groups-rings-fields

[steel pulsar]

To show that H = ⟨id,(12)(34),(13)(24),(14)(23)⟩ is a subgroup of S_4
I need to show stability by product
but I don't underwtand why
(12)(34)(13)(24) = (1234)
(12)(34)(14)(23)= (1324)
why (1234) for first and (1324) for second ?

[torn warren]

"So you can't necessarily realize the isomorphism by mapping alpha to beta no. You may have to map alpha to something else
"
sorry, when you say map alpha to something else, could you please give an example? do you mean to map generator to non-generator for the arbitrary extension field case? and keep F fixed at the same time?

[rocky cloak]

Like in the Q(sqrt(2)) = Q(2sqrt(2)) example. You can't map sqrt(2) to 2sqrt(2), but you can map it to sqrt(2)

[torn warren]

so we get this contradiction?

[delicate bloom]

it might help to think about how it maps the irreducible polynomial over Q, phi(x^2-2)=phi(x)^2-2 you can only map roots to roots

[torn warren]

this is due to there is only one isomorphism Q to Q, which is the identical map, right?

[delicate bloom]

you should try to prove it

[last spoke]

Ignoring well-defined is good for your mental wellbeing

[crystal turtle]

Ignoring well-defined-ness lets you prove nonsensical statements. Don't ignore it.

[torn warren]

do you mean this?

[untold cloud]

Hi, guys. I know that for a field extension K \subset L, the transcendental degree of L over K is well defined if the degree finite or countable infinite. Because of the swapping lemma

but how can i show the transcendental degree is still well defined if it is uncountable infinite?

[last spoke]

exactly, it means I don't have to worry as much

[rocky cloak]

I think you just use the same argument you would for vector spaces. Like replace K by its algebraic closure in L, and let B and B' be two transcendence bases.

If |B'| < |B|, then express each element of B' as a polynomial in elements of B. Since a union of |B'| finite sets is at most |B'| there is some b in B that doesn't appear in any of these polynomials. Then write b as a polynomial in elements of B'. Write each of the elements out as polynomials in B, boom that's a polynomial relation, so B is not a transcendence basis.

[rocky cloak]

Hmm, you may have to modify it slightly like

You know L is algebraic over K(B), so for each element of B' consider it's minimal polynomial over K(B). Then gather every variable that apears in a coefficients of such a polynomial into a set B''. Like before B'' is a proper subset of B, and K(B' u B'') is algebraic over K(B'') and L is algebraic over K(B'), thus L is algebraic over K(B''). Then take the minimal polynomial of an element of B and contradiction!

[untold cloud]

Thank you so much! I think i feel your idea!

[ocean charm]

my professor wrote in her notes that p-groups are not the semidirect product of two of their subgroups. It should be true for finite cyclic p-groups but doesn't D8 work as a counterexample? Am I going insane?

[lusty marlin]

D₈ isn't cyclic though

[ocean charm]

I know

[rocky cloak]

Don't know what your teacher meant, but there are loads of p-groups that are semidirect products

[ocean charm]

she probably forgot to add cyclic

[lusty marlin]

Wrong channel

[icy bear]

why?

oh yeah this is not really group theory it's just in a group theory book sorry

[lusty marlin]

Try #proofs-and-logic

[icy bear]

ty

[tardy hedge]

For people in grad school, doing masters or phds etc, how much more difficult does math get?

To understand the theory , working on problems etc

I really like learning math but im curious if im “good” enough to be able to do gead school

[delicate orchid]

Understanding the theory doesn’t get harder, sometimes it even gets easier because you’re more familiar with the objects the theory is about

For me, the main difficulty with working on problems is finding where to start/even knowing what questions to ask, cause unlike undergrad nobody knows what the answer should be

[rocky cloak]

Another new difficulty is that you have to choose which problem to work on. Hard to say how difficult a problem is if no one has solved it before.

But yeah, having some patience/perseverance is probably better measures for how you handle grad school than wheter you're "good enough"

[delicate orchid]

I’m doing ok in my phd not because I’m particularly intelligent (see any of my posts in here or #advanced-algebra) but instead because I am stubborn and refuse to give up

[tardy hedge]

Thats very insightful thank you

[tardy hedge]

Yeah thats what I admittedly need to get better at

If im stuck on a problem i get pretty stressed out and get obsessive over it

Im not very chill

I find it hard to “detach”

[white oxide]

yea i get that bro, i find that it helps to take breaks/force myself to take breaks

because when you get frustrated it's harder to think and it's not productive

[tiny jolt]

I'm confused about this problem, wouldn't there only be one elementary divisor that is a power of p since the primes must be distinct?

Or 0 if a is greater than the power of p in the elementary divisors

[rocky cloak]

What do you mean by "the primes must be distinct"

Like if M = (Z/2)^2 over Z for example, then n1 is 2

[tiny jolt]

We had this result from class

I meant that each p_j is distinct

[rocky cloak]

But M doesn't have to be equal to R/d

[tiny jolt]

Oh I see

We were first decomposing into invariant factor form and then into elementary divisor form, so there can be repeat primes

[vagrant zinc]

name of bookk ?

[tiny jolt]

these are class notes but we're using D&F

[white oxide]

here, would the content be (1 - i)?

also would an example of an ideal in Z[x] that can't be generated by 2 elements just be (x)?

<@&286206848099549185>

[abstract rock]

takin that back, yes thats a fine enough example

[white oxide]

lol these questions got me trippin

liek the answers seem too simple

the other one was give an example of a prime ideal in Z[x] which is not maximal

and i just said (1 + x) lol

[south patrol]

Yeah or even 0

Works for any integral domain which isn't a field

[south patrol]

Eh I don't think this is really fair lol

[white oxide]

wdym lol

[south patrol]

Generated by 2 elements presumably means generated by <= 2 elements

Or you can also do (x, -x)

[white oxide]

well maybe he meant something different by 2-elements with the hyphen im not sure

yeah....

[south patrol]

Well

I think he must mean you need > 2 elements to generate

[ivory trail]

if you remove the hyphen the question makes sense as a question

[south patrol]

(x) is an ideal which can be generated by 2 elements

[ivory trail]

wait yeah

[south patrol]

Otherwise this is trivial, again just take 0 lol

[ivory trail]

what is part a

my bad

[white oxide]

wait wat

i didn't know that

[south patrol]

I mean really when we say generated by n elements we only care about <= n elements

Unless they say smth about minimal generating sets

But yes it turns out for Z[x] there are ideals with minimal generating set of size n for all n

[white oxide]

hmm ok

and then for c i'm assuming it's a minimality argument

choose two elements of the maximal ideal with minimal degree

[ivory trail]

ah yes, the non-prime ideals

[white oxide]

division algorithim and then you're done or smt

wait hold up

wait yeah

[south patrol]

Hm I'd not do that but maybe it works fine lol

I would consider the preimage in Z, which will still be prime

For example

[white oxide]

oh huh ok

yeah division algorithiim doesn't seem to work

what does lang mean by reduce mod p for any prime p?

[ivory trail]

change the coefficients from being in Z to Z/p

[white oxide]

ah i see thanks

also what precisely does lang mean by "substituting elements of k for s_0, ... s_n - 1

[steel pulsar]

Hello

To show that H = ⟨id,(12)(34),(13)(24),(14)(23)⟩ is a subgroup of S_4
I need to show stability by product
but I don't understand why
(12)(34)(13)(24) = (1234)
(12)(34)(14)(23)= (1324)
why (1234) for first and (1324) for second ?

I need help to understand that

how to find the "=" in (12)(34)(13)(24)

[lusty marlin]

(1 2)(3 4)(1 3)(2 4) = (1 4)(2 3)

Start with an arbitrary element, say 1, and apply each cycle to it, starting from the right. The first cycle fixes 1. The second sends 1 to 3. The next sends 3 to 4. The last fixes 4. Hence 1 is mapped to 4 by this product of cycles.

Now we go through the same process for 4.

Since 4 is mapped to 1(work it out), we have that the RHS contains the cycle (1 4) in its disjoint cycle decomposition.

Now do it for 2

[steel pulsar]

I don't understand "The first cycle fixes 1. The second sends 1 to 3"

The first cycle is that ? (1 2)

and the second is that ? (3 4)

@lusty marlin

[lusty marlin]

From the right

[steel pulsar]

The right is (2 4) ?

I don't understand

[lusty marlin]

Yes, (2 4) is the first cycle from the right

[steel pulsar]

and why you said "The first cycle fixe 1"

We have no 1 in (2 4)

[lusty marlin]

Exactly, 1 isn't part of the cycle

Which is why applying it to 1 gives us 1 itself

So 1 is called a 'fixed point' of the permutation (2 4), and hence is said to be 'fixed' by it.

