groups-rings-fields

[maiden ocean]

lmfao the *q discourse

Yeah because insane 17 year olds like to ruin #discussion with debates on that topic

[rustic crown]

haven't read lee, but ig yea... spent a lot of time with covariant derivative stuff and defining different thigns like geodesics and curvature tensor from it

[maiden ocean]

makes sense

Not having like. a normal undergrad diff geo course is the weirdest thing about this place's curriculum

I think it used to be a thing and then whoever taught it retired or something but its still pretty inexcusable lol

[high cypress]

@rustic crown how does a silly billy like me learn topology?

[rustic crown]

what you doing in topology

our top1 is doing standard stuff with homology theory

[maiden ocean]

oh thats not even undergraduate material then

at least usually not

[high cypress]

No like

How do I learn it

[rustic crown]

make pictures in head

idk what else to say lol, unless you ask something more specific

[high cypress]

Like what book did you use

Did you use YouTube?

[rustic crown]

i read some stuff from munkres, but found it to be a pretty boring book

i might try tom dieck or peter may

[gilded gulch]

kanna!

[lethal dune]

Try bredon

[rustic crown]

okie

[next obsidian]

Try Chmonkey

[tender wharf]

chmonkey textbook? :O

[rustic crown]

[high cypress]

Fr?

[solar glacier]

Munkres is a classic text but mainly only good for point set top not alg top

Good stuff

[solar glacier]

Topology and Geometry by Bredon?

[lethal dune]

yes

[high cypress]

Wait, that is the difference in application of point set and algebraic topology?

[mighty spade]

algebraic topology involves a lot of algebraic invariants of spaces

I have no idea how to express what point-set topology is as compared to other branches of topology

It’s like your bread and butter definitions

Plus separation axioms stuff I guess

[warm wyvern]

algebraic topology is studying topological spaces using tools from algebra

pointset topology is what defines what a topological space is in the first place lel

pointset is very very bare bone

so you can't prove a lot using it alone

but it's the necessary formulation to do more advanced stuff

[hidden haven]

Algebraic topology is when homotopy and point set is when homeomorph

[elder wave]

Real

[ember field]

Where from in India?

How is f a unit here?

C(X) is the ring of all real valued continous function on X where X is compact hausdorff.

[elder wave]

if it doesn't vanish anywhere then you can define its inverse by 1/f(x) if it's real valued functions

more of a wild guess, i'm missing a bit of context here

[south patrol]

Indeed smth is a unit iff it vanishes nowhere ig

[nocturne bone]

Why can't the kernel be trivial in c?

[formal ermine]

what's gamma_g

[rustic crown]

conjugation

[rustic crown]

because if kernel is trivial then the map is injective. so |G| will divide |S_X|, i.e. 72 divides 4! = 24

[gleaming fable]

Hello
I was wondering about something
Now for an inteval A to be called a group it should have a law ◇ that is :

1. ◇ is associative
2. ◇ admits a neutral element
3. all elements from A are symetrisable by ◇
   • if ◇ is commutative then (A,◇) is a commutative group
   (Sorry if my terminology is bad I don't know how to say most of these terms in english)

•Now my question is
Does A have to be 'stable' with ◇ for it to be a group?
Like : ∀ x,y ∈ A , x◇y ∈ A
Like for example the interval 'N' isn't stable with the law '-', etc
• since all elements should be symetrisable, meaning symetrisable from the left and the right
x ◇ x' = e
x' ◇ x = e
Doesn't that mean that ◇ is commutative? Or is it possible for x to be symetrisable from both left and right while the law is not commutative??
Thank you for anyone who's gonna answer

[rustic crown]

'stable' is usually called the closure property in english

we say A is closed under the operation

[gleaming fable]

Oh thank you

[gleaming fable]

I'll remember that 🙏🏼🙏🏼

[rustic crown]

but people don't write that in the definition because that's already included in the definition of what it means to be a operation

when you say, ◇ is a binary operation on A, what you really mean is a function
(◇) : A x A --> A
where we write a ◇ b := (◇)(a, b)

[gleaming fable]

But - is an operation and while R is closed under - , N isn't
Doesn't that make it not an operation?..

[rustic crown]

so you see the target is assumed to be A already

yea so - isn't a binary operation on N

[formal ermine]

btw a commutative group is usually called an abelian group

[rustic crown]

you would only talk about closure under some operation if it's already defined in some bigger set

like if you see the different conditions for H to be a subgroup of G

one requirement is that H is closed under the operation

which is needed because we use the operation of G to get something on H

[gleaming fable]

Oh
I understand....
Alright that cleared things up thank you 🙏🏼

[gleaming fable]

Thanks !!

[rustic crown]

i.e. you use the inclusion map H --> G and use the operation on G to get H x H --> G x G --> G

[gleaming fable]

Alright I get that part

But what about the neuter element

Isn't having a neuter element from both left and right makes it commutative/abelian ?

[formal ermine]

(neutral element)

[gleaming fable]

Oh
Thanks again

[rustic crown]

yes we want the element e to be both left and right identity

and inverses should also be left and right inverses

[gleaming fable]

Wait no
Not the neutral element sorry i mistaken

I meant the symmetrical element

The element that :
x ◇ the symmetrical = the neutral element

[rustic crown]

people put this in the definition because that's how they think about it, it's enough to ask that there is an element e such that e * x = x for all x and for each x an element x' such that x' * x = e. it requires a little bit of work to prove this is equivalent and is a little boring

[gleaming fable]

Should the symmetrical be both from left and right?

[rustic crown]

yep

you want x to commute with its powers

x^m * x^n = x^{m+n} = x^{n+m} = x^n * x^m

but two arbitrary elements in your group may not satisfy that x * y = y * x

[gleaming fable]

Ok but if it was symetrisable from both sides
Doesn't this make it abelian ?..

[rustic crown]

nope, it's only abelian if that is true for all pairs of elements, not just the very special pairs (x, x')

[thick ledge]

$E_o$ is supposed to be a subset of an unknown extension field $\hat{E}$ acquired by evaluating all polynomials of degree less than that of a known irreducible polynomial $f(x)$ at a point $\alpha$. Is this proof basically saying that since $F(\beta)$ and $F'(\beta')$ mimic $n$-dimensional vector spaces over $F$, they're naturally isomorphic to $E_o$?

This is from Emil Artin's Galois Theory book

[cloud walrus]

Manifold

[gleaming fable]

Ummm so it have to be commutative for the pair x x' but not with other pairs?

[thick ledge]

also, sidenote - why does homological algebra go here and not in #algebraic-geometry? aren't homological and commutative very heavily tied or something?

[tawny magnet]

yes x and x^-1 commute but not necessarily everything

[rustic crown]

yep

[warm wyvern]

are you french by any chance?

[gleaming fable]

Ah thank youuu guys
I understand it now
That was kind of you

[gleaming fable]

I am

[warm wyvern]

I see

🫂

[gleaming fable]

🫂🫂🫂

[rustic crown]

aww

[tawny magnet]

how did you guys just recognise that both of yall were french? ahahah

[rustic crown]

yep looks like it

[gleaming fable]

Probably terminology?..

[warm wyvern]

ye

[formal ermine]

translating terminology can be messed up at times

[tawny magnet]

je vois je vois

[formal ermine]

a monic polynomial is called normed in german

[warm wyvern]

such a weird terminology

like

"l'inverse" is a french word too

just use that ffs

[delicate orchid]

if you call the inverse the symmetric element then how tf do you talk about S_n

[rustic crown]

inverse group on n letters

[coral spindle]

Admittedly, "symmetric group" isn't a great name

[tawny magnet]

😂

[rustic crown]

[delicate orchid]

general linears on the field with one element

[warm wyvern]

just "groupe symétrique" lmao

[thick ledge]

ah, was worried I didn't get it lol - thanks!

