groups-rings-fields

[night onyx]

and rings/algebras are quotients of tensor products

[alpine island]

I ended up in it in high school because I wanted to avoid linear algebra. Turned out I needed linear algebra anyways

[delicate orchid]

linear algebra is some of the nicest maths there is

[alpine island]

idk, it was fun, not widely supported enough to be useful though

If I ever get into game design / 3d modeling I might try it again

[night onyx]

ya I'm a Unity developer and linear algebra is 95% of the math I use

[alpine island]

I think the idea is you replace the lin alg with PGA formulas

and afaik it is pretty clean

but the problem is most lin alg has well-optimized algorithms due to its heavy use

PGA doesn't

wdym by this, out of curiosity? I know tensors as "multidimensional arrays", where 1d tensors are vectors and 2d are matrices

but not much more than that

[night onyx]

so if you think of a matrix ring, you can obviously take products, like A,B -> AB

in a tensor product of two vector spaces, you can take "products" but it's not an element of the same type, it's literally an "abstract product" of the two things, like v, w -> v \tensor w

[delicate orchid]

they don't need to be vector spaces! which is the cool part, they just have to be in modules

[alpine island]

Ah so like the symbolic addition 1+i in C, you're not actually evaluating that

[delicate orchid]

and you have nice properties like a(v tensor w) = (av) tensor w = v tensor (aw) and stuff

[night onyx]

yeah exactly, basically any sort of algebraic structure you want to "multiply" you can do so, but it's not a real multiplication, it's almost just a symbol for "v times w"

but then like I mentioned before, if you want to impose relations on that multiplication, you take a quotient

lol don't worry too much about it, there are much more intuitive ways of understanding tensors, but really it's like "a symbol for a product"

[delicate orchid]

yeah like factoring multilinear maps through it to get a linear one and stuff

[alpine island]

I don't know if this applies, but could you give a construction of say, R_2x2 from tensors?

of 2x2 matrices

[delicate orchid]

sure, a matrix over a vector space(?) is an element of V \otimes V* where V* is the dual space

now let me remember why

oh yeah

[alpine island]

dual is set of functions V -> K where K is the underlying field, yes?

[delicate orchid]

yeah exactly

each row of a matrix acts exactly like one of them

[alpine island]

Yeah

[delicate orchid]

I forgot how it goes... fuck

[alpine island]

In the case of vectors they would just be... sideways vectors

[delicate orchid]

I think you have to view it as blinears factoring into linears

which isn't useful for intuition

@south patrol do you remember lol

[south patrol]

V* (x) W is iso to End(V,W) by sending f tensor w -> ( v -> f(v)w)

It is fairly clearly injective and hence an isomorphism by dimension reasons

[delicate orchid]

god damn it that's what I was typing
"f : VxV* -> V given by f(x, g) = g(Tx), which is now a linear map" or something along those lines

[south patrol]

That is true

[alpine island]

I found this, but it kinda makes it seem like all matrices here would be linearly dependent

[south patrol]

That's for a specific construction of tensor prod ig

[delicate orchid]

yeah that's using the fact that, for vector spaces U, V (or free modules) with bases B_U, B_V, the basis of the tensor product V tensor U is {v tensor u : v \in B_V, U \in B_U}

[alpine island]

But like for a simple example V = W = R^2, then (a, b) tensor (c, d) = [[ac, bc], [ad, bd]]?

[south patrol]

Not sure what you mean by that

[delicate orchid]

uhhh

right hmm

[delicate orchid]

I think this is right, yeah

in general the resulting matrix x \otimes f = M would be of the form M_{ij} = f_j(e_i)

[alpine island]

Great, me neither :)

[alpine island]

Yeah that's what I ended up with, just seems odd that these are all linearly dependent

[delicate orchid]

why are they linearly dependent

N = 2M are clearly two linearly dependant matrices

[alpine island]

Sorry, the rows/columns are linearly dependent

[delicate orchid]

I don't think so

[alpine island]

Like [[1, 2], [3, 4]] couldn't be written as a tensor product

but [[1, 1], [2, 2]] could

[delicate orchid]

good point

this shouldn't be the case

[alpine island]

(1, 1) tensor (1, 2)

[delicate orchid]

oh yeah duh

they're sums of elementary tensors

not all tensors are of the form x tensor y

those are just the basis elements

[alpine island]

Ah, okay

that makes sense

So R_{2x2} = span(R^2 tensor R^2)

For a certain definition of span

[delicate orchid]

if you mean taking R linear combinations then yeah

[white oxide]

why is the codomain GL_n(C)? inputting g spits out a composition of linear maps, not matrices

or am i tripping

[delicate orchid]

what do you think a matrix is

[white oxide]

well a linear transformation isn't explicitly a matrix

[delicate orchid]

right you're one of them

WHO CARES?!?!?!

[white oxide]

if so then the notion of the set of all linear maps from V to W being isomorphic to the set of all m x n matrices would be useless

fine lol

[delicate orchid]

GL(V) is the group of invertible linear maps anyway

you're not given a basis with which to define matrices explicitly

but here, we are

line 3 2nd paragraph

[delicate orchid]

oh nevermind I just got the context

this is like the one time you do actually give a shit

of course it is

[south patrol]

Lol

No so like the point here is that we are choosing a basis, using that to identify V with C^n for some n

Then under that correspondence, a map into GL(V) corresponds to a map into GL_n(C)

[white oxide]

ohh that makes more sense thank you

[south patrol]

Oh okay I see

Well do I, one sec lol

Technically they are still abusing stuff a bit by identifying GL(C^n) with GL_n(C)

[long nebula]

🤫

[south patrol]

But that can be done canonically anyway

[white oxide]

yea that was what prolly fucked me up

[south patrol]

Like basically this is saying like

Take a basis of V and send g to the matrix of φ(g) with respect to that basis

That is the key takeaway anyway lol

[white oxide]

Ok I think that helps

Hopefully by the tenth reading I’ll understand it

[glossy crag]

Is this correctly formulated?

[delicate orchid]

Yeah this looks good

[south patrol]

Is there any nice expression for the characteristic of the ring W(A) of p-typical Witt vectors? If p is invertible in A then this is easy (same as A) but otherwise I'm unsure

I conjecture that if A has char p then W(A) has characteristic 0 from what I've seen though

Eh okay if A has characteristic p then we know that p.ais non-zero in W(A) if a is, so characteristic isn't a power of p, and further if n.1 = 0 in W(A) then n.1 = 0 in A (by considering the map W(A) -> A^N) and these two constrain W(A) to be characteristic 0 I think in that case

[south patrol]

I think you can make this a bit simpler though? Like considering
the projection maps and factoring through ought to allow you more easily to use the induction hypothesis

Like the important thing being that the isomorphism R1 \simeq S1 you have isn't just a random iso, but actually inducd by the isomorphism of products

[prisma ibex]

Yup

[south patrol]

Is my argument similar enough to what you'd give?

Lol was reading Bourbaki and couldn't find this actually stated

I wonder what the non char p case is like tbh hm

[prisma ibex]

essentially yes. There's a universal property that Witt vectors satisfy which is that they are the universal deformation from char p to char 0 with a lift of Frobenius

[prisma ibex]

if p is invertible in R then W(R)=R^N

[south patrol]

That's what I meant by this aha

I mean cases where perhaps A has p nonzero but not invertible

[prisma ibex]

ahh hmm

[south patrol]

Perhaps this just isn't of interest and I shouldn't worry lol

But idk

[prisma ibex]

yeah I mean generally speaking this case doesn't really come up

the main use of Witt vectors is to deform from char p to char 0

[south patrol]

Sure yeah it seems mostly for char p perfect fields right lol

Thank

[lusty marlin]

Is there a characteristization of all rings that contain a subring that is a field?

[lusty marlin]

Or some weaker results like necessary or sufficient conditions for a ring to contain a field as a subring.

[formal ermine]

F_1

[lusty marlin]

What's F_1?

Please don't tell me this is some 'field with one element' nonsense

[formal ermine]

if it's char p you can look at the prime subfield

[delicate bloom]

or just the prime subfield in general, no char restriction

[hidden haven]

If there is a subfield, there is a prime subfield

[lusty marlin]

Right, but my question is about whether any subfield exists at all.

[delicate bloom]

does your ring have 1

[hidden haven]

You check whether there is a prime subfield, which is easier

[lusty marlin]

Ok I didn't realise it would be easier. Why is this the case? Thanks.

[lusty marlin]

Not necessarily.

[hidden haven]

Then this is not equivalent

Nvm

[lusty marlin]

Ok, but for rings with unity it is, right?

