groups-rings-fields

[molten viper]

I definitely do not have it

[delicate orchid]

you have got it enough

[molten viper]

I think I'm just not understanding the definition of Krull dimension

[delicate orchid]

it's the longest dude you can make

[molten viper]

It's the length of the longest chain of prime ideals

[delicate orchid]

think of the longest dude imaginable. There can be no longer

[molten viper]

so A sub B sub C etc. where A, B, C, etc. are prime

[delicate orchid]

it's ascending or descending chains iirc

[molten viper]

so say our longest chain is A sub B sub C

then the ring has Krull dimension... 2?

[delicate orchid]

yeah

it's the number of subs not the number of doodads

bit annoying but w/e it has some geometric justification I'm sure of it

oh look it does

[molten viper]

So for an Artinian ring, we're claiming that every prime ideal is maximal, which means that these chains are all length 0

[delicate orchid]

yes

[molten viper]

Ahhhhh got it

[delicate orchid]

your rings are commutative right

[molten viper]

I should keep working through Atiyah Macdonald

Long way to go before I get to chapter 8 😅

[molten viper]

The only rings I'm worried about right now are k[x_1,...,x_n]/I, hence my asking about them earlier

[delicate orchid]

yur ok

[molten viper]

I am going to be in study hell this year

It's gonna be very fun

math GRE, actual uni courses, thesis work (feat. going through Atiyah Macdonald), all my other courses, grad svhool apps

Academic. Weapon.

[molten viper]

Not to mention all the time partying and attending football games 😉

[delicate orchid]

I'm at a party rn

[molten viper]

It's gonna be really fun relearning all of the calculus I've forgotten in the last 2 years

anyways

All I'm worried aobut it my meeting in 15 minutes

[white oxide]

Does $L_1 = F(\alpha_1)$?

[cloud walrus]

okeyokay

[white oxide]

also why is the polynomial invariant under action by any \sigma in G(L_1/F)? is it because sigma(alpha_2)^n_2 = sigma(alpha_2^n_2) and since alpha_2^n_2 is in L1 sigma(alpha_2^n_2) is in L1, and because the sigma_i in G(L_1, F) are a group the coefficients are the same except for order under action by all elements of G(L_1, F)?

[rocky cloak]

Not necessarily. Depends whether F contains roots of unity.

[rocky cloak]

Seems the action of G(L1/F) just permuts the factors in the product

[white oxide]

I have a couple of questions; in the case that the primitive roots of unity are not contained in F, would L1 = F(\alpha_1, \xi) where \xi is a primitive n_1 root of unity?
Is L1 a normal extension of F since it is obviously the splitting field of f_1(x) in F[x], but how do we know that it's separable over F? Is it because x^n_1 - \alpha_1^n_1 is irreducible over F and has all zeros of multiplicity one?

sorry you don't have to answer any of these if you want just wanted to put it out there just in case somebody knew

[white oxide]

why? i get that \sigma(\alpha_2)^n_2 \in L1, but why can't it send it to another element of L1 that wasn't a coefficient in the original polynomial

Hence not fixing f_2(x)

[rocky cloak]

I don't follow what you're saying. The polynomial is the product of (x^n - sigma(a^n)). Applying sigma' to this gives (x^n - sigma'sigma(a^n)) which is just another factor in the product

[white oxide]

yeah, my question is why it's another factor in the product sorry got my wording mixed up

[rocky cloak]

Because that's what we're taking the product over

[white oxide]

Ah ok

well I was trying to use the logic here

[rocky cloak]

Seems like just a convoluted way to say that a group permutes it's elements by left multiplication

[white oxide]

yeah I guess so

[charred crescent]

can someone help me understand the language used in defining the universal property on free groups

i kind of don't understand what "extends to a unique homomorphism" means

side note: i think it means that a unique hom. can be constructed

and then when i read the proof i don't really understand it

[paper aurora]

can you please explain how we are overcounting by p-1 a little more ? i take it that we are counting, up to isomorphism, the kernels of the p^2-1 homomorphisms

[hidden haven]

There is a homomorphism extending the given map, and any two such homomorphisms are equal

Extending the given map means that its restriction to the generating set is the given map

[charred crescent]

to probe at this; does this mean that for a given map f: S -> F(S), f(S) = S by the inclusion?