[chilly ocean]

symmetry group

[rustic crown]

but you really need to prove that they are isomorphic as fields, not just vector spaces

usually people would prove this
E iso to F[x]/(f)

[thick ledge]

that was proved earlier in the chapter

first he considers a known field extension and element of interest and then looks at a subset of polynomials (degree less than that of the irreducible corresponding to the element) then shows it's isomorphic to starting with the ambient field and extending it using the same irreducible

[rustic crown]

oh okie

[formal ermine]

automorphism group on n elements

[rustic crown]

.<

[delicate orchid]

Aut(X) with X being a set is correct

[elder wave]

tldr

[rustic crown]

just use permutation group >.<

for auto you need to tell me the category

[delicate orchid]

I will use the gosh darn permutation matrices

[coral spindle]

Permutation group refers to all groups actually

[delicate orchid]

I did! it's in set!!

[coral spindle]

Every group is (isomorphic to) a permutation group

[formal ermine]

what's a permutation group

[chilly ocean]

the category of categories

[rustic crown]

you only told me the object is set

[delicate orchid]

a group of permutations

[coral spindle]

A group consisting of permutations

[formal ermine]

ah

[elder wave]

[coral spindle]

jinx

[formal ermine]

ok that makes it clear

idk why I was asking that

it's just a subgroup of a sn

[chilly ocean]

permutation group is a subgroup of symmetric group

[delicate orchid]

how could the morphisms not be in Set if their codomain and domain are in Set

[rustic crown]

lol

[formal ermine]

beat you to it

[delicate orchid]

don't see what tin has to do with this

[chilly ocean]

beat me? I only confirmed what you was confused about

[rustic crown]

subcategory of set with only identity functions 😂

[coral spindle]

The monoid of all functions A->A is called the full transformation monoid. Maybe then a more consistent name would then be the "full transformation group" :) but ofc in the case that A is infinite, we may not want to consider all bijections

[chilly ocean]

full transformation semigroup

[delicate orchid]

my favourite!!

[chilly ocean]

is used more often tbh

[coral spindle]

I've only every seen it called a monoid 🤷

[chilly ocean]

no, go on like wikipedia

pretty sure they mention it there

it's a monoid, yes, but they still call it semigroup more often than not

[coral spindle]

OK

🤷

[chilly ocean]

everyone can edit wikipedia

[coral spindle]

Yup

[rustic crown]

you don't need abelian uwu

[warm wyvern]

I get that the idea is to repeated take quotients of elements with a prime order

[warm wyvern]

cauchy theorem yea

lel

that's 2 chapters from now

[rustic crown]

yea

if you find p, stop

[warm wyvern]

till you get an element with the order you want

[rustic crown]

else keep looking in the quotient

then bring this back to an element of order p in your original group

[chilly ocean]

my hot take is that there's really no reason to study monoids any more than it is to study semigroups
the identity only becomes relevant when we're doing ring theory or group theory

[rustic crown]

what about monoid ring >.<

[thick ledge]

Since f is irreducible (but with its root 's' in E), and the polynomials are a factorial ring with maximal ideals corresponding to irreducible polynomials, F[x]/(f) would be a field by applying Bezout's Lemma, and the homomorphism from F[x]/(f) to E given by evaluating polynomials in the latter at the root is an isomorphism

Is that how what you said would work?

[formal ermine]

are there group actions but for rings

so like a ring acting on a set

[chilly ocean]

but studying both seems pretty useless if unless you're studying them for studying sake

[rustic crown]

you act ring on abelian groups

[formal ermine]

I want ring on set

[rustic crown]

like you act groups on anything because Aut_{C}(A) is alwasy a groups

you can act ring on abelian groups because End(A) would be ring

[warm wyvern]

yeah, so that's what I'm a bit hazy on

[coral spindle]

How would this action behave wrt addition in the ring? If it has no bearing, you're just talking about monoid actions.

[formal ermine]

so I was thinking like a module but the underlying abelian group is just a set

so no adding moduli

[coral spindle]

So you're just talking about a monoid action

It is an interesting question to ask what monoids can also have a ring structure, but I doubt there is a nice characterisation

[warm wyvern]

my idea is if your element is h and the last quotient was by <k> (which has order q) then h^p<k>=<k> then either h^p=1 or it's an element of <k> which means h^p has order q which mean h^q has order p

does that make sense?

[coral spindle]

It does to me, DarQ

[rustic crown]

h^p could already be identity but yep

h^q will still have order p, as q is a different than p

[warm wyvern]

I did mention that

[rustic crown]

ah oops

my bad

[warm wyvern]

no worries

thank you guys for the sanity check

[analog zephyr]

Guys

The proof of characterization of all finite simple groups is in the internet as pdf?

And sorry for my English ( I'm Latin)

Mmm idk if it's the proof that I looking for, the ten thousand page proof

Or so they say

[tiny jolt]

I think the “10k pages” was over many papers, not just a single one

This book compiles them all I think though https://books.google.lk/books?id=lYMAg_Sj7hUC&printsec=frontcover&dq=finite+simple+groups&hl=en&sa=X&ved=2ahUKEwiJqr2Yl6T8AhXa1XMBHYbFC4sQ6AF6BAgCEAI#v=onepage&q=finite simple groups&f=false

[analog zephyr]

Really, I thought it was just about the proof

[chilly ocean]

whole 10 thousand pages?

in a single book?

[tiny jolt]

oh no, i just mean it compiles all the knowledge needed to classify them

[analog zephyr]

Thanks you, I'll save this link for later

[analog zephyr]

And even so is too much

[formal ermine]

latin?

[tribal moss]

Probably means "American with Spanish as their first language".

Usually that's said as "Latino/Latina" in US English, but if they feel insecure enough about their English to apologize for it, they might reasonably have assumed that the word would lose the gender ending in English.

[glossy crag]

how is the notation MR_S meant to be understood? is it meant to be the same as the product of ideals, i.e. MR_S is the set of sums of products of elements of M and R_S?

[thick ledge]

$MR_S$ means the set of all products of elements of $M$ and $R_S$, where $R_S$ means all the fractions that can be built using $R$ and $S$. The exercise is asking you to show that the fractions you can make using the ideal and $S$ are the same as the product of all ideal elements and ring fractions.

[cloud walrus]

Manifold

[glossy crag]

So $MR_S={\frac ms|m\in M,s\in S}$?

[cloud walrus]

Ocean Man

[glossy crag]

So MR_S happens to be the set of all products of M and R_S, but generally speaking, if R is a ring and X,Y are some subsets, then XY is the set of all sums of products, right (it's just in this case the sums can all be brought to the same denominator)?

Just to make sure I understand the conventions correctly.

[thick ledge]

as far as I'm aware, XY just means the set of products and not necessarily sums

sometimes they end up encapsulating sums but as far as notation goes that's the safest bet

[thick ledge]

To be a bit more precise,
$R \times S$ is the set of all pairs of elements of $R$ and $S$
Now, what we do is quotient $R \times S$ by an equivalence relation --- we want these to behave exactly like fractions, so we'll say that $(r, s), (r', s') \in R \times S$ are equal iff there is some $u \in S$ s.t. $u(rs' - r's) = 0$. The equivalence classes in this quotient are the fractions we meant, so $R_S := (R \times S) \slash \sim$

$MR_S$ is then ${m \frac{r}{s} : m \in M, r \in R, s \in S}$ and $M_S := {\frac{m}{s} : m \in M, s \in S}$

[cloud walrus]

Manifold

[thick ledge]

you'll need to show these are equivalent

I'm guessing setting up a bijection should do

[glossy crag]

What are, MR_S and M_S? They're equal considered as subsets of the fraction field of R (the context here is domains).