[steel pulsar]

What do you mean by "Hence 1 is mapped to 4 by this product of cycles."

Now we go through the same process for 4.
Since 4 is mapped to 1(work it out), we have that the RHS contains the cycle (1 4) in its disjoint cycle decomposition.
Now do it for 2

I don't understand

you say "now we go the same process for 4"

and after "do it for 2"

[lusty marlin]

Ok, I don't have time for this. Hopefully someone else can explain it to you.

[steel pulsar]

Ok thank you

[lusty marlin]

I suggest you go through whatever textbook/notes you were using again, and try to understand how these things work.

[steel pulsar]

(12)(34)(13)(24)

Start with (1 2)(3 4). This operation exchanges 1 and 2, then 3 and 4. After this operation, we have (2 1)(4 3)

Then I apply (1 3). This means I exchange 1 and 3.After this operation, we have (2 3)(4 1)

And (2 4), which exchanges 2 and 4. After this operation, we have (4 3)(2 1)

??

[faint fractal]

What's a good way to approach this problem?

[night onyx]

First thought is to consider R as a R-module, and look for the right subgroup to consider

see what it means for that subgroup to be an R-submodule (a left ideal)

[steel pulsar]

I don't find (1 2)(3 4)(1 3)(2 4) = (1 4)(2 3) ..... for permutation cycles

1 -> 3 -> 4
2 -> 4 -> 3

3 -> 1 -> 2
4 -> 2 -> 1

so (1 2)(3 4)(1 3)(2 4) = (4, 1)(2 1) no ?...

why ( 1 4 ) (2 3) ???

[faint fractal]

right, so subrings of R as an R-module are left ideals. So that means I*r is in I right?

I thought about this earlier, but I wans't sure how this was related to the problem

So this statement isn't true for R unless every subring is an ideal right?

and this is only true for Z or Z/n?

[prime kelp]

2 gets sent to 4, and 4 gets sent to 3 so (2 3) is product

[chilly ocean]

Can someone explain why C[Z/3] is isomorphic to C[x]/x^3-1?

is it because like x^3-1 is 0 if x \in Z/3? or something

More specifically I'm curious if anyone can explain how this works

[steel pulsar]

1 -> 3 -> 4
2 -> 4 -> 3
3 -> 1 -> 2
4 -> 2 -> 1
Where do you see 2 sents to 4 ????

1 sent to 4

2 sent to 3

3 sent to 2

4 sent to 1

[pliant forge]

2 to 4 to 3
4 to 2 to 1
3 to 1 to 2
1 to 3 to 4

compose and u got (2 3)(4 1)

[pliant forge]

look at where each loop starts and ends

(4 1)(2 1) would be simplified to (2 4 1) im pretty sure

[delicate orchid]

Let Z/3 be generated by x, then x^3 = 1, because Z/3 is cyclic or order 3. hence C[Z/3] is iso to C[x]/(x^3-1)

[delicate orchid]

You know that these are one dimensional as given in the question, so you have to consider maps from Z/3 into GL_1(C) = C*. If our map isn’t just sending everything to 1, then each non identity element of Z/3 must map to an element of order 3 in C*, aka a third root of unity

[chilly ocean]

Thanks, I figured this problem out eventualy.

Can you help explain how kronecker product is related to tensor product? Like how can you derive it from properties of tensor products?

This is the problem I'm trying to solve. I understand how this works using kronecker. And I have a vague idea of how the tensor product relates to it. Because for like A(e1) x B(e1) we get (a,c) x (e,g) which populates the first column of the matrix as ae, ag, ce, cg like the kronecker product does. But I don't understand why that's the case I guess

or maybe I do

I don't really know

[delicate orchid]

If you have matrices M, N in GL(V), GL(U) then M (x) N is the corresponding map in GL(V (x) U) such that the obvious diagram commutes

Specifically it’s the unique map out of the tensor product corresponding to the map (M, N) out of V x U

[chilly ocean]

for what I tried to do, (a,c) (x) (e,g) can be thought of as a vector of elements ae, ag, ce, cg right?

[delicate orchid]

Now that I’ve read the actual problem it’s far easier, Tr(A) = a+d, Tr(B) = e+h, Tr(A (x) B) = ae+ah+de+dh = (a+d)(e+h)

[chilly ocean]

why is Tr(A (x) B) = ae + ah + de + dh?

[steel pulsar]

It's what I said no ??

don't understand why you repeat

[delicate orchid]

Follows quickly from the definition of the kronecker product, A(x)B is a 4x4 matrix with the upper left 2x2 being eA, upper right being fA, lower left being gA, lower right being hA. So for the trace we don’t care about upper right/lower left as they don’t intersect the main diagonal, so Tr(A(x)B) = Tr(eA)+Tr(hA) = eTr(A)+hTr(A) = Tr(A)Tr(B)

[chilly ocean]

right, but we don't have kronecker product defined in my class, only tensor product

[pliant forge]

so what are you even confused about

[delicate orchid]

The tensor product of two matrices is the kronecker product, it makes no difference

[steel pulsar]

I have

1 -> 3 -> 4
2 -> 4 -> 3



3 -> 1 -> 2
4 -> 2 -> 1

[delicate orchid]

Because of this

[steel pulsar]

I ask you, why : (1 2)(3 4)(1 3)(2 4) = (1 4)(2 3)

[steel pulsar]

with that

[pliant forge]

because if you start at one you will end at 4. and if u start at 4 u end at 1 so (1 4) is a loop

[chilly ocean]

ok so with the end we were supposed to look at the image of the basis elements. e1 (x) e1 gets sent to (a,c) (x) (e,g) right? Why does this construct the first column of the kronecker product?

[delicate orchid]

The real reason I’m skirting around just explicitly computing everything is because I’m on my phone

[chilly ocean]

I'm trying to understand it a bit more tangibly, because I don't really undrestand the more abstract definition you gave

that's fair haah

thanks

[delicate orchid]

Because that’s how matrices work! The image of the first basis vector is the first column

Just because there’s a funny (x) doesn’t change that

[chilly ocean]

ah ok, so (a,c) (x) (e,g) is like equivalent to the column vector (ae, ag, ce, cg) by like distributivity right?

I was kinda just unsure if that was allowed

[delicate orchid]

Yeah so you can do

Gimme a sec

[chilly ocean]

all good

[delicate orchid]

Do you mean (x) or just x here

[chilly ocean]

I meant (x) because I got it from A(e1) (x) B(e_1)

I think

[delicate orchid]

(0,1)(x)(0,1) |-> (a,c)(x)(e, g) yeah I agree

[chilly ocean]

'i think I feel ok about this now

Do you mind explaining this proof? How does the base case work haha

[delicate orchid]

I think here you’re implicitly using the isomorphism between R^2 (x) R^2 and R^4 but that’s fine

[delicate orchid]

Proving maschke’s theorem is not something I can do off of the top of my head. I don’t think that proof is anywhere near sufficient to convince me

[chilly ocean]

oh ahah

I see I see

[delicate orchid]

It’s assuming indecomposable <=> irreducible which is not certain

[chilly ocean]

I didn' tknow the name so I can look more into it now haha

[delicate orchid]

Take for example, the F_2[Z/2]-module corresponding to the representation sending the generator of Z/2 to the matrix (1,1)(0,1) over F_2, this clearly has a subrepresentation(the first column of the matrix (1,0) is fixed) but cannot be decomposed as the sum of this subspace with another subspace

You won’t need to worry about this because you’re working over characteristic 0 but it’s still something that needs to be addressed

[chilly ocean]

oh interesting

I see

I have my midterm in 2 days, do you have any good advice on how to study algebra?

I feel like I didn't pick up good study habits in algebra 1 haha

[delicate orchid]

Nothing in particular that wouldn’t just be general advice

[chilly ocean]

okie

[rocky cloak]

The base case is just that a 1 dimensional vector space doesn't have any nontrivial subspaces. For the induction case they use that any submodule is a direct summand (which they presumably have proven earlier?)
As wew says, this is not something obvious. Typically you would prove it by defining a G-invariant inner product on your representation, and then taking orthogonal complement with respect to it

[chilly ocean]

is C* the conjugate transpose when looking at modules of k[G]?

because they are described by G -> C* right?

[rocky cloak]

The multiplicative group of complex numbers is what I would think.

[chilly ocean]

oooh

I see

so we have the example in our notes of C[G] where G is the dihedral group of odd order. How would I classify all of the simple modules of it?

what does it mean to map D_n -> C*?

[rocky cloak]

So a representation is a map G -> GLn(C) where n is the dimension. So a map Dn -> C* would be a 1-dimensional representation of Dn

[chilly ocean]

ah ok

So there would also be 2d reprsentations but those won't be G -> C*

just G -> GL2(C)

[rocky cloak]

So you can classify the 1-dimensional representations, just by computing these homomorphisms.

