and use that a field has no zero divisors to conclude that 1p = 0 for some prime p.

[pastel cliff]

it sleep tiem for mem

[wise igloo]

what's the notation for homomorphisms

[pastel cliff]

will do this in morning but merci

[wise igloo]

like how can I write "G is homomorphic to H"

[rustic crown]

what's homomorphic?

judging from your notation, i presume you want group homomophisms. there is at least one hom for any two groups, G --> 1 --> H

[wise igloo]

yes

[wise igloo]

is that not how you say it?

[rustic crown]

so you mean any two groups are homomorphic?

[wise igloo]

I just want to know how to say that two groups are homomorphic

[rustic crown]

i've only seen people use isomorphic

[wise igloo]

using notation

oh

so would I write it in words?

[rustic crown]

how are you defining homomorphic

[wise igloo]

idk

if there exists a homomorphism between them?

[rustic crown]

just give it a name if you wanna say there is one f : G --> H

[wise igloo]

alright ty

so could I say "f:G→H is a homomorphism"

and then continue on whatever I wanna say

[tawny pine]

theres no real reason to define such a term bc as said above theres always a homomorphism between any groups

[rustic crown]

well it depends on what you want.... there is always the trivial map between groups >.<

[wise igloo]

ohhhhhhhhhhh

[tawny pine]

namely map everything to the identity of H

[wise igloo]

so two groups can be isomorphic but it doesn't make sense to call two groups homomorphic

thats what im gathering

[tawny pine]

its not that it doesnt make sense to say homomorphic

[upbeat juniper]

you can say one group is a homomorphic image of another

[tawny pine]

but its just a bit meaningless bc any groups are homomorphic

[wise igloo]

it's just not useful?

alright

thanks guys

[tawny pine]

but smth like isomorphic IS very meaningful

[wise igloo]

I see

[thorn delta]

the only context ive heard "homomorphic" is in a statement like "H is the homomorphic image of G [under some map which is implicitly known]"

[barren sierra]

I'm going through this Naive Lie Theory book by Stillwell and I am stuck (idk where else to ask this)

"Show that a reflection in any hyperplane orthogonal to any coordinate axis has determinant -1"

wait nvm I got it I'm dumb

[weary terrace]

Thanks.
The reason I was asking is that I'm trying to figure out the binary polyhedral groups from the polyhedral groups (finite subgroups of SO(3) (rotations). e.g. the dihedral group).
The idea is that SU(2) is a double cover of SO(3), so polyhedral subgroups of SO(3) have double covers which are subgroups of SU(2). These are the binary polyhedral groups.

This should somehow give a complex 2-dim representations of these groups.
So I'm trying to find a way to understand the structure of these binaries from this description.

My current idea is to construct the following diagram:

2P ---------> P
| |
| |
SU(2) ----> SO(3)

Where the vertical lines are the representations: left is for the rep of the binary polyhedral and right is for the rep of the polyhedral.
and the horizontal lines are the surjective homomorphisms.

This way we can describe the rep of 2P in SU(2) by the other maps.
But it feels like I'm missing something...

SORRY FOR THE LENGTH

[upper pivot]

Anyone know an easy proof of cayley hamilton for either finite fields or F_p^alg. I know the normal ones are jcf or zariski dense, but i want to find something simpler for either class, as then i can use model theory to make a nicer proof of this for all fields.

[hybrid lichen]

can someone help me to finish this proof?

[delicate mauve]

Your picture is illegible

Post a pic with better quality

[hybrid lichen]

is it better?

[hidden haven]

is legible

[pastel cliff]

i saw a prof use cyclic generating functions but idk if this applies to those conditions…? i also dont know what zariski is so take that as you will

[upper pivot]

Oh interesting ill look into it

Heres the zariski proof if you are interested

https://johnds1234.github.io/2021/08/05/Proof-of-Cayley-Hamilton-using-the-Zariski-Topology.html

[hybrid lichen]

can someone help?

[delicate mauve]

Post legible image then I'll look at it

[delicate mauve]

it's still very bad

[south patrol]

The way we did it in general was just using triangular form which is easy to prove existence of w quotients

[hybrid lichen]

this is the original form of the question. I cant upload my proof nicely

[delicate mauve]

Can you show that it's a homomorphism?

I.e. do you understand why it's a homomorphism

This looks like a two part question. Which part are you unsure about?