[thick ledge]

they ought to be the sets defined above

$MR_S$ is the product of elements of $M$ and fractions, $M_S$ are the fractions with $M$ as the numerator

[cloud walrus]

Manifold

[glossy crag]

The "what are?" was directed to this (these are equivalent).

[thick ledge]

ohh

yeah you need to show those two are equivalent

[glossy crag]

I don't think that's necessary

They're clearly the same thing in the quotient field.

Where R and R_S are embedded.

I mean there's no question of "equivalence", they're just equal, plain and simple.

[thick ledge]

yeah definitely

okay, wait, no, dang it I've misread the question

yeah $M_S$ is just defined as $MR_S$

[cloud walrus]

Manifold

[thick ledge]

my bad for probably being confusing

[glossy crag]

Yeah my question was just how to interpret MR_S

I still think the general definition (for all manner of subsets) ought to be about sums of products (e.g. IJ would not be an ideal if it were just products), it's just in such cases it defaults to products because of the absorption of sums by one of the factors.

[thick ledge]

wait, but what about $(2)$ and $(3)$ in $\mathbb{Z}$

[cloud walrus]

Manifold

[glossy crag]

That's a PID, things are ultra-simple there.

[thick ledge]

you get $(2)(3) = (6)$, but $(2) + (3) = Z$

[cloud walrus]

Manifold

[glossy crag]

You mean (2)(3) should be interpreted as the set of products?

[thick ledge]

that's what it means when you usually state it like that

[glossy crag]

I'm 100% sure that for ideals my definition is the correct one (don't know about general subsets).

Open any textbook on algebra.

[glossy crag]

the ideal generated by products

that's the same thing as the set of all sums of such products

What you said is literally refuted in the line below

"Note that just the set of products would in general not be an ideal".

We are in agreement?

[thick ledge]

ah

my apologies once again

[glossy crag]

All good, discourse is fun (usually).

[thick ledge]

indeed

[rotund aurora]

uh bruh

https://www.ams.org/publications/authors/books/postpub/surv-40

I wanted to send this

[rotund aurora]

in response to this

like they are trying to make the whole proof "accesible". But idk if anyone cares tbh haha

[warm wyvern]

what does adding up ideals mean??

[rustic crown]

I+J

= {i + j | i in I, j in J}

[chilly ocean]

but IJ is not {ij}

[next obsidian]

It is if I or J = 0 or (1)

And that’s what matters most

[warm wyvern]

oh

that's it?

[next obsidian]

Yah

It’s the smallest ideal containing I and J

[solar glacier]

could I run a proof by you guys

[next obsidian]

No 😢

It has to walk by us

[warm wyvern]

[cloud walrus]

MyMathYourMath

[next obsidian]

What?

Maybe you mean phi(n) in phi(N)?

[cloud walrus]

MyMathYourMath

MyMathYourMath

[solar glacier]

then

[cloud walrus]

MyMathYourMath

[solar glacier]

But then

[cloud walrus]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[next obsidian]

Yee

[rustic crown]

Yee

[solar glacier]

cool

[next obsidian]

You might have already proven this tho

Without realizing it

If you have proven that subgroups of G/K are the same as subgroups of G which contain K and that it sends normal subgroups to normal subgroups

You can replace H with G/K by the first iso

[next obsidian]

And then this tells you that N is normal because N/K is

😮

[solar glacier]

woah, i just brute forced it lol

[next obsidian]

So eventually you have to brute force it

Because proving that the correspondence of like subgroups of G/K and subgroups of G containing K

Showing that this sends normal to normal

[rustic crown]

also if you wanna brute force, then use maps :p
N is the kernel of G --> H --> H/phi(N)

[next obsidian]

Requires you to brute force

[next obsidian]

True af

Very very true

[light iris]

When is the cartesian product not the same thing as the direct sum of two structures

Im confused after looking this up online

[south patrol]

Uh well whether it is the same depends on context ig

[light iris]

I guess ill see a case where its differnt when i need to

[south patrol]

The main case is groups in general

[glossy crag]

I don't think this ever happens for two structures, but for infinitely many it does.

[light iris]

Wouldnt the cartesian product of two groups always be a group if you define multiplication component wise

[solar glacier]

i think its the same for finite

[solar glacier]

yes

[rustic crown]

most of the algebraic structures you study in the start are just enrichment of sets, and in particular the forgetful functor happens to be a right adjoint functor (the left adjoint being the free functor). Now since forgetful is right adjoint it preserves products, so the underlying set of a product will be product of the underlying sets

[coral spindle]

In the category Set

if by "direct sum" you mean "coproduct" ofc :)

[light iris]

I dont think i know enough to understnd this yet but thanks

[light iris]

Im just starting to learn about category theory so hopefully i can learn about that soon

[rustic crown]

oh so basically if there is a notion of free objects, like free groups, free R-algebras, free monoids, free modules, then you'll have that the cartesian product naturally has the enriched structure

[tribal moss]

Another answer is "when you don't just have two structures but an infinite family of them".

[light iris]

Oh and this isnt necessarily true for whatever a "not free" object is

[light iris]

Ok

[tribal moss]

Even though this is true, I think it might be missing the mark a bit, since the question said "direct sum" rather than "product".

[light iris]

Hm

[rustic crown]

nu, having free objects would mean that looking at the underlying set of an object has some really nice properties, one such is that it preserves products

[light iris]

i understand how these are different things for vector spaces but not groups

or rings

mainly cause i dont know as much about abstract algebra

Whats wrong with having infinitely many that creates a difference

[tribal moss]

There's nothing wrong, but you get two different possible results. One is just the cartesian product of the familiy; that is usually known as the "product" of the structures. The other also starts with the cartesian product, but then discards all the elements that have nonzero values at infinitely many of the positions. This is called the "direct sum"; it is the smallest structure that each of the original structures inject into (such that the injections are, in a technically precise sense, independent from each other).

[light iris]

Sorry wrong was the ironically wrong word

I see, so its about the way the direct sum is defined over an infinite collection

[tribal moss]

Yes.

[light iris]

By definition you don't include anything thats nonzero for more than finitely many "places"

but in a cartesian product you would include that because it is every imaginable (ordered) combination of things from each of the infinitely many objects

[tribal moss]

Yes.

it is the smallest structure that each of the original structures inject into
(this is only true if all the operations of the the algebraic structures you're considering are commutative, by the way. Otherwise this characterization calls for a "free product", which is a much more hairy beast).

[light iris]

Does that have something in common with the idea that a linear combination in an infinite dimensional vector space must be zero ** for all but finitely many dimensions?

because the idea of an infnite sum is weird?

[tribal moss]

Yes, exactly. In the case of vector spaces or modules, a direct sum consists of linear combinations of everything from the incoming structures.

