groups-rings-fields

[coral shale]

ok its both internal and external, literally doesnt matter

cus the enclosing group is an external

[tender wharf]

alright

thanks for explaining, notation was confusing af

[coral shale]

i uh

just usually dont mind about internal vs external

ive not seen where it like matter matters yet.

[tender wharf]

i mean the book says that too so

[warm urchin]

Let G be a finite group of order $p^{n}-1$ where $p,n\in \Bbb{N}^{\ast}$ .Show that for each $x\in G$, there exists $y\in G$ such that $x=y^{p}$ \ \ Notice that $x^{p^{n}-1}=e_{G} \implies x^{p^{n}}=x \implies (x^{p^{n-1}})^{p}=x$ Set $y=x^{p^{n-1}}\in G$ and we are done.

[cloud walrus]

Susilian

[warm urchin]

am i crazy or does this work? feels too straight forward

[coral shale]

yep

[warm urchin]

so ,i'm trying to prove fermats little thereom using induction , lmn if my proof works.

Let p be a positive prime number and $b\in \Bbb{Z}$. Show that $b^{p}\equiv b$(mod p). \ \

Lets proceed by induction , for the case b=1 , $1^{p}\equiv 1$(mod p) trivially, suppose the claim holds till b=n and lets show it for b=n+1.

Lets prove that $(x+y)^{p}\equiv x^{p}+y^{p}$ (mod p) for p prime, first we have that \begin{equation*}
(x+y)^{p}=x^{p}+\binom{p}{1}x^{p-1}y+\binom{p}{2}x^{p-2}y^{2}+\dots+\binom{p}{1}xy^{p-1}+y^{p}
\end{equation*}
and since $\binom{p}{r}$ is a multiple of p for all $0<r<p$ we get the result required. \

finally $(b+1)^{p}\equiv b^{p}+1^{p}$(mod p) and since $b^{p}\equiv b$(mod p) by assumption of the induction , we get $(b+1)^{p} \equiv b+1$ and the induction is done.

[cloud walrus]

Susilian

[warm urchin]

and just out of curiosity , is there other ways to approach this ? yknow using groups

[cloud walrus]

potato

[warm urchin]

ohhh thats cleaner , thank you

[south patrol]

Np

[south patrol]

Another way is uh multiplication by a sends {1,2,...,p-1} to itself bijectively (well, mod p) and so in particular (p-1)! = a^(p-1) (p-1)! mod p by considering the product of all the elements

I think this is probably the nicest way that doesn't appeal to group theory lol to me at least

[coral spindle]

That's a cute proof that generalises to all groups

[delicate bloom]

ah yeah I forgot about that one, plus kind of nice to see (p-1)! appear, since it seems to pop up in other ways too, so kind of unifying in approach.

[formal ermine]

this is more of a meta question but do you guys have any tips on how to learn for my algebra final?

this will be my first uni exam

[delicate bloom]

for instance if p=1 mod 4, then you can split $(p-1)!$ into the first and last halves, reverse the second half, $$(p-1)! = \left(\frac{p-1}{2}\right)! (-1)^\frac{p-1}{2} \left(\frac{p-1}{2}\right)! \mod p$$
$$-1 = \left(\frac{p-1}{2}\right)!^2 \mod p$$
so you can think of $\left(\frac{p-1}{2}\right)!$ as a kind of canonical choice for $i$

[cloud walrus]

mOwOsity

[delicate bloom]

sorry didn't mean to bury your question

[formal ermine]

dw

[untold basin]

Hello

For a finite group G, is there always at least one subgroup that its order divides |G| ?

[chilly ocean]

G

:^)

see lagrange's theorem

[untold basin]

I'm talking about the existence

[barren sierra]

Trivial group also

[chilly ocean]

yeah, G exists

[untold basin]

Of course I meant the non trivial one

[chilly ocean]

G is a subgroup of G

[untold basin]

xd

[coral shale]

can you formulate your question to be precise

[untold basin]

Ok

[coral shale]

rather than us shooting catchems

[untold basin]

I'm doing an exercice

[barren sierra]

Are you asking for a criterion of whether non-trivial proper subgroups exist?

[untold basin]

Let G be a group of order 21

[barren sierra]

Sylow's theorems probably

If it's 21

[untold basin]

Show that G contains 6 elements of order 7

[barren sierra]

[coral shale]

Yeah I think you're asking about the existence of non-trivial proper subgroups

[untold basin]

I said that if H < G s.t |H| = 7 then all the elements of H (non-trivial) are generator of H

it means there exists 6 elements of order 7

[coral shale]

Isn't the answer always yes unless u have a prime cyclic?

[untold basin]

But I assumed that H exists

That's what I'm asking

Is there always an H of order 7

[barren sierra]

Do you know Sylow's theorems?

[untold basin]

I don't know

Perfect

[barren sierra]

Hm

[coral shale]

wait now you're asking existence for each order. not any

[untold basin]

Because I'm doing an previous exam of my course

From last year

[coral shale]

(is why I wanted a precise formulation)

[untold basin]

They saw it last year but I haven't

Ok perfect

thank you guys

[coral shale]

Specific then sylow probably yh

but not only way to show it

[untold basin]

I haven't formulated it well but it's ok I got my answer

[barren sierra]

Yea I think that's the intended answer

Pretty standard question is showing that there exist exactly two groups of order 21

And both will contain a subgroup of order 7

[rotund aurora]

Follow your heart

xd

[coral shale]

intuit everything

as in find an intuition for things u havent found one for

[delicate bloom]

yeah and then tell me your intuition for sylow thms thx

[rotund aurora]

like they are the axioms

idk what you want me to say

[coral shale]

i dont know them

came at the end of my g-t course and i was yh fckit by then

[rotund aurora]

SylOwO

[delicate bloom]

like I think there's an intuition for them that I'd be satisfied with in terms of group actions or something

I just haven't really sat down to roll up my sleeves to do it and want someone to just spoonfeed it to me 😛

[coral shale]

so when do they like be... useful

sylow

outside of exercises

[barren sierra]

primes r cool 🙂

[coral shale]

I didnt need them for galois, say.

or algebraic number theory iirc

[delicate bloom]

when you start classifying finite simple groups ||aka never ||

[coral shale]

which direction would u need them

[coral shale]

is that it

[rotund aurora]

Tbh, how the proof (of one of them at least) starts its pretty natural. You don't know if subgroups of order p^k exist, but you do know subsets of order p^k exist

I ought to revisit group actions soon, I'm still pretty noob

[delicate bloom]

I heard there's a way of proving it that makes proving them all in one go obvious and intuitive

like it's one theorem instead of separate ones

[agile burrow]

Yeah, the first time around I didn't get Sylow at all. The second time around I didn't get Sylow at all. The third time around I began understanding why the proof works because I had finally internalized group actions. I don't know if I'll ever have an intuition for the actual statement of the theorems but I do think that you can make the proof feel natural

[delicate bloom]

what sorta stuff did you do in the mean time that got you internalizing group actions just out of curiosity @agile burrow

[coral shale]

I haven't found a reason to bother with sylow, like its a dead end. Other than for the sake of doing it

i havent got very far with gt though ofc

[agile burrow]

I started studying some algebraic topology and group actions on trees and stuff

[delicate bloom]

ah I see

[agile burrow]

The more group actions I see the better 🙂

[rotund aurora]

Counting with group actions is probably nice too

[agile burrow]