To classify as irreducible ones, do you know about characters?

[chilly ocean]

not really

[cobalt heath]

Woah, Dn -> C* is a representation?

[rocky cloak]

Hmmm, do you know that every irreducible representation is a summand of the regular module?

[chilly ocean]

yea I think so

this is what I'm looking at btw

[rocky cloak]

Alright, then you start looking for representations, and then whenever you find them you mod out all the summands of it in the regular module. When you've killed everything you know you have all of them

[rocky cloak]

Well, there you have your classification

[chilly ocean]

how do you look for representations?

that's what I'm a bit confused about

[delicate orchid]

ohhhh there are many a whimiscal and mystical manner in which you may find a bounty of representations

[chilly ocean]

lol

[delicate orchid]

number 1 is to just consider any obvious group action associated with your group

dihedral groups for example have a very obvious group action on an n-gon, giving you lots of 2 dimensional representations

A_4 on the tetrahedron etc. etc.

[chilly ocean]

wow

[delicate orchid]

S_n on a set of size n

that one in particular gets you the permutation matrices

finding irreducibles is a bit more tricky and that's where I'd just pass to character theory

[cobalt heath]

Is this what you learn in representation theory?

[delicate orchid]

but there's methods like lifting and inducing representations

[chilly ocean]

So let's say I have G = Z/3 and I want to look for it's representations over Q

[cobalt heath]

I need to take the course

[delicate orchid]

over Q
oh NO

[chilly ocean]

I don't have roots of unity right?

because I'm over Q not C?

[delicate orchid]

you don't no

[chilly ocean]

sad

[delicate orchid]

well, you have 2nd roots of unity

not like they'd help us here

[chilly ocean]

I have the identity map?

[rocky cloak]

This is representation theory of (finite) groups yes

[delicate orchid]

you always have the trivial representation

[chilly ocean]

right

[delicate orchid]

and also basically compact groups

[chilly ocean]

I want the dimensions to sum to 3 or whatever

so that's 1

uh

[cobalt heath]

..maybe I should have learnt this in undergrad algebra course

[chilly ocean]

Like I have this theorem that I want to use

[delicate orchid]

yeah this is Artin-Wedderburn decomposition

[chilly ocean]

this tells me the dimension I'm looking for right?

[delicate orchid]

very standard

[rocky cloak]

If you have an undergrad algebra course called "representation theory"

[chilly ocean]

dim(Q[G]) in this case is 3

and we already have 1

so we need 2 more right?

does that mean it's 2 more 1-d representations? Since 2 can't be squared? Or does the e_i take account here

[cobalt heath]

This gives vibe of "introduction to algebra".

[delicate orchid]

we definitely don't have anymore 1-d representations

[chilly ocean]

ohreally?

[delicate orchid]

there aren't any elements of order 3 in Q*

what we gonna map to

[chilly ocean]

oh I see

so let's look at Gl2(Q) instead?

[delicate orchid]

yeah

[chilly ocean]

so now we need to find matricies with order 3?

2x2 matricies with order 3

[delicate orchid]

yup, then we can just map our generator to it

there should be one cause SL_2(Z) is a subgroup of GL_2(Q) and it has elements of order 6

what it actually is??? not sure

[chilly ocean]

I think it's in my notes

[delicate orchid]

very useful

lemme quickly check

yup, it works

[chilly ocean]

Why does the eigenvalue part matter?

is it some multiplicity thing?

[delicate orchid]

if it had a rational eigenvalue there would be a Q-subspace fixed under the action of G

[cobalt heath]

Diagonalizability?

Oh wait, so it is abt eigenspace

[delicate orchid]

hence the rep wouldn't be irreducible

hopefully you can see why

[chilly ocean]

I don't

hol don

oh is it about eigenspace?

[cobalt heath]

Eigenspace is precisely the subspace fixed under the action, if I recall correctly

As always, I can be wrong

[delicate orchid]

yeah, so the action by G is multiplying by our (0, -1)(1, -1) matrix

[chilly ocean]

rigrht

[delicate orchid]

if this had a rational eigenvalue, then there would be a subspace (the eigenspace, as you said) that would be fixed under that multiplication

hence a 1 dimensional subrep

[chilly ocean]

ah

So now we have dimension 2, but we need to divide it by e_i right, so e_i is the dimension of the endomorphism of this representation?

what does the dimension of the endomorphisms mean?

[delicate orchid]

well we can see that e_i should be 2 because we need 1+2^2/e_i = 3

[chilly ocean]

right

but is there a way to figure it out if I didn't know it equaled 3?

[delicate orchid]

the endomorphisms of a module are themselves a module

[chilly ocean]

right, I remember proving that. How is dimension defined in this case? Amount of elements in the basis?

I remember there isn't a definition of dimension for moduels right?

[delicate orchid]

just as it's always definied

[chilly ocean]

oh ok

[delicate orchid]

but you're correct, however

we take the dimension as a k-vector space

[chilly ocean]

ah

[delicate orchid]

I've gtg now but I hope this as helped

[chilly ocean]

ah ok thank you

yes it did help!

[low wyvern]

Is there ever an instance that a Cayley table can be used as a proof?

[open sluice]

would you accept a truth table as a proof in logic?

[low wyvern]

Idk, it has been like a year and a half since I looked at them... I didn't devote too much time to em anyway

[cobalt heath]

Wouldn’t you?

[open sluice]

idk I'd say they're on the same level of legitimacy here

I probably would
unless the specification was to not use one

[rocky cloak]

Sure, if you want to prove that two groups are isomorphic, and you write down both Cayley tables and they're the same. That would be a proof.

A very tedious proof, but yk whatever

[chilly ocean]

I have the solution but I don't really underrstand it

[rocky cloak]

Like you want to use the classification of the irreducible representations of Dn you just did, or you want it from first principles?

[abstract rock]

going off the idea that $\mathbb{C}[x]/(f(x))$ determines a $\mathbb{C}$-space of dimension degree $f(x)$ and extending the definition for $\mathbb{C}[[x]]/(f(x))$ for deg $f(x)=\infty$ , how would one determine what the infinite-dimensional vector space would look like? I was thinking about the example of $f(x)=exp(x)$ since then we can leave out questions about open sets.

[coral spindle]

Honestly I don't expect there's going to be any nice way to describe this

[cloud walrus]

nastydasty

[coral spindle]

yeah I'm trying to think of a clever way to do this

I mean C[[x]] is a complete archimedean ring right

or valued ring

shit that's not correct

C[[x]] is the completion of C[x] wrt a particular metric is what I mean

So I'm wondering if we can use this to say a particular thing about the quotients in general, but I can't see it clearly

However you should be aware that any series in C[[x]] with a nonzero constant term is a unit, so with f(x) = exp(x) you have the trivial quotient.

[abstract rock]

well a nicely tamed version of C[[x]] would be constructed with a metric in mind right?

what about exp(x)-1 then, to remove the unit

[coral spindle]

I really couldn't say 🤷

again I sincerely don't think there's gonna be a nice answer here

At least not an entirely algebraic one. This feels like you could tackle it via some complex analysis.

[delicate orchid]

the cosets themselves would just be offset copies of whatever surface e^x-1 makes I think?

I'm thinking of these like coordinate rings of varieties but like

bigger

cause they're not polynomials

[abstract rock]

i had an idea that maybe you oculd up with an inner product against these offset copies

[coral spindle]

Yeah it would be like coordinate rings of varieties with some sort of crazy topology. My idea is that we have to guarantee that these series converge

[abstract rock]

as a sort of bound between the zeroes

[delicate orchid]

then again, I doubt e^x-1 is maximal cause C[[x]] is local

so any zariski-analog would be boring

[delicate orchid]

but this sounds promising

[abstract rock]

i don't work with rings much so do you mind explaining what you mean

[coral spindle]

I need to go do work, but put simply my kernel-of-an-idea is that we could treat these formal series as functions if we have some kind of guarantee that they converge, which might be enforceable via a topology on a space.

But I kinda doubt 🙊

[abstract rock]

well there's a usual topology on the space of analytic complex functions, which should be our space

hmm, the only paper i find about this only shows there's a nice choice (sup seminorm) for nice domains (nice being pseudoconvex) (https://arxiv.org/pdf/0707.1876.pdf)

[rocky cloak]

In C[[x]] any polynomial with nonzero constant term is a unit, so any polynomial is just x^n times a unit. Hence every ideal is equal to (x^n) for some n.

So such quotients just look like C[x]/x^n

[next obsidian]

Or you can just say it’s a DVR

:dvacat:

[delicate orchid]

or just say that it's local :/

:\

;3

[next obsidian]

That is not equivalent to this

:dvacat:

[delicate orchid]

:/

u want ur ideal lattice to be a chain? Grow up

[abstract rock]

but the question remains not nice for polynomial with zero constant term without a topology then?