[light iris]

Got it, thats pretty cool

[rotund aurora]

https://math.stackexchange.com/questions/39895/the-direct-sum-oplus-versus-the-cartesian-product-times

You may be interested in that

[light iris]

Thanks for sharing that

i read it but after i saw the accepted answer and didnt understand a lot of the words they used i closed the tab. now i see that the 2nd answer is understandable for me

[rotund aurora]

Its very useful to look at diagrams

[light iris]

I saw the word categories in the first and assumed i wouldnt get it.. but now i see it makes sense lol

i agree, but i also find making visualizations for a lot of these things can be hard

especially when infinity is at play

[agile burrow]

To see that it's independent, suppose you have another set of coset representatives y_1, ..., y_m. Then you can write y_i = x_i h_i for some h_i in H. Substituting these into the product gives you something that should look nice, but in the end you will need to use the fact that H is abelian to cancel out relevant terms

<@&268886789983436800>

[tiny jolt]

For homomorphism– I assume you showed that $x_{\overline{g}(i)}^{-1}gx_i\in H$. To show that
$$T(g_1g_2)=T(g_1)T(g_2)$$ think about rearranging to get the $x_{\overline{g}(i)}^{-1}$ and $x_i$'s to cancel out (since $H$ is abelian)

[cloud walrus]

pramana

[agile burrow]

Yeah, something like that sounds right

[tiny jolt]

I meant you could rearrange the $x_{\overline{g}(i)}^{-1}gx_i$'s

[cloud walrus]

pramana

[tiny jolt]

since those are guaranteed to be in H

[agile burrow]

transfer is cool

[tiny jolt]

ive never heard of it, yeah that's sick

[agile burrow]

it's actually a specific case of a more general map in group (co)homology

you'll notice that all the above constructions carry through even if H isn't abelian by just replacing H with its abelianization

In group homology you learn that the first homology of a group is its abelianization, so transfer is just the composition G -> H_1(G) -> H_1(H). In particular, you need H to be a finite index subgroup for the latter map to exist

[next obsidian]

Is H_0 the group itself

[agile burrow]

H_0 is the coinvariants

[next obsidian]

Actually idk

[agile burrow]

M / (gm - m)

[next obsidian]

Chmonkey

[agile burrow]

or just tensor with Z over ZG

[next obsidian]

[agile burrow]

but really the way to think of it is as the first homology of a K(G, 1)

[next obsidian]

True

[agile burrow]

because then just use Hurewicz

[next obsidian]

This I know

Is a thing

Idk why but

It’s a thing

[agile burrow]

you don't know why hurewicz is a thing?

[next obsidian]

Why homotopy groups of K(G,1) are group chohomolgoy

[agile burrow]

homology groups

homotopy groups of K(G, 1) is just G in degree 1 and 0 otherwise lol

[next obsidian]

Oh yeah

I meant that

[agile burrow]

But yeah, universal cover of K(G, 1) is contractible, right?

because it's simply connected and higher homotopy groups are the same as higher homotopy groups of K(G, 1)

[next obsidian]

Yeah

Or cuz u just made it directly

And contracted it to a point

💪

[agile burrow]

Real

But yeah, the cellular chain groups of the universal cover are free ZG-modules

which kinda just follows from the cell structure on the universal cover

so you have a free ZG-resolution of Z

[next obsidian]

Walter dropping the group cochmology knowledge

[agile burrow]

when you mod out by the action of G, that's the same as tensoring with Z over ZG

call the universal cover X

[next obsidian]

I wanna call it EG

:(

[agile burrow]

ok fair

you get C(EG/G) = C(EG)/G

[next obsidian]

[agile burrow]

But EG/G is just K(G, 1)

so you recover the homology of the group

oops I guess I should've defined the group homology to be Tor^{ZG}(Z, M)

[next obsidian]

Waltermology

[agile burrow]

real

but yeah, homology is derived functor of coinvariants, cohomology is derived functor of invariants

and it all agrees with singular homology and cohomology of K(G, 1)

I am very glad that I actually understand this stuff enough to talk about it lol

[bleak abyss]

Alright

Let's do some of that algebra

@sturdy marsh get in here. Not sure why but do it

[sturdy marsh]

ok ;star

algebre

[bleak abyss]

So first off

Gonna understand restriction/extension of scalars for linear algebraic groups

[cloud walrus]

Sloth King Daminark

[bleak abyss]

Now schemes go the opposite way

[cloud walrus]

Sloth King Daminark

[bleak abyss]

This is a set-valued functor, but the question is, is it representable by Spec of some k-algebra?

Wellllllll

Sometimes

If k'/k is a finite degree field extension, or more generally k'/k is finite and locally free

[cloud walrus]

Sloth King Daminark

[next obsidian]

Mf decides to use the algebra channel, not the AG channel for this

[cloud walrus]

Sloth King Daminark

[bleak abyss]

This... doesn't look right

[next obsidian]

It’s maps A -> R’ over k

[bleak abyss]

Yeah that's what I expect

I'm just trying to get there from hom tensor adjunction but I think I fucked up

[next obsidian]

That Hom is just R’

Maps from k’ over k’ is just R’

[bleak abyss]

Oh

[next obsidian]

Lol

[bleak abyss]

Yeah so I was thinking k-algebra homs had to be ring homs for a sec

In the sense of

F(1) = 1

And I was like

How the fuck is this hom space not trivial

[next obsidian]

It is…

What

That is Hom_k’(k’,R’) as module maps

Hom-tensor is about modules

[bleak abyss]

Oh oh

[next obsidian]

You have to trace out the isomorphisms there which are only true as k’-modules and see they’re actually multiplicative

[bleak abyss]

Gotcha

[cloud walrus]

Sloth King Daminark

[bleak abyss]

So an example is the Deligne torus

[cloud walrus]

Sloth King Daminark

[knotty frigate]

for groups(and for rings and abelian rings) we have one for each positive integer
Isnt that false for fields though? i.e. arent there positive integers such that there is no finite field of that cardinality?

[cloud walrus]

Sloth King Daminark

[next obsidian]

Yes, there’s exaxtlt one field for every prime power

That’s all of them

[knotty frigate]

so 6 is the first one without a field?

[cloud walrus]

Sloth King Daminark

[next obsidian]

And 1 haha

[knotty frigate]

ig but i prefer the field axioms without the 1≠0 constraint

haha

[next obsidian]

Here’s a quick and dirty proof that each finite field has to have order a prime power.

The map Z -> F has kernel a prime number. So F contains some Z/pZ, and is then a vector space over Z/pZ, so it has order p^n

[knotty frigate]

is this like a version of the fundemental homomorphism theorem for rings?

[next obsidian]

It’s the first isomorphism theorem yeah

[knotty frigate]

oh okay

[next obsidian]

You can also see it like this:

A finite field has prime characteristic

So like, let n = 1 + … + 1 n times

First off, n = 0 for some n > 0

Or else it’s infinite

Then n has to be prime, cuz if not you’d have like n’n’’ = n = 0

Where n’,n’’ < n so they’re non-zero

But then it’s not a field cuz zero divisors

So p = 0 for some p

[knotty frigate]

whats n' and n'' here?

[next obsidian]

Then the set {0,…,p-1} is isomorphic to Z/pZ as a field

[bleak abyss]

(I'll move to #algebraic-geometry)

[next obsidian]

Just factors of n

If n wasn’t prime

So this is a proof that F contains Z/pZ without first iso

[toxic zephyr]

sorry to interrupt. i have a stupid question. if you have something like x^(ak)y^(bk) in something like the dihedral group, is there some way to simplify that/make it look nicer, since the powers have a common factor of k? i have an ugly expression, and it'd be nice to condense it a bit.