[rocky cloak]

So you have a map Z^2 -> Z/p. If you compose this with an isomorphism then it has the same kernel. Conversely if two maps f and g both have kernel K, then f induces isomorphism f' : Z^2/K -> Z/p and g induces g' : Z^2 -> Z/p, so f and g are related by the isomorphism g'f'^(-1)

[charred crescent]

well f(S) = S is obvious, but im asking if this is the so-called "restriction"

[rocky cloak]

So the number of kernels is the number of surjective homomorphisms divided by the number of isomorphisms from Z/p to itself

[hidden haven]

Yeah the map S → F(S) in the universal property is fixed, and I am abusing notation here by treating it as an inclusion to use the word "restrict". It is a consequence of this universal property that this map must be injective.

The universal property says that given a map g: S → G where G is a group, there is a unique way to "extend" this to a map h: F(S) → G. Again, by "extend", I mean that g = hf, where f is the map S → F(S).

[topaz solar]

Drawing out the triangle might help

[hidden haven]

(this is not the relevant triangle)

[delicate orchid]

the arrows are the same

[charred crescent]

it's a good example, maybe if i see it work on this on too then the abstract definition will pop out

[hidden haven]

No we have f g and h

[delicate orchid]

wow I see how it is

[topaz solar]

Well, intuitively

F(S) is just words on S with the appropriate reduction of like a a^-1 right?

[charred crescent]

yeah

[topaz solar]

So clearly sending each element in S to one in G

Tells you how to send words there

And it’s important that there are no relations on F(S) so it always works

[next obsidian]

All triangles are triangomorphic to one another

[paper aurora]

thank you 🙂

[hidden haven]

You might also have already seen the vector space version of this universal property. Given a set S, the "free vector space generated by S" is the vector space of formal finite linear combinations of elements of S. This has a basis given by the set S itself. The result I am referring to is that a describing a map from F(S) to another vector space V is equivalent to describing just the images of the elements of S, i.e. the restricted map S → V

[charred crescent]

@hidden haven so which arrows correspond? im guessing (f is to phi), (h is to pi), and (g is to phi_bar)

[hidden haven]

f is pi, g is phi, h is phi_bar

h is the dotted map

[charred crescent]

im sorry, i just don't understand

[hidden haven]

Also S is G, F(S) is G/ker(phi), G is H

Would be better to draw the entire diagram separately

[charred crescent]

okay im going to draw it

both of them

[hidden haven]

S → G
↓
F(S)

f is the vertical, g is the horizontal, and h goes diagonally up

[cloud walrus]

wew

[delicate orchid]

hgahahaha nice

[hidden haven]

🤓

[charred crescent]

dumb question, but does the dotted line represent the "extension" or what

[topaz solar]

For every g, there is a unique extension h

[charred crescent]

or the one that is "dependent" on the others, similar to the one for 1st iso

[hidden haven]

The solid arrows are the hypotheses of the result, the dotted arrow is the one that the result asserts exists

[topaz solar]

It’s the extension and is dependent

[hidden haven]

This diagram as it is now does not assert the uniqueness of h

[charred crescent]

!h

right

[hidden haven]

To assert that, you put an exclamation mark over the dotted arrow

Yes

[ivory trail]

think about a "flag" in a vector space: a point (dim 0 subspace) in a line (dim 1 subspace) in a plane in a ... and then like in the krull dimension definition, the supremum of lengths of a flag is the dimension of the vector space

[hidden haven]

And the assertion of the result is that the dotted arrow exists such that the triangle commutes, i.e. both ways of going from S to G are the same.

Which is what we mean when we say "h is an extension of g along f"

[delicate orchid]

ahhh I hate phrases like that just draw the triangle you lazy bones

[charred crescent]

is this what is meant exactly by the term "universal property" ?

[topaz solar]

Well, you know how every vector space map can be determined by where you map basis elements?

