groups-rings-fields

[fervent rock]

is it just the factor ring F[x]/<r> where r is irreducible

[prisma ibex]

if you have an extension of fields E/F you can talk about this extension being Galois

[smoky cypress]

And also exhibiting a polynomial with S_5 being its galois group but that isn’t too hard

[fervent rock]

https://tenor.com/view/depressed-desparate-smoking-sad-gif-19116342

all i need is a 70 on this exam to pass with an A

never gonna take an algebra course again

i love the people that do algebra

but not for me.

[prime sundial]

i just realized this is #groups-rings-fields my bad

[smoky cypress]

Bump

[sonic coral]

what would typically come after reducing polynomials and producing finite fields of order p^n in an undergrad aa course

[fervent rock]

dropping whatever courses come after

[sonic coral]

good one

[prime sundial]

have you already talked about field extensions and algebraic closures

[sonic coral]

i don’t believe so

is that like Q(sqrt(2)) and stuff. i’ve read ab stuff like that but havnt learned about it yet

[prime sundial]

yes that's the sort of stuff

my class went through finite fields then extensions, algebraic extensions, and now we're onto separability and galois correspondence (hopefully galois theory)

but there's not much time left so we can't really do much more

[sonic coral]

we’ve basically just done ideals and integral domain definitions and then proven some simple properties of both. and then the theorems to reduce polynomials over either Q or Z and produce finite fields

[prime sundial]

i think field extensions make sense then as a next step

[sonic coral]

okay. i ask bc i’m doing it as an independent study and he just handed me notes and problems mixed in with them and proofs ect. i finished them pretty quickly, in like 3 days. he said i’m good for the next test but like we have 3 weeks left and it’s our final test before the final exam

so i was just looking for a direction into what to study next

[prime sundial]

mood but also going to take more abstract courses cause i hate myself

[prime sundial]

i would consult the syllabus/professor, but field extensions are the next step in my limited experience

[sonic coral]

i will ask him on monday. but we’ve covered the whole syllabus

[prime sundial]

if you have 3 weeks (~2 with the exam) then im sure you'll get to extensions and algebraic extensions

[sonic coral]

well we’re done. he’s told me we aren’t gonna cover any more material in the class

[prime sundial]

oh

[sonic coral]

he’s covered everything he would cover in a standard class but since it was an independent study and i was motivated we got through things quick

[prime sundial]

then if you want to continue i'd look over what i mentioned above

[sonic coral]

but i wanna learn more so i will ask ab field extensions

also, for anyone with grad school experience, would you recommend jumping straight into graduate analysis or beginning with undergrad analysis at the new program? i’ve been told to start with undergrad.

[prime sundial]

have you done UG analysis at your current school

[sonic coral]

yes

i start grad school in the fall

[prime sundial]

i haven't heard much about taking UG analysis again

are you not confident in your ability to succeed in the grad version

[sonic coral]

not from reading the syllabus. a professors at the new program said it’s typical for their students to start at UG analysis again. i think i will start there but i wanted other opinions and if that is something somewhat standard to do

[prime sundial]

i've never heard of that but i am not a grad student so i can't really say what's standard

have you done grad analysis at your current uni

[sonic coral]

no just UG. i read the UG syllabus at the new school and they have stuff in there that we didn’t cover in my UG analysis so that should be enough of an indication to retake it right

[prime sundial]

i don't think it warrants a whole retake unless the syllabus is completely different. i'm guessing they'll cover similar stuff, and what is different you could review in your own time
but i have 0 experience so i'm not really qualified to suggest anything

[sonic coral]

okay. thanks your your direction regardless

[prime sundial]

can a surjective automorphism on a countable order group be not injective?

must it be an isomorphism

whoops did not mean automorphism

just meant a surjective homomorphism with domain and codomain equal

[celest furnace]

by "countable" do you mean finite ?

[prime sundial]

im working specifically over Z^2 so that might help

[prime sundial]

nah just countably infinite

if it was finite i can see it must be true, but here i can't justify it

[void cosmos]

are you guys ready for the worst proof possible?

its wrong

problem 12

now

[celest furnace]

what about something like phi(a, b) = (a, b/2) if b is even, and (a, b-1) is b is odd

definitely onto but not sure if it preserves +

Nope

[prime sundial]

not well defined sadly

oh

[celest furnace]

maybe now

[prime sundial]

didn't see the second half

[celest furnace]

Still no...

[prime sundial]

i don't think that'll preserve addition

it feels true

[void cosmos]

by induction suppose F has a basis of cardinality m for every m>=n , we wish to show that it has a basis of cardinality m+1 for every m>=n , suppose for the sake of contradiction it does not: consider the basis of cardinality m {x1,x2,....,xm}. we now have 3 basis : {y1,y2,..yn} , [z1,z2,..,zn+1] and {x1,x2,...,xm}. consider now the set {x1,x2,...,xm,y1}. this has cardinality m+1 so it cant be a basis but it spans the module so we must have its linear dependant --> there exists k_i not all 0 such that sum(kix_i) = 0. but each x_i is a linear combination of y_is so we arrive at a contradiction, contradicting that {y1,y2,...ym} is linearly independant

now ik this is trash but thats all i could do

i used the hypotehsis just for the base case , like the base case works just cuz the assumption but thats it

[void cosmos]

this is for problem 12.

[celest furnace]

True for Z -> Z since every additive map on the integers is just linear

same for rational i think

[void cosmos]

anyone 😦

[thorn delta]

im stuck on your induction hypothesis. It really should be something like: let F be a free module with a basis of cardinality n >= 1. Suppose (for the induction hypothesis) that F has a basis of cardinality m' for each n <= m' <= m. (Then go on to show that F has a basis of cardinality m+1)

[void cosmos]

okayy

Im inducting on m itsef

So start with m=n for base

or wait

[thorn delta]

right, the exact statement you want to show is like
P(m) = "F has a basis of cardinality m"
and you want to show P(m) is true for all m >= n

[void cosmos]

okayy

but i get to assume i have p(m)

so it doesnt matter rifht?

[thorn delta]

what doesn't matter?

[void cosmos]

the induction hypothesis

i dont see the difference 😅

[thorn delta]

by induction suppose F has a basis of cardinality m for every m>=n
what you said here is just asserting the conclusion

[void cosmos]

yes

I assumed the theorem for m and want to show yhe theorem for m*1

m+1

this + base case

[thorn delta]

if F has a basis of cardinality m for every m >= n, then it's vacuously true that it has a basis of cardinality m+1, because m+1 >= n. You don't need any of your assumptions to prove this (and this is a problem)

[void cosmos]

Yea mb

[void cosmos]

This works better

The rest is trash too tho right?

Or does it work

[thorn delta]

there exists k_i not all 0 such that sum(kix_i) = 0. but each x_i is a linear combination of y_is so we arrive at a contradiction, contradicting that {y1,y2,...ym} is linearly independant
i don't think you've said enough to conclude that that {y1,y2,...ym} is linearly dependent

err

this might work actually

its much more complicated than what i would've done

[void cosmos]

Lmao

cool

[thorn delta]

i think hungerford proves in this section that F has a basis of cardinality k iff F is iso to R^k

[void cosmos]

he doesnt

Its an exercise

oh

Yea its true

But i forgot that

and i see the proof using that

But i just cant aee anything aeonf

Wrong with my proof

And i wanted to ask

Other than the missing <m u pointed out

[thorn delta]

yea, im pretty sure it does work.