[next obsidian]

Unless you have a relation that lets you do it, no

[toxic zephyr]

$x^{m_0(1-n)}(sx)^{(m_1-d_1)(1-n)}$ in $D_n$ (where $n$ is odd)

[cloud walrus]

nilpotent nix

[toxic zephyr]

rip

[next obsidian]

For the dihedral group

Yeah I don’t think there’s a great one

Like you can reverse them

And pick up an inverse somewhere

Assuming like x is rotation, y is a flip let’s say

But there’s not a great way to like combine it and condense it

Afaict

[toxic zephyr]

i had some hope that maybe because the common power was 1-n, then maybe we could do something special

but i guess it'll just have to be ugly lol 😆

thanks anyway haha

[next obsidian]

I mean maybe there’s something

But it isn’t immediate

[delicate bloom]

it'd be cool if you made that render as latex

[glossy crag]

How can I show the inclusion $(MN)^{-1}\subset M^{-1}N^{-1}$ for (finitely-generated) fractional ideals $M,N$? I was thinking something along the lines of: if $M$ is generated by ${x_i}$ and $N$ by ${y_j}$, then ${x_iy_j}$ generate $MN$, so if $z$ is such that $zMN\subset R$, we get that $zx_i\in N^{-1}$ for all $i$ and $zy_j\in M^{-1}$ for all $j$. Now I'd be done if I could express $z$ as a sum of products of these somehow, but this I can't figure out.

[cloud walrus]

Ocean Man

[gleaming fable]

Hello
Just wanna make sure, am I correct here? :

[coral spindle]

This is not the right channel for this question. You're looking for #prealg-and-algebra. I'll say that although you don't need the part where you take square roots, your working is correct.

[gleaming fable]

Oh sorry I missed used the channel (since the subject I was aolving is algebra..)
+thank you :)

[glossy crag]

Anyone?

Actually I have a solution for when the base ring $R$ is a Dedekind domain, but I feel like $(MN)^{-1}=M^{-1}N^{-1}$ should hold for all integral domains (if only because the exercise shows up in the text before the prime decomposition of fractional ideals is proven).
Here's my solution for $R$ a DD: as mentioned above, if $z\in(MN)^{-1}$ and ${x_i}$ generate $M$, then $zx_i\in N^{-1}$. Because all fractional ideals over a DD are invertible we have $R=MM^{-1}$, in particular $1=\sum_k\mu_k\widetilde{\mu}_k$ where $\mu_k\in M,\widetilde{\mu}_k\in M^{-1}$. Rewriting this in terms of $x_i$ you get $1=\sum_ix_i\widetilde{\mu}_i$ (for some other $\widetilde{\mu}_i\in M^{-1}$), so $z=\sum_izx_i\widetilde{\mu}_i\in N^{-1}M^{-1}$.
Does this look right?

[cloud walrus]

Ocean Man

[knotty frigate]

Are there any infinite torsion groups?

[tribal moss]

Q/Z is one.

[delicate bloom]

if you're familiar with C, the roots of unity in there is a nice example

[tribal moss]

So is any infinite product of finite cyclic groups.

[knotty frigate]

But the set of the nth roots of unity only make finite groups no?

[delicate orchid]

yes, but they're talking about all the roots of unity

[chilly ocean]

Prufer groups

[knotty frigate]

Oh

[tribal moss]

"All the roots of unity" is coincidentally isomorphic to Q/Z.

[delicate bloom]

I sorta hesitated to say it because of that haha

[knotty frigate]

Oh

Ooooh

I think ik why

[formal ermine]

why doesn't A iso B and B iso C imply A iso C?

[coral shale]

isomorphisms are equivalence relations

so

[south patrol]

it does imply that, illum

[formal ermine]

Q/Z and R/Z are both isomorphic to the unit circle but Q/Z isn't isomorphic to R/Z

[south patrol]

Uhhhh

[coral shale]

what is 'the unit circle'

[solar inlet]

Q/Z is not isomorphic to the unit circle

[formal ermine]

multiplicative group of roots of unity

[south patrol]

that is not the unit circle

[solar inlet]

"roots of unity" here is the rational roots

[formal ermine]

ah

[south patrol]

e.g. not everything on the unit circle is algebraic ig

[delicate bloom]

order 2pi 🤓 (jk)

[formal ermine]

@elder wave shut

[elder wave]

i thought this was discussion

mb

[coral shale]

it is discussion

[delicate bloom]

ok now prove A iso B and B iso C imply A iso C

[formal ermine]

function composition of the isomorphisms

[delicate bloom]

a little too terse for my liking

but good enough I guess lol

[glossy crag]

Does this look right and does anyone know how to prove this for general domains (or whether it holds at all)?

[ruby sundial]

proof looks good to me

[karmic moat]

Aluffi claims that there are 7^4 set-functions from C4 to C7 (where C is the cyclic group), but there are only 7^3 group homomorphisms (because the homomorphism has to map identity to identity)

but where did they get 7^3

[formal ermine]

related: https://math.stackexchange.com/questions/273169/number-of-homomorphisms-from-bbbz-m-to-bbbz-n

[karmic moat]

ty

[next obsidian]

This is definitely not true

There’s no group homomorphisms C4 to C7

Besides the trivial one

[karmic moat]

maybe i read wrong

oh i read wrong

[next obsidian]

It says that alone rules out all but 7^3

[karmic moat]

yea oops

this is what i get for playing league of legends

ty

[next obsidian]

Lmfao

[chilly ocean]

uninstall

for ur own sake

[karmic moat]

but…

i already bought the pass for this event…

[ruby sundial]

aluffi high?

i proved this with a budddy btw

there are 7 choose 4 set functions

if thats wut u meant

no

im high

[warm wyvern]

Aluffi

[white grotto]

show that if f is an endomorphism (E->E) whose characteristic polynomial is irreducible over E which is a sub field of C, then rank(fg-gf)!=1 for all g

[formal ermine]

the only groups of order 21 are Z21 and Z3 x Z7 yes?

[tribal furnace]

yes

[formal ermine]

google shows semidirect product for the latter

so do I have to use that

because also by crt Z3 cartesian product Z7 iso Z21

[tribal furnace]

oh, sorry, it should be the semi-direct product

[south patrol]

Maybe you already know but like if a group has order pq where p > q and q doesn't divide p-1, it's cyclic and otherwise it's either cyclic or a specific semidirect product which you can find in general lol

[coral spindle]

Z_21 is isomorphic to Z_3 x Z_7.

[formal ermine]

what group homomorphism are we using for the semidirect product?

conjugation?

[warm wyvern]

wdym?

[formal ermine]

the outer semidirect product is always wrt some group homomorphism A -> Aut(B), no?

[wooden ember]

Look at all morphisms from Z7 to Aut(Z3)

You’ll see there are only two possibilities for the induced semidirect products

[formal ermine]

Aut(Z3) has two elements id and x -> x^2

so Z7 -> Aut(Z3) is either x -> (x -> x) or x -> (x -> x^2)?

[wooden ember]

By x^2 I assume you mean 2x

[formal ermine]

ohhh wait

I mean the square map

x^2 to x and x to x^2

e to e

[wooden ember]

That’s not an automorphism since 2 and 1 are sent to the same element

You’re working in Z3 with addition

But I think that’s what you mean anyways

[formal ermine]

hmm wait

what about x -> e

is it just id and trivial map?

[wooden ember]

Im having doubts on this semidirect product business though : Aut(Z_3) is isomorphic to Z_2 and the only morphism from Z_7 to Z_2 is trivial

You should be doing Z7 semidirect Z3

[formal ermine]

wat

that's what I'm trying to do though?

[wooden ember]

No

We want a morphism Z3 to Aut(Z7)

Not Z7 to Aut(Z3)

[formal ermine]

oh

lmao

it's the other way around

[wooden ember]

There are clearly non trivial ones here

[formal ermine]

yeah ok

[coral spindle]

You also have to argue why Z_7 is normal in the group of order 21

this is a necessary condition for a semidirect product

[formal ermine]

isn't that for the inner semidirect product?

[coral spindle]

Lol

That's for all semidirect products

The inner and outer semidirect product produce the same result!