[next obsidian]

Nothing they said involves anything about a constant or a topology

[rocky cloak]

What do you mean? There's no topology involved

[next obsidian]

The ideals are all just (x^n)

And 0

[abstract rock]

oh i misread, apologies

[delicate orchid]

yeah I forgot that fun fact earlier

I knew there was something about (x^n) but I forgot

[rocky cloak]

I guess if you introduce topology, like only consider the subring of those power series where things converge it might get interesting

[stark helm]

A definition of automorphism. Can I say automorphism is a definiton when there is a function f from G1 to G1, where we map g1 in G1 to g2 in G1, g1 may equivalent to g2 or g1 not equal to g2?

[delicate orchid]

what

an automorphism is an isomorphism from a group (I'm presuming) to itself

[abstract rock]

that was my hope, that when you only care about the functions that converge say uniformly you get atleast a reasonably wierd frechet space over all such f C[[x]]/(f(x))

for example exp(-x^2) would be nice

[delicate orchid]

although you're bringing up equivalences so you could be talking about automorphisms/autoequivalences of categories

[abstract rock]

but i dont know how to approach building up the space

[stark helm]

so I writes from G1 to G1, and I want to make sure if automorphism refers to map an element in G1 to another element in G1?I assume G1 to G1 means a specific isomorphism and also means a group to itself

[delicate orchid]

g2 does not need to be conjugate (which is probably what you mean by equivalent) nor equal to g1

take the automorphism of Z/3Z sending the generator g to g^2, for example

[stark helm]

so if both g1 and g2 are all in G1, then it finishes the automorphism right?

[rocky cloak]

Everywhere convergent power series are just entire functions. So you could look into the ring of entire functions, but it's probably a wild beast

[stark helm]

I actually feel kind of confused about what does conjugate mean, does that means like if a and b are in G1, then we can map into the result such as a^-1ba?

[abstract rock]

what's a good place to start learning about that ring? i've already some background complex analysis

[rocky cloak]

Then you're probably more equiped than me 😛

But yeah, idk. Maybe ask #advanced-analysis or something

[delicate orchid]

All it is is a isomorphism $\phi$ such that $\phi \colon G \rightarrow G$. There's nothing more to it

[cloud walrus]

WewGhostTbh

[delicate orchid]

I feel like you're confusing yourself with all this g_1, g_2 nonsense

if phi doesn't map into G, or isn't an isomorphism, it isn't an automorphism

[abstract rock]

hmm, i'll think on it after i gather some ideas

[delicate orchid]

$a, b \in G$ are conjugate if and only if there exists some $g \in G$ such that $gag^{-1} = b$ yes

[abstract rock]

thanks!

[cloud walrus]

WewGhostTbh

[stark helm]

I got it, really appreciate that!

[stark helm]

g1 and g2 are just element that I assume in G1, it can also be named as a and b. I just want to confirm if it is correct to map an element from G1 to another element in G1, like if G1 is D4, and it maps from r^2->r^3, is it still be considered as automorphism

[delicate orchid]

that specific map will not be an automorphism because it won't be a homomorphism

you're mapping an element of order 2 to an element of order 4

an actual example would be r -> r^3

and fixing s

lemme double check that actually

[stark helm]

because that doesn't satisfy isomorphism if the order is not consistent ?

[delicate orchid]

if f is a homomorphism then the order of f(x) must divide the order of x

[delicate orchid]

yeah that map works

another good example of automorphisms are conjugation maps, f(a) = gag^-1 for some fixed g

[stark helm]

If f(r^2) maps to r^3, then f(e)=r^2, which can't be correct, so I at least need to make sure that e maps to e here, is that correct?

[stark helm]

understood, it satisfies preservation and it is also bijective

[delicate orchid]

yeah, hopefully you can see why f(ab) = f(a)f(b) there

[stark helm]

f(ab)=f(a)f(b)=ga(g^-1g)bg^-1->gabg^-1

[delicate orchid]

yup!

[stark helm]

Another question is can I treat r1,r2,r3 the same rotation r while treating f1,f2,f3 as same reflection, I got the final answer f^3=f and it reaches reflection, does it make sense?

[rocky cloak]

Sort of. Like you have a homomorphism Dn -> Z/2 that maps rotations to 0 and reflections to 1. So applying this all rotations become "the same" and all reflections become "the same". But if course the final answer doesn't have to be f, it could be some other reflection

[stark helm]

In this question, is it possible to mean r1 is not same as r2 while f1 is not same as f2?

Because in that case I get r1r2r3^-2*f1f2f3, but it seems to have no answer of whether it is rotation or reflection, I even can not make sure how many times does the graph rotate in which direction, but reflection should always be the same, do you think this idea is correct?

[alpine island]

Given a ring R, is every subgroup S of the additive group of R an ideal of R?

[rocky cloak]

Usually not. That would have made the definition of ideal a bit redundant I guess.

[alpine island]

Every ideal is obviously a subgroup of the additive group, but I don't know about the inverse

[elder wave]

Think about a counterexample

[alpine island]

That fails for fields of char 0

[rocky cloak]

Like in very specific cases you can have every subgroup be an ideal. Like R =Z and R=Z/n

I think in every other case it should fail

[rocky cloak]

I'm not sure why you say there's no answer about whether it's a reflection or rotation. It's a reflection, by the argument from before. I'm not sure which idea your asking is correct either

[stark helm]

I mean if r1,r2,r3 are all different, f1,f2,f3 are all different, it can't give the exact answer, so I can regard all ri and all fi to be the same right?

[alpine island]

Do you know what n is?

[delicate orchid]

it should be 3 from context

[delicate orchid]

you just need it to be some f or some r

[rocky cloak]

Like the ri are rotations and the fi are reflections, so they're "the same" in the sense that their all rotations/reflections

Other than that I don't know what you mean

[rocky cloak]

You mean in Z/n? Then n is a (positive) integer

[alpine island]

I meant in Kingwisdom20's question

if he had a specific value for n it would help

[stark helm]

just about if r1=r2=r3, f1=f2=f3, if that's correct, then I get conclusion about f1,f2 or f3, which is just reflection, that's correct right?

[rocky cloak]

I mean the answer is the same no matter what n is

[alpine island]

oh, you're right, I misread it

[rocky cloak]

If their all equal then you get a reflection just like you would if they weren't equal, yes

[alpine island]

Imagine applying these rotations and flips to a paper cutout of the polyhedron. Specifically, one where you drew a big A on one side of the paper and a B on the other.

Every time you rotate, you stay on the same side. Every time you flip, you swap sides. The question is, in a sense, asking which side you end up on eventually. If it's a rotation, you should end on the same side you started on. If it's a reflection, you should end on the other side.

[white oxide]

in working in $\mathbb{Z}[x]$, why does this approach not work to show that any ideal generated by $n$ elements is principal? say we're considering the ideal $(f_1(x), f_2(x), \dots, f_n(x))$. take some element of minimal degree $h(x)$, and divide $f_1(x)g_1(x) + \dots + f_n(x)g_n(x)$ by $h(x)$. then you get $g_i(x)f_i(x) = h(x)q_i(x) + r_i(x)$ where $0 \leq \text{deg}r_i < \text{deg}g_i$, so all the $r_i$s are zero since $g_i$ is of minimal degree. hence each summand can be written as a multiple of $g_i$ and hence the ideal is generated by $g_i$

[cloud walrus]

okeyokay

[rocky cloak]

How do you know you can divide stuff by h

[delicate orchid]

wait but... Z[x] isn't a PID?

I am confusion

[rocky cloak]

Hence why the proof doesn't work

[delicate orchid]

yeah but why is he trying to prove this at all

cause I presumed it was an exercise?

[white oxide]

trying to show that every maximal ideal in Z[x] is generated by two elements and i wanted to try something lmao

like

i'm assuming that it's generated by n > 2 elements

and trying to derive a contradiction or smt idek lol

[stark helm]

The confusing part is that the question doesn't tell me if r1=r2=r3, so we can just considering the case when r1=r2=r3, f1=f2=f3, then we end up on reflection. Otherwise, we discuss if it's on the same side we started on, then it is rotation. If end up on different side, then reflection. Is that true?

[alpine island]

They may not be equal. They probably aren't.

if it's on the same side we started on, then it is rotation. If end up on different side, then reflection.
Yes.

[stark helm]

Ok, I think I got it, really appreciate that

[alpine island]

Just grab a sticky note, treat it as a square, and try this for yourself

[rocky cloak]

Is it like an exercise, or are you doing on your own little quest?