[hidden haven]

I am captioning the triangle you drew pewpyhead

[topaz solar]

This is literally the same thing

[delicate orchid]

https://cdn.discordapp.com/attachments/335352580777443338/1130637327979511898/doyfoj858tza1.jpg
you talkin to me with that tone of voice?

[charred crescent]

actually no, sorry im fairly new to all of this advanced stuff

[topaz solar]

It’s a property and holds universally

[delicate orchid]

that's not advanced, it's just linear algebra

[topaz solar]

Did you not do linear algebra?

[charred crescent]

not yet

[topaz solar]

ok that explains it

[charred crescent]

i mean i can multiply two matrices and i know about linear maps a fair bit

[delicate orchid]

why are you doing this before linear algebra

[charred crescent]

more interesting

im self-studying for the summer

[ivory trail]

linear algebra is pretty interesting if you approach it from the abstract angle

[delicate orchid]

google "representation theory" nerd

[ivory trail]

it's just very solved so you don't spend so much time on it

[hidden haven]

A universal property is something more precise and specific. Giving the exact definition of that would be difficult without going into cat theory, but a universal property is a statement that describes an object (up to isomorphism) by only describing what the maps out of that object look like

[topaz solar]

Well I doubt the fair bit since basis determining the map is unknown

[charred crescent]

well that's funny im actually trying to study that eventually

[topaz solar]

Then study linalg

[delicate orchid]

representation theory is studying maps into vector spaces (usually!) and you didn't think it prudent to learn about vector spaces

[ivory trail]

we did universal properties kinda in the abstract linear algebra i took

[charred crescent]

dang im sorry guys😂

[ivory trail]

like for tensor and wedge products

[topaz solar]

Anyway, shook, how is F(S) defined

[charred crescent]

ik what a vector space is but i was under the impression that i needed much more group theory

[hidden haven]

In this case, the universal property of F(S) describes F(S) based solely on what the maps from F(S) to any group G are like - they are in 1-1 correspondence with set maps S → G, and the correspondence is given by composing with the map f: S → F(S). This characterizes F(S) up to isomorphism

[charred crescent]

i have no guidance in this stuff

[delicate orchid]

oh yeah I suppose the wedge product does have a universal property

it's just a really lame one

[ivory trail]

no it has a straightforward one that is just a bit annoying to put in purely categorical language

[hidden haven]

Which wedge are you talking about?

[ivory trail]

exterior product

well actually just the exterior algebra of a vector space in general too

[hidden haven]

Oh the upside down wedge lmao

Damn algebrains

That has a pretty good universal property

Maps out of it correspond to alternating linear maps out of the product

[ivory trail]

anyway the "linear map determined by where it sends a basis" is very fundamental to linear algebra

so you definitely need a bit more time on that

[topaz solar]

anticommutative throwback

[charred crescent]

i have axler's and lang's text (taking linalg in the fall), which treatment do y'all think is more clear for this sort of thing? im an idiot so i know nothing

[delicate orchid]

yeah this is lame

"when u quotient out the thing with the universal property u get the universal property of the quotient + the other one whoohohoho"

[hidden haven]

Yeah it is the first isomorphism theorem

Just like the tensor product

[delicate orchid]

true, but that quotient has a particuarly nice doodad with multilinear maps

[hidden haven]

tf is a doodad

[ivory trail]

i didn't use a particular book so i'm deferring to others, but i am a bit inclined towards axler

[charred crescent]

okay thanks for the advice everyone, i will study some linalg

[delicate orchid]

whatever u want bby

[topaz solar]

But uhh, F(S) is just words in S (and inverses) with concatenation

So if you map S then clearly you map the inverses and words

By f(ab) = f(a)f(b)

so that defines a map to G that’s a group homomorphism right?