A friend of mine also began writing a book on group theory from the perspective of group actions rather than abstract groups. Honestly most of it is the same but it's an interesting perspective

[coral shale]

hmm hmm, time to do more math one day

[delicate bloom]

sounds fun

[agile burrow]

For example, you might instead derive Lagrange's theorem as a corollary of orbit stabilizer

[delicate bloom]

oh, sounds fun

[agile burrow]

Even though they are essentially the same result

[delicate bloom]

I guess I thinkk of lagrange's theorem as being so easy, idk how you'd go backwards

but I like the more pictoral way of thinking of group actions

[agile burrow]

Yeah that's fair. He mostly does geometric group theory so he's very interested in having geometric interpretations for most of these things

I should ask him about Sylow sometime and see if he has any insight

[delicate bloom]

yeah lmk if you do, sounds nice

[agile burrow]

Maybe this is kind of niche but I think the only time I've seen Sylow outside of classifying small groups is in group cohomology. You can reduce the computation of the cohomology of a finite group to some subgroup of the cohomology of the Sylow subgroups. This result relies on existence (Sylow I)

Actually I've probably seen it elsewhere too but I can't think of anything right now

[coral shale]

for something sounding so major it seems little used at early-mid level

say undergrad

[agile burrow]

I agree

Really I mostly just treat it as a capstone result

Existence in general can pop up here and there - I think there's a Galois theory proof of the fundamental theorem of algebra that uses existence of a 2-Sylow subgroup

[coral shale]

its not like isomorphism thm say idk

[delicate bloom]

I just want to know it better because I think it'll give me a better feel for what groups in general look like

[coral shale]

maybe its to do with category theory things

like sylow is irrelevant surely there?

you're talking about group structure but thats specific to groups

[agile burrow]

Yeah, I don't see why it would be relevant to any category theory

[coral shale]

sure, this i lack.

[coral spindle]

You can use them to do some weak classification results of groups in terms of their orders. You can e.g. say that if groups have particular orders, they must be simple/nonsimple etc

[coral shale]

its like it helps you zoom in on specific things

Structure of this group with this specific order

And maybe that's putting me off in finding it 'interesting'

I can't say I've ever been enthusiastic about hearing about finite group classifications

[coral spindle]

That's finite group theory 🤷

[coral shale]

maybe when i do more math past ug level

[agile burrow]

https://math.stackexchange.com/questions/1561986/different-applications-of-sylow-theorems

Top reply is the proof I was thinking of

[coral shale]

Oh damn

this is good

[agile burrow]

The linked MO post in the reply to the question also describes further results in group cohomology which I talked a little bit about earlier

[coral shale]

rip lol

[delicate bloom]

a few times while reading about elliptic curves some group thing would pop out and needed to figure out which one it was, and it ended up being handy

wish I could give a specific example but none comes to mind atm

[agile burrow]

That sounds neat

[delicate bloom]

to be fair I think I never needed the sylow theorems for them or just appealed to like "there are only such and such choices for this group of this order, and so we reason out this and that" without sylow, but

idk maybe I'll go back and see if I can find something tomorrow

[agile burrow]

Does this have to do with like Galois groups? Or the group law on elliptic curves? I don't know anything about elliptic curves lol

[delicate bloom]

I was thinking there were examples from both actually

[south patrol]

sylow is used in the fundamental theorem of algebra and so like linked to galois groups ig lol

[coral shale]

proof i saw was only relevant for R and C right

What about proper fields like Q >.>

[solar glacier]

for this exercise, is the forward containment the trivial one

cause if youre in N_H(A) then youre in H and youre in G such that hah^-1 \in A for. every a \in A

[coral shale]

im a bit confused - statement looks trivial

[chilly ocean]

both directions are the trivial one

the next part of the problem is probably more interesting

and part a is just there to call out that fact

[solar glacier]

is just says to define the center and centralizer

then part c ....

[chilly ocean]

[south patrol]

Tau is injective I guess so it sends roots to roots bijectively

(Since it already splits in L)

no problem sir

i was a bit unsure cause i typed that out instinctively before reading the proof lol

but yes that'll be it

owo

[warm urchin]

How do i show that any 3-cycles in An are conjugate? n>=5 ofcourse (Just need a hint) I've been thinking about this for a while with no luck ,tried testing conjugates by brute force to try to construct it for any two 3-cycles but little luck was found.

[south patrol]

you know they are conjugate in Sn. Use the fact that n >= 5 so that you can modify the element you conjugate to lie in A_n

[coral shale]

o = (1 ... n)
fof' = (f(1) ... f(n))

[barren sierra]

This is just one of those problems where you just basically write the definitions and the proof is immediate

[junior harbor]

Any tips for how to properly read through an AA book? Rn I'm on page 200-ish of Michael Artin's "Algebra" and I am having a harder time understanding each concept the more I read, its just too many definitions mixed together in one I'm having a hard time keeping up

[next obsidian]

Are you doing exercises

[junior harbor]

Nah but the issue is mostly with grasping everything the chapter throws at u at once

Idk if exercises will help but if they might, I will try it

[next obsidian]

They will

You aren’t using any of the concepts

[barren sierra]

Exercises will always help

[next obsidian]

So it’s hard to synthesize the info

If you’re just reading like a novel ofc it feels like way too many definitions

[barren sierra]

This goes for any math text. You retain the theorems and proofs more if you actually see how they're used

[next obsidian]

Because you haven’t had the time or experience working with the new concepts

[junior harbor]

I mean, it is. In the symmetry chapter they used like 5 definitions in a single corollary once iirc

[next obsidian]

Part of doing exercises is getting used to concepts, because the later stuff only keeps building on top of it

[chilly ocean]

you're on page 200 but still need advice on how to read the book?

[next obsidian]

Nah

[junior harbor]

Wasn't having trouble before

[barren sierra]

Go back and do the exercises

[barren sierra]

Seeing a theorem being used != Using a theorem yourself

[junior harbor]

Gotcha.

[coral shale]

||applying pure math is the easiest way to learn it||

[chilly ocean]

it's almost like you get better at things when you practice them

[warm urchin]

Ah , ye well i havent proved that for Sn yet actually lol , but here is my attempt

[cloud walrus]

Susilian

[warm urchin]

well , i guess that how i did it assumes the cycles are disjoint , so i need to do some modifications.

[warm urchin]

oh , wait that makes writing things much easier.

ok i think i got how to proceed now

we just take a cycle to send (a1 a2 a3) to (sigma(a1) , sigma(a2) , sigma (a3)) , defining sigma appropriately so they allign , then if the cycle is odd we add an extra transposition ( allowed by the fact that n>=5 ) and we are done.

[rotund aurora]

I would simply say that math is not a contemplative subject

[unique valve]

False

[rotund aurora]

its true but ok

[unique valve]

I have contemplated math many times

[rotund aurora]

ok

[unique valve]

Are most people here undergraduate students

[rotund aurora]

#discussion or #math-discussion

(mmh... unless you meant here in this channel, but then idk)

[junior harbor]

Ye I figured. Tho I got stuck on some exercises for long than I could care, so I just skipped them. Now in retrospect I can do most of them no problem, but others are just over my head

Like, idk how to prove that, being $a, b \in G$ and $G$ being a group, having $a^7 = 1$ and $a^3b = ba^3$ implies that $ab = ba$

[cloud walrus]

Duhon

[junior harbor]

It just looks super obvious but idk how to do it lol

[rotund aurora]

well that's how you see that reading doesn't give you that much

[junior harbor]

True. Well, any tips for what to do when I get stuck on exercises?