[void cosmos]

wow

okayy

Tysm

its fully original

Lmfaooo

[thorn delta]

it just bugs me a little bit tbh because there is an easy direct proof:
$$F \cong R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m + 1}$$

[cloud walrus]

kxrider

[void cosmos]

yeaa

I couldnt come up with tbis

It took me like 35 mins

To come up awith what i sent

Or even more lmao

sad ig

[thorn delta]

eh yea that's learning anything new for me. getting stuck for a ridiculously long time on (in hindsight) trivial things.

also that example in q13 goes hard. a lot of people don't like this about noncommutative rings, but I think it is cool they can have properties like this.

[void cosmos]

Ugh im sleeping now

Will check it tmrw

Ty

[thorn delta]

nothing to worry about. just pointing out a fact i find neat

[solar shore]

im having such a tough time struggling with understanding the proof for the fundamental theorem of abelian groups

[rustic crown]

finite?

[lethal dune]

[plucky flicker]

Hi! Can you guys give me some hint about this problem? I have to prove the following statement: if G has only one irreducible representation of dimension > 1, then the commutator subgroup of G is elementary Abelian p-group.

[plucky flicker]

I can bring G' in the game knowing that the number of 1 dimensional irreps is |G/G'|

[wooden ember]

I think I might have smth? Let c(H,G) denote the number of conjugacy classes of G contained in a subset H. Then if N is normal I’m pretty sure that c(G/N,G/N) <= c(G,G)-c(N,G)+1. If this is true then write G’ as a union of conjugacy classes and observe that it must then by made of exactly 2 classes. Show that it’s thus elementary abelian

Pretty neat fact gotta say

Rep theory is so cool

[untold cloud]

Hi, guys, why these vectors form a basis?

[wooden ember]

What have you tried in showing these are linearly independent vector?

[south patrol]

I think it's easiest just to show that $\mathbb C \cdot v_{\zeta^i} \not \cong \mathbb C \cdot v_{\zeta^j}$ as $\mathbb C[G]$-modules

[cloud walrus]

potato

[south patrol]

Though this uses more theory lol

[untold cloud]

i tried to put them in matrix

show the det is non-zero

[south patrol]

Yeah this is presumably a vandermonde moment

[wooden ember]

Yeah you should get a vandermonde matrix

Sniped again

[south patrol]

lel

[delicate orchid]

not only is it a vandermonde matrix, isn't it just straight up the character table of C_n?

[south patrol]

Which presumably is what we are proving lol

[delicate orchid]

right ok lol

[south patrol]

More or less

[delicate orchid]

anyway, definitely non-zero determinant!

[south patrol]

TIL the proof every field automorphisim of R is the identity

[delicate orchid]

this is true

[south patrol]

LIke I heard that whether C admits non-trivial field automorphisms depends on foundations but never actually thought about R lol

[delicate orchid]

C always admits i -> -i even without AOC

[wooden ember]

How does it depend on foundations

[south patrol]

...besides conjugation

lol

[wooden ember]

Oh okay

[untold cloud]

Thanks! i see

[delicate orchid]

it's just that with AOC that C_2 becomes an unconstructable uncountable mess

[south patrol]

lol

[wooden ember]

Every day I’m made to reconsider my current stance on AC

[south patrol]

Lol

[delicate orchid]

I only work with finite sets so I really don't care about AOC

[south patrol]

Basiert

[wooden ember]

DC is clearly an easier alternative

[south patrol]

Yeah I basically just trust that every time I employ choice I could probably just do it by hand in that specific example anyway lol

[delicate orchid]

there's a C in there actually but I just take vector spaces over it and then stop caring

[south patrol]

just choice lets me prove smth more general

¯_(ツ)_/¯

Anyway I am do Galois Theories

are there any fun questions

that i can do lol

[delicate orchid]

no

[south patrol]

understandable

tryna show x^4 + 1 reducible over Fp for all

p

Got it down to p = 3 mod 8 lol

So perhaps time to just consider p = 3 and 11 to see what happens

[lethal dune]

[south patrol]

Why 😭

[lethal dune]

you want hints or ...?

[south patrol]

Oh just wasn't sure what that sticker was meant to mean lol

[lethal dune]

idk either, looks cool

[south patrol]

Nah I'll keep going but like is considering quad residues good so far to like rule out roots of 2 and -1 etc

And do you have any interesting galois theory questions

Lol

[lethal dune]

I do

[wooden ember]

I think I have a broken way to do this with alg num

[south patrol]

Bruh

[wooden ember]

But im probably using a result that doesn’t satisfy the required hypotheses or smth as usual

[lethal dune]

@south patrol

[wooden ember]

Okay yeah I think it works

[south patrol]

Okay this is more advanced than my galois theory course lol

but thank ointeresting

[lethal dune]

Maybe these then

[wooden ember]

||x^4+1 is the 8th cyclotomic polynomial. Look at the extension $\mathbb{Q}(\zeta_8)|\mathbb{Q}$. The galois group of this is not cyclic so no primes remain inert, thus $x^4+1$ must split into factors over $\mathbb{F}_p$ for each p||

Aah

How do you spoiler latex

[lethal dune]

,texsp spoiler

[summer path]

[lethal dune]

W/o ||

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

Thanks

[south patrol]

Okay so my idea is basically like

[formal ermine]

is there any way to understand tor functors without derived functors

[wooden ember]

You don’t really have to explicitly mention the whole derived functor machinery but in essence it’s a derived functor yeah

[south patrol]

\texsp sps $K = F_p(\epsilon)$ w $\epsilon$ a root of $x^4 + 1$. if irred then $[F_p(\epsilon) : F_p] = 4$ so that $# F_p(\epsilon) = p^4$ and $# F_p(\epsilon)^\times = p^3(p-1)$ so $8 \mid p-1$ and $p \equiv 1 \mod 8$. But then $(2/p)=1$ and we can factor $x^4 + 1$ anyway lol

[cloud walrus]

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

[formal ermine]

I can't find a resource online that explains them without derived functors

[wooden ember]

I mean it’ll be explained by using the same machinery, just without using the word derived functor

[lethal dune]

Think of measuring the failure of injectivity

[south patrol]

Well they are derived functors

[agile burrow]

You could take the properties that they satisfy on faith, but if you want to understand why they work then you might as well learn derived functors lol

[south patrol]

Is this ok lol

like i'm doing a lagrange moment

[formal ermine]

what properties

[agile burrow]

Long exact sequence, how to compute, etc.

[south patrol]

oh very cool

[formal ermine]

[lethal dune]

@south patrol

[south patrol]

oh thankies

thanks for flexing the fact you did them

jk

[lethal dune]

past assignments

[wooden ember]

I know I struggled less with tor because I saw ext first

And ext feels a little more natural as a derived functor to me

[formal ermine]

ext?