[white oxide]

it's an exercise

a stupid stupid exercise which i'm too stupid to figure out

[delicate orchid]

there's a swag way to do it

what else do you know about maximal ideals

other than they are the head honchos

[white oxide]

lol

i mean now i'm quotienting out by it

and i know that it's a field

which has no nontrivial proper ideals

[delicate orchid]

yur

the way I'm viewing it is, we need to quotient Z by something to get F[x] for some field F, and then quotient by an irreducible polynomial to get some field F'

[white oxide]

hmm ok

sorry i'm dumb, how will that help?

i'm looking at $\mathbb{F}_p/(x - 1)$

[cloud walrus]

okeyokay

[white oxide]

i forgot all of my field theory and stuf fholy shit

[coral spindle]

Can you please stop prefacing your questions with some variation of "I'm stupid"

[delicate orchid]

that's a pretty boring example

this is just F_p

assuming you meant F_p[x]/(x-1)

[white oxide]

oh yea meant that my bad

[delicate orchid]

degree 1 polynomial => degree 1 field extenstion so you don't go anywhere

ok wow this proof was nowhere near as easy as I remember it being wtf

this shit is kind of cracked

[white oxide]

lol

not a good sign

[topaz solar]

What y’all trying to do

[delicate orchid]

krull dimension of Z[x] is 2

[topaz solar]

Maximal ideal has 2 gen?

[white oxide]

yea

[delicate orchid]

aka maximal ideals are of the form (p, f(x))

say we have some polynomials f_1, ... f_n and a maximal ideal I = (p, f_1, ..., f_n) , then Z[x]/(p, f_1, ..., f_n) = F_p[x]/(f_1, ..., f_n) = F_q for some q by maximality. So F_p[x]/(f_1, ..., f_n) = F_p[x]/(g) for some irreducible polynomial g, as F_p[x] is a PID.
I got this far and tried to use the corrispondence theorem to shoehorm (g) backwards but it didn't work

[white oxide]

i thought c wasn't going to be so bad judging off of a and b

[topaz solar]

Well, if your thing just has a polynomial and no integer, you’d be able to stick some integer in there and violate maximality

[delicate orchid]

ok but that doesn't invalidate the cases of (p, f, g)

[chilly ocean]

why is algebra so hard 😭

[delicate orchid]

this guy on stack exchange is taking localisations which is insane

[topaz solar]

So gotta be a prime so it stays integral, then you do def get F_p[x]/(f’), but there’s maybe something like

f’ = g’

[chilly ocean]

my other classes need to stop having "modules" as a section on the canvas, it's giving me PTSD

[delicate orchid]

like by correspondence I can get that (f_1, ..., f_k) is a subideal of (g) in Z[x] which isn't very useful

[topaz solar]

Can’t you get there’s a sub ideal in the opposite direction

[delicate orchid]

not by correspondence

and I can't see of any other way of doing it

[topaz solar]

indeed less trivial

[delicate orchid]

ok this localisation looks completely unneeded. Passing to Q[x] to show that (f) isn't maximal is deranged

[rocky cloak]

Why doesn't it work? Doesn't this just show your ideal equals (p, g)

[delicate orchid]

if you could explain how I'd be very happy

[topaz solar]

No ideal proper containing (f_1, …, f_n) because field?

I mean, it do be maximal

[delicate orchid]

did I get the correspondence theorem backwards again

ideals of R/I are J containing I

(g) maximal

yeah ok...? still not following

[topaz solar]

So (g) = (f_1, …, f_n) over da F_p[x]

[rocky cloak]

It's the correspondence theorem. The ideals containing p are in one two one correspondence with ideals of R/p

[topaz solar]

I think

[delicate orchid]

yeah so there are no ideal that contain (f_1, ..., f_n) and (g) in F_p[x], completely agree there

how does this show that they're equal

soceity if F_p[x] was local

[topaz solar]

Well, (f_1, …, f_n) <= (g)

I mean, g is literally the thing that makes it a principal ideal inF_p?

[rocky cloak]

So (p, g) is an ideal, which corresponds to the ideal (g) in Fp[x]. (p, f1, ...) is an ideal which corresponds to (g) in Fp[x]. So by the correspondence theorem, they are the same ideal

[delicate orchid]

I need to think about this lol

gimme a min

[rocky cloak]

The tricky part I guess is showing that an ideal that doesn't intersect Z can't be maximal

[delicate orchid]

well no that's the easy part

[topaz solar]

The flames of Babylon

[delicate orchid]

(p, f, ..., f_n) strictly contains (f, ..., f_n) and is proper

[rocky cloak]

Is it obvious that it's proper?

[delicate orchid]

yeah it's pretty obvious

the (f, ... ,f_n) part can't access the 0th degree term

and the (p) part is obviously proper

probably something to do with Z[x] being a valuation ring

[rocky cloak]

I mean it's clearly not for all p, but I guess since there are infinitely many primes it's fine

[delicate orchid]

actually yeah, it's because Z[x] is a valuation ring

[delicate orchid]

it's this fuckin subtle bullshit that's the hard part

[topaz solar]

The isomorphism theorems strike again

[delicate orchid]

I still don't really understand how we're concluding R/I \cong R/J => I =J but I'll just take ur word on it

my ring theory is shit anyway

[topaz solar]

I don’t think we’re quite concluding that

[delicate orchid]

anyway okeyokay I hope u can read all of this stuff and understand it now

[white oxide]

yeah thanks i'm going to look back at this when i return to this problem

lol i thought it was an easy question and was getting mad at myself but turns out it's not ig

[rocky cloak]

I mean it's just "there is a bijection between ideal of R/p and ideals of R that contain p"

So looking at ideals that contain p is the same as looking at ideals of R/p

[delicate orchid]

yeah, that's the correspondence theorem, I'm aware

and then R/p is a field so the only ideals of R that contain p are p and R, which can also be seen by maximality

[rocky cloak]

Right, but there's nothing else going on

[rocky cloak]

No R/p is Z/p[x]

[delicate orchid]

yeah, sorry

I'm just using p to be a general maximal ideal

[stark helm]

I feel pretty confused about a collection of 12 integers. In my understanding, phi(56)=phi(2^3*8)=24, so we have 24 integers as a group st we can find closure, identity, and inverse. Could anyone explain what does this question mean?

[delicate orchid]

ohhh I got it now

thanks

the only ideals that contain (p, f_1, ..., f_n) is R and (p, g) and the only ideals that contain (p, g) are (p, f_1, ..., f_n) and R, both ideals are proper so we have to have (p, f_1, ..., f_n) = (p, g) as they contain one another

[lusty marlin]

These 12 integers are a subgroup of the group of 24

*more precisely, these 12 residue classes

[stark helm]

I can pick 1, (5,45), (3,19),(9,25),15,(11,51),(17,33), just make sure we can get identity and inverse and find 12 integers out of all 24 integers, is that true?

[stark helm]

12 residue classes, I understand it as any 12 integers out of 24 integers have different residues mod 56, is that true?

[rocky cloak]

Hmm, okay so how do you pick this prime p (or argue that it exists) do you like at like the gcd or the lcm of the coefficients of the polynomials or something?

[white oxide]

how are we able to view B = A[\alpha] by considering this tower?

[delicate orchid]

yeah so I'd just focus on the constant terms of each of these polynomials, then pick a prime coprime to all of them

this doesn't generalise nicely to R[x] with R a PID but I don't care

ah no wait hmm

[rocky cloak]

Well I mean for the ideal (2x + 1) you can't add 2 and still have a proper ideal, so you have to look at the other coefficients as well

And without some restrictions on which generators you use, things like (x^2 + 1, x^2 - 2x) can cause trouble as well

[delicate orchid]

even just looking at the constant terms I'm not sure my idea works if any two of them are coprime

[white oxide]

because if B is an A-algebra, doesn't that mean that B is an A-module?

[delicate orchid]

this is probably why the geezer on stack exchange used a localisation to move up to Q[x]

[rocky cloak]

They're using induction

Like adjoin one element at a time

[delicate orchid]

is that the full proof?

[white oxide]

yes

[rocky cloak]

Yeah if you pass to Q[x] then your ideal is generated by just one polynomial, and then it's not so hard to find a p

[delicate orchid]

exactly

[delicate orchid]

yeah dw I don't get it either then.

[white oxide]

at this point i don't even know what an A-algebra is, something like a module over A with a bilinear product

but i don't even know what a bilinear product is

[white oxide]

ah ok lol

[delicate orchid]

we're gonna do induction

absolutely no inductive step
great.

[white oxide]

love lang

[rocky cloak]

The inductive step is B = A[alpha]

[delicate orchid]

but then they say you can assume that by considering a tower?

is it just poorly written?