[delicate orchid]

burrrppp

[topaz solar]

And it’s gonna be the only one that agrees with your original mapping of S

Since it’s just extending the original map by the group homomorphism property that f(ab)=f(a)f(b)

[charred crescent]

one more thing; does anyone mind giving me a very brief outline on the topics in lin alg i would need for rep thry, idk where to start honestly

[topaz solar]

Start with linear algebra

[charred crescent]

i know vectors spaces and some very basic things about linear maps

yeah but like what next

[elder wave]

just go through an entire lin alg course

[delicate orchid]

lin alg, group theory, comm alg, rep theory

[elder wave]

you'll need all of it

[ivory trail]

algebra as in groups rings modules

[topaz solar]

Worry about the rest later, might even be covered partially in a lin alg course what to do next

[delicate orchid]

depends on what type of rep theory you want to do btw

sorry for the typos I'm intoxicated

[elder wave]

your lin alg should be solid

[hidden haven]

They should make representation theory 2 with less group actions

[charred crescent]

they're types?

[elder wave]

no matter what you end up doing

[delicate orchid]

*fewer

[delicate orchid]

finite group and infinite group rep theory are very different

[hidden haven]

britchip

[delicate orchid]

not to mention shit like quiver reps

[charred crescent]

there are only finite groups

[delicate orchid]

BASED

ok you're in the club

[topaz solar]

Ok Wittgenstein

[ivory trail]

did wittgenstein talk about infinity

[charred crescent]

okay finite rep it is

[hidden haven]

ayo what about the best group U(1)

[delicate orchid]

or.... [checks the room for cameras].... the uhhh.... [no feds outside the window]..... modular reps....

[ivory trail]

seems like too much metaphysics for him

[charred crescent]

1x1 unitary matrices?😅

[delicate orchid]

S^1 for the normal folk

I think

[hidden haven]

Yes

[topaz solar]

Conjectural category of mixed motives

[ivory trail]

i.e. the unit complex numbers

[topaz solar]

Or ya know

Z

[delicate orchid]

S^1 has lots of torsion tf you mean Z

[charred crescent]

aren't they the same

[topaz solar]

I remember teasing him saying something dumb but I do not care about philosophy

[delicate orchid]

no?

S^1 is R/Z

the cardinalities don't even match

[topaz solar]

An infinite group

[delicate orchid]

Z is dog water

[charred crescent]

i thought the fundamental group of S^1 is the integers

[hidden haven]

Maybe shook means that they are the same in that they don't exist

[delicate orchid]

well, yeah

but S^1 is a group?

[topaz solar]

S^1 is the group

[delicate orchid]

oh my golly gosh le wholesome K(Z, 1) oh my goodness gracious

[charred crescent]

yeah im too dumb for this, i will come back when im smarter

[topaz solar]

Must be the same if they exist (principle of explosion )

[delicate orchid]

what

I'm going to demonstrate the principle of explosion on everything you love

[hidden haven]

Falso truther

[topaz solar]

Only 1 axiom

[hidden haven]

[ornate vessel]

R is commutative ring with 1
I and J are ideals of R
image phi R - > R/I x R/J is defined as phi(r) = (r+I, r+J)
Is phi surjective?

[topaz solar]

I do remember a rather simple type theoretic presentation for \omega-categories (in the sense of contractible globular operad algebras)

[ivory trail]

often no, take R = Z and I = J = (2), then (0,1) isn't in the image

[topaz solar]

Ok you me, but I mean it’s way simpler to see that than “contractible globular operad algebra”

[ivory trail]

phi is an isomorphism if I \cap J = (0) and I + J = (1) by the chinese remainder theorem

[hidden haven]

How tf did you reach globular operad algebras from what we were discussing

[topaz solar]

Idk

[hidden haven]

Mental illness

Anyway what's the globular operad

[topaz solar]

Ok so uhh, give me a sec

You know globular sets

[ornate vessel]

so its isomorphism when I+J=R ?

[topaz solar]

Imagine an operad globular set

[ivory trail]

AND I and J have trivial (i.e. the zero ideal) intersection

idk about "only if"

[topaz solar]

I forget the exact definition setup though unfortunately

[hidden haven]

Is there is a simplicial or cubical version

[topaz solar]

Take an operad definition and apply it to gSet

I think that’s equivalent

[hidden haven]

Oh it is an operad over gSet?

[topaz solar]

I think equivalently?