[hidden haven]

Ye

[lusty marlin]

So how is checking for a prime subfield easier?

[hidden haven]

Then you just have to check whether all the non-zero integers are invertible or not

So you have a concrete set of elements to check

[lusty marlin]

non-zero integers?

[hidden haven]

Elements of the form 1+...+1 and their inverses

[lusty marlin]

Ah. I didn't know those were called integers in rings. Thanks

[hidden haven]

Where subfield is assumed to be a unital subfield i.e. must have the same 1

[lusty marlin]

Right

So in a ring with unity, if all non-zero integers are invertible, then the ring contains a subfield?

[hidden haven]

Yes, the integers and their inverses will generate a prime subfield

[lusty marlin]

Hmm, right.

[formal ermine]

wait but what field is inside e.g. Z

it has char 0

[hidden haven]

Prime subfields also include Q, in case that is the confusion

That is the smallest char 0 field and is contained uniquely in any char 0 field

[formal ermine]

they were asking for fields contained in rings, not fields though

[hidden haven]

Ye so you check whether the ring contains a prime field

[orchid iron]

Hi, people..
I have a doubt here...please resolve

Q.} Why is the bilinear form being non-degenerate,makes the map injective......see the last proposition please

[coral spindle]

If y =/= 0 then B'_y =/= 0

In other words, the kernel of the map y |-> B'_y... is {0}.

Does that solve it for you?

[orchid iron]

Yeah if kernel is trivial the map is injective.....

Thank you thank you bro/sis very much,i see it now

[glossy crag]

I want to show that if V is an infinite-dimensional vector space over a division ring D, then End_D(V) is not a left artinian ring, does this work:
Since V is not finite-dimensional there is a countable set of independent vectors {x_n}. Take V_0=0, V_1=span(x_1), V_2=span(x_1,x_2) and so on. Define I_n as the set of endomorphisms whose kernel contains V_n. These are left ideals of End_D(V) and form a descending chain End_D(V)=I_0\supset I_1\supset\cdots. The inclusions are proper: for any n take the projector p onto Dx_n+1, then p(x_1)=\cdots=p(x_n)=0, but p(x_n+1)\neq0, so p\in I_n, but p\notin I_n+1. Thus the chain never stabilises and End_D(V) is not a left artinian ring.

[lethal dune]

sounds logical but I haven't checked the details

one thing is End becomes an infinite dimensional vs over D so can't be noetherian as D module

[glossy crag]

the question was about artinianness

[lethal dune]

Over division ring they are equivalent (I think)

Your argument holds nonetheless

[glossy crag]

but are D-subspaces of End_D(V) and left ideals equivalent? i'm not sure

[lethal dune]

You want artinian as D module or as ring?

[glossy crag]

i meant as a ring, should've specified that perhaps

[lethal dune]

They aren’t equivalent, my argument is as D module

[glossy crag]

but my argument works for rings, right

[lethal dune]

Yes

Here's something else you can try as well, try to show the ring is not Noetherian and that'll show it's not Artinian as all Artinian rings are Noetherian.

[rocky cloak]

Is there any step of the argument you're unsure about, or are you just doing a sanity check?

Seems like a perfectly good argument to me.

[slim kayak]

I have to show that the map from the HNN extension A * (phi_1, C , phi_2) with A = stab(alpha(e)) [starting vertix] and C = stab(e) for a specific edge e and phi_1 = id and phi_2 (c)= s^-1 c s into G, which acts without inversion on a tree, is injective.

Someone familiar with their Bass-Serre theory/Cayley graphs etc. have some tips?

[glossy crag]

sanity check

Another sanity check (tensor products of vector spaces, U\otimes T meant to embed naturally into V\otimes W).

[ebon heron]

IF YOU NEED HELP WITH ASSIGNMENTS HIT MY DM

[delicate orchid]

lmfao

alright buster

[glossy crag]

This might be splitting hairs, but shouldn't this be an isomorphism, strictly-speaking?

[coral spindle]

Yeah

[formal ermine]

,, = = \iso

[chilly ocean]

so true illum

[lethal dune]

Not true

[barren sierra]

Let em cook

[white oxide]

halp

im stuck on showing that every integral domain that is artinian is a field

need hint

lost

[coral spindle]

Let x be a nonzero element. Consider the ideal generated by x. What's a natural chain you could build with this

[white oxide]

yea i was considering that

hm i tried powers of x

[coral spindle]

OK so what does that tell us

We know that <x^n> = <x^{n+1}> eventually

So can we turn that into a statement about elements in the ring now?

Remember in a commutative ring R, the ideal <x> is equal to xR.

C'mon okeyokey my children are starving!!!!!!!!!!

[white oxide]

go feed your kids lol i'm too slow to get it instantly

don't starve the children

starving children is bad

i wrote a^nr = a^n+1

cmon i got this

[coral spindle]

they're picky eaters! They will only eat solutions to this problem !!!!!

[coral spindle]

OK try the other way. We already know there is an element r such that x^n r = x^n+1, namely r=x, so this tells us nothing.

[white oxide]

sorry went to go take a piss

ok ill try that

[coral spindle]

🧿👄🧿 they're so hungry, okey

[white oxide]

ah

$a^n(ar) = a^n$

[cloud walrus]

okeyokay

[white oxide]

implies an = 1

[coral spindle]

[white oxide]

oops

ar = 1

[coral spindle]

Let's gooooooo

[white oxide]

WWWWWWW

thank youuuuu

gg

feed your children now

[coral spindle]

My children eat well tonight 🙏

[white oxide]

glad to know i'm responsible for that

[coral spindle]

See that wasn't so bad eh

[white oxide]

true true but let's be honest you did most of the heavy lifting

[coral spindle]

Your name is weird to say in a sentence so I'm going to call you keyo

because I decided to

Also no

:(

[white oxide]

yes >:)

keyo

interesting

i like that

rhymes with mayo

nvm

[coral spindle]

sounds like K.O.

[white oxide]

but it's pronounced o-kee o-kay

:(

[coral spindle]

O.K.

[white oxide]

if R is a division ring but not a field and A is a R-module, is there any reason why we refer to A as a vector space?

because i thought vector spaces are always over fields

i'm being picky but just curious

[coral spindle]

Yeah honestly I would never call it a vector space personally

But fwiw, they have bases just as vector spaces do

I.e. every R-module is free if R is a division ring

[hidden haven]

That makes them resemble vector spaces way more than they resemble modules so seems like a fair name

[white oxide]

hm ok i see

[white oxide]

bro hungerford trying to be different writing n in N* instead of n in Z+ 😹

😹

[cloud walrus]

pikachupikachu

[white oxide]

nah bro's trying to seem COOL

[coral spindle]

But Z^\times = {+1, -1}

Wait wrong thing

[chilly ocean]

-1

[delicate orchid]

-1

[coral spindle]

If you interpret N as a monoid under multiplication 🤓 then the group of units is just {1}

so either way

Fucked up to write N^\times

[ashen heron]

i still like it tho

[open sluice]

$\mathbb{N} = {1, 2, 3, \dots}$

[cloud walrus]

bladewood

[open sluice]

😎

[ashen heron]

0 in N

[white oxide]

0 in N

[coral spindle]

If you don't stop sharing opinions on 0 I will be forced to algebrapost

[ashen heron]

wouldn't it hurt your feelings if i write "f abstract algebra" in my about me too

[white oxide]

yes

fuck analysis tho

[ashen heron]

so...

[summer path]

Algechillpost?

[barren sierra]

0 in N

[ashen heron]

blocked

[chilly ocean]

low quality channel

[barren sierra]

#adv-algebra when

[coral spindle]

Repeat after me: #groups-rings-fields is not #chill

[summer path]

Or #1100503863586455632 I guess

[warped fulcrum]

I’m posting this in both #groups-rings-fields and #elementary-number-theory as my question relates to both.

does anyone know of a short pdf document (or perhaps an appendix of some book) that sums up all the number theory ‘facts’ (theorems and definitions) needed for algebra? Preferably it should include proofs, but I’d be fine with accepting many ‘obvious’ statements without proof like the division algorithm and whatnot.

[south patrol]

what does "needed for algebra" mean

You don't need any to get started on algebra

[celest furnace]

The two algebra textbooks I’ve used (Dummit and Foote and T.W. Hungerford abstract algebra) has like a chapter in the beginning cover everything you might need to know

But also try not to double post as it is against the rules

[warped fulcrum]

I am going through Aluffi and he seems to assume a lot of number theory background in the exercises (and doesn’t have an appendix on it or anything of the sort). I can do most of the exercises, but I struggle when it comes to ones that involve number theory in a direct way

[warped fulcrum]

thanks, will check them out

[celest furnace]

What sort of exercises? It might be helpful if you could post one here

[long nebula]

The only number theory you need to learn abstract algebra is to understand what divisibility is, what prime factorization is, and how the division algorithm works tbh

I don't recall really knowing any number theory before starting abstract algebra

(And I still don't!)