[wooden ember]

But all in all it’s still pretty weird stuff imo

[wooden ember]

Derived functor of Hom

[formal ermine]

[wooden ember]

Measures « failure of exactness » of an ses if you will

[formal ermine]

how can a ses fail to be exact

[wooden ember]

Uhm

Lmao

[lethal dune]

of you go deep into homological algebra, you'll see they are just the natural functors to define in the derived category

[wooden ember]

Nvm me

[lethal dune]

homology

[wooden ember]

Should have said failure of exactness of the last term in a not so short exact sequence

[formal ermine]

idk homology

[wooden ember]

If you get what I mean

[elder wave]

maybe follow the advice

and skip those AM exercises for now

[agile burrow]

ah

I understand the context now lol

[formal ermine]

es kribbelt mein gehirn

[elder wave]

you really don't need this machinery to do comm alg for now

[wooden ember]

Honestly as much as I enjoyed the homological algebra stuff I did it felt pretty unmotivated and I didn’t get a whole lot out of it so I’d also recommend skipping

It’s probably better to see that stuff after having seen (co)-homology in topology first anyways

[elder wave]

idk how people learn hom alg without having seen homology in alg top

[agile burrow]

I just did it because it was the next thing in the book I was reading

[elder wave]

like i know people do it but it'd feel so random to me without it

[agile burrow]

But then I learned the topology 🙂

[south patrol]

i more just like

don't see how you could have the willpower to learn hom alg by itself

[elder wave]

yeah

[south patrol]

Anyway i am do that question from you ryu like

[wooden ember]

They introduce hom algebra before homology at my uni and it’s so dumb

[south patrol]

the irred factors of f(g(x)) one

Here they introduce them at the same time basically lol

[wooden ember]

Especially since we end up repeating a bunch of stuff in algebraic topology afterwards

[south patrol]

oh lol i think i misread 2) frick

Okay ryu is it like

uh

[lethal dune]

they are nice problems

[south patrol]

Well

[plucky flicker]

But wait. G’ must be abelian, so the #of conj. classes in it is the order of G’

[south patrol]

Let $h \mid f(g(x))$ be irreducible and $\alpha$ any root of $h$ in some extension lol. Then $f(g(\alpha))=0$, so $[F(g(\alpha)) : F] = n$ and this divides $[F(\alpha):F]= \deg h$

[cloud walrus]

potato

[lethal dune]

ye

[south patrol]

noic

3 looks spicy

and important lol

[lethal dune]

in 3 assume the factors are monic btw

[south patrol]

Lol

Ok

Ngl I am stuck seeing how normality (but not separability) comes in

but i shall think

Obviously it is needed or otherwise e.g. x^3 - 2 lol

hm

[sly storm]

Hello. Can anyone here explain in a simple and intuitive way and in detail why in the hyperbolic plane it is generally impossible, starting with an arbitrary circle of area less than 2\pi, to construct a quadrilateral with all sides and angles congruent having the same area, and vice versa, that is, starting with an arbitrary quadrilateral with all sides and angles congruent, it is impossible in general to construct a circle having the same area?

[formal ermine]

what does this have to do with abstract algebra

[sly storm]

It can so.

[formal ermine]

huh?

[sly storm]

It can have to do with abstract algebra.

This question can be deleted if abstract algebraists here see no relationship between it and abstract algebra.

[fair quartz]

is basic functional analysis on-topic enough for this channel?
if so id like to ask something about proving properties of a predefined scalar product in a complex hilbert space

[formal ermine]

I think this belongs in #real-complex-analysis

[summer path]

Or #advanced-analysis

[fair quartz]

ah yeah i think adv is the right one, thx!

[south patrol]

Any hints for 3 btw? Done the rest of them except 5 I think

[fervent rock]

i have a question

i have the question

[delicate orchid]

THE question??

the one and only?

[fervent rock]

wait im writing fuck off

wait

[south patrol]

Yes

[delicate orchid]

waiting

[south patrol]

The answer is 42

[fervent rock]

Let G be a group and supposed [G:Z(G)] = 4. Prove G/Z(G) ismorphic to Z_2 x Z_2

my approach is this

attaching photo

[delicate orchid]

yup!

[south patrol]

Sure

[fervent rock]

is it poggers?

[delicate orchid]

I'm pogging irl rn

[south patrol]

I mean the hardest bit is the bit you already know

[fervent rock]

thx i didnt cheat

doing my practice final

[south patrol]

That if G/Z(G) is cyclic then G abelian

[fervent rock]

trying to just load up my cheat sheet

i just need an 80

never gonna algebra again

[south patrol]

Lol

[delicate orchid]

unwise....

u think ANALYSIS is RIGOROUS??? FOOL!

[silent oxide]

Dunno how you can avoid basic group theory forever

[fervent rock]

for me, up to kernel theorems im fine, but isomorphism theorems make no sense

[south patrol]

Wdym kernel theorems

[fervent rock]

the kernel of group homomorphisms.

also the rings isnt that bad

[delicate orchid]

if you mean G/ker\phi \cong img \phi then that is an isomorphism theorem

[fervent rock]

but like some of these theorems are fucked

[delicate orchid]

not sure about that one boss

the 2nd and third iso theorems are just special cases of the first and the first just says "if u say two things are the same if they map to the same thing in an isomorphism u still have a group"

*homomorphism

[fervent rock]

was able to do 3/5 of the questions on the practice final in like 25 minutes

so we chillin

literally a genius

[lethal dune]

α, β be roots of g(x) and h(x) respectively then there is an automorphism fixing base k sending α ↦β. Show this automorphism is enough

[south patrol]

hm

uh so how do we know they have roots

or are you extending to a Galois closure

[lethal dune]

Alg closure

[south patrol]

Yee sure

[lethal dune]

got it?

[bitter shard]

can i make a question or is a discussion on going ?

[lethal dune]

you can ask

[lethal dune]

galois closure may not exist as K is not knows to be a spearable ext

[bitter shard]

thanks, ok so i want to show that for a r-homomorphism f from M to N which is surjective, if exist K submodule of M such that M = kerf + K (kerf intersection K = {0}) then f splits (exist g: N -> M such that gof=id)
what i have done until now is :
from second isomorphic theorem ... K isomorphic to K/kerf
from first isomorphic theorem ... K/kerf isomorphic to N
So K isomorphic to N
And if these are correct there is a h from N to K which is r-isomorphism
how i will define my g ? probably something trivial with h but i am blind right now

[lethal dune]

sorry K/kerf? you meant M/kerf?

[bitter shard]

no i think K/kerf is isomorphic to N

[lethal dune]

no that's not correct

[bitter shard]

oh

[lethal dune]

K is isomorphic to N

is what you'll get

infact K ∩ kerf ={0}

that's what you said as well

have you seen exact sequences and splitting of exact sequences?

[bitter shard]

yeah

but wait i got confused... is K isomorphic to N ?

[lethal dune]

0 → kerf → M= kerf ⊕K → N → 0

N ≅ M/kerf = (K ⊕ kerf)/kerf ≅ K

[bitter shard]

oh ok

and how the g will be ?

[lethal dune]

think a bit

[bitter shard]

btw cant we say this (K ⊕ kerf)/kerf = K/kerf ⊕ kerf/kerf ?

[hot lake]

ker f isn't a submodule of K so K/ker f makes no sense

[dreamy chasm]

can anyone help me understand this question? I know that the order of an element g is the smallest integer n such that g^n = e. and one element of N is the identity matrix and the other is rotation by 180 degrees

just not sure what the order of the given element of G/N might be. is it 2?

[bitter shard]

oh yeah you are right

[delicate orchid]

keep multiplying it by itself until you get N

[dreamy chasm]

what is "it"

[delicate orchid]

(0 -1)
(1 0) N, as that's what you're trying to find the order of

it is 2 though you are right

[dreamy chasm]

what I don't understand is that the identity element just keeps giving (0 -1) (1 0 ) which isn't in N

[delicate orchid]

yeah, that's a coset of N

[dreamy chasm]

but I noticed that multiplying it by (-1 0) (0 -1) twice gives the identity

[delicate orchid]

if that matrix was in N you'd just have the identity of G/N straight away

[bitter shard]

can i say i have a r-isomorphism h from N to K such that h(n)=k and an φ from K to M such that φ (k) = k1 + 0 (k1 in K and 0 in kerf) such that f(k1) = n. So g from N to M is g=φοh ?

[dreamy chasm]

so do we just need to test out different elements of N until the product is in N? that'll give the order?

[delicate orchid]

no?