@hybrid lichen

[hybrid lichen]

I'm not sure about the second part

I can show that it's a homomorphism

[delicate mauve]

What do you know about homomorphisms?

What is the kernel of this homomorphism?

[hybrid lichen]

identity?

[delicate mauve]

Do you know what the kernel means?

Every homomorphism has a kernel

It's the set of elements that map to identity

You need to go back and review the fundamentals

[half anvil]

I am writing a Geometric Algebra library in C++. My reference text is "Linear and Geometric Algebra" by Alan Macdonald.

I need to represent a general $k$-vector using an array of elements corresponding to the coefficients attached to the blades in the general $k$-vector form whose array size is computed in terms of the dimension of the geometric algebra $n$ and $k$. For example, in $\mathbb{G}^{3}$ the $2$-vector contains an array of $3$ elements $[ a, b, c ]$ corresponding to the general $2$-vector form $a\mathbf{e}{1}\mathbf{e}{2} + b\mathbf{e}{1}\mathbf{e}{3}+c\mathbf{e}{2}\mathbf{e}{3}$.

On the Wikipedia page for "Blade (geometry)", the following statement is made: "In a vector space of dimension $n$, there are $k(n-k)+1$ dimensions of freedom in choosing a $k$-blade, of which one dimension is an overall scaling multiplier."

This array sizing of $k(n-k)+1$ appeared to work initially, but this is not the case when considering a $2$-vector in $\mathbb{G}^{4}$ for example: $2(4-2)+1=5$, which suggests that the basis consists of $5$ blades. This contrasts with the actual basis which consists of $6$ blades: $\mathbf{e}{1}\mathbf{e}{2}, \mathbf{e}{1}\mathbf{e}{3}, \mathbf{e}{1}\mathbf{e}{4}, \mathbf{e}{2}\mathbf{e}{3}, \mathbf{e}{2}\mathbf{e}{4}, \mathbf{e}{3}\mathbf{e}{4}$.

How can I correctly determine the number of elements to be contained within the array of a general $k$-vector given $n$ and $k$?

[cloud walrus]

Casey

[delicate bloom]

I think the problem is blades aren't closed under addition, you can take $e_1e_2+e_3e_4$ and this is not a 2-blade even though it's the sum of two 2-blades, because you can't factor it into a product of vectors.

[cloud walrus]

Merosity

[trim grove]

This question was asked to me in an interview i replied identity ,but the they said it is zero(0).

[delicate mauve]

What is identity?

What was the question you were asked?

[trim grove]

Elements in kernal are mapped to?

[delicate mauve]

It depends on the context

In the context of vector spaces, kernels map to zero

Same with rings

But in the context of groups it maps to identity

Zero is the additive identity in vector spaces and rings

[trim grove]

That makes sense

[delicate mauve]

Also the identity in abelian groups is usually called zero

[half anvil]

Hmm, do you perhaps have an idea as to how to determine the number of blades defining the basis of a k-vector then for that matter? Perhaps I ought to scavenge other GA libraries and see what they do.

[delicate bloom]

I think they cover it, but I can't guarantee if they do, in this book called Geometric Algebra: An Object Oriented Approach

I know they discuss the problem in there and probably implement it so I'm pretty sure it does

[half anvil]

Aha, thank you kindly, I will certainly give it a look.

[delicate bloom]

yeah good luck, feel free to ping me I'll probably look in there later today out of curiosity myself

looks like they discuss it in chapter 2, section 2.9.3

ah paragraph 5 they say there is a necessary and sufficient condition for a k-vector to be a k-blade and cite the paper, "but it's not easily summarized"

they might go into more detail on programming it since it's a programming book in part

[half anvil]

What is the cited paper called? I'd like to read that if possible

[hot tinsel]

If $A_p$ is the localization of $\mathbb{Z}$ at the prime ideal $p\mathbb{Z}$ then is $A_p$ complete local ring?

[cloud walrus]

K零ꓘ

[delicate mauve]

Complete in what sense?

[hidden haven]

wrt m-adic topology

where m is the maximal ideal

[stoic rose]

No it is not

Its completion is the p-adic integers Z_p

[trim grove]

Is S4 x A3 a normal subgroup of S4 x S3? I am confused with this , anyone can please help?

[delicate mauve]

What about this confuses you?