[south patrol]

Lol I learnt that stuff in the context of algebra basically

Idk why you'd need it in algebra usually

Hm

[warped fulcrum]

1.14 is an exercise I just went back to (from months ago, currently I’m at chapter 3) after having stumbled upon Bezout’s identity, and now I have been finally able to solve it. I constantly feel I’m missing little facts from number theory and what I’m operating with is not enough.

[long nebula]

Lagrange's theorem for the first one, classifying finite groups for the second one, learning about Euclidean domains or something for the third one

Oh that's true, Bezout's identity is kind of useful

But it's just a result of the Euclidean algorithm

Which you should know

[coral spindle]

Then it seems you should read an intro book on number theory

[warped fulcrum]

tbh I don’t find number theory all that interesting so I don’t want to do that haha… i just want to gather an arsenal of number theory facts

[coral spindle]

Bézout and the CRT are probably the most useful results in elementary number theory in application to basic algebra

[long nebula]

^

[coral spindle]

So I guess read up on those.

[long nebula]

Yeah I don't think you need a ton of number theory apart from that

[coral spindle]

Maybe if you read it you'd find it interesting I hope it doesn't disappoint you, but a LOT of abstract algebra is just number theory

[south patrol]

Lagrange doesn't need number theory right lol

[warped fulcrum]

PIDs, Euclidean domains, and whatnot, right? I’m planning on skipping these parts of Aluffi as well

[south patrol]

and we proved bezout's identity within my group theory course

Oop

[long nebula]

No I just said to understand what divisibility is

Lol

[warped fulcrum]

sort of just wanna get into the exciting linear algebra and hom algebra at the end

[south patrol]

Oh sure

lol

That isn't so much a number theoretic thing i guess but sure lol

[long nebula]

exciting

hom algebra

[coral spindle]

Yeah this is a funny thing to say

[long nebula]

you terrify me

[south patrol]

Hom alg is fun

[coral spindle]

You might as well have said that you don't plan on doing much commutative algebra

[warped fulcrum]

commutative algebra requires number theory?

[south patrol]

other way round mostly lol

[coral spindle]

It certainly helps lmao

[warped fulcrum]

i mostly found drawing commutative diagrams and thinking of ways to stick them together fun

defining things like pullback and whatnot

that’s why I’m reading aluffi

i don’t suppose I’ll need that much number theory for that sort of thing

i could just skip these exercises (the body of the text doesn’t have much number theory in it) but then I’d feel a bit guilty, so I guess I’m going to look at the sources you guys recommended. Thanks!

[paper aurora]

why is this true? i think its something related to group theory and polynomials that i forgot . m is non-negative integer here

i mean the 0 part, not the 1 part

[coral spindle]

If you multiply by e^{2*pi*i*m/d}, then the sum remains the same, right?

And certainly this multiplier is not equal to 1

Therefore, the sum must be zero, since only zero remains the same after multiplication by a number other than 1.

[paper aurora]

wow i thoguht it had something to do with something being a root of 1+x+...+x^{d-1}

[coral spindle]

uhuhuhhhhh sure that works too

Yeah that does work

But I think my way is more swag.

[paper aurora]

yeah yours is more swag and i dont even see how to do the approach i mentioned

[coral spindle]

Well I mean e^{whatever/d} is a dth root of unity right

and if d doesn't divide m, then it's not 1

so it must satisfy the polynomial you mention. Donezo!

[paper aurora]

oh right you factor x^d - 1

[coral spindle]

Uh-huh

[paper aurora]

but this is cool i feel like i learned a new trick 😎

[coral spindle]

Tbh this is something I have been using a whole lot in my own research

Bc I am summing a lot of roots of unity

so this is my instinct immediately

[delicate orchid]

sums of roots of unity gang

[coral spindle]

gang

[south patrol]

Representation theory

[coral spindle]

Me when I compute a Frobenius–Schur indicator after doing approximately 16 billion sums

"Wow it was 1 nice"

[south patrol]

Actually this happens to generalise lol

Well

[coral spindle]

It generalises a whole lot. It's really wonderful.

[south patrol]

Well you can view this as like

Oh wait nvm lol

the (mb/d) is weird lol

Actually like why is the b there lol

Is that a typo? otherwise we can take b = d and break it lol

[coral spindle]

To slip into character brain for a bit...

No the b is the index

[south patrol]

Oh I misread it

Lol

[formal ermine]

what are you researching

[coral spindle]

So for example this could be seen as the inner product of a certain character of Z/dZ with itself

[coral spindle]

groump... .theory... :D fsrepresentaion.........s

If I didn't know better I'd tell you I research tedious calculations

[formal ermine]

iSnT tHaT a sOlVeD fiELd

[coral spindle]

Well ok let me tell you more exactly.

[south patrol]

I mean my way of thinking of it was like let C_d act on R^2 by multiplication by e^{2pi i m / d} and this is computing the # of occurneces of the trivial rep in that lol

[coral spindle]

My supervisor wants me to verify a conjecture for a handful of finite groups. Nice ones.

[coral spindle]

Yeah I was gonna say

[south patrol]

which ig is exactly what you meant right

[coral spindle]

Uh-huh

[formal ermine]

can't you automate that using a computer program

[south patrol]

I was gonna say as well that it links to the nice theorem that uh

[coral spindle]

Yeah I just didn't mean to say "with itself" that was way off

[coral spindle]

No, it's an infinite handful. And we would like to know why

[south patrol]

if the average of n roots of unity is an algebraic integer then they're all the same or the sum is 0 lol

[formal ermine]

what is the conjecture

[coral spindle]

Well let me tell you

The McKay conjecture says that if we take a finite group G and a Sylow p-subgroup S, then the characters chi in Irr(N_G(S)) such that p does not divide chi(1) are in bijection with characters with the same property in Irr(G)

This is not the conjecture

[formal ermine]

what's Irr(...)?

[coral spindle]

Irreducible complex characters

[formal ermine]

ah

[coral spindle]

The McKay–Isaacs–Navarro conjecture says that the McKay conjecture holds, and furthermore this correspondence is equivariant with respect to the absolute Galois group of bar Q_p over Q_p

This is not the conjecture

The McKay–Isaacs–Navarro–Turull conjecture says that the McKay–Isaacs–Navarro conjecture holds, and furthermore the correspondence preserves the p-adic Schur indices of the characters.

[formal ermine]

equivariant?

[coral spindle]

g . f(x) = f(g . x).

[formal ermine]

ah

[coral spindle]

This is the conjecture that I'm looking at

[formal ermine]

tbh this one looks like it could be a problem on a homework sheet lol

[coral spindle]

Sure does

But it's not proven yet.

[formal ermine]

definitely sounds interesting

[south patrol]

Saw sylow subgroups today in alg top

Bruh

[coral spindle]

Some homology stuff?

[formal ermine]

oh

mckay died last year

rip

[coral spindle]

Indeed

[south patrol]

Cohomology but yes sure

[coral spindle]

potato otatop

[south patrol]

It was like uh

Changing a G-bundle to an H-bundle for H a sylow subgroup of G lol

[formal ermine]

powotato

[tough raven]

Let K/F be a field extension and V a vector space over K.
What is the connection between the exterior powers Λ^k V over K and over F? Are there natural maps between them in either direction?

Actually, the same question for tensor powers.
But in that case IG the tensor product over K is a quotient of the tensor product over F (quotienting to force the tensor product of vectors to be K-linear).

Maybe the same is true here?

https://kconrad.math.uconn.edu/blurbs/linmultialg/extmodbaseextn.pdf
Almost supiciously convenient.

Hmm. If Λ^n V = 0, for V a module over a ring R, does that imply that V is spanned by some n elements?

[coral spindle]

Well no, because if we just take R to be a vector space field, it means that it is of dimension strictly less than n

But I don't know how it might turn out for a general ring immediately.

[rustic crown]

isn't Λ^2 Q = 0?
each (a/b)Λ(p/q) = apbq(1/bq)Λ(1/bq) = 0

[coral spindle]

Yees

[rustic crown]

(and Q is definitely not f.g.)