[dreamy chasm]

I think I misunderstood you

[delicate orchid]

you need to multiply that element with itself until you get the identity, the number of times you do that (+1 ;3) is the order

by definition of order

[dreamy chasm]

right. I think I see what you mean now. we're taking the product of sets. my bad, I kept thinking about each element of N

[delicate orchid]

yeah think about elements of G/N instead

[lapis trail]

So I've been learning Galois groups. To try to explore the mechanics, I took the polynomial x^4-2 and found the roots z, -z, zi, -zi. I already know that the Galois group of these roots is D4 (square), but to see why I listed all possible permutations of those roots (which gives S4), then I started checking those permutations to see if applied to a generic splitting field element if it reproduces the splitting field. For instance, take X= {1,z,z^2,z^3,i,iz,iz^2,iz^3}, the splitting field being spanX. Acting on X by a permutation k in S4 will either give the whole field back (by which I mean k acting on X equals X), or it will destroy the field by "doubling up" some elements in X and removing others (in other words k applied to X is a proper subset of X). My hypothesis is that the permutations in S4 that preserve X are the ones in D4.

Tl;Dr the Galois group of x^4-2 is D4 and not S4. I'm trying to figure out through brute force why this is so. In other words, I'm trying to figure out what permutations in S4 will give a valid automorphism, which together compose D4. Hope it makes sense!

[formal ermine]

what

[lapis trail]

Am I barking up the wrong tree

[formal ermine]

your automorphisms are uniquely determined by what they send z or i to

[lapis trail]

I know but why is that

[formal ermine]

definition of an automorphism...?

[lapis trail]

Ok

[fleet pelican]

they are generators and an automorphism is a homomorphism

[formal ermine]

write a table for all possible automorphisms using this

[lapis trail]

But why can't I show using permutations on the roots directly is my question

[formal ermine]

who says you can't

[lapis trail]

I know what you're talking about I'm asking if the way I'm going about it, though bashy, is also correct

[formal ermine]

I have no idea what you're doing there

[bitter shard]

ahh i guess is not well defined cause f isnt injective

[wraith cargo]

you can
But I think you are forgetting that the automorphisms in the galois group have to keep the field you're extending fixed

Here you have like Gal(Q(4rt(2),zeta_4)/Q) so every element of this group has to keep Q fixed

[lapis trail]

Here's an example of a permutation of roots that I think does not belong, i.e., doesn't represent an automorphism

[lapis trail]

I think this is what I was trying to say

[formal ermine]

you can more easily see this by checking whether it is a homomorphism

-iz = phi(-1)phi(iz) = phi(-iz) = z

[wraith cargo]

Also note that $[\mathbb{Q}[i,\sqrt[4]{2}]:\mathbb{Q}]=8$ so it can't be $S_4$

[cloud walrus]

Irony Incarnate

[lapis trail]

Right I know that. I was just exploring how to derive D4 from automorphisms on the roots only and no other techniques

@wraith cargo I appreciate your help

[devout slate]

hello! I'm quite confused if all division rings are integral domain (or are there some cases that are not?). We know that integral domain is a nonzero commutative ring. But there are noncommutative division rings such as the one shown on the screenshot

[elder wave]

not all division rings are integral domains since, like you said there are non commutative ones like the quaternions but division rings are domains in the sense that they don't have zero divisors

this is kinda tricky because it depends on conventions and stuff

[formal ermine]

TIL domains have to be commutative lol

what do you call a non commutative domain

[elder wave]

integral domains do

[south patrol]

Domains don't

[formal ermine]

[elder wave]

yeah

that's why i said it's tricky

[south patrol]

In the context I've been reading up on (normed) division algebras needn't even be associative

Lol

[elder wave]

some authors use these interchangeably

[south patrol]

So yeah depends on context

[formal ermine]

just like integritatsring and integritatsbereich

[elder wave]

yes

awesome if you have two profs who both use integritätsring where for one it's necessarily commutative and for the other it isn't

🙂

[south patrol]

Die haben keine Integrität

[elder wave]

true

[chilly ocean]

how do I do this?

[summer path]

Use the following result

[chilly ocean]

mx + ny = 1

implies m and n are coprime

i need to show closure: gcd(a,b) = ax + by = 1

idk how to show the four axioms

[formal ermine]

I was gonna ask something

but I forgor what I was gonna ask

OH I remember again

nvm it answered itself

[lapis trail]

The group acting on the set of roots of the polynomial has a single orbit iff it is the Galois group, is that right?

[formal ermine]

uh

what restrictions are you putting on the polynomial

[lapis trail]

Yes or no question

[formal ermine]

no

orbits correspond to irreducible factors

[lapis trail]

All I needed

[ripe basalt]

Suppose that GL(n,R) has a continuous representation V (a continuous homomorphism from GL(n,R) to GL(V)) Show that the stabilizer of any vector v in V is a matrix Lie group. I'm not sure where to start this besides saying that the stabilizer subgroup is a closed subgroup of GL(n,R)

[ripe basalt]

it's a subgroup of GL(n,R) because it's the stabilizer

but i need a way to know that every sequence in the stabilizer converges to something in the stabilizer

maybe something about the stabilizer has to lie in a connected component of GL(n,R)?

is it true that all the matrices in the stabilizer will have determinant 1?

hm i guess not

[agile burrow]

Are you trying to show the stabilizer is closed?

[formal ermine]

what does the u represent

like for the field of fractions we have a similar relation but without the u in this case

which seems logical because then it's just what it means for two fractions to be equal

[fleet pelican]

It's for zero divisors.

[formal ermine]

huh?

[fleet pelican]

Do you know what a zero divisor is?

[formal ermine]

yes

[fleet pelican]

Good, it's for the case that a ring has those.

[formal ermine]

how? what?

[fleet pelican]

What is your question?

[formal ermine]

what does the u represent

[fleet pelican]

The zero divisor.

[formal ermine]

why not just at - bs = 0

then we'd get the obvious fraction

why do we add a u

[fleet pelican]

Because there might be a zero divisor.

[formal ermine]

I don't see the connection

[void cosmos]

suppose x is a zero divisor

--> there exists y such that xy=0

but then inside the localization u would get someething like x/1 = xy/y = 0/y=0/1

not cool

[tribal moss]

They're saying "a/s = b/t if at-bs is a zero divisor", not just "a/s = b/t if at-bs=0".

[formal ermine]

idk localization yet

this is the text leading up to localization

[formal ermine]

ah

[fleet pelican]

That relation IS localization

S is the complement of a prime ideal in general.

[formal ermine]

S is any multiplicatively closed subset here

[void cosmos]

the equivalence classes of the equivalence (a,s) ~ (b,x) iff u(ax-bs) =0 for some u is localizing with respect to S

ur basically inverting S

the field of fractions is when S is the whole ring except 0

so u get a field

cuz every element is now invertable

[tribal moss]

Well, not quite, only if it's a zero divisor with the other factor being in S.

[fleet pelican]

In particular, the complement of a prime ideal.

[formal ermine]

why is at - bs always a zero divisor

[tribal moss]

It isn't necessarily.

[fleet pelican]

Zero divisor is just a generalization of equaling 0

[tribal moss]

For random choices of a,b,s,t there's no particular reason to expect at-bs to be a zero divisor.

However, suppose we have a,b,s,t,u such that (at-bs)u = 0 and s,t,u all in S.
Then if we take a/b and s/t and multiply each of them by 1/u we get a/(bu) and s/(tu) which definitely ought to be the the same, since cross-multiplying and subtracting yields 0/(btu). But then we can't have both (a/(bu))·(u/1) = a/b and (s/(tu))·(u/1) = s/t -- unless we make sure a/b and s/t are the same from the beginning.

[ripe basalt]

pretty much

do you know why it's true?