Just skipping them sounds bad

And so does asking for hints lol

[rotund aurora]

You are just facing problems

there is no recipie

try to think hard, be perseverant, etc. Also, you should be strategic and know when to give up.

[barren sierra]

you try them

get stuck

maybe come back and try them again

and if you're still stuck, a hint isn't bad

but give them an honest effort

[rotund aurora]

but like

being stuck is normal

it's in fact ideal

[junior harbor]

Ok. Thanks

[warm urchin]

alright folks , follow up exercise

[cloud walrus]

Susilian

[warm urchin]

looks like it should work , but im always suspecious im missing a detail when dealing with these pesky permutation exercises

[rotund aurora]

this is correct

But I wouldn't say "sigma is generated by 3-cycles", I would simply say that sigma is a product of 3-cycles (as generated, is usually to talk about sets/subgroups)

[warm urchin]

ohh , true true , i see what you mean

[next obsidian]

Yeah it would be like “as A_n is generated by 3-cycles…”

Then you can write it as a product of them

[warm urchin]

yep , gotcha , ile be more careful next time , thank you

[next obsidian]

It’s just wording, it was totally understandable

So no real error

[warm urchin]

[rustic crown]

[next obsidian]

[rustic crown]

i miss moldi here

@hidden haven

[south patrol]

Btw so like

[rustic crown]

uwu

[south patrol]

I've got a spectral sequence with mostly Z/2Z (and a single Z) on the E2 page and apparently I can show that there are no elements of odd order (besides 0) in any group on the E_{infty} page too - is there a particularly easy way to see this?

I imagine the point is just like

The Z must map into Z/2Z or 0 and so like the differentials from Z must have kernel 0 or 2Z or Z?

And so on the E_{infty} page you can only possibly end up with 0,Z/2 and at most one Z

[rustic crown]

me wanna learn about spectral sequences too >.<

[south patrol]

Run whilst you can

jk

[rustic crown]

where you reading it from uwu?

[south patrol]

Oh I mean this is from a topology paper that is using it lol

I learnt a bit about spectral sequences from random sources oof like practising with the actual machinery

otherwise i think it can get fairly daunting lol

but e.g. McLeary's book is nice ig

[rustic crown]

oh okie

i think the only place i was close to encountering spectral sequence was while computing the composite of two derived functors or something

i should do more cat theory and read that again

[ruby sundial]

same

i had the total complex introduced to me in AT for something i don’t remember

but not for spectral sequences

[rustic crown]

maybe kunneth's theorem?

idk AT

[ruby sundial]

yeah

i was gonna say for lunneth

because i remember we introduced diagonal map around then

[lethal dune]

Ah you prove Kuneth using spectral seq

[barren sierra]

ugh this is a throw back to a conversation earlier in the day

I am reviewing my algebra notes in prep for next semester

going over the proofs of the Sylow theorems again and I'm still like "yea these sure are theorems. They sure do the thing. I have no intuition for why they're true but they are"

[prisma ibex]

they're various applications of orbit-stabilizer/class formula

[barren sierra]

yea the proofs use that alot

and I mean they make sense

just I could never come up with these proofs lmfao

[next obsidian]

Yeah who cares

Like honestly

That’s why we read things other ppl do

[prisma ibex]

if you absolutely have to remember them, then it's maybe easier to try to remember what G-set you use in the class equation proof for each part

[barren sierra]

I mean it's preferable to see "hm yes, I can see why they did the proof this way"

[prisma ibex]

but also yeah you probably will not need to actually regurgitate the proof

[next obsidian]

And once you know the technique I think you could prove it

Like

[barren sierra]

but I cannot see how one makes the leap from p-subgroups to using actions

[next obsidian]

The dude didn’t prove this like

[barren sierra]

oh of course not

[next obsidian]

Sitting at his desk for 4 hours

For his Algebra 1 class

[barren sierra]

fair fair

and when taking Algebra, I never had to regurgitate the proof

just apply it

[next obsidian]

Like if you just sat there and tried to orbit stabilizer on literally everything you can think of

I’m sure you’d eventually stumble onto the proof

[prisma ibex]

for Sylow III it's not so hard to guess how the proof goes

since you're trying to say something about the number of Sylow subgroups

[barren sierra]

Proof of Sylow 3 definitely makes the most sense

[prisma ibex]

let G act on this set of Sylow subgroups by conjugation and apply class formula

yeah

[next obsidian]

Bro I remember the proof better than I remember what part of Sylow Sylow X is

For any value of X

Lmfao

[prisma ibex]

definition: Sylow subgroups
Sylow I: we didn't make an empty definition

For Sylow II showing that all the Sylows are conjugate to one another, you already have a Sylow P in G so let G act by left translation on G/P

Sylow I is the hardest to remember I think

[barren sierra]

yea fs

[vocal patrol]

How does 1 generate the group of all integers?

I mean every positive integer can be seen as 1+…+1 right but

How do I get to the negative integers using only 1

[coral shale]

-1...

[vocal patrol]

So if I have a single generator 1 I can use its inverse?

[coral shale]

lookup what generator means i think.

[vocal patrol]

I thought I’m only able to use that generator and the operation of addition of the group

Oh ok thanks

[formal ermine]

@vocal patrol $\gen{g}$ means $\Set{g^n | n \in \bZ}$

[cloud walrus]

rectangle cube

[vocal patrol]

g to the power of n in this case means I use the operation of the group on the amount of n elements of the generator g, right?

[formal ermine]

yes

and it's also defined for negative n

[delicate bloom]

it almost becomes tautological if they wrote it additively haha

[formal ermine]

g^(-2) means (-g) \circ (-g)

and of course for 0

g^0 = e

[coral shale]

in all cases.

There is only 1 group operation

[vocal patrol]

I see

Thanks a lot!!

[formal ermine]

how would I find the galois group of $\bQ(\sqrt[3]{2})/\bQ$? this extension is not galois because it's not the splitting field of anything

[cloud walrus]

rectangle cube

[formal ermine]

I can't really go by the root of the minimal polynomial, can I?

because that will also have zeta_3 in it

[delicate bloom]

yeah I'd start by trying to find the automorphisms of Q(cbrt(2)) that fix Q

[formal ermine]

wait

oh

yeah

[formal ermine]

isn't it only the identity

cuz every auto is determined by where it sends cbrt(2) to

[delicate bloom]

well I was thinking you wanted to prove that somehow

[formal ermine]

and we only have one possibility

because cbrt(2) -> anything in Q wouldn't make it an auto anymore

so cbrt(2) -> cbrt(2) is our only possibility

therefore the galois group is trivial?

is that correct

[delicate bloom]

sure

I guess it depends on what you're willing to accept or not

[delicate bloom]

like you asserted this, but I didn't really see a proof, but maybe that's sufficient for you

[formal ermine]

if sigma(cbrt(2)) = x for x in Q but also sigma(x) = x (because sigma fixes Q) then we'd have sigma(cbrt(2)) = sigma(x) but cbrt(2) \neq x, so sigma wouldn't be injective anymore

[delicate bloom]

what's Q \ { cbrt(2) }

[coral shale]

wuh

[formal ermine]

wait lol

[delicate bloom]

why couldn't sigma(cbrt(2)) = a + b cbrt(2) + c cbrt(2)^2

[formal ermine]

I mean Q

kek

[delicate bloom]

for some a, b, c rational

b, c not both 0

[formal ermine]

can't seem to figure that out

do you have a hint for me

[delicate bloom]

well I was gonna say you could make that subsitution and cube it,
$$\sigma(\sqrt[3]{2}) = a + b \sqrt[3]{2} + c \sqrt[3]{2}^2$$
$$2= (a + b \sqrt[3]{2} + c \sqrt[3]{2}^2)^3$$
then equate coefficients to get a system of 3 equations in rational numbers but that seems like a terrible way to do this

[cloud walrus]

mOwOsity

[hot lake]

You could try and find all the real solutions to x^3 = 2

[formal ermine]

cbrt(2)

[coral shale]

roots have to map to roots

do u have a thm saying this?