[trim grove]

How they showed this ?

Any solid proof.

[delicate mauve]

By producing a homomorphism whose kernel is S4 x A3

[trim grove]

Not again

[delicate mauve]

Dude just learn what homomorphisms and kernels are

These are fundamental concepts

[trim grove]

Can you please tell what homomorphism will work here?

[hidden haven]

Do you know about the index of a subgroup?

[trim grove]

144/72=2

[delicate mauve]

The signature homomorphism on the second component of the product

[trim grove]

So normal

[hidden haven]

Every index 2 subgroup of any group is normal

yes

[trim grove]

I know what homomorphism is ,but it is very difficult to think about that particular homomorphism that will work for given situation.

[hidden haven]

That comes with practice, with such problems its good to try to find homomorphisms. Any thought you put into that will help you in the long term

[trim grove]

I will try 👍

[delicate mauve]

Did you understand what homomorphism I was talking about?

[trim grove]

Looks like f(a,b)= sign(b) wich will always give kernal S4 x A3?

[delicate mauve]

Yes

Do you understand why that is the kernel?

[trim grove]

Yes beacuse signature of even permutation is always one

[trim grove]

So kernal will be S4 x A3

[untold cloud]

Hi, guys, what does this corollary 3.6 mean?

[hidden haven]

It links representations of g with those of U(g) so you can study either to understand the other

Or are you asking for the literal meaning

[untold cloud]

Oh, i see, so this is like constructing a injection between representation of g and representation of U(g)?

[untold cloud]

yeah, can't understand what it wants to prove

[hidden haven]

not just an injection

it is a bijective correspondence (and also an equivalence of categories)

given a representation pi of g, you can define a corresponding representation pi~ of U(g)

and you can also go back

and this is a bijective correspondence

[untold cloud]

Thanks

[trim grove]

I have to show that S4 x S3 have no normal sylow-2 subgroup

Order of given group is 144=2^4 x 3^2

So we have posabilites for 2 sylow subgroups 1,4,9
After that i am getting confused how to proceed further ,i also searched this on google but the answer was very confusing ,can some please help me with this?

[hidden haven]

There can't be 4 sylow 2 subgroups because their number has to be 1 mod 2

All sylow 2 subgroups are conjugates of each other

So sylow 2 subgroups are normal iff there's only 1 of them

So you have to prove that there's more than 1 sylow 2 subgroup

The sylow 2 subgroup of the product would be a product of sylow 2 subgroups of S4 and S3

See if you can figure anything out from this

[hidden haven]

This might be false

Is not true in general at least

[wooden ember]

Maybe the fact that if there are 9 sylow 2 subgroups they must be self normalizing?

Since you have such a nice caractérisation of conjugation in symmetric groups there might be smth you can do with that

I’m at the cinema rn but I’ll think about it later

[hidden haven]

We know 1 of the sylow 2 subgroups

Which is the product of the sylow 2 subgroups of both groups

And this is not normal

Because the sylow 2 subgroups of S3 aren't normal

[hidden haven]

I mean we already know multiple lol

So that proves that the sylow 2 subgroups aren't normal in the product

@trim grove

[wise igloo]

can someone explain to me wtf a group action is

[delicate mauve]

An action of a group G on a set X is a homomorphism G-> Sym(X)

Sym(X) = the group of bijections of X

[wise igloo]

Sym is the symmetric group right

[delicate mauve]

Yes

[wise igloo]

oh alright

[delicate mauve]

The group of symmetries of X

[wise igloo]

yeah

alright thats not too hard to understand

ty

[frank fiber]

A G-action is a homomorfism G->Aut(X) where X is an object of a category

[wise igloo]

im not familiar with Aut

nor categories

[delicate bloom]

basically a group action is something that takes some set and moves it around

[subtle ivy]

u can use feit thompson ;)

[wise igloo]

what does "moves it around" mean?

[delicate bloom]

simplest example I can give is take a set of elements like {a,b,c} then permute the elements around in the set, to make {a,c,b}. That's an example of a group action by an element of the permutation group S_3 on that set

[wise igloo]

oh okay that makes sense

ty

[delicate bloom]

other examples are like taking a shape, like say a square and labelling the corners, faces, or edges

and then using the dihedral group D_4 to move it around

all the different configurations are the elements of your set that your group is acting on

[wise igloo]

oh okay

ty for the examples, that really helps

[delicate bloom]

yup you're welcome

[proud bear]

might be a dumb question but why can't n11=3 * 5^3?