[tribal moss]

If 0 notin S and you don't have zero divisors in your ring, "(at-bs)u = 0 for some u in S" is the same as "at=bs", and in that case it's easier to think of it the latter way.

[agile burrow]

Do you see how the map GL(n, R) -> V which maps g to gv is continuous?

[ripe basalt]

not really

[agile burrow]

Well as a continuous representation, you can view it as a continuous map G x V -> V

[ripe basalt]

like it makes sense that it is i guess

ok

[agile burrow]

Perfect, now what is the preimage of v under this map?

[ripe basalt]

the stabilizer of v and some of the orbit (which has v as image)

[agile burrow]

Oh I'm referring to this map

[ripe basalt]

it's still the stabilizer of v

all matrices A with Av=v

(A in GL(n,R))

[tribal moss]

Once you have fixed v, consider the map G -> V that tells you what the group action does on that v. That is continuous, right?

[ripe basalt]

i think so

all weve said is "it's continuous" so sure

[tribal moss]

And {v} is a closed subset of V, right?

[ripe basalt]

ah yes

so the preimage is closed

this would also imply that any sequence of matrices in the stabilizer converges to a matrix in the stabilizer too then right?

(or something non-invertible)

[tribal moss]

Yes, since a closed subset is sequentially closed.

If the sequence converges at all, of course.

[ripe basalt]

nice

lie theory is so weird

by continuous representation though it says that theres a cont homomorphism from GL(n,R) to GL(V)

but youre saying g*v which is an element of V, not GL(V)

i guess it's the same thing with some different wording though

like you can map g to whatever its image is in GL(V)

so if the homomorphism is p you get p(g) * v or something

thanks for the help though guys. i will think on it because i don't have it fully internalized but you got me what i needed to know

[formal ermine]

what are some local properties of rings? a+m only lists a couple of local properties for modules

[wraith cargo]

All the sources on localizations I've read never really talk about those
Maybe there's something in Eisenbud

[glossy crag]

I just saw this incredibly basic proof of "G finite subgroup of K* for a field K => G cyclic" and I just want to check there isn't some subtlety I'm missing:
Let |G|=n and p a prime dividing n. Then since x^{n/p}-1 has no more than n/p roots, there exists an element b_p in G with b_p^{n/p}\neq 1. Let a_p=b_p^{n/p^k} where k is the exponent of p in n, then a_p has order p^k, since a_p^{p^k}=b_p^n=1 and a_p^{p^{k-1}}=b_p^{n/p}\neq 1. Let a=\prod_{p|n} a_p, then a has order n (x,y of coprime order and commute => ord(xy)=ord(x)ord(y)) and we're done.
Makes me wonder why textbooks typically use the classification of finite abelian groups when such an easy proof exists.

[void cosmos]

yo

[shell brook]

NO AI MATHEMATICS

PROVE YOUR OWN RESULTS

READ A BOOK!!

[void cosmos]

yo the proof is correct if the ring has IBN right?

[shell brook]

i’m not reading the AI proof and i think no one else should either

[void cosmos]

its the standard one

dw

[bleak abyss]

So, chatgpt is obviously not reliable here so don't take the stuff wholesale

[void cosmos]

yea it was for fun

and also guess what was the above question

[bleak abyss]

But if you're just shooting the shit

[void cosmos]

u can guess by reading some of the text

( it answered it incorrectly too )

[bleak abyss]

If it answered incorrect then those 2 lines are underdetermined

[void cosmos]

it showed that something (called it F i wonder what did i mean by F ) was not a projective module

[bleak abyss]

Because nothing specific about F was mentioned other than the fact that it's not projective

[void cosmos]

but yet its incorrect

well mr detective

[bleak abyss]

So F can be anything that is projective

[void cosmos]

i used F to denote a free module next question

nvm

i didnot

xd

[bleak abyss]

I was gonna say

[void cosmos]

but yea the question was are free modules projective

[bleak abyss]

Lying mf

Equivalent characterization of projective is, direct summand of a free module

😛

[void cosmos]

wait really

col

[bleak abyss]

Yup

[void cosmos]

i am still at the beginning of the section

the diagram definition

and i can see why free implies projective easily

ig u would use the category definitino of free

and then u would have ur homomorphism from the free diagram

check this out

this was a problem i struggled with and two from here did too

maybe yyou could do it @bleak abyss

give an example of a ring that doesnt have IBN

[bleak abyss]

Lmfao

WHat's IBN again?

[solar inlet]

||limit of matrix rings as dimension goes to infity||

[void cosmos]

inside any free-Rmodule any two basis would have same cardinality

R is your ring

[bleak abyss]

Wait what

Can't you just like

[solar inlet]

||(embeddings are upper left corner)||

[bleak abyss]

Reduce mod a maximal ideal

Unless you mean noncommutative ring

[void cosmos]

no

commie rings

will always have that

i proved why up

so i always mean

noncomm rings

[bleak abyss]

I mean I also proved it just now

😛

[void cosmos]

hmmmmmmmmmmmmmmmm

[bleak abyss]

Like if M is an R-module and R is commutative, take a maximal ideal m

[void cosmos]

sounds mega complicated

lets try gpt's answer

[void cosmos]

yes thats how the proof goes

ur so good wow

[solar inlet]

oh wait silly example lol

zero ring

[void cosmos]

hmmmmmmmmmmmm

[bleak abyss]

uh you sure?

[solar inlet]

yes there is only one module over the zero ring

[bleak abyss]

And the only basis is empty set

[solar inlet]

well the zero module is free of dimension n for all n if you're silly enough with your definitions

[bleak abyss]

What

[solar inlet]

well actually id argue it's not silly at all

R^n = R

for all n

[bleak abyss]

No

That's not how you define it

[void cosmos]

ig we would have to resort to our trusty chtgpt

[bleak abyss]

Doesn't fly

[void cosmos]

ofc it knows the right answer

[bleak abyss]

Tru it did confirm to us that free modules aren't projective

[void cosmos]

yea exactly

[solar inlet]

the free module generated by the set S is always 0

[void cosmos]

its great with algebraic topology tho , a friend of mine told me that

it computes "K theory groups" (idkw hat are those) correct

[bleak abyss]

Free module of dimension n means a free module with an n-element basis

[void cosmos]

aktshwally dimension here is bad wording

[bleak abyss]

0 has a basis of n elements for exactly one n, namely n=0

[void cosmos]

it should be rank

dimension when ur talking about a module over a divsion ring, a vector fiedl

cuz u would have multiple dimensions and it would suck

but rank is better

[bleak abyss]

I'm saying commutative for now 😛

Rank is a bit more general

[void cosmos]

:angry

are all left notherian rings right noetherian rings?

[solar inlet]

I'm using the universal property definition

which is the definitively better definition

bc it's the one I'm using 🙂

[bleak abyss]

What even is that exactly

[void cosmos]

the free object in the category

i g

on a set X

[solar inlet]

yeah I think this is true

idr tbh

[void cosmos]

it is yea

given any module homo from X to S u can find a homo from F to S

commie diagram

[solar inlet]

I'd be fine discounting this example, but I want to mess with dami

[void cosmos]

i:X-->F the inclusion

[void cosmos]

u can miss with him but u cant miss with gpt

mess*

[bleak abyss]

I mean I'm just deciding whether to reject tox's definition or not

[void cosmos]

they are all equivalent

[bleak abyss]

That's the thing it's hard to mess with me when I reserve the right to simply discount what anyone's saying at any time

[void cosmos]

module has basis--> direct sum of R copies --> free object in the category --> go back

[solar inlet]

except in the zero ring ig lmfao

[void cosmos]

hmmmmmmmmmm

yo gpt is fire with combinatorics

[solar inlet]

there's more interesting examples in like Leavitt path algebras but these are pretty obscure and difficult to construct

[pastel cliff]

i cant help but feel a little dirty using gpt

besides maybe like

"explain ___ in laymans terms"

[solar inlet]

you should always feel dirty using chatgpt

[pastel cliff]

quick link to this, ignore

[lament dawn]

quit posting gpt in advanced channels

(or maybe anywhere)

[pastel cliff]

(also the fact that's it tends to be wrong lmao)

[summer path]

#chill

[pastel cliff]

can someone help decipher this

it has to do with the question i posted before

but i dont see how it carries over

ok i do see how it carries over obv not how

[thorn delta]

what's ur goal here? to write down a matrix expressing the tensor product of two matrices?