[coral spindle]

I did make a typo

So clearly for R = Z the same statement doesnt hold. Nice one

[rustic crown]

[delicate orchid]

holy shit someone else has heard of the McKay conjecture

[coral spindle]

I'm forced to

[delicate orchid]

it's like, where my research is heading

[coral spindle]

Bruh

wew

are we the same person

[delicate orchid]

it's ok boytjie, because I lied

[coral spindle]

NOoooooo

[delicate orchid]

I'm studying the modified version "The Alperin-McKay" conjecture

which is for fusion systems

well, associated more with fusion systems

it's the block version

[coral spindle]

Ah okay well I can only imagine that it's a strictly stronger statement

[delicate orchid]

it's actually incredibly similar

[coral spindle]

Well Turull's version is cooler because Schur indices are there

and that's cool

so take that

[delicate orchid]

oh woowwwww I'm really impressed

higher schur indicator functions my beloved

[coral spindle]

Oh is that some Hopf algebra stuff?

[white oxide]

how does this equality hold? wouldn't that imply that ni = nivi

[delicate orchid]

no clue! I just like making the funny number bigger

[coral spindle]

I have obviously been sweating my arse off trying to use the Frobenius–Schur indicator well

[delicate orchid]

you literally take the g^2 in the schur indicator and turn it into a g^m for whatever m you please

[coral spindle]

Whassat do then

[delicate orchid]

the m = 1 one case is especially useful, it's non-zero if and only if your representation is trivial!! How handy!!

[coral spindle]

They're using different delimiters: [] vs {}. It's unclear to me what the second one means, so you should go back and check the definition.

[white oxide]

nvm

[delicate orchid]

dunno lol I just like the concept

[white oxide]

i can't read

didn't see that lol

[coral spindle]

I knew that one, keep up

[coral spindle]

easy to miss

[white oxide]

read it both as [] makes sense

[coral spindle]

so did I the first time

[white oxide]

gracias

[delicate orchid]

the zeroth schur indicator is the dimension of the rep

[coral spindle]

My mind is being blown wew

But like ok

n=2, this is some sick stuff

what happens with n=3, what info do we get there

[delicate orchid]

the negative first schur indicator is the dot product of the corresponding algebra idempotent with the vector (1,1,1...,1) \in Z^{|G|}

[delicate orchid]

I've said I don't actually know what they've used for "n=3" times now

I just think it's a cool idea

[coral spindle]

:(

Shit did I just not read

No wait you made it look like you were replying to someone else >:( it's actually your fault now

[delicate orchid]

perhaps this could be an avenue for you to peak down

[coral spindle]

I found a paper https://www.cambridge.org/core/journals/bulletin-of-the-australian-mathematical-society/article/on-higher-frobeniusschur-indicators/2F39C70E9852E2BD2F814F8996F8FD90

[delicate orchid]

yeah I looked at this paper before

didn't qget a word of it

also I cna't VIEW IT

[coral spindle]

those colours................ these authors are psychopaths

[delicate orchid]

found the preprint :epicwin:

[coral spindle]

access issue?

nice

[delicate orchid]

:uponthewitnessing:

[coral spindle]

Vile

OK, whatever, I need to learn Brauer character schtuff

[delicate orchid]

well well well

[coral spindle]

three in a row wow

so much water

[delicate orchid]

I LOVE WATER

[coral spindle]

Me too.

Isaacs' book doesn't cover Brauer characters, wow

I guess I'll have to look elsewhere

[delicate orchid]

fusion systems in algebra and topology by Bobby fusions et. al covers them

it's a bit uhhh

inaccesable

though

[coral spindle]

Is this like, your bible or something

do u pray to it every night

[delicate orchid]

yes

yup

[coral spindle]

Oh ok I respect that

I might go for just a regular ol' book on modular rep theory but idk maybe I'll dive into fusion systems

[delicate orchid]

I don't know of any modular rep theory books lol ur on ur own

[coral spindle]

There's a couple of modular rep theorists at my uni so I'll just annoy them until they throw a book at me

(read: ask them nicely for recommendations)

[white oxide]

how useful are totally inseparable extensions? fraleigh lists it as a chapter that is not used for the remainder of the text, and i'm tempted to skip to go straight to the expository section of galois theory 😊

[coral spindle]

Well uh not gonna lie I haven't heard of them

but

after looking up the definition

every finite field is a totally inseparable extension, so I guess it's helpful to know? Someone with more ken of Galois theory than I can feel free to comment.

[white oxide]

ooh ok

[rocky cloak]

I think maybe you misread the definition. Perfect fields have no inseparable extensions.

[coral spindle]

Different definition to the one that I saw then

In fact yeah, the one I was looking at was a "purely" inseparable extension, so in the first place something completely different.

[rocky cloak]

No, that's the same thing

Perfect fields have no purely inseparable extensions (except the trivial extension)

[rocky cloak]

So if E/F is a totally inseparable extension, then any the only automorphism of E fixing F is the identity. So if you wanna do Galois theory you have to deal with them somehow, or restrict yourself to separable extensions.

Luckily, most nice fields are perfect, e.g. finite fields or fields with characteristic 0. Which means all algebraic extensions are separable.

[coral spindle]

I hate to quote wikipedia, but perhaps the definition here is simply different to the one you have in mind: https://en.wikipedia.org/wiki/Purely_inseparable_extension

This is quite plainly defined exactly for fields of positive characteristic.

It seems to me that this simply means something different from extensions which are not separable.

[rocky cloak]

Yes, everything you're saying here is correct. But finite fields don't have any inseparable extension, and therefore not any purely inseparable ones either

This goes perfectly well with Wikipedia

[white oxide]

ok good to know than kyou

[rocky cloak]

An extension is purely inseparable if every (nontrivial) element is inseparable. An extension is inseparable if at least one element is inseparable. This leads to the funny thing where the trivial extension is purely inseparable, but not inseparable.

Slight correction: there are inseparable elements whose irreducible polynomial has several roots.

Whereas for it to be purely inseparable it needs to have only a single root, so being purely inseparable is slightly stronger.

[white oxide]

got it got it thx

is tau inverse onto Fbar because it's an isomorphism?

[rocky cloak]

Yeah tau^-(tau(x)) = x, so it's onto

[chilly radish]

you should learn brauer groups instead

[coral spindle]

Make me, nerd

[white oxide]

i'm still struggling to see how all of that implies that t[Fbar] = Fbar'

why are we talking about tau[F] and not tau[Fbar]

in their proof

or is that a typo

[glossy crag]

I want to show that for algebras "A left artinian, B finite-dimensional => A\otimes B left artinian", does this work: if n is the dimension of B, then A\otimes B\cong A^n as A-modules (if y_1,\dots,y_n is a basis of B, then every element of A\otimes B has a unique expression as \sum_ix_i\otimes y_i, map that to the tuple (x_i)_i), so A\otimes B is artinian as an A-module. A chain of left ideals of A\otimes B is in particular a chain of A-submodules (since a(x\otimes y) is the same as (a\otimes 1)(x\otimes y)), so the ring A\otimes B is left artinian.

[glossy crag]

Learning that now, it's a pain

[next obsidian]

It is different yes, it means the separable closure of F inside of E is E again. Perfect is the same as saying you have no inseparable extensions (so no purely inseparable ones other than the trivial one), and is equivalent to being char 0 or having the Frobenius be surjective. This is evident for finite fields by the pigeonhole principle

[glossy crag]

How does similarity (isomorphic to matrix rings over the same division ring) imply Z(C)=Z(B)?

[delicate orchid]

is there not a proof of this theorem directly below this

[glossy crag]

It doesn't explain how similarity implies the centres are equal.

[delicate orchid]

ok just to double check, B^o is the opposite algebra right

[glossy crag]

yes

[delicate orchid]

Z(A) = K,
Z(Z_A(B^o otimes A)) = Z(Z_A(B^o) otimes Z_A(A)) = Z(Z_A(B^o) otimes K) = Z(Z_A(B^o)) = Z(B)

does this work

= is iso or equals I cba differentiating them

[next obsidian]

The center of a matrix ring is just that ring, embedded as diagonal matrices

No?