[rocky cloak]

No, they're saying because of the tower we can use induction, so we may as well assume B = A[alpha]

And that case is apearantly proven earlier

[delicate orchid]

I'm going to go back to doing something that makes sense. Like cohomology of finite categories in arbitary modules

[rocky cloak]

That's what it is yes. Bilinear just means linear in both arguments. So

x * (ay + bz) = a(x*y) + b(x*z)

And similarly in the other argument

Sometimes you might also assume the multiplication is associative or has a unit

[white oxide]

i see

what's the intuition behind it? like what special properties does it have

[delicate orchid]

it's like you can multiply vectors in a vector space

[rocky cloak]

It's like a ring, but also a module

[white oxide]

ooh

that does seem handy

[delicate orchid]

wanting them to be rings are why some weirdos don't require rings to have units

[rocky cloak]

Rings in general can be a little tricky, so it's nice to have some nice base ring (or field) that can restrict things a bit

[open sluice]

bilinear forms

[delicate orchid]

that reminds me jagr you said a really nice way of thinking about modules a while back that I can't remember the details of

you said something like "a module is a ring with a map" or something like that

[rocky cloak]

Hmmm, like a module is a ring homomorphism R -> End_Ab(M)?

[delicate orchid]

yeah that's the one

very obvious but still nice

[rocky cloak]

I prefer "a module is just an additive functor"

[delicate orchid]

a module is something that transforms like a module

[white oxide]

here what does B[C] mean?

[rocky cloak]

They say it in the parenthesis

[delicate orchid]

it says immediately

yeah

[white oxide]

oh oops

lmao

thanks

[delicate orchid]

the ring algebra B[C]

[white oxide]

oh yea i meant more like

[delicate orchid]

or as I like to call it, the ring B[C]

[white oxide]

what algebraic structure is it

oh ok

[delicate orchid]

homedawg...

[white oxide]

so BB[C] would be another ring algebra

😏

[delicate orchid]

ok maybe I shouldn't have joked about "ring algebras" cause you took me seriously

[stark helm]

Could someone answer this question?

[delicate orchid]

<5,15> is a subgroup of phi(56), find its elements

[stark helm]

Could you explain it in depth, I think <5,15> includes identity, and 5 has inverse 45 while 15 has inverse itself? But there are only 4 elements here?

[delicate orchid]

5*15 mod 56 is 19

[stark helm]

so it generates 1,(5,45),(19,3)(23, 39),(13), (9,25),(27),15

i think that's exactly 12 elements here

[stark helm]

it seems that I ignored closure...

[delicate orchid]

it seems so!

[white oxide]

ok i have a couple questions about this proof 😹1) how does showing that a has a factorization show the uniqueness portion of a is not any arbitrary element of A, and is instead chosen to be the generator satisfying the maximal condition? 2) wtf is he even doing logically at the bottom paragraph with contradiction and stuff

[delicate orchid]

huh, another source using "factorial", strange.

[white oxide]

lang's just tryna be different

like he says "we note that a cannot be irreducible" then proceeds to show that a is a factorization of irreducible elements?? even though earlier on he said that (otherwise it has a factorization)??? like wtf is he even contradicting here

[delicate orchid]

I'll give it a read but no promises

I'm skill issuing heavily today

[white oxide]

lol ur good i'm reading the lang supplement rn and it's helping

[delicate orchid]

what does the supplement say

also... if ur book needs a supplement to understand it I think something has gone wrong

[white oxide]

LMFAO i mean it's lang

page 17

he has commentary on almost every page which is nice

[delicate orchid]

:uponthewitnessing:

holy fuck he uses "entire ring" instead of integral domain

who does this guy think he is

[white oxide]

na fr what is he doing

[coral spindle]

I expected to see something like the talmud

[delicate orchid]

wait yeah this is just the proof of PID => UFD

this is why language should be standardised

[white oxide]

i did not know that noetherian induction is a thing

[delicate orchid]

yeah inducting on lattices other than (N, \leq) comes up sometimes

[white oxide]

cool

‘‘To prove that every nonzero nonunit a∈A has a factorization into
irreducibles, assume by Noetherian induction, applied to the nonzero principal ideals, that every nonunit
that generates a larger ideal than (a) can be so factored. Now if a is itself irreducible, there is nothing
to prove, while if there is a nontrivial factorization a = bc, then b and c each generate larger ideals
than a, so by our inductive hypothesis, each may be factored into irreducibles. Multiplying together these
factorizations, we get a factorization of a into irreducibles, as required.’’

[delicate orchid]

this supplement is written way better than the book

[white oxide]

yeah this is way more clear lol

[stark helm]

Is there only 2 subgroups here? i only figured out {<r>} and {<sr>} starting by assuming element x=s^i*r( meaning to figure out it must have r as cycle and discuss it into whether we have s here, then by closure)

[crystal turtle]

(sr)(sr) = s (sr^{-1}) r = 1. So that is order 2.
There are two more subgroups of order 4, which is generated by two elements

Hint but not full answer: ||Some power of rotation + reflection for one of them|| and ||some power of rotation + (rotation and reflection)|| for the other

[stark helm]

subgroups of order 4, I understand it as a subgroup with four elements actually

Do you think this question mean all elements with order 4 besides identity?

[alpine island]

Isn't D_4 itself 4 elements?

or is this D_8

[stark helm]

D4 has 8 elements I think

it is about r,r^2,r^3,e, and as we multiply s, we get 8 results

I think sugroups of order 4 means a group containing 4 elements right?

[stark helm]

So I acutally doubt why we see sr has order 2 here, because it seems irrelevant to how many orders of an element

[alpine island]

Yeah some people would call it D_8 instead. Just clarifying.

[stark helm]

I actually regard D_8 as r has smallest order 8.

[stark helm]

How would you think of the answer to this question? Like how many kind of subgroups do you think here?

[rocky cloak]

Some hints:
||The product of two reflections is a rotation, so any subgroup of order 4 contains a rotation||

||If you contain a reflection you contain r^2, which generators a normal subgroup||

||D4 / r^2 = (C2)^2, how many subgroups of order 2 does it have?||

[stark helm]

does D4/r^2 means D4 without r^2? And could you tell subgroup of order 2 means a group containing 2 elements or each non-identity element has order 2?

[crystal turtle]

If you take an element of order n, the subgroup generated by it has n elements, so it's kinda relevant yeah

[rocky cloak]

By D4/r^2 I mean the quotient group of D4 modulo the subgroup generated by r^2 .

Group of order 2, means group with 2 elements.

[delicate orchid]

Not sure if they’ve done quotient groups yet

The way to find them at your level would to just bash elements together

[stark helm]

I know where is the problem, i assume srsr as s^2r^2 here...

[stark helm]

I didn't learn quotient group yet

[rocky cloak]

Alright, then I'll modify my hint to say that the subgroup contains r^2, which other element could you add to make a subgroup of order 4?

[stark helm]

I can conclude additional subgroup{e,sr^2,r^2, s}

and another one is just e, r, r^2, r^3

[rocky cloak]

Yeah, that's 2. Any other elements of D4?

[stark helm]

r with other exponent like r^3 will generate r with all exponents here, which can not work, but if there is no rotation, then only s and e in a subgroup, so I think none of them

[rocky cloak]

So there are 2 elements that don't apear in the subgroups you've listed

[stark helm]

You mean sr and sr^3 right?

[rocky cloak]

Yeah, maybe try creating a subgroup containing them

[stark helm]

e, sr, sr^3, r^2

[rocky cloak]

Not sure where you got s from

[stark helm]

No s here

[delicate orchid]

<sr, sr^3> = {e, sr, sr^3, r^2} so yeah that’s all of them

[alpine island]

Well, if it has r, it has all the powers of r, which is 4 elements.

So without r, there is <s, r^2>. That’s 4 in a klein group setup.

If it has two reflections, you can combine them for a rotation. <s,r^2s> is the one above, and <s,rs> and <s,sr> include r, so too many elements.

Then yeah, <sr,rs>

[delicate orchid]

Consider the subgroup lattice of a semidirect product…

[stark helm]

I haven't learned that actually

we are just covering up to isomorphism and homomorphism

[delicate orchid]

Yeah I know, you obviously haven’t

[rocky cloak]

Is there an easy way to think of the subgroups of a semidirect product using the factors?

[delicate orchid]

No it’s usually quite awful!

We’re just extending a very nice group by a very nice group here so it works

[stark helm]

I am thinking m would be the multiple of 6 in order to get 5-cycle here because alpha are disjoint cycles, and the purpose is to eliminate 2-cycle and 3-cycle by letting them be identity, is it true or is there anything that I miss here?

[delicate orchid]

Yup that’s true

[stark helm]

OK, really appreciate the response

[south patrol]

Yeah that's a good argument

If you'd like you can also give a sufficient condition on m - not much harder

[stark helm]

You mean mine? which sufficient condition you would give on m?