I’d have to reread it

[hidden haven]

What do you mean apply the definition of an operad to a category

[topaz solar]

Its got some wacky composition stuff

[hidden haven]

Only interpretation I can think of is that it would be an operad that acts on objects of that cat

[topaz solar]

I mean, take an operad in set, but make it globular sets in there instead

[hidden haven]

Yeah I think that's an operad over gSet lol

[topaz solar]

Yeah, I forget the exact setup unfortunately

[hidden haven]

You can define an operad over any closed symmetric monoidal cat

[topaz solar]

That’s probably equivalent to the way it was described in terms of like the explicit set way

[hidden haven]

Probably

[topaz solar]

And contractible meant if we had some pasting diagram in there with parallel top arrows or smth (since globular setup) then you can fill it in

[hidden haven]

I see

[topaz solar]

f, g: •=>• thing

[hidden haven]

Ye

[topaz solar]

Then we can get a f->g lying over it

[hidden haven]

Got it

[topaz solar]

In like P(•=>•) or smth

Where P is the operad

So when we take an algebra over the operad that leads to composition iirc

[hidden haven]

[topaz solar]

There’s some boundary condition on like

The top boundary & bottom boundary have to be compatible

But •->•->• has two “parallel” 0-cells

Something like that

Look homie it’s been a while since I read it

[hidden haven]

Lol

[topaz solar]

But the algebra needn’t be nonempty on all of them and that’s why it’s not a groupoid necessarily

[hidden haven]

I stopped following around 10 messages ago

[topaz solar]

But uhh concretely

If there’s a pasting diagram

You can paste em

gg

(As in you get a pasted one in the operad)

So the algebra will get it and behave nicely because algebra definition

[hidden haven]

Ok

[topaz solar]

I forget why it’s not a groupoid

[hidden haven]

This is just the monad for an operad construction

[topaz solar]

It’s probably “just” something categorical ofc

[hidden haven]

Ofc

[topaz solar]

It’s an algebra for a monad based on P

And if P is contractible then it’s an \omega category because everything that should be coherent is guaranteed to be because of the contractibility condition on how things lift to pasting diagram coherators

Which ends up giving you identity and such terms too

Here it was

And T1 being all the pasting diagrams, so any pasting diagram that can exist in P will exist

[hidden haven]

Don't know omega-cats

[topaz solar]

It’s just one of the infinity category variations

But these are globular-y, as opposed to, say, quasicategories as simplical sets with the horn filler condition things

[south patrol]

hm interested what the advantages of other "shapes" are

[topaz solar]

(Though that’s (\infty, 1) iirc, I don’t believe these \omega categories have that all higher cells are trivial?)

[topaz solar]

Uhhh opetopes have some “unbiased” composition kinds of benefits and simultaneously describe more uhh complex composition types at once

Like think how you can glue two homotopies together in different ways based on like, f=>f’ and g=>g’ to get gf => g’f’ versus f=>g=>h to get f=>h

[south patrol]

Sure

[topaz solar]

And in composing as binary vs ternary vs quarternary etc operations and being compatible with the relevant associativity coherences

So not just the two ways to get (fg)h and f(gh) but also immediately composing fgh

(All homotopic yada yada what have you)

[topaz solar]

Probably also useful in the differences of how the maps and such in the shape category work out combinatorially? Combinatorics has some black magic

[south patrol]

Hm sure interesting

Thank

[topaz solar]

Globular are particularly simple

Since they’re like

“Yep, you have two maps, source and target, and the source of source = source of target, and t o s = t o t”

think uhh f, g: A -> B and h: f -> g

(These are maps of the globular set not the globes themselves, since those are reversed)

Important to note though is we can take like globular nerves of topological spaces and do realizations but gSet doesn’t have the right structure for that adjunction to be good for homotopy theory (not a test category so no Quillin equivalence @south patrol, I do not know about cubes or opetopes in this area though)

[delicate bloom]

sick, thank you checkin it out

[hidden haven]

Ye quasicats are also infty,1

[topaz solar]