[formal ermine]

my lecture script says:

Let H and N be groups and theta : H -> Aut(N) a group homomorphism. Then (N x H, *_theta) with (n_1, h_1) *_theta (n_2, h_2) = (n_1theta(h_1)(n_2), h_1h_2) will be a group. We denote N x H with this group structure as N semidirect_theta H. In this case we call G = N semidirect_theta H the outer semidirect product

[coral spindle]

OK let me make my point clear

If G is the inner semidirect product of N and H, then it is isomorphic to an outer semidirect product N x_phi H with some phi. Similarly, if N x_phi H is some outer semidirect product, then setting G = N x_phi H, we have G is the inner semidirect product of N x {1} and {1} x H.

So they are the same.

In normal parlance, we don't even use the words "inner" and "outer".

Now your task is to find all groups of order 21. If you want to find them via semidirect products, you must first show that they are a semidirect product in the first place. For this end you must show in particular that Z_7 is a normal subgroup in every case.

[coral spindle]

Elaborating on this further: not only are they isomorphic as groups, but they are isomorphic as group extensions, which is to say that the isomorphism preserves the normal subgroup N and the quotient H in a nice way. In fact, semidirect products characterise special extensions called 'split' extensions.

[formal ermine]

ah ok

isn't another requirement that N and H are disjoint

as in

$N \cap H = \Set{e}$

[cloud walrus]

rectangle cube

[south patrol]

I'd not say disjoint but rather "have trivial intersection" or similar

But ye

[formal ermine]

but $\bZ_3 \cap \bZ_7 = \bZ_3$?

[cloud walrus]

rectangle cube

[formal ermine]

or wait

no

wait

I'm confused

elements of Z3 are 0 + 3Z, 1 + 3Z, 2 + 3Z

elements of Z7 are 0 + 7Z, ..., 6 + 7Z

[south patrol]

Uhhh

They aren't subsets of one another like

(This doesn't make sense algebraically ig, you'd just get the empty set)

But

If a group G has subgroups of order 3 and 7, then their intersection is trivial

Think about why

[formal ermine]

they are both cyclic so every element has order 3 bzw 7

and an element of order 3 cannot be an element of order 7

(not taking the identity into account)

[formal ermine]

how would I show that Z7 or Z3 is normal in Z21

[tiny jolt]

Sylow theorems

[formal ermine]

ah because the 7-sylow group is unique

[tiny jolt]

(or just subgroups of abelian groups are normal)

[formal ermine]

lmao

[south patrol]

Well also like

If p is smallest prime dividing order of G then any subgroup of index p is normal

Oh I didn't see you said Z21

[rustic crown]

or use that Z21 is nilpotent group, so is a direct product of its sylow subgroups

[small ingot]

Bumping this because I think I solved it using linear algebra (assume rank 1; eigenvector implies commutator is scalar; etc.) but I feel someone more familiar with abstract algebra can use galois theory to solve this

[lethal dune]

endomorphism over what?

[white grotto]

its an endomorphism over a subfield of the complex numbers

but nvm idc about the quest

[lethal dune]

End_E(E)?

[white grotto]

endomorphism of E, where E is a subfield of C

[wooden ember]

You could also use your classification from earlier

[lethal dune]

okay

I was thinking of it as a linear space

okay whatever don't care anymore

[white grotto]

yeah it doesnt matter, its way above my level

i thought i was supposed to know

but turns out we didnt do it in class

[small ingot]

Ok fine ill review galois theory and try to find the solution w/ it

[small ingot]

Yeah its actually really simple you just need galois theory to justify some roots things to formalize the argument i gave above

Kind of lame thought itd be something sick

[rustic crown]

okie need help with some homological alg

ik it shouldn't be hard to verify but idk how to show that you can find an injective resolution of any bounded below complex A* --> E*, in the sense that this cochain map is a quasi-isomorphism

[agile burrow]

If I'm not mistaken, Aluffi does this explicitly for projective resolutions somewhere in chapter 9

[rustic crown]

oh it does?

lemme check

[agile burrow]

Can you just dualize it?

Yeah, I forget which section

Maybe 5 or 6

[rustic crown]

i never fully read the last chapter of aluffi

[agile burrow]

It's ok, me neither

I actually quite like most of it up to section 8. I never finished double complexes because I got scared

But now I think I am prepared :)

[rustic crown]

yee i see something nice in theorem 6.6

i was literally wondering that :p

[agile burrow]

Yeah, that's the one

Took me a long time to work through this proof, and I've probably forgotten how it works by now

[rustic crown]

i think i just want the existence, the stuff about uniqueness upto homotopy equivalence i think follows easily from stuff i've proven

[agile burrow]

I would think that there are alternate constructions using mapping cones or something, but I don't know of any explicit references off the top of my head

Just inductively building the map so that the mapping cone is exact sounds reasonable

[rustic crown]

i don't have much intuition about the mapping cone, but the algebra with it is always so clean

thanks walter

[white yoke]

How can I prove that the Hurwitz quaternions are a maximal Z-order in the real quaternions? I would suppose O is an order containing the Hurwitz quaternions and aim to prove that it is equal. Does anyone have any ideas?

I'm looking for a more algebraic proof, not really analysis

[rustic crown]

aluffi's proof is a little tricky, but super clean even without using any mapping cones. it's tricky because it sets up a stronger induction hypothesis. we find a map A* --> E* such that at each degree it's a mono and moreover the map between cokernels of the differential is also a mono.

[agile burrow]

yeah, I've just looked at it again and I think it makes more sense now

that's a neat proof

[rustic crown]

yea really neat, but hard to come up with :p

idk why aluffi does so much work for proving the uniqueness and lifting of morphism

there is a really cute argument for this you know

[agile burrow]

how does it go?

[rustic crown]

the idea is to not shy away from Hom-complex and mapping cone :3

so in an abelian category, say you have two complexes A and B, then define the complex Hom(A, B) in Ch(Ab) where nth degree thing is product of Hom(A^i, B^n+i) where i varies in Z

there is a simple way to define the differentials, and with that the Z^0 cocycles correspond to cochain maps, and B^0 coboundary corresponds to null-homotopic maps, so H^0(Hom(A, B)) = Hom_K(A, B)

where K is the homotopy category of your abelian category

once you have this Hom-complex the argument is very cute

so by hand you first show that if C is bounded below and exact, and E is a complex of injective, then any f : C --> E is null-homotopic

using this you can easily prove that if A --> B is a q-iso between bounded below stuff and A --> E is a map to a complex of injectives, then you can find a unique map B --> E in K such that the obvious triangle commutes

so if C is the mapping cone of A --> B, then 0 --> B --> C --> A[1] --> 0 is exact. and moreover, it is split exact at each degree

therefore you can apply the functor Hom(-, E) : Ch(the abelian category) --> Ch(Ab)

since the thing is split exact at each degree, you get a short exact sequence
0 --> Hom(A[1], E) --> Hom(C, E) --> Hom(B, E) --> 0

now H^n(Hom(C, E)) = H^0(Hom(C, E[n])) = Hom_K(C, E[n]) and since C is bounded below exact and E[n] is a complex of injectives this is 0

so the middle term is acyclic

this gives you an iso
H^0(Hom(B, E)) --> H^1(Hom(A[1], E)) = H^1(Hom(A, E)[-1]) = H^0(Hom(A, E))

and so the map
Hom_K(B, E) --> Hom_K(A, E)
given by composition with A --> B is an iso

[agile burrow]

that is very neat

now i am actually very confused why aluffi proves uniqueness the way he does lol

[rustic crown]

like this gives all the stuff so quickly

both the uniqueness and the lifting

i've seen people show some horseshoe lemma

but that's not needed at all with this

[agile burrow]

yeah this is a really cool way of showing lifting which is what tripped me up

Maybe I'm being silly, but I think uniqueness also follows from the second part of this exercise too

which can definitely be done by hand

or at least, I did it with some mapping cone stuff

but it seems like most of these proofs are simplified/illuminated by hom-complexes

[rustic crown]

mapping cones and hom-complexes are really op

[agile burrow]

out of curiosity, why are you looking at injective stuff?