[normal heron]

is there a special name for fields that are isomoprihc to a subset of the real numebrs

[trim grove]

Thankyou @hidden haven

[dense phoenix]

Yeah: F

[normal heron]

?

[tribal moss]

I think "Archimedean ordered field" comes close, but that implies that a particular embedding into the reals has been chosen.

[trim grove]

Sorry for that, it should be 3.but how to eliminate this posability?

[proud bear]

you don't have to eliminate that possibility to show that S4 x S3 has no normal sylow 2-subgroup. you just have to show that there is more than one sylow 2-subgroup (which moldi showed by taking products of the sylow 2-subgroups of S4 and S3)

[trim grove]

thank you , i got it

need some help with this question, i discarded option a by taking n=4, and second option is direct a result because of cyclotomic polynomials , but i am confused in 3rd and 4th , any hint?

[hidden haven]

4 is a number of the form p^e lol

[trim grove]

what for last one?

[rustic crown]

You can rule out 3 the same way you ruled out 1.

[hidden haven]

Maybe cheese it with e=1 lmao

Units aren't considered irreducible are they

I always forget this

[rustic crown]

nope

[hidden haven]

Epic

Oh wait

It just becomes f_p(x) then

Nvm

[rustic crown]

one way to say this is like this

$f_n = \frac{x^n - 1}{x - 1}$

So if $d \mid n$ then $f_n = \frac{x^n - 1}{x^d - 1}\cdot \frac{x^d - 1}{x-1}$

[cloud walrus]

det

[rustic crown]

the first terms looks like (1 + x^d + ... + x^{n-d})

[trim grove]

by doing this we are just proving it is true, but this will be for a particular case (e=1)?

[hidden haven]

I thought that it becomes f_p(1)

But that was wrong

[broken stirrup]

go

[proud bear]

\[
\frac{x^{p^e}-1}{x-1}=\frac{x^{p^e}-1}{x^{p^{e-1}}-1}\frac{x^{p^{e-1}}-1}{x-1}\implies f_{p^e}(x)=f_p\big(x^{p^{e-1}}\big)f_{p^{e-1}}(x).
\]
Then I think you can use
\[
f_{p^e}(x+1)\equiv f_p\big((x+1)^{p^{e-1}}\big)f_{p^{e-1}}(x+1)\bmod p
\]
to show that you can apply Eisenstein to $f_p\big((x+1)^{p^{e-1}}\big)$.

[trim grove]

I got the first line, but not getting the second line.

[proud bear]

well if $f_{p^e}(x)=f_p\big(x^{p^{e-1}}\big)f_{p^{e-1}}(x)$ then $f_{p^e}(x+1)=f_p\big((x+1)^{p^{e-1}}\big)f_{p^{e-1}}(x+1)$ so they will also be equal mod $p$

[proud bear]

,texsp
||

\[
\frac{x^{p^e}-1}{x-1}=\frac{x^{p^e}-1}{x^{p^{e-1}}-1}\frac{x^{p^{e-1}}-1}{x-1}\implies f_{p^e}(x)=f_p\big(x^{p^{e-1}}\big)f_{p^{e-1}}(x)
\]
We have $f_p(1)=p$ so the constant term of $f_p\big((x+1)^{p^{e-1}}\big)$ is $p$. Also
\begin{align*}
f_{p^e}(x+1)&\equiv\frac{(x+1)^{p^e}-1}{(x+1)-1}\equiv\frac{x^{p^e}+1-1}x\equiv x^{p^e-1}\bmod p\\ f_{p^{e-1}}(x+1)&\equiv x^{p^{e-1}-1}\bmod p.
\end{align*}
Since
\[
f_p\big((x+1)^{p^{e-1}}\big)=(x+1)^{p^{e-1}(p-1)}+(x+1)^{p^{e-1}(p-2)}+\cdots+1=x^{p^{e-1}(p-1)}+\text{lower degree terms},
\]
this means that
\[
x^{p^e-1}\equiv f_{p^{e-1}}(x+1)f_p\big((x+1)^{p^{e-1}}\big)\equiv x^{p^{e-1}-1}x^{p^{e-1}(p-1)}+\text{lower degree terms}\equiv x^{p^e-1}+\cdots\bmod p
\]
so $p$ divides the coefficient of every term of $f_p\big((x+1)^{p^{e-1}}\big)$ other than the coefficient of $x^{p^{e-1}(p-1)}$. Since $p^2$ doesn't divide the constant term, Eisenstein says $f_p\big((x+1)^{p^{e-1}}\big)$ is irreducible.