[coral shale]

if yes then this

[formal ermine]

no

[coral shale]

well thats unfortunate

[delicate bloom]

can you show that, like with some calculus maybe

[coral shale]

???

[delicate bloom]

I'm just following up with Zef's hint

[coral shale]

dont think i follow but ok

[delicate bloom]

how do you know cbrt(2) is the only real* root of x^3-2?

[hot lake]

Yeah we are relying on proving that there is no other real root

[formal ermine]

x^3 is strictly increasing so x^3 = k has at max only one real solution

[hot lake]

Yeah

Then you have to explain that an automorphism have to send a root to a root

[formal ermine]

I think I figured it out?

[south patrol]

just factor it over C to see there's only one real root lol

no calculus 😎

[formal ermine]

so for an f in K[x] with a root at x = a and a sigma in Gal(L/K) we have that f(sigma(a)) = sigma(f(a)) = 0 because sigma fixes K?

this seems so trivial why did I spent the last hour trying to figure it out

so sigma(cbrt(2)) must send it to another real root of x^3 - 2, but the only real root is cbrt(2) so sigma(cbrt(2)) = cbrt(2)?

[delicate bloom]

yeah you got it

or if you don't like R you could use some other field that contains both Q and Q(cbrt(2)) like the 5-adics, Q_5. Since x^3-2 factors as (x+2)(x^2+3x+4) mod 5, and lifts to a factorization in Q_5, but the quadratic has no roots, then it means we can only map cbrt(2) to itself in the same way.

just kidding

[formal ermine]

do you have an article or w/e on p-adic numbers

I still don't understand them and the youtube videos aren't helping

[delicate bloom]

I have many books on them, articles not so much I'm afraid lol

I just was kinda feeling rambunctious and thought, "why use R here when I can use some p-adic field just as well?"

nothing particularly special except that most people are only familiar with that one completion of Q so most arguments get threaded through it that don't necessarily need to be

[delicate bloom]

Actually got to thinking about it more, seems we could generalize this idea further of how $x^n-a$ is surjective on $\mathbb{R}$ when $n$ is odd to p-adic scenarios too to prove it.

[cloud walrus]

mOwOsity

[delicate bloom]

Suppose $\gcd(a,p)=1$ and $\gcd(n,p-1)=1$. Then $\mathbb{Q}(\sqrt[n]{a})$ has trivial Galois group.
Proof: $f(x)=x^n-a$ is surjective on $\mathbb{F}_p$ because of the gcd condition on $n$, which means it has only one root $r$. $f'(r) \ne 0$ because of the gcd condition on $a$. This means we can lift the factorization to $\mathbb{Q}_p$ and there are no other roots of $f$ there. Furthermore, $\mathbb{Q}(\sqrt[n]{a}) \subseteq \mathbb{Q}_p$ so we're done.

[cloud walrus]

mOwOsity

[delicate bloom]

one of the few times where p=2 is the only good case, since odd primes give a strictly worse theorem than the one you can get over R lol

idk something doesn't smell right in p=2 case so maybe it's time to go to sleep, I just proved $\mathbb{Q}(\sqrt{1+2n})$ has trivial galois group

[cloud walrus]

mOwOsity

[delicate bloom]

oh I just need the extra condition gcd(n, p)=1 too... that fixes it at least, but makes p=2 not so great nevermind lol, nite nite

[stone stag]

An old T.A. of mine got a doctorate and lists "algebra" in her research areas; do I understand correctly that after calculus you wrap around again and Algebra means something quite different?

[chilly ocean]

lol

that tends to be the case

"algebra" means fucking around with equations and polynomials until you learn what a group is

then it means groups, rings, modules, etc.

[chilly ocean]

of course, all this abstract nonsense is inspired by the algebra we learn before calculus

some people use the term "college algebra" to refer to the algebra you're referring to, i've seen

[formal ermine]

I like the term hs algebra better

[rustic crown]

can't wait to cat-pill new kids learning algebra uwu

[stone stag]

Alright, I've been afraid to even guess because it felt like a trap

[formal ermine]

merry christmas det

[rustic crown]

mewwy chwistmas uwu

[chilly ocean]

"algebra" changes meaning many times through one's math career

solving polynomials (hs) -> groups, rings, fields, etc. (university) -> solving polynomials (ag)

(half-serious post)

[stone stag]

^ More than half-serious relief getting this sorted

[formal ermine]

what's an example of a captive group

(captive as in non-free)

[agile burrow]

I'm not familiar with the terminology, but if you're looking for any group that isn't free then there are plenty. Take any finite group, for instance

[formal ermine]

what about any finite cyclic group

[agile burrow]

those are not free groups

[formal ermine]

does free group and free abelian group mean something different

[agile burrow]

they are different in general, yes

a free abelian group is isomorphic to Z^|S| for some set S

[formal ermine]

ah ok thanks

[rustic crown]

uwu

[tribal furnace]

uwu

[rustic crown]

[tribal furnace]

merry xmas det!

[rustic crown]

mewwy xmas uwu

hewwo walter uwu

[agile burrow]

hi det

merry christmas 🎅

[rustic crown]

[lament dawn]

Wrong channel? Lol

[rustic crown]

i gave me the studying role, so no more #chill >.<

[warm urchin]

any hints not sure where to even start

[chilly ocean]

start from the definition of "solvable" maybe?

[elder wave]

Merry Christmas friends

[chilly ocean]

merry christmas timo

(and all others)

[tribal furnace]

Frohe Weihnachten!

[rustic crown]

mewwy chwistmas tewwa

[cloud walrus]

Susilian

[warm urchin]

actually i have an idea using last one

[chilly ocean]

think about what you're given

sure

[chilly ocean]

i was just thinking, you want a chain of nice subgroups for HN, and you have one for H and one for N, so you can try to get one for HN using them...

[warm urchin]

that could also work let me go give these 2 ideas a try

[junior harbor]

Can someone help me with a hint or 2 on how to prove fermat's little theorem? (The one where a^p = a mod p if a is coprime to p and p is prime). It for some reason came up as an exercise in the vector fields chapter of a book I'm reading and I just want to know what I really need to use here

[formal ermine]

do you know euler's theorem

[junior harbor]

Which one of them

[formal ermine]

$g^{|G|} = e$

[cloud walrus]

rectangle cube

[junior harbor]

e being the identity here?

Or what?

[formal ermine]

yes

g is any element in G

and G is finite

[junior harbor]

Hm right, so this would be the perfect tool here

[formal ermine]

yeah

[junior harbor]

But idk if it was shown

Lemme check

[formal ermine]

do you know basic group theory (specifically lagrange's theorem)?