Or is this for noncomm stuff

Like Z(C) ≈ Z(M_n(A)) ≈ A ≈ Z(M_n’(A)) ≈ Z(B)

I dunno

[delicate orchid]

we may never know

[next obsidian]

Oops I sent 2 photos

[delicate orchid]

chmonkey we're not in ivory

[next obsidian]

#groups-rings-fields is my original stomping grounds

I get to make the rules here

[delicate orchid]

understood sarge

[glossy crag]

Oh shit, I forgot A is central, now it works: Z(B^\circ\otimes A)=Z(B)\otimes K\cong Z(B) and similar algebras have isomorphic centres, so Z(C)\cong Z(B). However I still don't see why equality should hold

[delicate orchid]

You are way too hung up on being pedantic about equality vs iso in my opinion, does this book not just use = for isomorphic

Or is it important that they’re literally equal as sets for some reason

[glossy crag]

i don't know, hence me asking

[delicate orchid]

Lemme think though my explanation again

[glossy crag]

I don't like calling isomorphic things equal unless e.g. they are literally the same subset of a set. You wouldn't write Q(\sqrt{3}{2})=Q(\omega\sqrt{3}{2}), now would you.

[delicate orchid]

First and third = might be an iso

[delicate orchid]

No I wouldn’t because I don’t work with field extensions

I would write Z/nZ = C_n though

[glossy crag]

I would too.

I feel like the distinction here is more tangible, though.

[delicate orchid]

But they’re not equal as sets :uponthewitnessing:

[glossy crag]

True, but here it's sort of irrelevant.

All the lines in R^2 are isomorphic, but not literally equal.

[glossy crag]

cba?

What I was wondering about just now: what if Z(B) happens to be central, does that make C central? We have K\subset Z(C)\cong K, does that necessarily mean Z(C)=K? Here knowing Z(C)=Z(B) might be helpful.

Although since K is contained in both centres and they're isomorphic I suppose it's enough for one to be equal to K (i expect the isomorphism preserves K).

Am I correct in understanding that if B\subset A are CSAs, then automatically A\cong B\otimes Z_A(B)?

Is this useful in any particular way (e.g. some kind of multipliactive property)?

[frigid lark]

Could someone look over this proof (I'm self studying and haven't had any algebra checked in a while, I just want to know if I'm not writing absolute baloney)

Also I apologise for the bad handwriting

[frigid lark]

Oh, I was rereading this and I noticed that I didn't state that I want to assign sigma_i, such that if j < i, sigma_i restricted to k_j is sigma_j

[elder wave]

consider latexing it

[frigid lark]

Sure, tommorow

[teal vessel]

hey guys, a little help understanding what the question is asking about specifically? 10 is asking about actions of the symmetric group on a set of subsets, but the notation has left me confused about whether I'm supposed to be checking the faithfulness of the action for all members of the symmetric group, or only for the given ones; also about whether I'm checking for the k values of an arbitrarily large set A in this structure, or only the size 4 set. I can't tell if the subscript changing from A to n is meaningful here or just a typo.

I've already completed 8 and 9, so I have some idea of what I'm doing

[delicate orchid]

I don't know what you mean "only for the given ones"

faithfullness of an action isn't dependant on the element doing the acting

[teal vessel]

idk, I'm just dumb. I'm more concerned about whether it's asking for the k values associated with an arbitrarily sized A, or if it's specifically the set given in 8.B

[frigid lark]

I think it's just the values of k

[teal vessel]

that doesn't help me much, because the values of k depend on the size of A

so I don't know if I'm checking a particular A or checking if it's something like "if A is a multiple of K then it's not faithful"

[night onyx]

that sounds like what they're looking for, the values of k will probably depend on both |A| and n

[teal vessel]

I'm not sure what n is though, it appears nowhere else in the problem

[frigid lark]

I think n = |A|

Or A = Z/nZ

[teal vessel]

I think I'll check for cases where (|A| k)=1. That seems like a reasonable guess.

[delicate orchid]

Yes, top line

[frigid lark]

Golden phoenix spoilers don't read ||isn't part a just k < n?||

[teal vessel]

wait a sec... my gut is telling me that it's faithful on all sets with cardinality k<|A| because if it's always missing at least one element, then somewhere there has to be some set which gets an element replaced by the missing one.

unfortunately I need to make that notion slightly more rigorous

[delicate orchid]

||yeah||

[frigid lark]

|| If k=n don't you just have the whole set, so every action is trivial?||

[teal vessel]

the ordered tuples will be slightly trickier but I think the order mattering makes it faithful throughout.

[delicate orchid]

||yeah I realised that and fixed it ||

ok thinking done

[frigid lark]

Haha

[delicate orchid]

no more thinking will be scheduled for today

[frigid lark]

Good thing today ends in 43 minutes

[delicate orchid]

what wack ass timezone are you in

[frigid lark]

and you can live without thinking, politicians do it

[delicate orchid]

bro's in kamchatka

[frigid lark]

Australia

[delicate orchid]

got the latitude basically right :pack:

[frigid lark]

Don't you see my pfp

[teal vessel]

no, actually, hold on. I think that my reasoning is properly rigorous for part a. yeah, I submit that for all k<|A|, the action is faithful on B={b|b in P(A) and |b|=k}

[frigid lark]

Well I did tell you my time zone

[frigid lark]

Do you mean |b|=n?

[delicate orchid]

|b| = k is right

they're saying B = the set of all subsets of size k

[teal vessel]

I crashed through an entire ZFC set theory textbook, I gotta break out the proper notation at some point or else I'm not properly snooty

[delicate orchid]

the real snooty thing would be writing 2^A instead of P(A)

but I understand the restraint in being overly snooty

[frigid lark]

Fuuuuu

[teal vessel]

well... that would be the right cardinality but not necessarily the right set from which to take subsets

[frigid lark]

I've double negatived myself for the k'th time today

[teal vessel]

especially considering that the set A here seems to be unit indexed instead of 0 indexed, which is a total noob move

[delicate orchid]

irregardless, see if you can do part b

[teal vessel]

but regarding part B, since order matters, and all members of the symmetric group permute order (except the identity permutation), then it seems nearly trivial to assert that, even for the k=|A| case, it is faithful.

[frigid lark]

Part b reminds me of doubly transitive actions

[delicate orchid]

how so

[frigid lark]

Acting on tuples

Well pairs specifically

[delicate orchid]

right, fairs

I suppose you could say that if S_n has some number of orbits on the set of subtuples of size k then it's k transitive on the set of size n elements

not a very useful construction (I can't remember the number)

I'll think about it later

[delicate orchid]

it's k orbits lol

[frigid lark]

Is it just S_n or any group?

[south patrol]

UwU

[frigid lark]

Cause I think S_n is n transitive on n elements.

[frigid lark]

Mathematics server try not to be a furry challenge, difficulty: impossible

[coral spindle]

uwu has transcended furries

[frigid lark]

But we have not

[summer path]

detuwu

potuwu

[spice whale]

the server is owned by a furry

[teal vessel]

sometimes I miss the days when I could use a slide rule for my mathematics, because slide rules are cool. Then I remember that I teach math to students almost daily, so I still get to use numbers in my daily life.

[frigid lark]

With algebra you can miss those days no more, we've got counting galore

[delicate orchid]

yeah S_n is definitely n-transtitive, but my idea is for any group

basically the orbit representatives are tuples where m for (0 <= m <= n, m \neq 1? lol) of the entires are the same

[frigid lark]

Nah m can be 1

m is 1 orbit corresponds to the thing you need to show for a k transitive action

And m = 0 is silly

[rustic crown]

tubuwu

[south patrol]

uwu

[delicate orchid]

"one of the entires is the same" is equivalent to saying "there is an entry" and "zero of the entries are the same" is equivalent to saying "they're all different"
m = 1 is the silly one

[frigid lark]

Ok m = 1 is silly, m = 0 is silly which implies that 1 = 0

Problem solved

[delicate orchid]

true

[glossy crag]

I'm trying to understand maximal abelian extensions of (not necessarily finite) Galois extensions.

Let $L/K$ be Galois with group $G$.

First off, if $N\lhd G$ and $F=\operatorname{Fix}(N)$, then $F/K$ is Galois (it is enough to show it's normal and that's equivalent to $gF=F$ for all $g\in G$: if $x\in F$ is fixed and $n\in N$ is arbitrary, then $g^{-1}ng\in N$ and thus $gx=g(g^{-1}ng(x))=n(gx)$, thus all $n\in N$ fix $gx$ and $gx\in F$). The restriction $G\to\operatorname{Gal}(F/K)$ is surjective and $N$ is contained in the kernel (it might be larger than $N$), thus there is a surjective $G/N\to\operatorname{Gal}(F/K)$. In particular, if we take $N=[G,G]$, then there is a subfield $K^{ab}$ such that $K^{ab}/K$ is abelian with $G^{ab}\to\operatorname{Gal}(K^{ab}/K)$ surjective.