[stark helm]

I think more than one elements implies the order is at least 2, when the order is prime, then done, otherwise, all numbers are product of prime numbers, so (g^k)*m must exist where m is the component of product prime, do you think it is true?

[south patrol]

Well, just 6|m and 5 doesn't divide m

should be necessary and sufficient

[south patrol]

Hm this is a bit unclear

Like what do you mean "(g^k)*m must exist"

[stark helm]

i guess you mean adding that it is necessary and sufficient for 6|m

[south patrol]

no like you need 5 does not | m as well

[stark helm]

I see, but actually 6 and 5 co-prime, so i didn't consider that at that time

[stark helm]

i mean say if the order of g is not prime, it should be composite, then composite is the product of prime say n=a*b where a is a single prime or a product of prime numbers while b is a prime here, so if we have smallest order n, then we have g^n=(g^a)^b here so there exists element that has prime order

[lilac mango]

Why are simple groups so important in group theory? I know that by the Jordan-Hölder theorem you can in a sense "decompose" a group into simple groups, but I really don't see how this decomposition is enlightening in any way

What I mean is what does a composition series of a group tell you about the group that actually makes you understand it better in some way

[barren sierra]

It's more the other way around. If we better understand simple groups then Jordan Holder may allow us to reduce other problems to simple groups

[lilac mango]

Do you have some example?

Of when can you reduce some bigger problem into a simple group problem

I think another question I have is why are normal subgroups so important, because I really haven't trully grasped that yet so I guess I'd need to understand that better in the first place

[crystal turtle]

This one is a little bit easier to motivate: They are precisely the subgroups which we can quotient by, and get a group again. There are a variety of reasons why this is important. For example, there are several important properties that, if it holds from a group G, then it holds for any quotient group G/N (a rather silly one is being simple >.<).
It can also give connections between two groups that we would like to understand, but one of them might be easier to understand. In particular, this is seen by the first isomorphism theorem: for a group homomorphism f : G --> H, we have (1) ker(f) is normal and (2) that G/ker(f) ~= im(f).
Now, along those lines, one purpose of knowing that a group G is simple is that it actually classifies what the group homomorphisms out of G look like. Since ker(f) is normal, it must either be that ker(f) = G, and hence f is the constant identity map, or that ker(f) = {e}, and hence f is injective. Or in otherwords, G is (isomorphic to) a subgroup of H.

Among, of course, many other reasons to care about normal subgroups, as well as simple groups.

[next obsidian]

I like normal subgroups because I HATE deviants

I want things to be NORMAL

[sly rain]

Vague question, but what are good ways to think about tensor products in the realm of commutative algebra?

[coral spindle]

The hom-tensor adjunction, or more simply, the fact that they define bilinear maps.

[inner needle]

How do we prove this simple fact? If a and b are mth and nth primitive roots of unity where m and n are coprime, then ab is a primitive mnᵗʰ root of unity.

[night onyx]

Outside of the normal algebraic way of thinking about them, that they describe bilinear maps as linear ones, tensor products are the right algebraic structure to talk about functions on the intersection of varieties, in that C[x,y]/(f, g) is the tensor product of C[x]/(f) and C[x]/(g)

Or I guess more generally that functions on product spaces are tensor products of the individual function spaces

[inner needle]

I think I got it. But it's kind of convoluted.

[delicate orchid]

The roots of unity form an abelian group under multiplication

[inner needle]

Well, I just did it but it isn't elegant. I'm sure there is some theorem of group theory which gives the result quickly.

Considering μₘₙ as a group containing μₘ and μₙ as subgroups

[south patrol]

Yes this is great

[inner needle]

But their orders are ϕ(m) and ϕ(n)

[south patrol]

Oh yes sorry lmao

[inner needle]

So this doesn't really work

[south patrol]

Sorry yes

Lol

[inner needle]

Can I share the argument I wrote?

[south patrol]

Sure

[inner needle]

I'm not that pleased with it but it works

[south patrol]

The way I would do it is to show that mu_m and mu_n intersect trivially

Which you can do Galois-theoretically

for example

if memory serves

[delicate orchid]

Moderators!!!

[south patrol]

lol

[inner needle]

As you can see, this isn't very elegant.

[delicate orchid]

It’s also obvious that they intersect trivially when m, n coprime, just think about the orders of elements

[south patrol]

Wait lol

[south patrol]

No they aren't

Well

the orders of the groups of primitive roots are that

But you can just consider the orders of the elements, which was what my original argument was

Sadge

[delicate orchid]

Wow I wish I’d thought of that….

[south patrol]

Yes

[inner needle]

Hmm

Oh. Are you saying the element ζₘ has order m?

Oh jeez. I am so confused today.

The number of primitive roots is ϕ(m), not the order of the group.

I'm sorry. You were absolutely right.

[south patrol]

Dw.

[inner needle]

Can we use this to prove that ϕ is multiplicative or something? That would be cool!

[crystal turtle]

φ is multiplicative on coprime numbers: if m and n are coprime then φ(mn) = φ(m)φ(n). But this is not true in general (take m=n)

[lusty marlin]

In the context of number-theoretic functions, a function f is said to be multiplicative if f(mn) = f(m)f(n) whenever m and n are coprime.

[delicate orchid]

Yur. If it’s just straight up a homomorphism it’s totally multiplicative

[crystal turtle]

Ah, I did not know that

I don't like numbers.

[lusty marlin]

Let p and q be distinct primes. Then is there a characterisation of all positive integers n for which there exist simple groups of order pⁿq?

[delicate orchid]

There will never exist any simple groups of those orders, by burnsides all groups of order p^aq^b are solvable

[lusty marlin]

Burnside's lemma?

The one about counting orbits?

[delicate orchid]

Oh wait abelian groups

[delicate orchid]

There are no simple abelian groups of non-prime order. There we go

[delicate orchid]

No. The one that states what I said it states about p^aq^b groups

[lusty marlin]

I didn't mention abelian groups

[lusty marlin]

Ah ok. Could you provide a link or something to what it is?

[rocky cloak]

https://en.m.wikipedia.org/wiki/Burnside's_theorem

This one

[delicate orchid]

??? It was a specific case burnsides wouldn’t have covered

Because simple abelian groups are solvable

[lusty marlin]

Thanks!

[karmic moat]

burnsides has nice sideburns

[delicate orchid]

But all of them are prime order so we don’t care!

[rocky cloak]

So the characterization is n=0

[delicate orchid]

0 is positive now… I will inform the field theorists…

[karmic moat]

is there an analogous statement for non abelian groups? bc A5 is not abelian but simple with nonprime order

[lusty marlin]

Yeah sorry, I had gotten confused

[delicate orchid]

Yeah it’s called the classification of finite simple groups

[ivory trail]

can you explain the proof to us?

[ivory trail]

also this is a very funny statement to read out of context

[karmic moat]

oh true

[karmic moat]

eagerly awaiting your five volume proof wew!

[south patrol]

Saur good

[delicate orchid]

237985 years of character theory with ONE real world use

[south patrol]

Fr

Oh yeah my dissertation will involve character theory btw heh

[delicate orchid]

what ur fucked up alg top homology computations?

involving rep theory makes sense cause uknow Z[G] and what not but character theory specifically

[south patrol]

Nah that was my summer project

[delicate orchid]

ah

[south patrol]

Uhh this is like

So the starting point is like equivariant topology stuff which involves the representation ring

But then it gets harder and idk how

[delicate orchid]

I'd be surprised if character theory is involved

it's very uhhh K[G] with K a field ykwim

[summer path]

As opposed to not very K[G]?

[delicate orchid]

it's the vibe

[rocky cloak]

x^n + 1 = (x^m + 1)^2^k

Check that x^m + 1 is seperable by taking the derivative, thus has m distinct roots.

And I guess you mean your field to be algebraically closed or something, since if not you can of course face fewer roots of unity

[inner needle]

Sorry, I got it.

Yes. That is what I used.

[inner needle]

Oh yes, over an algebraic closure or a splitting field.

[delicate orchid]

the derivative result is so cool

[inner needle]

1 = -1, I completely forgot that is true, haha.