[junior harbor]

That showed up before here yes

Tho I found no mention of Euler's theorem in the short time I looked

I mean it does make sense that |G| divides itself

But it does not restrict it to being only |G|

[cloud walrus]

Susilian

[warm urchin]

here is my attempt using the 3rd definition

which i mentioned above

[junior harbor]

Tho ig even then, the theorem does make sense. At least if the group is <g^n> and we are talking about g^1 being raised to |<g^n>|

[warm urchin]

hopefully it works , but i will try to do it the other way as well once i figure out how to do so

[formal ermine]

ok so proving euler's theorem could be a fun little exercise

anyway

so to prove the original thing you want to look at $\bZ_p^*$

[cloud walrus]

rectangle cube

[junior harbor]

So multiplicative modulo p?

Well yeah assuming Euler's I managed it I think

Only one issue

The order of the integers mod p is p

So a^p ≡ 1 (mod p)

Which is wrong

So idk what I'm missing

[warm urchin]

you can argue by induction.

[south patrol]

No, it is p-1

If you're doing them under multiplication

(0 doesn't have a multiplicative inverse)

An elementary method is to consider (Z/pZ)^x (which is just 1,2,...,p-1 under multiplication mod p) and compute the product of all the elements before and after multiplying everything by a

Because multiplying elements by a will just permute the elements, those two numbers are the same (mod p)

If you get stuck just lmk

[cloud walrus]

MyMathYourMath

[solar glacier]

btw thats integers modulo p not p-adic lol

clear from statement though i feel like

i had a prof who hated ussing Z_p for integers moudlo p

[chilly ocean]

it's usually clear from context if Z_p means Z/pZ or p-adic integers. the only people that get mad about that are graders who are looking for a reason to deduct points

[formal ermine]

Z/p

[chilly ocean]

i'm not too sure how that fact will be useful. G is abelian, so all of its subgroups are normal

[solar glacier]

good point

[chilly ocean]

that also serves as a hint of sorts

namely: what do you do to normal subgroups?

[solar glacier]

quotient out by them>

?

[chilly ocean]

right

and does that have anything to do with the index?

[solar glacier]

when you quotient out by H you get p elements? i.e., the order of size of the idnex is the order of the quotient group

[chilly ocean]

the order of the quotient group is the index of the subgroup you quotiented by

[formal ermine]

(a bit off-topic, but @chilly ocean for these kind of problems, do you already know them or did you just immediately 'see' it?)

[solar glacier]

yeah thats what i meant

[chilly ocean]

order(G/H) = index(H)

[chilly ocean]

what's the difference?

[formal ermine]

already know as in you've done this problem in the past

[chilly ocean]

neat. i think you can use this

[solar glacier]

does the first iso thm get used here at all

[formal ermine]

see as in this is the first time you've seen this but you already know the solution

[chilly ocean]

i can't remember if this is the first time i've seen this or not

i just "see" it then

[chilly ocean]

possibly

it tends to come up when you quotient

if H were the kernel of some onto G -> Z_p, then you'd have G/H = Z_p, for example

[solar glacier]

ok i think i should be able to figure it out with these hints ..i think lol

[chilly ocean]

i think you can

[chilly ocean]

chances are that i've probably seen enough similar problems that i just know how it ought to go

like

"onto group homomorphism" just screams QUOTIENT

[formal ermine]

yeah I mean I "saw" the one direction immediately too but I'm still trying to figure out the other direction for myself right now lmao

[chilly ocean]

and the index of a subgroup is related to the quotient

[solar glacier]

yeah supposing it has prime index then showing its the kernel

[formal ermine]

(my course actually defined the index as the size of the quotient group)

[solar glacier]

supposing its the kernel seems like the easier direction imo

[chilly ocean]

if H has prime index then G/H has prime order and is abelian...

that should remind you of a certain group

[solar glacier]

cyclic groups Z_p

[chilly ocean]

i'll leave the details to you

[solar glacier]

ok thanks !

[chilly ocean]

but there is indeed an easier direction and a harder direction

[formal ermine]

||so we know that G/H is isomorphic to Z_p, how do we know that there must exist a surjective homomorphism from G to Z_p?||

[chilly ocean]

you should think a bit more about that

lol

give it 10 minutes and if you still dunno ping me

[chilly ocean]

hint

@formal ermine ||the quotient map G -> G/H is surjective, compose with the isomorphism to get G -> Z_p surjective||

[formal ermine]

||ah that seems so obvious||

I sometimes wonder if math isn't right for me

there are so many of these blatantly obvious things

that I just don't 'see'

[chilly ocean]

it's not like you'll be able to see these things immediately

you're literally in your first abstract algebra class

[formal ermine]

yeah but a bunch of my course mates seem to do

[chilly ocean]

i've been studying math for a good 5 years now and it's been like, 3 or 4 since i was first exposed to the first isomorphism theorem, for example

[formal ermine]

they were even able to figure out really non obvious things during the lecture, right after the topics were introduced

[chilly ocean]

you would do well to learn early that it's pointless to compare yourself to your classmates

there's always going to be someone better than you

that's just life

also you're a highschooler

[formal ermine]

you're right

btw do you have any tips for my algebra exam?

this will be the first uni exam I'm taking

a bit scared

[chilly ocean]

mental tips or studying tips

[formal ermine]

both

[chilly ocean]

get enough sleep the night before

for real

[formal ermine]

that will be hard

[chilly ocean]

do it anyways

[formal ermine]

I usually get like 4-5 hours during school days

[chilly ocean]

studying tips are harder to give because what works varies between people, but good general advice is to look for past exams and try to do problems you think/know will be on the exam.

this is kinda getting off topic though

[formal ermine]

sure maybe we can move to #advanced-lounge

[formal ermine]

thanks walter

wait

these are the same problems again

no?

[agile burrow]

no problem - I think the last two problems on 8 and then hw 9 cover module theory and hw 10 covers some field theory

I've reuploaded the rest so that they can be in one post and deleted the original

If someone wants to repin this that would be great

[formal ermine]

thank you!

[formal ermine]

@uneven folio pwease? 🥺

[agile burrow]

I said this the first time too, but I'm pretty sure most if not all of these are just from Dummit and Foote. Since there isn't much module theory in these exercises, you might get some extra practice from looking at relevant sections in those

[formal ermine]

Yeah

[agile burrow]

thank you ryc

[solar glacier]

question for showing N_G(H)=C_G(H) for a subgroup of order 2 does this work

let H={e,h_0} where h_0 \neq e

clearly we have C_G(H) \subset N_G(H)

[south patrol]

This seems quite unclear to me like

[solar glacier]

for the other direction

let g \in N_G(H)

must show gh_0g^-1=h_0

clearly gh_0g^-1 \in H as g \in N_G(H)

then gh_0g^-1 is either e or h_0 but if its e you get that h_0 is e

thus weve shown ghg^-1=h for every h \in H

[solar glacier]

proving the normalizer is equal to the centralizer for a subgroup of order 2

more specifically:

is this correct

[south patrol]

Yeah seems fine

Or slightly more briefly note conjugation by g sends H to itself and sends e to e, so it must send h0 to h0

(But that is basically what you are arguing)

[solar glacier]

yep true

[south patrol]

Oh interesting

Apparently N_G(H)/C_G(H) always embeds in Aut(H)

(N/C theorem)

And Aut(H) is trivial here so that gives this

[solar glacier]

thats an interesting way of seeing it!

yeah since H has only 2 elemetns

Aut(H) is trivial

whats N/C theorem never heard of this

[south patrol]

Well I just stated it aha

For H any subgroup of G

[solar glacier]

the normalizer modulo the centralizer embeds into the auto group of the subgroup?