[pastel cliff]

this

maybe im just getting lost in the notation though

[thorn delta]

so, you can think of T_1 \otimes T_2 as a matrix with 4 columns

If $T = T_1 \otimes T_2$, then the matrix for $T$ is just $$[T(e_1 \otimes e_1) \quad T(e_2 \otimes e_1) \quad T(e_1 \otimes e_2) \quad T(e_2 \otimes e_2)]$$

[cloud walrus]

kxrider

[thorn delta]

(the matrix which has T(e_i \otimes e_j) as columns)

So for example,
\begin{align*}
T(1,0,0,0) &= T(e_1 \otimes e_1) = T_1e_1 \otimes T_2e_1 = e_1 \otimes (3e_1 + e_2) \ &= 3 e_1 \otimes e_1 + e_1 \otimes e_2 = (3,1,0,0)
\end{align*}

[cloud walrus]

kxrider

[thorn delta]

so the first column of T is (3,1,0,0)

rinse and repeat with the rest of your basis vectors

[pastel cliff]

this should be a square matrix though no?

oh wait uhhh

wait no yeah that should be a square

[thorn delta]

ye

4x4

[pastel cliff]

that's the square right

im lazy lol

[pastel cliff]

that second equality is what i was failing to see though

[thorn delta]

nope, each T(e1ej) is a column

wait which one?

[pastel cliff]

T(e1 \otimes e1) = T1e1 \otimes T2e1

[thorn delta]

that's the definition of T

T = T1 \otimes T2

[pastel cliff]

i mean more that like

the tensor product is T1e1 \otimes T2e1 repeated 3 times

for corresponding vectors ofc

[thorn delta]

hmm not sure what ur getting at, T1e1 \otimes T2e1 is computed once tho?

[pastel cliff]

it's okay i got that part

my only question i think then is like

hmmm let me think more and i'll phrase an actual question

[lethal dune]

Use the following definition of tensor instead:
A tensor is something that transforms like a tensor

[fleet pelican]

Tensors are tbh just kind of odd. The best construction is the universal property they satisfy, but that’s lowkey just abstract nonesense. Tensors just be tensoring.

[woeful sage]

oohh tensors, my favorite

I actually dont understand the general tensor product I only understand the tensor products between vector spaces

[rustic crown]

[round sandal]

Can someone help me simplify (C[x]/(x^2-x))_{x-1} = C[x]_{x-1}/(x^2-x)_{x-1}? I'm trying to localize C[x]/(x^2-x) at x-1 using that localization commutes with quotients, but can't get any further with the terms C[x]_{x-1} and (x^2-x)_{x-1}.

I think that (x^2-x)_{x-1} will just be (x), but C[x]_{x-1} is mysterious

[rustic crown]

yea you could do that and again use commutativity to get that's (C[x]/(x))_{x-1}
or just directly show C[x]/(x^2-x) = C x C and then (x-1) corresponds to the idempotent (1, 0) so inverting it will just kill off the second component in the product. giving you C.

[round sandal]

How do you use commutativity again here? If we do it first we get (C[x]/(x^2-x))_{x-1} = C[x]_{x-1}/(x^2-x)_{x-1} = C[x]_{x-1}/(x), but do we have C[x]_{x-1}/(x) = (C[x]/(x))_{x-1}?

[rustic crown]

umm i mean the principal ideal (x^2-x) in C[x]_{x-1} is the principal ideal in C[x] localized at (x-1) right. that's what you used earlier.

so (x) in C[x]_{x-1} is same as (x)_{x-1} when you think of (x) as an ideal in C[x] instead

[round sandal]

I'm a bit confused with this haha. I think I need to read more about this on atiyah mcdonald or some other comm alg book

[rustic crown]

okie lemme say it like this

you have an exact sequence
0 --> (x^2-x)C[x] --> C[x] --> C[x]/(x^2-x) --> 0

then you invert x-1

which gives you the exact sequence
0 --> xC[x,1/(x-1)] --> C[x, 1/x-1] --> what you want --> 0

but we can get to this sequence in a different way
0 --> xC[x] --> C[x] --> C --> 0
now invert x-1
0 --> xC[x, 1/x-1] --> C[x, 1/x-1] --> C[1/-1] --> 0
so what you want shoudl be same as C

[round sandal]

Thanks I'll try to digest this. Is the notation C[x, 1/x-1] equivalent to C[x]_{x-1}? I've seen this C[x, 1/x-1], C[x]_{x-1} and C[x][1/x-1] and wondering whether they all mean the same thing.

[rustic crown]

yep, the first and third are pretty much the same. the second can be confusing as when you localize at a prime ideal then you're actually inverting its complement

so when you localize at a multiplicative set, the common notation is S^-1A or A[S^-1] and when you localize at a prime p, it's A_p = A[(A \ p)^-1]

[glossy crag]

Anyone?

[rustic crown]

yea looks correct

there are many easy proofs lol

a second one uses that order is multiplicative when coprime thingy to say that there exists an element of order lcm(d, d') if there exists elements of order d and d'. now you have an element of maximal order d, so all elements satisfy x^d = 1, so d >= n and d is order of some element so d | n giving d <= n.

[glossy crag]

Neat.

[ivory trail]

whoever just posted and deleted that free group free product commutator subgroup thing, i was kind of attempting an answer, but it involves covering space theory

or actually F_2 * F_3 = F_5 and then you can proceed from there lol

and then if you just get a covering space you can show that commutators gives paths that don't end where they start

[broken stirrup]

it was me

its a problem from D.J. Robinson's book A course in group theory

[south patrol]

Lol i saw F_2 * F_3 = F_5 and was immediately like wtf

[broken stirrup]

and we are not supposed to use covering space i guess

[ivory trail]

ah yes the field product

[broken stirrup]

actually what I'm trying to show is that derived subgroup of PSL(2,Z) is a free group of rank 2

and it's isomorphic to free product Z_2*Z_3

[ivory trail]

oh not free groups, my bad

i misremembered

i honestly don't know how to deal with free products besides covering spaces so i can't help too much

[lethal dune]

[broken stirrup]

hello ryu sama

[lethal dune]

Who’s ryu

[broken stirrup]

no clue

[lethal dune]

No clue either

[broken stirrup]

let Z_2 = <a> and Z_3 = <b>, then appearantly derived subgroup of PSL(2,Z) is freely generated by x=abab^2 and y=ab^2ab

sorry i forgot to delete a section

[lethal dune]

[ivory trail]

this particular thing is actually an example in hatcher page 78

you can see how looking at this thing hard enough kind of solves your problem

[smoky ivy]

has anyone ever done reperesentations of compact groups

[summer path]

if you have a question, just ask it

[smoky ivy]

i don’t have a particular question, but ik it has a lot of applications in physics, diff geo, but is it also interesting for people that go into pure algebra subjects

[summer path]

i don't think i know enough about reps of compact groups to answer that :(

[south patrol]

I mean sure I've used them and many here will have

[smoky ivy]

ok so the thing is there’s a whole class just abt that, but idk if it’s worth taking it if i alr know the basics of it

idk if there’s much more i need to know abt it

[delicate orchid]

it's a very deep and far reaching area of study, if there's an entire class on it I'm sure they'll move past the basics

[south patrol]

Ig it depends on what you mean by the basics ig

Anyway wew knows much more about this than me lmao

[summer path]

If your entire class is on some topic, surely it will go deeper than the basics right?