||

[proud bear]

i gotta go sleep. i think those are full details^

[trim grove]

Thank you

I will try

[trim grove]

i have to find the prime ideals of $\frac{\mathbb{Q}[x]}{\langle x^4-1\rangle}$\

so i can write it as $\frac{\mathbb{Q}[x]}{\langle x^4-1\rangle}=\frac{\mathbb{Q}[x]}{\langle x-1\rangle} \times \frac{\mathbb{Q}[x]}{\langle x+1\rangle} \times \frac{\mathbb{Q}[x]}{\langle x^2+1\rangle} $\
but now i'm not getting how to find its prime ideals, any hint will be appreciated.

[cloud walrus]

TheStudent

[hidden haven]

Use the correspondence theorem and the fact that ℚ[x] is a PID

[blazing parrot]

Hi. I don't know if it belongs here.
Let G be a finite group., assume we have in G exactly t elements of ord 2 while t>=0.
Prove: |G| - t is odd

[hidden haven]

It belongs here

[blazing parrot]

oh okay good to know 🙂

I don't get it, if they said exactly t elements so isn't |G| = t??

probably not because this way the proof doesn't make sense haha

[hidden haven]

exactly t elements that have order 2

[blazing parrot]

yes

[hidden haven]

not that G has t elements in all

[blazing parrot]

so we know that there is a that belongs to G. a^2 =1.
And there are t elements like a

[hidden haven]

yes

[blazing parrot]

but how do I know what is |G|?

haha sorry

[hidden haven]

No I mean idk lol

[blazing parrot]

hahaha

so |G| = t + x.
We need to proof |G| - t = odd, so basically t+x-t = x = odd.
So need to proof there is odd number of other elements that aren't of order 2

[trim grove]

Let order of G is odd ,then it will never have elemnet of order 2 ( because 2 will not divide odd number) so |G|-0= odd

I am trying to prove for even one

[blazing parrot]

I found it somewhere on the internet. Maybe it helps?
A group of even order must contain an odd number of elements of order 2.
So t = odd
then wat

[hidden haven]

lol that is what we have to prove

[blazing parrot]

really

?

[hidden haven]

yeah because if |G| is even

then t is odd iff |G| - t is odd

[blazing parrot]

yea youre right

haha

thank you two

[trim grove]

I am trying to figure this out but still not getting this , can you please give me a little more hint?

[hidden haven]

The prime ideals of the quotient correspond to the prime ideals of Q[x] that contain (x^4-1)

A prime ideal (p(x)) of Q[x] contains (x^4-1) iff p(x) divides x^4 - 1

so you just need to factorize x^4 - 1 as you already did

and the generators you get are the ones that generate the prime ideals you need

you just have to then prove that they all generate distinct prime ideals

for which you need to show that they don't divide each other

[trim grove]

I will try my best

[trim grove]

Can we use the fact that in PID, every irreducible is prime so it they all will be prime ideal,?

[hidden haven]

yes

I mean

you tell me if you can use it lol

Idk what facts you are allowed to use

[crystal lion]

Hey! Does any of you happen to know whether a finite simple group with an involution with an elementary abelian centralizer is necessarily alternating?

[crystal lion]

Provided my simple group is non-abelian

[trim grove]

the definition of ring corrospondence is "Let A be a multiplicative ring with identity and I an ideal of A. There is a one-to-one correspondence between the ideals of A that contain I and the ideals of the quotient ring A/I." this , but how you are using this here , can you please explain?

[hidden haven]

This correspondence restricts to a correspondence of prime ideals

As well as to maximal ideals

It's not hard to prove once you know what the correspondence is

Ideal J containing I corresponds to J/I

And if J is prime

Then A/J ≅ (A/I)/(J/I) is an integral domain

So J/I is prime in A/I

[trim grove]

now things getting clear to me

[hidden haven]

Similar reasoning works for maximal ideals, or you can use the fact that the correspondence is also an order isomorphism

Between the partially ordered sets

[trim grove]

just one more question , is this line applicable to any ideal ,or just prime ideals?