[south patrol]

But I think it is clear enough here because there is a clear map N_G(H) -> Aut(H) since the former acts on H by conjugation

And the kernel is just C_G(H)

Oh okay so it is easy lol

[languid moss]

are these graduate or undergrad abstract?

[chilly ocean]

wrong reply?

[formal ermine]

undergrad

[coral shale]

Is it possible to try find galois group of splitting field of x^3-x+1 without finding the roots

i recall this but seems like roots needed to compute discriminant

[south patrol]

You can calculate discriminants using the coefficients

[coral shale]

oh.

[south patrol]

There are formulae which are a bit of a pain but basically just expand it out and use the Vieta formulae

[coral shale]

The formula for discrim

[south patrol]

Yeah

[coral shale]

this generalises to higher?

what happens to the a^4

[south patrol]

Yeah it generalises but the formulae get nastier I guess

[coral shale]

not the same one?

i mean using the roots

product of root difference squared

[south patrol]

Wait wdym by this here

[coral shale]

times a^???

[south patrol]

Or like the definition of discriminant

[coral shale]

,,a^4\prod(a_i-a_j)^2

[coral shale]

this one

[south patrol]

Usually just assume a = 1 and then it is that product

[coral shale]

sure im just not sure how we got an a^4 in the first place

[south patrol]

I imagine the formula doesn't depend on scaling basically

Which is why the a is in there

[coral shale]

like this is cubic, i dont see how we got a^4 if u see what i mean

is it still a^4 for higher dim

ah maybe i should lookit up
i didnt realise this discrim is more generap

ok ok

ok i wasnt completely aware this is the same discriminant used everywhere

Still curious about the 2n-2 but I guess its a bit like degrees of freedom

[nocturne bone]

Why is the rank 4 and not 3?

[coral shale]

whats rank

this is pure group theory right?

[nocturne bone]

Ye

[coral shale]

nvm got it

[formal ermine]

15Z is isomorphic to Z

[coral shale]

This defn coincides with 'smallest cardinality of generating set'?

oh right u choose ((0,...,1,...,0), 1, ..., 1)

loosely notated

[rotund aurora]

Z x Z/2Z has rank 1 but is not generated by 1 element

[plush wasp]

The Counting Formula reads:

|G| = |N(H)| • (number of conjugate subgroups)
N(H) is normalizer of a subgroup H
How did they conclude the above statement? Is |N(H)| same for the normalizers of all subgroups?

[south patrol]

No, because different subgroups may have different numbers of conjugate subgroups.

But yes, this is gonna be a special case of orbit stabiliser if you've seen that?

I can explain further if you'd like

[plush wasp]

never seen. I was also perplexed when I read the order of conjugate classes also divide the order of the group

[south patrol]

Ah okay

[plush wasp]

can you explain on both rq

[south patrol]

Do you know anything about group actions?

[plush wasp]

nope

[south patrol]

Ah okay hm I'll try to write up an explanation that bears that in mind then lol

It's definitely worth knowing about them so I might as well explain them

[cloud walrus]

potato

potato

[south patrol]

Is this clear so far? @plush wasp

[plush wasp]

Got the first one! the second one's loading

understood!

[south patrol]

Okay, awesome, I hope the examples have been helpful

[plush wasp]

so centralizer and normalizer are stabilizers, right?

[south patrol]

So that's what I'm coming onto, yup aha

They are stabilisers for appropriate actions

Okay so also worth pointing out that if G acts on X then 1) the orbits partition X and 2) the stabiliser of any element is a subgroup of G - these are both easy checks but are important

[cloud walrus]

potato

potato

[south patrol]

So e.g. unioning over all elements in G.x we get |G| = |G.x | |Stab(x)|

Lmk when you're up to speed with this / if you need more owo

(also that is a typo lol, i meant h in G, not h in H of course)

[plush wasp]

sometimes I forget functions can be treated as group operations.

can you explain why {h | g^{-1}h ...} is of size Stab(x)?

the thing you wrote at the end

[white grotto]

find the proper values of this

so im solving for phi(X)=lambda X

any hint would be welcome

[sage spruce]

Maybe apply Tr to the equation

You get 2Tr(A)Tr(X) = lambda Tr(X)

[vocal patrol]

Why are the real numbers under multiplication only a monoid and not an abelian group?

[coral spindle]

Think about 0

[sage spruce]

Because zero does not have an invert

[white grotto]

nvm

[white grotto]

so the only lambda is 2 tr(A)?

[south patrol]

Sure, so ${ h \in G \mid g^{-1} h \in \text{Stab}(x)}$ is equivalently $g \text{Stab}(x) = {gh \mid h \in \text{Stab}(x)}$

[cloud walrus]

potato

[south patrol]

Like, there's a bijection between that set and Stab(x) by multiplying by g^{-1}

[sage spruce]

Yes and because you necessarily have at least a proper value in C

[south patrol]

Does that clear it up Arya?

[white grotto]

so just 1 proper value?

nice

thanks a lot

[plush wasp]

Yes it does. Beautifully explained tyvm

can you help cram school me for my upcoming exam? >_<

[south patrol]

Thanks aha

[south patrol]

Lol gl but i can probs help with the occasional q ofc

Anyway, so now we have orb stab in hand uhhh probably good to show you a couple of examples to get the feel for it lol and then we can do your question?

or we can go straight to it lol

[plush wasp]

Do you know Discrete Math? Incidence structures and stuff?

[south patrol]

No lol sorry

But yeah UHHHH so let's do a particularly nice example which will motivate orb stab lol

[cloud walrus]

potato

[plush wasp]

orbit of G/H?

[south patrol]

Well the orbit of H, which is an element of the set G/H (sorry if that was unclear)

or are you saying the orbit is G/H

[plush wasp]

ohh

no, okay.. H is an element of G/H right ✓

[south patrol]

Yup

(It's also eH i guess if e is the identity if that makes it clearer)

[plush wasp]

so orbit is G/H and stabilizer is H?

[south patrol]

Yup!

[plush wasp]

tysm

[south patrol]

So if G is finite, what can we conclude?

[plush wasp]

|G| = |G/H| • |H|

[south patrol]

Yeah so I guess that relation holds regardless of size

But in particular in the finite case this implies Lagrange's theorem

[plush wasp]

hmm

[south patrol]

Is there anything u r confused about like

I mean the infinite case like dw about multiplying infinite cardinalities if that is confusing lol

[plush wasp]

not at all. Actually, the topics I'm supposed to cover are Sylow Theorems, Nilpotent groups and solvable groups

[south patrol]

But yeah it's cool that this spits out lagrange

Like pretty much immediately

[plush wasp]

mhmm understood it!

[south patrol]

nice

Okay, now with this machinery in hand, the question you asked is now very straightforward.

$N_G(H) = { g \in G \mid g^{-1} Hg = H}$, so which group action are we going to consider and which set should G act on? i think there are a couple of reasonable choices and it doesn't matter lol)

[cloud walrus]

potato

[south patrol]

The definition basically gives away the action

[plush wasp]

hmm, H -> g^{-1}Hg

:c but that's not an action is it?