Next, if $L\supset E/K$ is Galois, then $\operatorname{Gal}(L/E)\lhd G$, and $G/\operatorname{Gal}(L/E)\cong\operatorname{Gal}(E/K)$ (the fixed field of $\operatorname{Gal}(L/E)$ is $E$, so the kernel is $\operatorname{Gal}(L/E)$). In particular, if $E/K$ is abelian, there is a surjective $G\to\operatorname{Gal}(E/K)$ and necessarily this descends to a surjective $G^{ab}\to\operatorname{Gal}(E/K)$ with kernel $\operatorname{Gal}(L/E)/[G,G]$.

So far, is this correct? Furthermore, I'm trying to understand the "meaning" of $K^{ab}/K$. Is this maximal in the sense that every abelian $E/K$ is contained in it or that there is no larger abelian extension containing it?

[cloud walrus]

leave_no_norm

[barren sierra]

Anyone got a good reference on wtf central and wreath products are

like a text or something

[delicate orchid]

wreath products are like little baby... wahhhh

[barren sierra]

oh cool now it makes sense /s

wdym little baby 😭

I mean idk I'm looking at the definition of wreath product in this paper and I understand the words but have 0 intuition for why people care about this

[delicate orchid]

you have a group G acting on some set X
let H be some other group, then you can define an action of G on H^{|X|} via the action on X
this is the wreath product*

*the unrestricted wreath product, which you don't have to worry about if |X| is finite

the classical example is taking G wr S_n for some n

and the group action of S_n on {1, ..., n} obvs

the wreath product* would be G^n \ltimes S_n with S_n permuting copies of G in G^n

[teal vessel]

Question: "pairs of" =/= "ordered pairs of" correct? the rotational symmetry of an even polygon is much easier to describe if the ordering of the pairs doesn't matter. question for context.

[delicate orchid]

try it with unordered pairs and ordered pairs

get two exercises out of one

[long nebula]

xD

[delicate orchid]

I mean, why not?

[long nebula]

It probably means ordered pairs

[coral spindle]

But like actually, why not

But fwiw, no I think it means unordered pairs.

[teal vessel]

the only real difference is that with unordered pairs, the kernel has an extra element in it because of the half-turn symmetry, but the notation for the description of transforming the ordered pairs is just so much more nasty

[coral spindle]

Think of it as referring to the line that connects two opposite vertices and you get the unordered action

This is actually a trick that comes up again when you look at the symmetries of the cube, which is why I think this is referring to unordered pairs.

[delicate orchid]

ah good point

I'm unordered pairs gang now

[teal vessel]

without ordered pairs I can call that set P={p|p={x,k+x}} with x being a one-indexed version of Z mod k

[delicate orchid]

what

[coral spindle]

what is that notation

[teal vessel]

shorthand cramming too much information into one line. That's what

[coral spindle]

Also no, in any case that's not the right set

it's not mod k, it's mod 2k

We require n = 2k

Just write P = { {x, x+k} | x in Z/2kZ } smh

[delicate orchid]

oh my golly gosh

[teal vessel]

having it be mod 2k overspecifies

[coral spindle]

No it doesn't, it specifies it precisely!

if you're saying this is mod k means that {x, x+k} = {x}.

[teal vessel]

no

[coral spindle]

Unless you mean you are working mod 2k, but you only want 0 <= x < k, in which case yes

[teal vessel]

that would be if we were taking (x+k) mod k

[delicate orchid]

geezah it's just {{x, y} : 0 < x < y < 2k} why are you overcompliacting this

[coral spindle]

Those not opposite tho wew

[delicate orchid]

since when do they have to be opposite

oh I see how it is

[coral spindle]

s'what the exercise says

[delicate orchid]

going back in time to change the timeline are we

not falling for it

[coral spindle]

TRUE mandela effekt so true

[teal vessel]

my notation aside, having unordered pairs means that {x, k+x}={k+x, x} which implies that the kernel includes r^k as well as just 1.

[delicate orchid]

yeah I buy this

[teal vessel]

so I was just checking because I started with ordered pairs, then realized it didn't say ordered, so swapped my notion so I wouldn't have to do wonky shit describing how to write some extravagant transformation performed by s in the form (x, k+x)

[delicate orchid]

I think the kernel with ordered pairs is trivial

[teal vessel]

me too, hence leaning towards the unordered exercise

[delicate orchid]

cause it's gonna be a proper subgroup of the kernel with unordered pairs

and that is just <r^k> if I'm visualising a square correctly

[teal vessel]

I used an octagon because the previous exercise was writing out all the permutations on a square, so I wanted two concrete examples to work from, but yeah, it would be {1_D, r^k}

[delicate orchid]

fairs

[teal vessel]

(haven't actually seen notation on how to write out kernels so I'm just bashing together a set of them)

[delicate orchid]

sets are fine

[south patrol]

UwU

[delicate orchid]

weird thinking about dihedral groups of anything other than a 2^n-gon

like t^3 is central?? WTF??!?!

[teal vessel]

especially since the kernel is just the fiber over 1, which is itself a set (thanks, Enderton). iirc another name being the "pull-back?"

[south patrol]

lol

[delicate orchid]

what did you just say

[teal vessel]

which part? I said many things

[delicate orchid]

all of it

why do you know any of those words

thinking about it, the pullback of fibres on sets might induce some sort of correspondence

[teal vessel]

1. am I correct in my statement, or at least close?
2. I did this same "do every exercise of a textbook in a new area" process previously with Elements of Set Theory by Enderton. It's what I do for fun. The term "pull-back" is the term my brother uses to describe fibers according to the textbook and class he took in abstract algebra. Whether this is just semantic trickery or is a genuine distinction is unknown to me.

[delicate orchid]

what the fuck is this pedagogy bro

[teal vessel]

autodidactics

[delicate orchid]

no the pull back is not another name for a fibre, it's a certain object in a category with a universal property

what happened to "preimage" dawg these kids growing up too fast

[teal vessel]

image/preimage terminology was always more confusing for me for no particular reason

[delicate orchid]

lemme think about this

[teal vessel]

like, I know what it is, but it wasn't part of most of my lexicon during early math and I had other terminology to describe it, so it didn't have much place

[delicate orchid]

the pullback of the diagram of sets A -> C <- B with f, g being the maps is the set {(a, b) : f(a) = g(b)}

ok yeah sure, so then you take the fibre of some c in C that's in the image of both functions and then this set is all of the (f^-1(c), g^-1(c)) so it is kind of like a product of fibres

but I would never ever use it to describe the fibres themselves

[teal vessel]

dunno. I'm just trying to navigate competing notations and nomenclatures between the texts I'm using and the ones other people use, and my brother is my primary point of contact for that lol.

[delicate orchid]

you're not at a uni I'm guessing

[rocky cloak]

The pullback is also called the fiber product though. Since the fiber of the pullback is the product of the fibers. Also the pullback along a subset is the primage, so the pullback along a point is a fiber.

[delicate orchid]

right, that makes sense

I would not use this in an intro group theory course that's covering group actions though

[rocky cloak]

No, me neither

[delicate orchid]

The pullback is also called the fiber product though.
I feel like I've heard this before and just forgort

[rocky cloak]

I feel like a lot of topologists call it fiber product.

Not sure if thats true or just my prejudice

[delicate orchid]

those guys smell a little funny though

[delicate orchid]

it's ok as long as they call pushouts fibre coproducts

[rocky cloak]

pushout (also called a fibered coproduct or fibered sum)
-Wikipedia

I like fibered sum

Though its not clear what this has to do with fibers anymore...

maybe it should be cofibered sum 🙂

[hidden haven]

Cofibered coproducts

[delicate orchid]

you can't co ur co

the cofibre of x, or as I like to call it, f(x)

[hidden haven]

This is good and healthy

[teal vessel]

No, all self teaching above calc 2. I'll go back eventually but my uni is yanking my chain and stringing along my first degree (music composition) so I'm taking a break.

[delicate orchid]

yooo music gang

[teal vessel]

I'll need to talk to my brother about his use of words when I see him next.

He might have just been using cat theory words because he was in cat theory at the time and cognitive bleedthrough

[delicate orchid]

right

that explains A LOT

[glossy crag]

Matrix ring over left artinian ring is left artinian, right?