[hidden haven]

Any ideal, for prime ideals you look at irreducible p(x)

For any other ideal you look at all factors

[fierce perch]

Very stupid question. For L/k finite separable and k_s the separable closure, when is Hom_k(L, k_s) ismorphic to Hom_k(L, L) = Aut(L/k)?

[oblique river]

Always, no?

[fierce perch]

I think so yes, but why?

[oblique river]

Isnt that kinda the definition of a separable extension?

At least thats how I think of the definition

Actually, wait, im sorry

Im wrong.

What you wrote is equivalent to L/k being normal as well

For example consider Q(2^(1/4))/Q

Has 4 homs into the algebraic closure

But only two automorphisms

[fierce perch]

ah

Are you thinking of this def?

Every embedding of L in \overline {K} induces an automorphism of L.
Why is it interchangable if we use the algebraic or separable closure?

[oblique river]

If L is separable then all embeddings into the algebraic closure will factor through the separable closure

If L isnt separable then it’s not even contained in the separable closure so there are no embeddings there

[chilly ocean]

lejoon come vc

[chilly ocean]

Could someone recommend me a good textbook for studying groups and rings?

dummit foote

[chilly ocean]

Hi, I’ve gotten parts (i), (iii), and (iv) just by picking arbitrary elements in S3 and Z[x1,x2,x3]. I can’t seem to get part (ii)… Any help?

[chilly ocean]

<@&286206848099549185>

For example, this is how i did parts (i) and (iii)

[small bison]

try looking at the products of x1 x2 and x3

[chilly ocean]

I computed that the orbit of x1x2 has 3 elements

Wouldn’t the orbit of x1x2x3 be the singleton {x1x2x3}?

[small bison]

Yeah

You can also add these together

[chilly ocean]

I’m looking for a two element orbit

[small bison]

Add products together**

[chilly ocean]

Ohh

I’ll try it out right now

Yeah I’m getting another 3 element orbit, unless I’ve made a mistake

[small bison]

what element did you try?

[small bison]

i was thinking more along the lines of something like x1x2 + x1x3

hmm actually i think that might have 3 elements also

[chilly ocean]

I could try that

I just did x1x2+x1x2 and got a 3 element orbit

[proud bear]

i think (x1-x2)(x2-x3)(x3-x1) should work

[chilly ocean]

Whoa hmm I’ll give that a shot @proud bear

@proud bear It was so close to being a 2 element orbit. It is a 3 element orbit.

[proud bear]

for the second rotation, when 2->3, 3->1 and 1->3, shouldn't you get (x2-x3)(x3-x1)(x1-x2)?

i don't think you permuted all of the indeterminates

[small bison]

I think stain’s thing works

Should be fixed by any 3-cycle

But not by any transpositions

[chilly ocean]

Okay I will do it again, thank you @proud bear

and @small bison

Yes it worked! @proud bear @small bison

@proud bear @small bison How did you know this choice of f would work? Was this just through experience?

[proud bear]

yeah kind of

[chilly ocean]

Interesting

[small bison]

hmm well you can sort of use the orbit-stabilizer theorem to help

but coming up with an example is prob done with experience or a lot of time

[chilly ocean]

Ahh I see. Thank you @small bison. This level of abstract algebra is the hardest class I’ve taken. I’m still an undergrad.

Bump

[small bison]

hmm Z(G) is solvable cause it's abeliean

and you have Z(G) -> G -> Aut(G)

[long obsidian]

If the product of two ideals in a domain is principal, does that mean the two ideals are finitely generated?