I am unsure if these \omega cats are infty, 1, but they’re at least that powerful, probably described somewhere in that text I can dig up

[hidden haven]

Ye when I looked at nlab the globe cat was described as one of the shapes for higher structures

and another page described shapes for higher structures as cats on which the presheaves model infty,1 cats

Might be misremembering though

[topaz solar]

Probably pretty similar anyhow

[opaque tartan]

I don't know why these two equalities hold. What happened here?

this circle dot-thing is an operation

[topaz solar]

Consider how it is defined

[opaque tartan]

OH, I forgot it was defined before

My bad

thank you

[median pawn]

y'all what's a free basis

is it just a free generating set of a free group?

[hidden haven]

Yes

[median pawn]

ah okay then! for the rank-n free group G, if S is a generating set of cardinality n, how do we know it's a free basis too?

[topaz solar]

it generates the free group?

[median pawn]

that it does, i feel like we need to show the univ property holds

[topaz solar]

well, if G satisfies the universal property, then like

S->a free basis for G

which I assume you know exists?

otherwise I question how you know it's rank n

@median pawn

[median pawn]

rank n just means that there's a generating set with n elements, not a free generating set necessarily

[topaz solar]

"rank-n free"

[hidden haven]

I think you are misinterpreting the problem. S is given to be a generating set, not a free generating set. The problem asks to prove that a generating set that has the cardinality of a free generating set must be free.

[topaz solar]

right, I'm saying if you know a free generating set of that cardinality exists

S -> that set

[hidden haven]

But that may not extend to a map from G

Because S may not be a free basis

[topaz solar]

bijection

because n = n

[hidden haven]

So?

[topaz solar]

then we get generating->S -> G

[hidden haven]

I don't get it

What are you trying to do with the bijection S → any free basis

[topaz solar]

lemme draw it rq

[hidden haven]

ok

[ivory trail]

you most likely need an automorphism that sends S to the given free basis

[hidden haven]

To say that S → G satisfies the universal property of the free group

The map S → G is fixed

[topaz solar]

ah right L bozo

I am 0 brain

[rocky cloak]

So if I understand great sharp, we have S' < F(S). So we can define a map F(S) -> F(S) by mapping S to S'. Since S' is a generating set, this is surjective, hence it splits. Now the problem is that the splitting doesn't a priori have to map S' to S...

[ivory trail]

it splits?

[topaz solar]

Well if we have that the free group has no nontrivial relations that kills kernels

[rocky cloak]

Right, but that's just the statement we're trying to prove

[rocky cloak]

Yeah, just map S to it's preimage

[topaz solar]

right right makes sense

[ivory trail]

this makes a right split exact sequence 0 -> ker -> F(S) -> F(S') -> 0 which doesn't mean it splits, no?

[rocky cloak]

The map splits. Which gives you a semidirect product

The kernel will probably not split no

[ivory trail]

oh i see

[topaz solar]

it's fixed but is it the inclusion necessarily?

I mean obviously injective but I mean

[hidden haven]

Yes, a generating subset means that it is in particular a subset

[topaz solar]

bijective inverse seems to satisfy a fixed map there

[ivory trail]

i don't know how to do anything with free groups except via bouquets of circles

[rocky cloak]

I found a comment on MSE that said this was a well known theorem without giving any reference

[median pawn]

I'm a bit lost how did we prove the thing

[ivory trail]

we didn't

😦

[median pawn]

ah okay i'm not so lost then

[hidden haven]

It might just be some annoying symbolic work with induction

[rocky cloak]

https://math.stackexchange.com/a/3045466

Found MSE

[topaz solar]

tbh I'd honestly just like

[hidden haven]

That's a fucked up proof

[topaz solar]

not care about this theorem

[hidden haven]

How is this an exercise

Unless you are doing some combinatorial group theory

I am sure that makes this easier

[topaz solar]

I'd just say S <-> free basis -> G is a fixed injection and satisfies the universal property

[ivory trail]

This is essentially the statement that finitely generated free groups are Hopfian.
do i need to have studied finite group theory???