[south patrol]

Yeah so that's how g should act on H, though you need to define it slightly more generally lol

i.e. the set it acts on can't (in general) just be {H}

[plush wasp]

hmm, G acts on G/H, action being H -> g^{-1}Hg?

[south patrol]

g^{-1}Hg needn't be a left coset of H i guess

But yeah you can more simply just consider the set of subgroups of G

Or even the set of all subsets, I suppose

Or more restrictively, the set of subgroups conjugate to H

[plush wasp]

Yeah I was thinking that but I recalled G/H partitions G so ...

[south patrol]

It doesn't really matter since the orbit and stabiliser will be the same

Yeah sure

Uhh let's take the set of subgroups of G then, say

Hopefully the orbit and stabiliser of H are clear

[plush wasp]

hmm! sorry I confused myself. Of course G/H is not the same as the set of subgroups of G

[south patrol]

And that answers your question

:)

[south patrol]

Yeah dw

[white grotto]

if A^q=1 show that the proper subspace of lambda=1 has dimensions this

quick hint

[south patrol]

interesting

[plush wasp]

so umm, yeah I see now, stabilizer would be {g | gHg^{-1} = H} and orbit would be the set of conjugates of subgroups of G.

right?

[south patrol]

Yup

So the stabiliser and orbit are both what you wanted (!)

[plush wasp]

hmm

[south patrol]

The stabiliser is just N_G(H)

[south patrol]

oh sorry

Orbit of H is not the set of conjugates of subgroups of G

the orbit of H is set of subgroups of G conjugate to H

[white grotto]

shouldnt the dimension be 1

[south patrol]

No

I mean so like

[white grotto]

since the characteristic polynomial has (x-1) as a factor

[south patrol]

If A = I, then the dimension should be the same as the dimension of your space

Which isn't necessarily 1

[white grotto]

but if the power of my factor in the charac polynomial is 1, isnt the dimension necessarily 1?

[plush wasp]

ooo that's what it meant ✓ conjugate subgroups of H are all g^{-1}Hg 🤦‍♂️

[south patrol]

The power in the minimal polynomial is 1, not necessarily in the characteristic polynomial

and this is ofc assuming q is invertible in the field or else A-I is nilpotent oop

[south patrol]

Yeahh dw

But yes this is cool hopefully now?

Group actions are super cool and you'll use them for the Sylow theorems, for example

[plush wasp]

yeah it's cool now

[white grotto]

wdym

[south patrol]

Wait i haven't tried that case with characteristic

[white grotto]

q is in N*

[south patrol]

But I imagined it'd be important here

So yeah I was kinda joking with characteristic cause I imagined you're using matrices over R or smth but like

[white grotto]

the matrix is over any field

[south patrol]

e.g. if you consider $A=\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \in M_2(\mathbb F_2)$ then $A^2 = I$ and the eigenspace corresponding to $1$ is of dimension $1$, but $\text{tr}(A) = \text{tr}(A^2) = 2$ and so the formula you wrote spits out $\frac{1}{2}(2 + 2) = 2$

[cloud walrus]

potato

[white grotto]

A^2 != I

[south patrol]

I'm pretty sure it is

[white grotto]

mmm

I am not getting it then as I

[south patrol]

Are you getting $\begin{pmatrix} 1 & 2 \ 0 & 1 \end{pmatrix}$

[cloud walrus]

potato

[white grotto]

yes

[south patrol]

Yeah so 2 = 0

cause we're in F_2

[white grotto]

ah

whats F_2

mod 2?

[south patrol]

Yeahh

{0,1} with 1+1 = 0 and everything else defined in the normal way

[white grotto]

what youre saying is that my formula is wrong?

[south patrol]

I'm saying your formula only holds under sufficiently nice conditions

But it should be fine with the field being R or C

That's why I crossed it out lol sorry

[white grotto]

Ah

[south patrol]

ANYWAY so this formula uh there is a fancy way I know to do this but I'll try to translate it into more standard stuff

[white grotto]

sure then Mn(R)

[white grotto]

make it as intuitive as possible i want to learn how you thought about it more importantly than the actual result

[south patrol]

Okay sure, there are a couple of ways I think and I'll write it up ig

[white grotto]

cause i got nada

[cloud walrus]

potato

[south patrol]

You can show that B is a projection of V onto W and note for any projection operator, the trace is just equal to the dimension of the space you project onto

[cloud walrus]

potato

[south patrol]

Okay, just came back to it and realised perhaps an easier solution: over C, A is diagonalisable. Show that B acts the same as A on the eigenvectors of eigenvalue 1 and acts as 0 on everything else

[white grotto]

this is beautiful

yeah i shouldve thought of some kind of average

As to the projection thing. God

i wouldve never thought of that

how do you go from average to projection

[cloud walrus]

potato

[south patrol]

oh hm

Well I guess yeah the point is that this just acts as the identity on W so it is a projection

Just like this is all using the fact that BA=A=B

It's worth pointing out this is part of a more general theorem in representation theory (here the important thing being basically that the {1,A,A^2,...} form a group of size q under multiplication)

[white grotto]

how did you know that

[south patrol]

Uh like well this is just from BA=A=AB but also like

if you take any vector v and apply A lots of times, you'll get {v,Av,A^2 v, ...., A^{q-1}v} and loop that

So if you add up everything and apply A again, you'll just get back what you already had

[formal ermine]

why do the field extensions need to be normal

[white grotto]

why ://

is it obvious that i am gonna get the same thing

[south patrol]

Well this is like

$A(I + A + \dots + A^{q-1}) = A + A^2 + \dots + A^{q-1} + A^q$ and $A^q = I$

[cloud walrus]

potato

[white grotto]

wait BA=A?

or B

[south patrol]

sorry yes

BA=B

lol

[white grotto]

okay i didnt see that before

thank you

soo, how does B act like the identity on W again?

[south patrol]

Well if Av = v, then A^k v = v for all k

So Bv = 1/q (v + ... + v) = v

Come to think of it, this does fit into more general linear algebra too, uh

[chilly ocean]

[delicate orchid]

🍿 _ _

[south patrol]

So like

[formal ermine]

hi wew

[south patrol]

Well i mean you have a decomposition of any vector space into A-stable subspaces and you can find a formula for the projection onto the 1-eigenspace

i am sleepy

[white grotto]

lol

i am always sleepy

[white grotto]

okay

should i ask why

[delicate orchid]

Completely lost track of wtf is going on

[formal ermine]

?

[south patrol]

i mean they don't have to be normal in general right like they're saying you need this for them to be normal extensions

since then (since you're probably working over char 0 anyway) you'll get finite galois extensions

[chilly radish]

You want your minimal polynomials to split

Otherwise things don't act nice in the automorphism groip

[celest cairn]

Merry Christmas guys! I have a question. Does ζ and ω mean the same thing when dealing with roots of unity?

[barren sierra]

Depends on context but iirc I see both those variables being used

[formal ermine]

this is kinda like asking "what does the letter m represent". it heavily depends on the context and the definitions you're working with

[celest cairn]

Ah ok. Also, what’s the difference between letting $\zeta$ be a nth root of unity and $\zeta_{n}$?