[sturdy marsh]

Z(G) ---> G ---> Inn(G)

but Inn(G) is solvable as Aut(G) is solvable

oh ig you didnt say that it was short exact

sorry

[small bison]

well i meant for it to be short exact 🥲

[sturdy marsh]

lol

[proud bear]

first one is the inclusion of Z(G) into G. second one is the map taking g to the automorphism of conjugation by g

it's called a short exact sequence because the first map is injective, the second one is surjective and the image of the first map is the kernel of the second

[hidden haven]

how proof

[next obsidian]

Tensor product distributes over direct sums

And a direct sum of chain complexes is exact iff each one is

Oh wait

Oh, this is about E-flatness

The simplest answer is that Tor is additive

If you can’t use Tor… uhhh you have to do some shenanigans i guess, i dunno

[hidden haven]

Yeah haven't covered Tor yet, but I will try to prove that it is additive

wait additivity just follows from adding resolutions doesn't it

and the fact that tensor is additive

by adding I mean direct summing

We have done some snek shenanigans for flatness proofs, maybe tensor an injective map into the direct sum with F and argue somehow but I just don't see how do that

so given E → dirsum E_i injective, for each i we have exact sequence 0 → K_i → E → E_i

nah we want E_i in the middle

😵‍💫

[next obsidian]

Wait what’s E-flatness again

Is it that if E is in the middle?

[hidden haven]

yes

[next obsidian]

Oh Hurb

[hidden haven]

ses with E in middle is preserved

[next obsidian]

Tor won’t even directly solve your problem

[hidden haven]

[next obsidian]

Oh maybe

try this

Direct sums are exact and commute with tensor

For something like

0 -> K -> (+) E_i

Denote K_i = K\cap E_i

Then you should be able to write K as a direct limit of K_i

Over the same index set as (+) E_i as the direct limit of the E_i

Then take direct limit and boom

Or something like that

Oh that won’t quite work

[hidden haven]

this is direct sum though

[next obsidian]

You have to do something like

Direct sum is a direct limit of all finite direct sums

[hidden haven]

right makes sense

[next obsidian]

So maybe you can try something like this for all finite direct sums

Then try that idea

Or something

Like if you can induct

To handle the finite

Doing this trick except

Not K_i

You have to index via the finite subsets of I the index set

If that makes sense

[hidden haven]

right

[next obsidian]

And intersect K with that

[hidden haven]

I shall try it

[next obsidian]

But you need to handle the finite case

Which maybe you can induct

Or use snake lemma

Or both

¯_(ツ)_/¯

At the very least this reduces it to finite

[hidden haven]

ye we have some theorems I might try applying 😵‍💫

[next obsidian]

I was thinking about induction

And I feel like

This works with Tor

Nicely

[hidden haven]

[next obsidian]

But without Tor the proof is a lot harder

The point here is that

You do something like this

[hidden haven]

I feel like this was given as an exercise because prof didn't have an easy proof

this is like the only exercise

in all of the lecture notes

[next obsidian]

0 -> E_ 1 -> (+)>=1 E_i -> (+)>1 E_i -> 0

Then you get the SES

Oh hmmm

Actually idk

This doesn’t quite work either

Lmfao

[hidden haven]

F

[next obsidian]

Maybe you can like

Make this 2-D

Put the arbitrary

0 -> K -> (+) E_i -> F -> 0

Down the middle

Then try to fill in the missing spots

Apply Snek

[hidden haven]

wait put what down the middle of what

[next obsidian]

Take this

Then you need arbitrary SES’s with the direvt sum of E_i in the middle

So put that vertically down the middle

Then you have some missing spots

[burnt sparrow]

👀 💯 🙏

[next obsidian]

The diagram is like a +

[hidden haven]

oh I see

[next obsidian]

¯_(ツ)_/¯

[hidden haven]

lol I will try doing that

[next obsidian]

I don’t think Tor is particularly helpful here anyway

[hidden haven]

You just brought it up to scare me

[next obsidian]

Idk about trying what I suggested either anymore

I mean I guess you can fill it in with push outs or pullbacks I think

But things get kind of dicey I think

¯_(ツ)_/¯

[hidden haven]

oof

Anyway I am planning to leave this till the end, haven't finished preparing for the exam tomorrow 😵‍💫

[broken stirrup]

I know it's simple question but i need validation. If A is subgroup of B, it implies A' is subgroup of B' ( where A' and B' are commutator groups of A and B) right?

[broken stirrup]

Good luck mate

[hidden haven]

yes

Ty

[trim grove]

i have check whether the given polynomial is irreducible or not on given field , the given polynomial is of three degree so i can just check its roots in the given field , and it has root x=3, but is there any other method to check its irreducibility , like checking each and every element will be difficult?

[next obsidian]

There’s various tools sure, but because you have just F_7 plugging in all the numbers is the easiest way

In general there’s a wide array of things you can try to do to determine if a polynomial is irreducible, Eisenstein’s criterion for example, but in general it is a hard problem

So there’s not going to be some magic formula you can follow, you have to try various things

[lethal dune]

Is it tedious or is there a better way?