[rustic crown]

because ag

[agile burrow]

not enough projectives 😵‍💫

[rustic crown]

yea >.<

also here the functor of interest is left exact

so we need injective resolution to compute the right derived functor

[agile burrow]

sheaf cohomology?

[rustic crown]

yea >.<

[agile burrow]

good luck lol

[rustic crown]

hehe :3

injectives are not so bad actually

at least the homological algebra is completely dual

[agile burrow]

yeah, i guess that's nice

makes sense to reprove theorems using injective resolutions just for comfort

[rustic crown]

i was happy with injective once i was playing a little with cofree objects. now i wanna read more about them

so like if you have an abelian group A, then you get a surjection Z^(⊕Hom(Z, A)) = Z^(⊕A) --> A

similarly, you have a complete dual to this

if A is an abelian group, you have an injection A --> (Q/Z)^Hom(A, Q/Z)

where you send a --> (f(a))_f

the injectivity follows from Baer's criterion :3

if a is non-zero, then you can easily find f' : <a> --> Q/Z such that it doesn't send a to 0. now extend this to f : A --> Q/Z by injectivity of Q/Z

so this f shows that (g(a))_g is non-zero

[agile burrow]

oh yeah, i think i've read about this a little bit before

[rustic crown]

i think i discovered this in this channel lol, shamrock helped a lot

[agile burrow]

or is this just necessary to show that module categories have enough injectives?

maybe i'm misremembering

[rustic crown]

yea once you show that Ab has enough injectives, it easily follows that R-mod has enough injectives as well

as co-inducing injectives gives injectives

[agile burrow]

that makes sense

i haven't thought about injectives in a while lol

[rustic crown]

Hom_Z(R, Q/Z) should be cofree R-module :3

[agile burrow]

for finite groups, projective ZG-modules are also injective lol

and group rings are all i think about these days

[rustic crown]

woah

me will try to read some group cohomology in summer

[agile burrow]

we can talk about it !

maybe i'll learn some AG eventually too

[slender hamlet]

I don't think that's true

Take G=Z/1Z

Z isn't injective as a Z-module, right?

[lethal dune]

Yes

[agile burrow]

ah wait, yeah I misremembered

something about relative injectivity, i should brush up on the details

[slender hamlet]

I guess you meant like acyclic

In group cohomology

[agile burrow]

yeah, that's the context

but this showed up when constructing complete resolutions of Z over finite group rings

[rustic crown]

finite (group rings) or (finite group) rings?

[agile burrow]

(finite group) rings

like you can make these backwards resolutions 0 -> Z -> P_0 -> P_1 -> ... where each P_i is projective and finitely generated

so if you stitch this together with a normal projective resolution you can apply functors and take the cohomology of the whole thing

and that's how you get tate cohomology

[lethal dune]

Yeah remember reading something similar

Using coinduction to find acyclic resolution

[slender hamlet]

Ok I haven't seen this

[agile burrow]

there's an extension of these ideas to infinite groups too called like farrell-tate cohomology or something

but idk that yet lol

[slender hamlet]

Tate cohomology was like joining together the group cohomology and homology

And tweaking the 0th and 1st terms a bit

[agile burrow]

yeah, you have the norm map from the coinvariants to the invariants so you look at the kernel and cokernel of that

[slender hamlet]

Yes

Also how do we get periodicity in tate cohomology when we are not in cyclic group case?

Like I saw certain non-cyclic groups showing periodicity in Tate cohomology

[agile burrow]

iirc the criterion is that every abelian subgroup is cyclic

you first prove it for p-groups and then show that every Sylow subgroup has periodic cohomology iff the group has periodic cohomology

and there's a criterion for groups where every abelian subgroup is cyclic, namely the Sylow subgroups are all cyclic or generalized quaternion groups

so really there aren't that many groups with periodic cohomology, which is kind of interesting

[slender hamlet]

Oh I thought the sylow subgroups should should be all cyclic

Didn't know generalised quarternions also work

[agile burrow]

yeah, you can see that having every abelian subgroup be cyclic is equivalent to saying that the elementary abelian p-subgroups have rank <= 1. That is, the Sylow subgroups have unique subgroups of order p. The only groups satisfying this are cyclic and generalized quaternion (because everything about 2 has to be weird)

[rustic crown]

2 is uwu

[tiny jolt]

any ideas on how to prove this?

[lament dawn]

Can you do it if everything is a finite extension?

[tiny jolt]

I don't see how

[cloud walrus]

pramana

[next obsidian]

I don’t see why you’d ever have a polynomial in E

[tiny jolt]

oh i wrote that wrong

If $f\in E$ and the minimal polynomial of $f$ in $F[x]$ is $p(x)$, and in $K[x]$ is $q(x)$, then showing that $p(x)\mid q(x)$ would finish the proof, but I'm not sure if that's true

[cloud walrus]

pramana

[next obsidian]

This is definitely true

By definition of minimal polynomial

[tiny jolt]

how so?

[next obsidian]

The minimal polynomial of a divides any polynomial a is a root of

[tiny jolt]

oh i see

thanks

[drifting plover]

If x is an element of an integral domain R and I is an ideal of R such that x*I is principal, must I also be principal?

[chilly ocean]

x = 0

[drifting plover]

Let's say x is non-zero haha

[chilly ocean]

I was thinking x=0 too but isn’t that too trivial?

Yup

[rustic crown]

yea in that case it should be principal

[drifting plover]

How do you prove this?

[rustic crown]

just do it by hand

say x*I = (y)

intuitively, you could guess that I is generated by something like y/x, so just try to formalize this

since y lives on the right, you can find i in I such that x*i = y

[drifting plover]

Then I is generated by i

[rustic crown]

now given any arbitrary element in I, say j then x*j lies in (y) so equals y*r,
xj = yr = (xi)r
canceling x (as it's non-zero) j = ir

so j in (i)

[drifting plover]

Great, thanks

[celest cairn]

Is the minimal polynomial of
$4 + 3i$ over $\mathbb{C}$ equal to $x - 3i + (2i)^2$?

[cloud walrus]

Sapphire Gaming

[drifting plover]

That's just x - (4+3i)

Which is correct, yeah

[celest cairn]

Ah ok, thanks 🙂

[rustic crown]

woah so many sully reactions

[elder wave]

make it 10 det

[rustic crown]

11 now uwu

[delicate bloom]

im usually not first to sully, but when i do...

[formal ermine]

let $G$ be a group of order $40$. sylow $1$ and $3$ say that the $5-$sylow-subgroup exists and is unique. how does sylow $2$ tell us that this subgroup is normal in $G$?

[cloud walrus]

$rectangle~cube$

[coral spindle]

All sylow 5-subgroups are conjugate. Since it is unique, what does this tell us about its conjugates?

[formal ermine]

it's conjugate to itself

[coral spindle]

Every subgroup is conjugate to itself. That means nothing.

A subgroup is normal if it is its only conjugate.

Do you see why?

[formal ermine]

wdym

[coral spindle]

What's the definition of a normal subgroup?

[formal ermine]

gNg^-1 = N for all g

[coral spindle]

So gNg^-1 is a conjugate of N. All conjugates are (by definition) of this form.