[lethal dune]

how many did you solve?

[south patrol]

Oh those problems you sent?

All but two

[lethal dune]

nice

which 2

[sly storm]

Hello. Can anyone here explain in a simple and intuitive way and in detail why in the hyperbolic plane it is generally impossible, starting with an arbitrary circle of area less than 2\pi, to construct (by straightedge and compass) a quadrilateral with all sides and angles congruent having the same area, and vice versa, that is, starting with an arbitrary quadrilateral with all sides and angles congruent, it is impossible in general to construct a circle having the same area?

[ebon pine]

Let $G, H$ and $K$ be groups then, $G \times K \cong H \times K \implies G \cong H$?

[cloud walrus]

numbpy

[ebon pine]

I thought that this was true in general but apparently not. Can someone give a counterexample for it?

And more importantly an intuition for why this is not true in general (I think this is true for finite groups). In particular if phi is an isomorphism between (G x H) and (H x K) then why wouldn't a projection of phi, be an isomorphism between G and H?

[lethal dune]

not necessarily

[pastel cliff]

something something finitely generated abelian groups

[broken stirrup]

Let K= Z_p x Z_p x ... , G = Z_p, H=Z_p x Z_p

this should work

sorry, fixed it

[south patrol]

What I'm interested in is uh

If you take K= Z

I think this is true for abelian groups but not in general

[ebon pine]

Is K an infinite product?

[delicate orchid]

I think it's true for f.g. abelian groups

backpack's example is a counter example for non-finitely generated abelian groups

[ebon pine]

Is there no simpler counterexample say for non-abelian groups

[delicate orchid]

wait, this might be true for finitely generated groups

[south patrol]

No what I mean is like

The more obvious counterexamples come from taking K to be smth like that

What I mean is like

[broken stirrup]

yes

[south patrol]

"G x Z iso to H x Z => G iso to H" seems true for G,H abelian but not necessarily if they aren't abelian

Iirc this was a problem I was given

Which is kinda inch resting

[ebon pine]

I think I saw a proof in an expository paper by Conrad

[south patrol]

Nice

[ebon pine]

My bad, it was supplementary material of some course. Turns out it is called cancellation of finite groups

[delicate orchid]

I no longer believe this (drew some diagrams)

[pastel cliff]

can i get a hint for this pls

i think it makes sense

[south patrol]

Just try to do it and check out the system of equations lol

[pastel cliff]

okie i be back

need to refresh

[ebon pine]

[R. Hirshon, On Cancellation in Groups, The American Mathematical Monthly, Vol. 76, No. 9 (Nov., 1969), pp. 1037-1039] proves that finite groups are cancellable. In other words, that if G is a finite group and A and B are groups such that A×G is isomorphic to B×G, then A is in fact isomorphic to B.
Found this while searching

[delicate orchid]

yeah it's trivial for finite groups, and I thought I could extend that idea to finitely generated groups by using the fact that the direct product of two groups <X|R> <Y|S> is generated by the coproduct of X and Y and then doing some diagram nonsense utilising the fact X and Y are finite

didn't get anywhere though

wait, if the coproduct of two finite sets is isomorphic then the two sets must be isomorphic (as sets) right

hold on

[ebon pine]

what does isomorphic as sets mean?

[delicate orchid]

ok there's a slight wrinkle in that we need show the relators of G and H are the same but surely (I pray) that's just the subsitution test

[dusty verge]

same size

[wooden ember]

Yeah this is just counting

[dusty verge]

Since morphisms on sets are functions, an isomorphism is a function with an inverse, i.e. a bijection

[lethal dune]

[wooden ember]

As long as everything is finite

[dusty verge]

Wait, if the coproduct of two sets is isomorphic to what, exactly?

[wooden ember]

They mean if A u C = B u C then A=B

Which is true for A,B,C finite

(Not equality here, isomorphisms, and union is disjoint)

[dusty verge]

That's a blocky u, right?

Yeah okay

Nevermind, it doesn't hold for infinite sets

As a counter example, consider {1}uN vs {1, 2}uN. They're clearly bijective, but {1} isn't with {1, 2}

[solemn acorn]

hey anyone smart to help

i got a quiz rn

i just need this

[pastel cliff]

this isnt algebra

[dusty verge]

Also, don't cheat it's lame

[solemn acorn]

yh it isnt

[solemn acorn]

its physics

[pastel cliff]

then why post it here

#old-network

[glossy crag]

I feel stupid for asking, but how does p being prime imply ab>p? I know that if p is prime, then there can't exist ideals a,b>p such that ab<=p, but the negation of ab<=p is not ab>p.

[rotund aurora]

is this notation standard?

[dusty verge]

I've always seen it as GX

[pastel cliff]

ive seen it with the little circle arrow

[dusty verge]

I'm a little rusty on rings, what's the ordering here?

[south patrol]

Never seen that notation lol

Bit weird given colons have meanings already

[delicate orchid]

I've seen [G : X] exactly ONCE to mean the number of orbits

but never G : X

[delicate orchid]

I'm so close it's unreal

bit word-salady lol

[rotund aurora]

🥗

[lethal dune]

Where algebra

Don’t really see any easy way to see this. Any help?

[glossy crag]

Just regular containment, lol. > means it's proper

[smoky ivy]

i’m sure of that, but how interesting is it as an algebraist

or let me reformulate that: in what other fields of algebra would you need to know a deeper understanding of it

[dusty verge]

Are there any extra facts about A? I'm assuming it's not necessarily a pid

[delicate orchid]

vague question. Just try learning it and see if you like it rather worrying about "t"he "d"eep "p"hilosophical "r"amifications of le funny root systems

[smoky ivy]

hmm, alright

[glossy crag]

None.

[delicate orchid]

someone check this lol

excuse the fact that it's appallingly written I was typing fast

[lethal dune]

no

[dusty verge]

What is A?

[wraith cargo]

A prime ideal is such that if ab in P then either a in P or b in P
Since here both a and b are in P we can't have that ab is bigger than P
So hence we require that ab is in P

Also the negation of ab<=p is ab>p

[dusty verge]

Let (x) denote the ideal of Z generated by x. Then (2)(3) = (6)!<=(5), but also (5)!>(2)(3).

[delicate orchid]

yeah the ideals of a ring are (not generally) a totally ordered set, just partially ordered

[dusty verge]

^

Whenever I read totally ordered, I hear it in a valley girl accent

[wraith cargo]

oops yep true

but I think that the rest of what I said still holds

(a)(b)<p => (a)<p or (b)<p by definition
So if the RHS is true the LHS also has to be

[dusty verge]

I don't believe you

[wraith cargo]

?????

[dusty verge]

Okay, so we have ideals a and b, such that p is contained in a, and p is contained in b.

The ideal ab is the set {xy| x \in a, y \in b}. That means that to show p is contained in ab, we need to give sufficient argument that for z \in p, there exists x in a, and y in b, such that xy = z.

You're mistaking the contrapositive for the converse.