[hidden haven]

Then you are just proving that a free bases is a free basis

By that logic any subset of cardinality n is a free basis

Doesn't even need to generate G

[topaz solar]

true L

[ivory trail]

where's the falso person

[hidden haven]

[median pawn]

yeah ok that proof is a bit wild

but at least we have a proof lol

[ivory trail]

i didn't read it

[topaz solar]

Can I just say

Oh yeah chief, F(basis) ~= G, and words satisfy no nontrivial relations because they are systematically reduced to get rid of a a^-1 looking pairs, then each s_i of S is represented by a word and since they generate G every word can be made from them: in particular, each e_i of the basis. So if some s_j = T for some word in (s_i) nontrivially, but since the e_i can be generated in terms of the s_i we must have s_j appear in at least one e_i (as it is clearly impossible to get n free generator terms from n-1 generators) then that means there will be a nontrivial relation among the e_i basis, which cannot happen?

the "clearly impossible" meaning we can't just throw out s_j and still cover the e_i since if <S-s_j> is the whole group still, that poses an issue when doing e_i |-> s_i, but e_j |-> anything else in <S-s_j>-S, since that's never iso because nontrivial kernel involving e_j

definitely surjective because it hits all generators

[rocky cloak]

Can you conclude that s_j = T, just from a nontrivial relation existing?

Moving away from free groups you could have a^2 = b^2 without either a or b being multiples of each other.

[median pawn]

Could someone explain what the shift action means here? I know wreath products in a slightly more particular context, i.e., when the first group is the direct product of G taken n times, and H is a subgroup of the symmetric group S_n on n letters

Then the action of H on the direct product is obvious, it shuffles the coordinates in the same way that it shuffles {1,2,...,n}

Ah do we just send the h^th coordinate under the action of x in H to the xh^th coordinate?

That's the obvious choice for me, it's a bijection too

[solar vessel]

yeah

maybe invert something

make sure the orders line up

[median pawn]

what do you mean?

[coral spindle]

Well this is just an incorrect definition

Let me describe the correct one.

Let $G$ be a group, let $X$ be a $G$-set, and let $H$ be a group.
The wreath product $H \wr G$, where the $G$-set $X$ is implicit, is the semidirect product $H^X \rtimes_\phi G$ where given $g \in G$ and $f \in H^X$ we define $\phi(g)(f)(x) = f(g \cdot x)$.

[cloud walrus]

Borty

[coral spindle]

So in particular note that we can do this for any group action of G.

[uneven jackal]

shift as in shifting indices, so the action you described seems alright

[coral spindle]

It's typical to see wreath products such as $C_2 \wr S_n$, which is the group of so-called `signed permutations'. These are permutations paired with $n$-tuples of entries in $C_2$, where $S_n$ acts on these $n$-tuples by the permutation action!

[cloud walrus]

Borty

[uneven jackal]

the definition of Borty seems more general but idk how the other is "incorrect" lol

[rocky cloak]

Guess one could distinguish between a wreath product and the wreath product.

[coral spindle]

It's incorrect because it only describes one particular wreath product.

That's like saying "A group is something that looks like Z/nZ"

This definition is incorrect despite Z/nZ being a group.

[coral spindle]

Maybe so

[uneven jackal]

eh

I guess the definition he sent is the special case where the G-set is G itself

[rocky cloak]

Wikipedia seems to call it the regular wreath product

[coral spindle]

Yes, hence it being incorrect

[rocky cloak]

Because regular representation I guess

[uneven jackal]

it's not incorrect, it's a special case

[coral spindle]

.

[uneven jackal]

I know how to read thank you

[coral spindle]

It is incorrect to say that it is the definition of a wreath product.

[rocky cloak]

Sometimes people use slightly different definitions, and that's okay

[uneven jackal]

yeah

also you can't compare definitions that don't involve the same objects to begin with

if you're not specifying the data of a specific G-set, it makes sense that the definition he sent would be a natural one

it's saying "if you don't specify the action, here's the one we're using"

nothing inherently incorrect imo

[median pawn]

same thought

[coral spindle]

This conversation is very silly in my opinion, but I will repeat that when people refer to e.g. the Coxeter group C_2 \wr S_n, they are not referring to the definition that your book gives.