[wise igloo]

Proof. Exercise.

please burn the author

[lethal dune]

QED even

[coarse storm]

Just did some star jumps. I am none the wiser about the proof!

[lethal dune]

[weary bane]

Hello i need help with a question in the book Linear algebraic groups by humphreys

Let G be a connected algebraic group, $x\in G$ is semisimple. Must $C_G(x)$ be connected?

[cloud walrus]

kh

[weary bane]

and $C_G(x)$ is the centralizer of $ x$ in $G$

[cloud walrus]

kh

[weary bane]

i tried using that in a connected algebraic group the centralizer of a torus is connected

can i prove that in a connected algebraic group the closure of the power set of $x$ is a torus?

[cloud walrus]

kh

[kind temple]

what’s [g,h]

[coarse storm]

Their commutator, I assume.

[cyan marten]

I don't think it's really tedious. Generally, to compute g^n h^m we need higher-order commutators like [g, [g, h]] etc., but in this case they vanish. For the first part, try to express [x, yz] in terms of commutators and conjugation.

[boreal lion]

Is it alright to ask for proof verification here? Small problem in intro group theory

[tawny pine]

yes

[boreal lion]

Does something like this work? I can't help but feel as though I'm skipping details, or not being general enough somehow

(This is only for the first part)

[tropic spade]

If x has order n and i,j are distinct elements of {0,...,n-1} such that x^i=x^j I'm pretty sure you can find a positive integer k less than n where x^k=1?

For ex take j>i wlog and consider x^(j-i) maybe?

[boreal lion]

Yeah I looked at the solution and it uses that argument exactly

I understood that one and am convinced by it, I just wanted to know if mine works or not since I couldn't come up with that one

[tropic spade]

Why can't x be 1 and how do you know n is not 1? (I could be forgetting basic group theory stuff lol)

[boreal lion]

Wait, does this not work if n = 1? Since x = 1?

I guess I assumed n was greater than 1, and then x wouldn't be equal to 1 since its exponent would be less than the order of x

Wait how does this actually work if n = 1...? They're not distinct anymore, since x^1 = 1, right?

[tropic spade]

Yeah I'm confused on that myself.

But ignoring x=1 specifically I think your proof will work.

[tropic spade]

It seems to me like an inductive version of this argument

Where you consider the set {1,...,x^k} for k<n.

And show all elts of that set are distinct.

[boreal lion]

Oohh I didn't think of it like that

[tropic spade]

You can apply the same trick with i,j at every step of the induction proof so it should work.

Though you'd be picking i j from a subset of 0,..,n-1.

[boreal lion]

I'll try and tackle it that way, and see if I can construct something. I really need to take another look at induction lol, I only know how to use one version of it, I think

In any case, thanks a bunch

I was feeling iffy about it since it felt a lot less "clean" than the one you suggested

[tropic spade]

For the case x=1 I guess n=1 so you can maybe argue x^0,...,x^(n-1) is just the list/set or whatev containing only 1?

Then that case is obvious enough that you could consider only x not equal to 1 maybe lol?

[boreal lion]

Wait it's actually n = 2 where the problem arises, isn't it

[tropic spade]

I think the x matters also

[boreal lion]

Oh wait my bad, no n = 2 is fine

Got confused

[tropic spade]

I'd say the induction arg is less clean, but it doesn't hurt to prove a result you already know in a diff way.

[boreal lion]

I think you're right then yeah, probably saying n = 1 gives the list containing only 1 is to the way to go

Yeah it definitely feels a lot more messy, but I guess it wouldn't hurt to get some practice in for me, I desperately need it

Appreciate your help a bunch, thanks man

[woven solar]

silly question about notation, if you would please. let R be a [commutative] ring and M and N be modules. What are some common meanings for the notation "N < M".

that's "<" as opposed to "\subset". Is it suppose to mean that N is a submodule (as opposed to just a subset) of M (which was given in my assumptions...)

[tawny pine]

ye submodule. generally in algebra < can be used to denote substructure. subgroup, subring, etc

[woven solar]

thank you, much obliged.

[trim grove]

can we make every finite field from ring Z[x]? (don't know is this question to worth it ask)