[delicate orchid]

it's the additive group generated by that set, but yes

[dusty verge]

Ah dammit, my in my head check said they were the same :(

[wraith cargo]

Wait ah yes OK I got it turned around in my head lol
I thought False => True has False truth value fuck

ok seems I am too tired for math today

[dusty verge]

that can't be possible. I feel like most of the math ever done was done by people who were really tired

[past temple]

is it true that for a given n, there exists only one partition of n that's equal to its transpose?

drawing out the young diagrams for n = 3, 4, 5 seems to suggest this is true, but i'm not sure how to prove it

[delicate orchid]

that's definitely not true

one mo lemme get ms paint up

[past temple]

tyty

[delicate orchid]

two partitions of 9 that are transpose invariant

[past temple]

ahh ok..

[delicate orchid]

there's a sequence somewhere one sec

https://oeis.org/A000700

[dusty verge]

Let's see, if you define a distance function that says how many times you have to move (in the above picture) squares to get from one partition to the other, then the distance between any partitions satisfying this will be at least two

This would definitely make it at least slightly unlikely for smaller numbers

[past temple]

i see

this definitely shakes up the strategy that i've been pursuing for my problem then lol

[dusty verge]

You can probably get better bounds to, since it seems like the distance would be at least three or more

[delicate orchid]

I also think the number of transpose invariant permutations has something to do with even/odd conjugacy classes of S_n

[dusty verge]

yeah it feels "parity"-y

[delicate orchid]

it's either #even conjugacy classes - #odd conjguacy classes or the other way around

I think

research time

[past temple]

right this is exactly what i'm trying to get at

so classifying the irreducibles of Sn isn't hard because the partitions of n give exactly its conjugacy classes

but it's a lot harder for An

[delicate orchid]

more over they give the uhh irreducibles directly

via le funny alternating/symmetric space right?

for A_n you can restrict irreducibles of S_n and since A_n is both index 2 and normal (realised that's redundent, lol) there is a boat load of nice theory for you to use

[past temple]

yeah ive constructed the irreducibles already

i just need to show that they're exhaustive, so this is purely combinatorial now

but i have absolutely no combinatorial background

[delicate orchid]

ah so you're finding the number of conjugacy classes?

[dusty verge]

Leave it as an exercise to the reader

[past temple]

well, more so trying to relate the conjugacy classes to young tableau

i guess i dont rly have a formula for the number of conjugacy classes either

[delicate orchid]

doesn't clifford's theorem guarantee that all irreducibles are consituents of a restriciton?

[past temple]

uhh that's probably a little too advanced for me

[past temple]

but do u think u could point me to some literature having to do with this proposed fact?

[delicate orchid]

I'd have to find where I read it from first

[rotund aurora]

how is it m to 1

tf discord dumb af

[past temple]

i'd appreciate it a lot haha

[delicate orchid]

it could be way simpler, one moment

[rotund aurora]

nothing is fixed by G_p besides p so no two elements can be in the same G_p-orbit

idk what I'm missing

[delicate orchid]

how uhh "accessible" is your formulation of the irreducibles? Specifically how well can you produce their dimensions

because we can use the fact that the sum of the dimensions of all the irreducibles is equal to the order of the group to show we have them all

and if that is tractable then maybe it's the way to go

[past temple]

hmm it's hard to say

they're all obtained by taking restrictions of the irreducibles of Sn

but sometimes those restrictions break into two irreducible summands

[delicate orchid]

yeah that will happen

[past temple]

in that case i'm not sure what the dimensions are

[delicate orchid]

when exactly that happens is clifford theory which we want to avoid

[runic hemlock]

Take an element x in the ring A. The multiplication by x map A->A is given by some integer matrix (by choosing a basis for A over the integers). Now take the characteristic polynomial of this matrix

[delicate orchid]

I've found a formula for the number of conjugacy classes of A_n in terms of a generating function

https://oeis.org/A000702/a000702.pdf

old ass paper

[past temple]

hmm ok

yeah i really dont have much information about the dimensions

i have some information about multiplicity though

namely, every restriction of an irrep corresponding to a transpose invariant representation splits into exactly two irreducible summands

[delicate orchid]

oooooo

that's nifty

[past temple]

i feel that if we take N = # partitions of n a priori then no sophisticated combinatorics is required

but even counting the number of conjugacy classes of An from N is weird

[delicate orchid]

this is what I'd expect, there's a vague correspondence between conjugacy classes that split and irreducibles that split

[past temple]

for conjugacy classes, they split when the cycle type consists of distinct odd integers

[rotund aurora]

wait I was speaking complete nonsense

[delicate orchid]

is that true? (12345) does split in A_5 and that has cycle type (5,1,1)

or this implication only one direction

[past temple]

hmm this is just a fact i found on wikipedia, hold on

https://planetmath.org/conjugacyinan

it's stated here

[delicate orchid]

perhaps I should have read this more in depth

[past temple]

wait what?

the cycle type is 5 + 0 + 0

[delicate orchid]

oh yeha DUH ahhahaha

ok I believe you now

so the number of conjugacy classes is #number of even partitions + 2*#paritions of distinct odd numbers

[past temple]

well not quite

i think you're double counting there

you take the subset of even partitions that consist of distinct odd integers, multiply that by 2

[delicate orchid]

I've also found this, Prop 12.17 in Reps and characters of groups by Liebeck, James:
"the conjugacy class of x in S_n only splits on restriction to A_n if x does not commute with any odd permutations in S_n"

[past temple]

then add the rest of the number of even partitiosn

[past temple]

yeah this is how this splitting is obtained to begin with

the splitting if distinct odd integers that is

[delicate orchid]

right ok so it doesn't help... darn

[lethal dune]

[delicate orchid]

the fact the ONLY formula I've found is a generating function does NOT bode well

oh wait lol

the splitting will be into two characters of equal dimension I think

again, by clifford's theorem

it's a sum of two conjugate (or one, in which case it's irreducible and we DO NOT CARE) characters by clifford's theorem, which have equal dimension as $\chi^g(1) := \chi(g1g^{-1}) = \chi(1)$

[cloud walrus]

Wew

[past temple]

ok so when they split they're of equal dimension, cool

[delicate orchid]

this should help you do this, if you chose to take that path

I gtg now but I wish u luck

[past temple]

ty for the help

ohhhhhhh i think i got it

[white oxide]

can somebody give me a hint for this question?

[void cosmos]

do it directly

[glossy crag]

It sounds like you've got it all turned around.

[glossy crag]

Firstly a and b contain P, are not inside P.

[glossy crag]

Secondly, the negation of A<=B is not A>B, it's A\cap B^c is non-empty (here <= means \subset, not \leq).

[white oxide]

oh wait i can just divide both sides of the equality a0 + a1x + ... anx^n by a^n oops

[kind temple]

simplify f(1/x)

[glossy crag]

Anyone?

[lethal dune]

what does> mean here?

[white oxide]

why can we not have both of the degrees of f(x) and g(x) less than the degree of p(x)? for instance, if p(x) is of degree n, why can't we have deg(f(x)) = n - 2 and deg(g(x)) = 2?

by the degree rule

[silent oxide]

What can we say about the ideal generated by f(x) if deg f < deg p?

[lethal dune]

because one of them is a multiple of p(x)

[white oxide]

ohh ok thanks got it

[glossy crag]

proper containtment

[rotund aurora]

but $a>p$ means $a\subset p$ no?

[cloud walrus]

Croqueta

[glossy crag]

Let me check the notation they use, but unless I'm taking crazy pills it should mean $a\supset p$ and $a\neq p$.

[cloud walrus]

Ocean Man

[rotund aurora]

So you are asking $a\supset p$ and $b\supset p$ implies $ab\supset p$

[cloud walrus]

Croqueta