[delicate orchid]

I think it holds in general that finitely generated algebras over left/right artinian rings are left/right Artinian

could be finitely presented

go with yes what could possibly go awry :pack:

[wooden ember]

honestly i often say "pulling back into ..." to talk about preimages

similarly to how someone might say contracting into

having full knowledge it's not the pullback im talking about

[delicate orchid]

pull:pack:

[wooden ember]

it's just a terminology that makes sense imo

[delicate orchid]

it does

[wooden ember]

you saw nothing

[delicate orchid]

I just don't like the possible confusion it could cause later on

[wooden ember]

yeah ofc

[delicate orchid]

but then again maybe it doesn't cause confusion at all and is instead a good way to think about the pullback intuitively

"the object where you can 'pull back' through two functions at the same time" kinda deal

[wooden ember]

i mean for my case i use the two uses of the word completely disjointly

[night onyx]

https://en.wikipedia.org/wiki/Hopkins–Levitzki_theorem tl;dr yeah, if R is artinian and M is a left R module then artinian <=> noetherian <=> finitely generated <=> has a composition series

[glossy crag]

I was just thinking about this again, I think we can claim equality. We already have a ring isomorphism Z(C)\cong Z(B) (as centres of similar rings), and this can be handwaved to an isomorphism of K-algebras (it's kind of tedious, but I checked, and I'm pretty sure that's correct), so they have equally finite dimensions (B is finite-dimensional), but Z(B)\subset Z(C), so literal equality follows.

[white oxide]

if H and K are subgroups of a group G and K is normal in G, what's the canonical inclusion map again from H --> HK

is it just h \mapsto hk for a fixed k in K

[chilly ocean]

bro

[coral spindle]

Well that wouldn't be a homomorphism

When we say an inclusion map, typically we mean a map that restricts to the identity

So............

h |-> h.

[white oxide]

ohhhhhhhhhhhhhhhhh

right

oops thanks

[quiet pelican]

Suppose I want a longer/more project-y question on abstract algebra (I don’t particularly mind wrt sub area), how would I find such a question/questions?
(For reference, my knowledge is roughly “average just-graduated Western European student”)

[coral spindle]

Are you asking for an open research question?

[glass thorn]

It may be good to find a researcher and ask them for a question

Because they can give you context, reading material, suggest approaches, etc.

[quiet pelican]

I literally just came back to explain why I can’t really - give me a few seconds

[coral spindle]

You should be more specific about your question.

[glass thorn]

I have an interesting open abstract algebra problem if you want

(specifically a finite group theory problem)

[quiet pelican]

Dogpiling me while I’m still trying to write is just going to slow down answers (and it’s not like this is not pretty stressful for me as is)

[white oxide]

bruh lmfao

[quiet pelican]

And the answer to this is basically the same as above, brain gets very stressed when emailing/communicating with when it’s not “supposed” to
And “supposed to” has no concrete definition
It’s just whatever my brain says

[glass thorn]

Googling "Open group theory problems" should come up with some decent results

[coral spindle]

The algebra section here is decent https://en.wikipedia.org/wiki/List_of_unsolved_problems_in_mathematics

[glass thorn]

In my experience the hard part is understanding the context well enough to feel motivated to try to solve it

[coral spindle]

You will need to consult current research in detail in order to make any kind of progress in these problems. This will involve at the very least reading papers, and likely making contacts with experts in the field.

[glass thorn]

Unfortunately mathematics is very collaborative, it may be possible to work on problems without regular communication with other people but I certainly wouldn't want to try

Basically because the rate of knowledge that can be transferred through direct communication is much faster than can be learned through just text

[barren sierra]

https://kourovka-notebook.org/

However some of these are like

"this question has been open for 3 decades"

[coral spindle]

only 3

[glass thorn]

Here's one I like: https://math.stackexchange.com/questions/4295186/which-pairs-of-groups-are-quotients-of-some-group-by-isomorphic-subgroups

[delicate orchid]

it's going to be a product of infinite copies of a gorup isn't it

[glass thorn]

The question is, given two finite groups $P$ and $Q$, can we tell whether there is a finite group $G$ with isomorphic normal subgroups $N$ and $M$ such that $G/N \cong P$ and $G/M \cong Q$?

[cloud walrus]

flopster0

[celest furnace]

What font is this? (Sorry to bother you)

[glass thorn]

Comfortaa

[delicate orchid]

oh nevermind

[glass thorn]

You're welcome, I know it is a pleasing sight

It can be shown that P and Q must have composition series with isomorphic factors. This is necessary but probably not sufficient, although no one has been able to prove a counterexample

C12 and A4 seem to be not compatible, but we can't prove it

[delicate orchid]

oh now I'm interested

that's fascinating

[rocky cloak]

Are Z/6 and S3 compatible? Or do you have any interesting example of compatible ones

[formal ermine]

what does it mean for two groups to be compatible

[glass thorn]

Sorry, compatible means they satisfy the properties of P and Q above

[rocky cloak]

I guess Z/3 x S3 will make Z/6 and S3 compatible

[glass thorn]

ie. They are quotients of a group by isomorphic subgroups

I believe Z/6 and S3 are compatible

It's a nice exercise to show that abelian groups are compatible

[delicate orchid]

invoke le structure theorem I presume

[rocky cloak]

It's a little more work than that I guess, but you should use the structure theorem yes

[formal ermine]

wtf is the structure theorem? I only know the ftofgmopids

[delicate orchid]

yeah I don't expect it to follow immediately

[glass thorn]

Sorry, I mean that abelian groups of the same order are compatible

And yeah, structure theorem is the bulk of it really

[delicate orchid]

well, it's not possible for two groups to be compatible if they're different orders

so I presumed

[rocky cloak]

The classification of finite(ly generated) abelian groups.

[glass thorn]

I think also that $P$ and $Q$ are compatible if they have an isomorphic subgroup $K$ such that $P/K \cong Q/K$?

[cloud walrus]

flopster0

[formal ermine]

you didn't get the joke

[delicate orchid]

cause the joke was shite

[rocky cloak]

No sorry, still don't get it

[glass thorn]

And I think it can be shown that P and Q are compatible if they are central extensions of compatible abelian groups by isomorphic groups

[delicate orchid]

funny long abbreviation of "fundemental theorem of finitely generated modules over principle ideal domains"

[delicate orchid]

lemme think about this

I lied I'm gonna think about split central extenstions instead

[glass thorn]

Ie. $0 \to A \to P \to K \to 0$ and $0 \to B \to P \to K \to 0$, where $K$ is abelian, and $A$ and $B$ are compatible

[cloud walrus]

flopster0

[glass thorn]

Ahh I typo'd

$0 \to A \to P \to K \to 0$ and $0 \to B \to Q \to K \to 0$, where $K$ is abelian, and $A$ and $B$ are compatible

[cloud walrus]

flopster0

[glass thorn]

Ok I'm changing the font why did I do this

It basically comes down to constructing N, M, and G as tuples of elements of A, B, and K, and figuring out how conjugation etc. should work

It's kind of a pain really and I don't know how it would generalise

[delicate orchid]

conjugation you say

do you have a paper on this result

[glass thorn]

Nah I know the guy working on it but I never got around to checking with him whether my approach works

[delicate orchid]

I'd love to see the proof of this if you've got a pdf lying around somewhere

it's fine if not

[glass thorn]

I've got a picture of my whiteboard scribblings

[delicate orchid]

I just hear the word "conjugation" and I move in for the kill

[glass thorn]

Lol it's more like conjugation is mostly trivial

[delicate orchid]

well yeah it's a central extenstion lol

anyway thanks for giving me another rabbit hole to fall down

[glass thorn]

Ah shit the photo was on my old phone that drowned

Oh well, I'll have to write it again

I'll let you know if I get around to it

Wait somehow I have a backup???

Ignore the matrices on the left

Also I called the groups J and K in the image

Also a "witness" is the corresponding G for the compatible groups

[rocky cloak]

Following up on this:
Are you looking for unsolved problems or just something that would be a fun project?
Do you have a more specific area in mind, since abstract algebra is pretty wide?
And it's not clear to me what an "average just-graduated western European student" knows

[prisma ibex]

learn more stuff first, there is a lot of stuff between the basic abstract algebra you learn in courses and modern themes in research

learn some flavor of representation theory for instance, there are lots of things to do there

[delicate orchid]

le wholesome maximal torus of GL_3 though

ah nvm there's 1s

E_8 got me like

[glass thorn]

Lol

[delicate orchid]

anyway I'm guessing a "witness" is this subgroup dodad that you quotient by to get P,Q

[glass thorn]

Technically it's an A_2 building I think

[delicate orchid]

nowt clue what's going on with the \bar{} geezers

[glass thorn]

Ye

I don't remember I think you can ignore those

[delicate orchid]

they're like half the proof

[glass thorn]

The bars I mean

[delicate orchid]

right ok

yeah basically the same

[glass thorn]

Oh that's right it's because those are isomorphic as a set but I'm defining a different group action

So I wanted to write a different symbol to represent that

[delicate orchid]

sure

I'm gonna trust you on the relators

[glass thorn]

Yeah I convinced myself that they work but I haven't actually proved it

[delicate orchid]

are the ones at the bottom independant of choice of representative?