[junior stump]

I'm not sure you actually need sylow 2 for this but maybe i'm being dumb

[coral spindle]

And it being = N means that all of the conjugates of N are just N

I.e. the only conjugate of N is itself

[formal ermine]

yeah but only for a specific g and not for all no?

[coral spindle]

No.

[formal ermine]

oh

[coral spindle]

That is for all g.

[formal ermine]

then I misinterpreted the theorem

[coral spindle]

Now let's look back at this: you have shown that there is only one sylow 5-subgroup

knowing that all sylow 5-subgroups are conjugate, what do you conclude?

[formal ermine]

idk what you want me to say

[coral spindle]

You want to know why the sylow 5-subgroup is normal

follow on from this argument to prove this

[junior stump]

I don't think sylow 2 says anything about this

[formal ermine]

it's conjugate to itself so it's normal?

[coral spindle]

Again, every subgroup is conjugate to itself.

[formal ermine]

why

[coral spindle]

1N1^-1 = N.

Let me fill in the rest of this proof, then.

Let H be the unique Sylow 5-subgroup. Note that, for any g in G, gHg^-1 is also a maximal 5-subgroup, i.e., it is a Sylow 5-subgroup. Therefore it must be equal to H, since we know that H is the unique such subgroup. Hence gHg^-1 = H for all g in G.

adronsw is correct in saying that the 2nd Sylow theorem isn't really necessary here: we can actually infer it blindly from the fact there is a unique such group

[junior stump]

yeah it's a bit of a red herring exercis

[formal ermine]

this was just me tryna make up an example to understand the 2nd sylow theorem

[formal ermine]

but that's what I was saying

it's conjugate to itself for every g

[junior stump]

ah ok

[formal ermine]

so it's normal

[coral spindle]

You did not use this terminology

I'm a little frustrated by this

[formal ermine]

me too

sorry

so sylow 2 says that if H is a p-sylow-subgroup then gHg^-1 is also a p-sylow-subgroup for any g?

[coral spindle]

No, the 2nd theorem states that all Sylow p-subgroups are conjugate

[formal ermine]

could you put that into notation please

[coral spindle]

If H, H' are sylow p-subgroups then there is some g for which H' = gHg^-1

[formal ermine]

Note that, for any g in G, gHg^-1 is also a maximal 5-subgroup, i.e., it is a Sylow 5-subgroup.
why can we use "for any g in G" here

[coral spindle]

OK

Do you agree that conjugation is an automorphism?

[formal ermine]

yes

[coral spindle]

In particular, it will send subgroups to subgroups

so we know that gHg^-1 is a subgroup of G.

Now we also know it is a p-subgroup, because it is an automorphism of G (in particular a bijection)

We can finally conclude that it is a maximal p-subgroup, since if there were a larger p-subgroup containing gHg^-1, it would contradict the maximality of H after conjugation

[formal ermine]

what's a maximal p-subgroup

[coral spindle]

That is the definition of a Sylow p-subgroup

A p-subgroup being maximal means there is no strictly larger p-subgroup containing it

It concerns me that you haven't seen these terms. What definition of a Sylow subgroup have you been referring to?

[formal ermine]

Let p be a prime. A group whose order is a power of p is called a p-group. Let G be a finite group, |G| = p^m k, m in N and p doesn't divide k. A subgroup whose order is p^m is called a p-sylow subgroup.

[coral spindle]

Right, OK. These definitions coincide once you've proven a particular theorem. Minor differences!

[formal ermine]

it would contradict the maximality of H after conjugation
can you elaborate on this pls

[coral spindle]

Suppose K is a larger p-subgroup containing gHg^-1. Then g^-1Kg contains H, which is contradictory.

[formal ermine]

why is it a contradiction

[coral spindle]

H is maximal

[formal ermine]

why is H maximal

[coral spindle]

That is the hypothesis

Are you asking why Sylow p-subgroups are maximal p-subgroups?

[formal ermine]

I only know that normal p-sylow subgroups are maximal p-subgroups

[coral spindle]

All Sylow p-subgroups are maximal. Try proving this.

Oh n.b.

maximal p-subgroups

Definitely not maximal subgroups haha

[formal ermine]

ah

H is the only subgroup of G whose order is |H|

gHg^-1 is also a subgroup of G

and |gHg^-1| = |H|

but because H is unique wrt its order it must follow that gHg^-1 = H

[coral spindle]

Yeah this is the proof we did above

There is a proof not assuming that H is unique wrt its order

[formal ermine]

yeah I was kinda confused about the maximality thing you were using

[coral spindle]

Yes, we were using different definitions

[formal ermine]

that's what threw me off

thanks tho!

[coral spindle]

No worries

[formal ermine]

if $p$ doesn't divide $|Z(G)|$ and $p^m k = |Z(G)| + |C_1| + \ldots + |C_s|$ how does this imply that there must exist an $i$ such that $p$ doesn't divide $C_i$

[cloud walrus]

$rectangle~cube$

[rustic crown]

if p divides every |C_i| and m >= 1, then the equation forces that p divides |Z(G)|

[formal ermine]

ah

because then you could factor out a p from the conjugacy classes

right

ok thanks

[rustic crown]

yee

[formal ermine]

why does $p^m$ divide $|Z(G)|$

[cloud walrus]

$rectangle~cube$

[agile burrow]

in what context?

certainly this isn't always true

take Q_8 for instance

[formal ermine]

in the context above

[agile burrow]

where p doesn't divide |Z(G)|?

[formal ermine]

yes

|G| = p^m * k with p doesn't divide k

[agile burrow]

how could p^m divide |Z(G)| with p not dividing |Z(G)|

[formal ermine]

ah wait I think I got it

,, |G| = |C(x)| \cdot |Z_G(x)|

[formal ermine]

by orbit stabilizer

left side is p^m k right side is something times Z_G(x)

the something isn't divided by p

[agile burrow]

oh, you meant the centralizer of an element

not the center of the group

[formal ermine]

oh wait yeah I got confused

mb

I was originally gonna ask about the second sentence where something with Z(G) was done but then I got it

[south patrol]

Group actions yummy

[formal ermine]

I hate them

[agile burrow]

but you like galois theory

[south patrol]

They r cool I hope u will like them more soon

lol

[formal ermine]

they feel unnatural to me

[south patrol]

How you prove every theorem in finite group theory

Bruh

[formal ermine]

another question

[south patrol]

Group actions are basically the main reason for considering groups

they are more natural than the groups themselves arguably in some cases

[formal ermine]

how does induction over |G| with this whole thing show that there exists a subgroup S subseteq Z_G(x) with |S| = p^m

[formal ermine]

what's an intuition behind group actions

for me they're just "group acts on a set"

which feels weird

[south patrol]

I guess one point is that sets can come with certain symmetries, which can be encoded using a group

[formal ermine]

like

what is the induction step

and induction beginning

or whatever it's called in english

[agile burrow]

Do you need induction here? Isn't the hypothesis that p doesn't divide |Z(G)|?

[formal ermine]

I'm trying to understand the proof of sylow 1

[agile burrow]

so if |G| = 1, then it trivially contains Sylow subgroup

assuming it holds for all groups with order less than |G| = p^r m, you show that p^r divides the order of some proper subgroup of G

then the inductive hypothesis implies that said proper subgroup contains a p-Sylow subgroup, which must also be a subgroup of G

[formal ermine]

but if $S \subseteq Z_G(x) \subsetneq G$ then how can $S \subseteq G$

[agile burrow]

.

If S is a subgroup of the centralizer of x, then it's certainly a subgroup of G

[formal ermine]

yes

but the centralizer of x is a proper subgroup of G

so how could S be the entire group G

[agile burrow]

we're not claiming that S is the entire group G