[karmic moat]

how is Nakayama's used here

R is an integral domain

[coral spindle]

But R_m is local

[karmic moat]

ye

but how do we conclude that mQ_m = 0

does that just hold true in general

[delicate orchid]

R_m is local, so J(R_m) = m

[karmic moat]

oh

yeah real okay

i forgot about that

[delicate orchid]

it's ok I can never remember nakayama's either

[karmic moat]

there are like

one kajillion versions of nakayamas

:(

[delicate orchid]

yeah ;3

the jacobson one is the useful one though

[karmic moat]

the only one i learned was 4

[delicate orchid]

I don't recall ever seeing that one being used

[summer path]

I learned it as 1 and 3

[karmic moat]

rats

[rocky cloak]

Very useful to show the existence of projective covers

[delicate orchid]

projective
this dude does cocohomology!!! ho ho ho he he he ha ha ha

[south patrol]

It is actually over a field

Like in K-theory one often works with like RO(G) and R(G)

[delicate orchid]

oh you're doing K-theory

can u commit to something it's hard to keep up PLEASE

then yeah they appear in like, the Aitayh-Segal theorem

which is a pretty fundemental K-theory result

[south patrol]

yes the atiyah-segal theorem is the starting point of the diss

hehe

[delicate orchid]

there's an analog for fusion systems

[south patrol]

Another thing is that you can define the adam operations on k-theory using the corresponding things for representations

oh really how come

[delicate orchid]

the situation just straight up lifts to fusion systems

you can't do like, BG/EG you need the nerve of some related category, but can't just use the fusion system itself

so it's little bit more annoying

[tardy hedge]

No proper subgroup of S4 can contain both (1,2,3,4) and (1,2)

Why?

[delicate orchid]

because they generate S4

tada!!

specifically I think it's because (1234)(12) = (134) which is a 3-cycle

order of S_4 is 24 = 2^3*3, the group generated by (1234) and (12) is at least order 8 (powers of (1234) and then those powers of (1234) times (12))

so you have a 3 cycle on top of that you have to have the whole group by lagrange

[tardy hedge]

Not sure on this part

So any subgrp containing (1234) and (12) will have at least order 8. There is also an element of order 3 in there cuz (1234)(12) is a 3 cycle

And the facts we r using is that orders of elements divide order of group and also lagrange thm right

[delicate orchid]

yeah so 3 also divides the order of <(1234), (12)>, so <(1234), (12)> has order 8*3 at least

but it can be at most 8*3, so it has to actually just be 8*3

[south patrol]

Hm okay sure

[tardy hedge]

Oh

[south patrol]

Nice

[tardy hedge]

What about order 12

[south patrol]

Wew has a nicer argument than what I was thinking about

Unsurprisingly lol

Only order 12 subgroup of S4 is A4 kian

This isn't a4 as it contains a 2 cycle

[tardy hedge]

What about if we didnt know what a4 was

I might be being stupid rn cuz im really tired

[south patrol]

I think the way I'd do this is uh

[delicate orchid]

without appealing to that, we have the identity element, (1234)^n for n = 1,2,3 and then (12) on it's own

so there are at least 5 elements all of order either 4 or 2, so again by lagrange...

[south patrol]

By conjugation we get (12),(23),(34), (41) and those generate S4

I believe actually for any n, (12),...,(n-1,n), (n,1) generate S_n

[delicate orchid]

yeah, S_n is rank n

[eager willow]

Is argument by order the best way here? A transposition (s,t) and an n cycle (s, t, ....) will always generate S_n, as it is easy to see how to get a bunch of other transpositions by conjugation.

[delicate orchid]

yeah ok I could generally prove that fact

[south patrol]

You can do this by induction on n - basically take a permutation f, multiple by these guys so f fixes one point, then induct

[delicate orchid]

but we're working in S_4 specifically so why not just use low level machinery

[south patrol]

Sure

[south patrol]

Hehe

Beat you to it ig lol

[delicate orchid]

consider the uhhhhhhhhhh standard representation

[south patrol]

But yeah for this thing probably wew's is nicest

By a similar argument you can show any p cycle and any 2 cycle generate Sp for prime p

Nice exercise

[delicate orchid]

stinky exercise...

[tardy hedge]

I have to look at this later when im not as sleep deprived

[delicate orchid]

I'm actually kind of warming up to this method instead of mine if you know beforehand that these generate S_4

[tardy hedge]

If you have (1234) and (12) in the grp then the subgroup needs to be at least order 12 cause (1234) has order 4 and (1234)(12) has order 3

Thts true right

[tardy hedge]

I dont get this

[delicate orchid]

so the group is at least order 8

cause 4 has to divide the order of <(1234), (12)> and there's at least 5 elements

[delicate orchid]

do you know this result btw, if you know this result we won't have to worry about this nonsense

[tardy hedge]

I dont

[tardy hedge]

Ok yea

But i think im still stuck on why cant it be order 12

4 3 and 2 divide 12

[delicate orchid]

because 8 doesn't divide 12?

we have a subgroup of order 8

[tardy hedge]

U said at least order 8

Is there an element of order 8?

[delicate orchid]

no, there are no elements of order 8 in S_4

we've established that <(1234), (12)> is at least order 8

it also has an element of order 3 in it, and so is at least order 3*8

it is contained in a group S_4 of order 24 = 3*8, so it must be the entire S_4

[tardy hedge]

I get what u saying but why cant this be order 12

[delicate orchid]

becuase. 8. doesn't. divide. 12.

[tardy hedge]

<(1234),(12)> is at least order 8 because that has order 4 so there are 4 distinct elementd and multippy each one by (12) u get a different element

[delicate orchid]

sure

[tardy hedge]

Ok so any subgroup that contains those 2 must be at least order 8

[south patrol]

Le Grange

[tardy hedge]

Is that just equal order 8? Why are we saying at least

If <(1234),(12)> is order 8 then any subgroup containing those elements has <(1234),(12)> as a subgroup so the subgroup needs to be divisible by 8 so it cant be 12

[delicate orchid]

yur

[tardy hedge]

So im pretty sure i was hung up on the fact that we said at least

When it was really = to 8

Like we know for sure <(1234),(12)> order is 8

[delicate orchid]

NO! It's order 24!!!

[tardy hedge]

Fail

[delicate orchid]

that's the entire point!!!

[tardy hedge]

Imma just walk out

🚶🏽‍♂️

[delicate orchid]

(1234)^n(12)(1234)^-n = (n+1,n+2 mod 4), this is a generating set of S_4
yeah there we go

ergo entire group is S_4

[tardy hedge]

Loolll yeah i just said it before

[delicate orchid]

so you had this fact

[tardy hedge]

I just said “IF it was order 8 then it wouldnt work cuz …”

Then i said its order 8

Lol

Ok yeah wtf it cant be order 8 cuz of that so it has to he order 24

Oops i cant say the r word

I was being r lol

Thx

So for dihedral groups, what are the main things to keep in mind?

I know they are subgroups of Sn

[delicate orchid]

D_2n = <a,b| a^n = b^2 = 1, bab = a^-1> and that's it

[tardy hedge]

That represent rigid motion symmetries of n gons

[tardy hedge]

Yeah

And that thing

[delicate orchid]

uhh centre is trivial for n odd, iso to C_2 for n even uhhh

[tardy hedge]

Yeah the book just also did some subgroup diagram of D4 and stuff

[delicate orchid]

they're metacyclic

[tardy hedge]

And did it based on orders and lagrange

My midterm is tmr brw

Btw

I feel like a lot of early material is based around orders

And lagrange

Cyclic stuff and all that

[delicate orchid]

maybe, I couldnt' comment

it's been 5 years since I was introduced to groups

[tardy hedge]

I see

[delicate orchid]

this is true of every finite group btw

[tardy hedge]

Only 5 yrs actually seems surprising to me tho

[tardy hedge]

Lol oh yeah

Cayleys theorem

[delicate orchid]

yurrr

[tardy hedge]

We covered that but im not rlly sure how we will be tested on it

Seems kind of like a “oh ok cool” kind of thing

[delicate orchid]

you might be asked to compute an explicit embedding

for example, can you find an embedding of C_2 x C_2 into S_4

[tardy hedge]

What is C?

[delicate orchid]

cyclic
I'm not using Z/nZ because these aren't rings

[tardy hedge]

Oh yea ok

I like how all cyclic stuff is just integer stuf

That was cool

[delicate orchid]

yeah all abelian stuff is basically just Z/nZ

[tardy hedge]

I dont think we got there yet

[delicate orchid]

just usually more than one Z/nZ

yeah you won't see the structure theorem for a while

[tardy hedge]

I thought this was just cool

Like basically finding formula for how many coprime ints less than n but using all this stuff

Like its cool the answer to a “simple” question can be solved using this stuff

[delicate orchid]

yeah that corollary is like, a specific application of a much larger ring theoretic result called the chinese remainder theorem

[tardy hedge]

Oh cool

[abstract rock]

something i remember from my intro to groups course that might be on a test will be showing that there are multiple cayley tables for the same group

[tardy hedge]

Hm wdym by that?

Cayley table is just a mult table right

A group should just have one multiplication table right?

Or are u saying groups of same order could have different tables

Like group order 6 is either cyclic of S3

[rocky cloak]

You can shuffle the rows and columns I guess

[tardy hedge]

Or