[glass thorn]

Yes because representative are up to a choice of p and q, which commute

[delicate orchid]

fair enough, I can't see anything else immediately wacky that might go awry

[glass thorn]

As you can see this really only works because of the central extension thingy

[delicate orchid]

yeah you get all the nice commuting

[glass thorn]

If you relax that at all you have to deal with how elements should commute which gets really icky

[delicate orchid]

here's one of my many random nonsense whiteboard pictures

giving back to the community and whatnot

[glass thorn]

Is that motion blur or is it reflecting off the back of the glass

[delicate orchid]

both probably

[glass thorn]

Very nice

brb plagiarising your work

[delicate orchid]

too bad, two random spanish guys did that months ago

ok ok

they didn't plagiarise

they just scooped me

for all the lawyers in chat

[glass thorn]

Ahh it's always the random spanish guys

[glossy crag]

If A and B are similar CSAs (or Brauer equivalent or whatever you call it), are A\otimes C and B\otimes C similar for C simple artinian central? I know this works for C a CSA, but I'm not sure if finite-dimensionality can be dropped.

[south patrol]

okay idk if i'm being dumb or if this is a non-standard definiion - if a (p,q) shuffle sends i -> 1, don't we necessarily have i = 1 or p+1?

Like i thought a (p,q) shuffle has σ(1) < ... < σ(p) and σ(p+1) < ... < σ(p+q)

so like, for example, consider σ_{24} - this sends 2 -> 1 and 4 -> 2, so where can it send 1 and 3?

I guess the only reasonable interpretation is that we do stuff cyclically

[frigid lark]

If K is a p-group, where every proper subgroup is contained in H, such that [K:H]=p, can we say that K is cyclic?

[frigid lark]

I think I got it

[hollow fjord]

Hi guys!, I want to know why the condition of H being a closed subgroup

(I'm trying to define the induced linear representations for compact groups )

[night onyx]

I think it could have to do with the fact that H being closed in necessary for G/H being Hausdorff, that could be one of the main reasons, it's common when dealing with quotients of topological groups do always work with closed subgroups

[frigid lark]

Hey I've latexed my response

[rocky cloak]

looks good to me at least

[chilly ocean]

Howdy, I reading a book on abstract algebra and I'm reading this table for z4 that shows its a group under addition but not under under multiplication. I noticed the table looks like a symmetric matrix. Also a subset of the multiplication table of z4 looks like it would be a group since it also looks symmetric...

Is this an incorrect thing I noticed?

[coral spindle]

The table is symmetric because both operations are commutative. It looks like a matrix because a matrix is a table.

You have observed that a + b = b + a and that ab = ba.

Being commutative is not enough to be a group, not even close. Many groups are not commutative.

We call groups that are commutative Abelian.

[chilly ocean]

After the Norwegian lol, I live in norway

[coral spindle]

Indeed, after Niels Abel.

[chilly ocean]

I'll name my next boy that

[rocky cloak]

All the cool Norwegians have groups named after them. Abelian groups, Lie groups, Sylow subgroups

[coral spindle]

Do you want him to die young and empoverished like Abel

[chilly ocean]

So if a group is abelian, will its table be symmetric

[coral spindle]

Yes

[frigid lark]

26>21 at least

[chilly ocean]

That's really pretty

[coral spindle]

Uh sure, I would call it boring but different strokes I guess

[chilly ocean]

He is my son after all, born unlucky lol

[rocky cloak]

It is true that a subset of Z/4 forms a group under multiplication. Specifically those elements that have a multiplicative inverse.

This is called the unit group, and can be defined for any ring (if you know that rings are)

[chilly ocean]

Rings are fields?

[rocky cloak]

Fields are rings

[chilly ocean]

I just started reading the book yesterday

[rocky cloak]

Not all rings are fields, for example Z/4 is not

Hvilken bok leser du?

[chilly ocean]

Contemporary abstract algebra

Has chegg solutions

So good for self learning I think

[chilly ocean]

Er du dansk eller

[frigid lark]

Don't trust published solutions

[rocky cloak]

Norge baby

[chilly ocean]

Ah I see, er du norsk

[rocky cloak]

Jupp

[chilly ocean]

Kool, jeg kommer fra usa, men jeg har bodde I norge I 5 år

[coral spindle]

Discussion goes elsewhere, folks.

[chilly ocean]

Jeg sente deg en melding

[tender wharf]

gallian has the instructors manual, lol

you can probably find it on certain websites

[chilly ocean]

Whats the difference between chegg and the instructors Manuel?

[chilly ocean]

It's on those website haha

[tender wharf]

The chegg seems to have some wrong answers lol

[chilly ocean]

Ah okay i see, well I'll use the Manuel thank you 🙂

[frigid lark]

But that's where the learning comes in

[tender wharf]

Well I know the chegg is wrong because my answers are right

[frigid lark]

Are they as egregious as this?

[tender wharf]

I remember it being subtly wrong

[frigid lark]

Idk if that's better or worse

[tender wharf]

answers subtly wrong could actually be good pedagogically (?)

encourage you to really check your logic

[frigid lark]

Mmm or to stop cheating

[tender wharf]

good for spotting cheaters tho

since they can't spot the subtleties

[chilly ocean]

Is it cheating using it to self study?

Or self learning?

[dim widget]

Learning is cheating

It gives you an unfair advantage over those who don’t learn

[tender wharf]

If you fail the course you just gotta pay a bunch of money to retake it lmao

[chilly ocean]

I'm having to do my thesis when the class is offered so I'm trying to learn now before so I just show up to the exam

And take it

[elder wave]

Why would that be cheating

[delicate orchid]

because you have an unfair advantage?

[teal vessel]

define cheating. What rules are being broken by teaching yourself? The notion that reading ahead, or learning a subject for oneself is somehow a bad thing stems from lazy teachers who can't be bothered to actually engage with the needs of students. There are a couple dangers to be aware of when engaging in autodidactics: you may end up learning unusual or incorrect methods to engage with material that are a result of the quirks of the material you learned from, or you may do things in a strange or suboptimal order for learning. These risks, however, are minimal compared to the actual understanding you'll get from honestly sitting down and teaching yourself. That's what I'm doing, and I have spaces like this to check my understanding and offset any eccentricities I may develop along the way.

[slim kayak]

It can also become a real bad habit. If you have the looming option of just getting cheap answers you often won't try hard enough on your own

[delicate orchid]

bro actually wrote an essay

[teal vessel]

OH using a website for self study. misread that whole situation lol

don't worry, I can't read

[slim kayak]

Lowkey cringe to write more than 3 sentences

[delicate orchid]

you need to bold the first letter to stop my adhd from distracting me

[chilly ocean]

I respect he took the time to deliver a though and thoughtful idea.

[delicate orchid]

*thorough

[dim widget]

No, it is not unethical to teach yourself material in a class. In fact, self-directed learning can be a valuable and effective approach to education. Many individuals choose to supplement their classroom instruction with independent study to deepen their understanding of a subject or explore topics of personal interest.

Here are a few reasons why teaching yourself material in a class is not unethical:

1. Taking initiative: By teaching yourself, you are demonstrating a proactive attitude towards learning and taking ownership of your education. This level of initiative can lead to a more comprehensive understanding of the subject matter.

2. Personalized learning: Everyone has different learning styles and paces. Self-study allows you to tailor your learning experience to suit your individual needs and preferences. You can explore additional resources, set your own goals, and focus on areas where you need more practice or clarification.

3. Supplementing classroom instruction: Classroom teaching is often designed to provide a general understanding of a topic, but it may not cater to every student's specific needs. Teaching yourself can fill any gaps in your understanding, provide extra practice, or delve deeper into the subject matter.

4. Lifelong learning: Self-directed learning cultivates a sense of curiosity, critical thinking, and a passion for learning beyond the confines of a classroom. These qualities are highly valued in today's rapidly changing world and can contribute to personal growth and professional development.

While self-teaching can be beneficial, it's important to strike a balance. Engaging with your instructor and classmates, participating in discussions, and seeking clarification when needed can still be valuable in the learning process. Collaboration and interaction can enhance your understanding and provide different perspectives on the subject matter.

[delicate orchid]

MODSSS

MODDSSSS

[elder wave]

Oh I didn’t read the chegg part

[delicate orchid]

I didn't read any part

[elder wave]

I thought it was just about self studying

[slim kayak]

This reads just like how it feels to drive over speed bumps