[cloud walrus]

Sapphire Gaming

[barren sierra]

Same thing

Again, look at the context

[celest cairn]

Ok, thanks. 🙂

[barren sierra]

You may see notation for the latter if like

You're talking about multiple roots of unity

[cloud walrus]

Spamakin🎷

[barren sierra]

But yea it's just notation

[solar glacier]

hey is showing image of cyclic groups under homomorphism ic cyclic a super easy proof? you just use properties of the homomorphism to show phi(x^m)=phi(x)^m

so if \phi:G \rightarrow H and we want to show phi(G) is cyclic let h \in \phi(G) then \phi(g)=h for some g \in G which is cyclic so we can put g=x^m for some m

merry xmas btw

[next obsidian]

Yah

[solar glacier]

ok sweet just makin sure

thx

[next obsidian]

Or use 1st iso

It’ll be a quotient of Z/nZ which is some Z/mZ

[solar glacier]

ah didnt think of it like that!

[next obsidian]

This is like

Harder to be clear

Hahaha

But it’s a different way to view it

[solar glacier]

i see, thanks!

[rustic crown]

thinking of them as image of Z should be even easier :p

[next obsidian]

uwu

[rustic crown]

uwu

[white grotto]

if this, then the gcd of characteristic polynomials for M,N has degree >= the rank of P
is there some making sense of this
why it would make sense

[cloud walrus]

MyMathYourMath

[coral spindle]

yes

[solar glacier]

thanks!

[agile burrow]

Here's an open problem I find neat: a deep result of Stallings and Swan characterizes (non-trivial) free groups as the only groups with cohomological dimension 1. It's conjectured that the only groups with homological dimension 1 are non-trivial locally free groups (that is, groups in which every finitely generated subgroup is free).

A potential counterexample would have to satisfy certain properties. For instance, a non-abelian counterexample necessarily has trivial center

[formal ermine]

the degree of the splitting field $\bQ(i, \sqrt[4]{2})$ of $x^4 + 2$ over $\bQ$ is $8$ because we have $[\bQ(i, \sqrt[4]{2}) : \bQ] = [\bQ(i, \sqrt[4]{2}) : \bQ(\sqrt[4]{2})] \cdot [\bQ(\sqrt[4]{2}) : \bQ] = 2 \cdot 4$ as the minimal polynom of $i$ is $x^2 + 1$ and the one of $\sqrt[4]{2}$ is $x^4 - 2$. is that correct?

[cloud walrus]

rectangle cube

[formal ermine]

x^2 + 1 is irreducible by eisenstein with 2, right?

I mean when looking at f(x + 1)

x^2 + 2x + 2

[tribal moss]

It's also irreducible over Q because we know its only factors over C are (x+i) and (x-i) and neither of those are rational. 😛

[formal ermine]

o

yeah that's simpler lmao

[tribal moss]

Eisenstein wouldn't work, because the prime has to divide the constant term too.

[formal ermine]

it does tho?

2 divides 2

[tribal moss]

Sorry, missed which polynomial you were applying it to.

[formal ermine]

np

[formal ermine]

@chilly ocean the reason why the grader deducted me a point was because I didn't show that it's an involution iff a_0 isn't 0

[chilly ocean]

sucks

but is understandable

it may be very easy and straightforward, but it's not trivial

i never like doing homework where i have to include every tiny little argument out of fear of losing marks

[formal ermine]

I mean

he said he'll give me back the point if I email him the proof

[chilly ocean]

oh lmao

kind of based

do it

[formal ermine]

yeah

[chilly ocean]

nice grader

[formal ermine]

but there are these other two things I don't get tho

[coral spindle]

What a cool prof

[formal ermine]

grader

[barren sierra]

This shit is so annoying

[coral spindle]

same difference

[barren sierra]

And then the prof releases the answer key and it's half the length

Or less

[formal ermine]

I wrote "Let f be an irreducible polynomial but f* = gh (with g,h non-constant) not irreducible. As the constant term of f* is not zero, the constant terms of g and h are also not zero, therefore g* and h* are both non-constant"

he said that "As the constant term of f* is not zero" should be "As the leading coefficient". why is this though?

f* = x^n f(1/x) fwiw

[barren sierra]

💀

[chilly ocean]

lol

[formal ermine]

heh

so why is it "leading coefficient" and not "constant term"?

I don't get it

[barren sierra]

Is the leading coefficient of f* the 1/x^n term or the actual constant constant term

[tribal moss]

The leading coefficient of one of the polynomials becomes the constant term of the other, and vice versa.

There may be confusion or ambiguity about which polynomial you're talking about the constant term in.

[barren sierra]

They said "the constant term of f*"

[formal ermine]

btw I did a typo here

in the first sentence

[formal ermine]

what's a primitive element of a field extension

[tribal moss]

https://en.wikipedia.org/wiki/Primitive_element_theorem#Terminology

[formal ermine]

ah

I was looking at the german wikipedia and it wasn't really helping lol

"Determine infinitely many primitive elements of the field extension Q(sqrt(2), sqrt(3))/Q"

so we know that Q(sqrt(2), sqrt(3)) = Q(sqrt(2) + sqrt(3))

so can I just answer "any element of the form k(sqrt(2) + sqrt(3)) for k in Q" lol?

[tribal moss]

Yeah -- though that does feel a bit cheaty. It might feel more substantial to say something like sqrt2 + k sqrt3 for any nonzero rational k -- then they're at least not all proportional over Q.

[formal ermine]

proportional?

[tribal moss]

I mean, any two of elements of your form would be linearly dependent over Q. There doesn't seem to be any prohibition against that, though.

[formal ermine]

ah ok thanks

another question

Q(thirteenth root of unity)/Q is a galois extension because x^13 - 1 has a nonzero derivative and so it is separable, it is normal because it's the splitting field of the positive roots?

I feel like my argument for normality is flawed

[coral shale]

Wait I just realised. U said ur 15? how are u learning all this shit already

[formal ermine]

they let me take one university course per semester

last semester I did numerical analysis

[coral shale]

sounds whack

[formal ermine]

this semester I'm doing abstract algebra

yeah

I'd love to do more courses

[coral shale]

very cool though, gl.

[formal ermine]

but my school grades are too bad for that I believe

I mean I still pass every class

but they're not that good

[formal ermine]

cuz I'm not sure if the definition of a splitting field allows it to only contain a partial amount of the roots?

[tribal moss]

What does "positive roots" mean here? It is the splitting field of the cyclotomic polynomial (x^13-1)/(x-1) = x^12 + x^11 + x^10 + ... + x +1.

[formal ermine]

oh sorry

typo

I was confusing it with something else

I mean it's normal because it's the splitting field of x^13 - 1 wrt to the thirteenth root of unity?

idk how I would write it using language

[tribal moss]

I would just say the splitting field of x^13 - 1. I'm not sure which additional meaning it would give to say "with respect to" a particular root -- the idea of a splitting field is that all the roots are there.

[formal ermine]

what about the other thirteenth roots of unity?

[coral spindle]

They are all powers of any primitive 13th root of unity.

[tribal moss]

13th.

[coral spindle]

what a typo!

[formal ermine]

right

is the argument for separability correct?

[coral spindle]

I don't recall what separability means

If you have that result you're using re: formal derivatives, I don't see why that wouldn't work

[formal ermine]

an algebraic field extension L/K is called separable if for every a in L the minimal polynomial of a only has simple roots in the algebraic closure of K

ah wait