[median pawn]

ahh, i see it now, thanks

[coral spindle]

:)

[median pawn]

more eye-opening than silly really

[uneven jackal]

wouldnt the permutation action be exactly the action described in his definition

[coral spindle]

No. The group described there would have $2^{n!}\cdot n!$ elements, whereas in reality $C_2 \wr S_n$ has $2^n \cdot n!$.

[cloud walrus]

Borty

[uneven jackal]

you said "S_n acts on these n-tuples by the permutation action" how is that different than the "shifting" action

[coral spindle]

The shift action of S_n on C_2^{S_n} is very different from that of S_n on C_2^n

I don't know how else to explain that

[uneven jackal]

oh right. so my point is even more true

there's a G-set here being acted on that isn't specified in his definition

[coral spindle]

No, it is specified

It is H

The H-set H

[uneven jackal]

lol

[coral spindle]

Why are we still talking about this as if this hasn't been solved?

[uneven jackal]

because I can and you're free to leave the conversation

[median pawn]

calm down y'all

[uneven jackal]

yeah it's all fine

[coral spindle]

If you have any questions about clarifying the definition hausdorff, feel free to ask here

[median pawn]

the definition is clear to me! thank you

[topaz solar]

Well yeah, you’d have at least one thing in the word, so T = s_i T’ = eor s_i^-1 T’ = e, either flip it or flip T and boom s_i = expression in s_i’s, and in this case we could have a = b^2 a^-1 yes, but not for the e_i since we can map them to groups where this doesn’t occur

[rocky cloak]

Right, but s_j can apear in T, so then I'm not sure I follow your argument

[topaz solar]

It can indeed

[rocky cloak]

So then we can't conclude anything about <S - sj>

[topaz solar]

It clearly can’t be written in terms of just the s_i otherwise you’d have n-1 generating n things, but it’s sufficient to know that S has a relation involving s_j, so S-s_j does not and will critically require s_j in order to remain generating for all G

*n free things

T must include s_j or s_j ^-1 in it somewhere

But by assumption we can’t pair things off and reduce them

So by carrying this along and that you cannot maintain independence, we can arrive at <S-s_j> is not allowed to be iso to G under swapping e_i to s_i and e_j to anything else

Because it would satisfy a nontrivial relation by virtue of being in the span of the other n-1 elements

[rocky cloak]

I still don't quite get this.

Like I agree <S - s_j> cannot be the whole group, but how does that tell us that S is a basis?

[topaz solar]

So, it needs s_j, and <S> covers the e_i basis

Since well it covers everything ofc

We can get e_i = word in S

[rocky cloak]

What do you mean by covers?

[topaz solar]

As in basis \subset <S>

Generated by oop

We cannot generate it with S-s_j since earlier argument about lack of n independent points, so at least one e_j has s_j in the word

[rocky cloak]

Sure sure

[topaz solar]

e_j = nontrivial expression involving other e_i is the next goal

[rocky cloak]

Yeah, if we can get there that would do it

[topaz solar]

We can take the s_j term in the e_j expression and it expand it out

[rocky cloak]

Expand in e_is you mean?

[topaz solar]

No, particularly in e_j because it’s the one depending most definitely on s_j

e_j = word in s_i looking like t s_j t’

Now = t T t’

[rocky cloak]

Sure sure

[topaz solar]

These can all be expanded out in terms of the e_i

[rocky cloak]

Indeed

[topaz solar]

Since obviously those generate it

[rocky cloak]

But then we're just back to square one right

[topaz solar]

This implies T is trivial in terms of e_i and would amount to e_j

But it’s nontrivial in terms of S

[rocky cloak]

Yeah tTt' simplifies to just e_j

[topaz solar]

And every e_i can be rewritten in terms of them

This however implies if we take e_j and expand out s_j -> T however many times, rewrite in terms of e_i, then rewrite in terms of s_i

We can reduce it

[rocky cloak]

Not sure I follow

Why does it imply this?

[topaz solar]

Well taking the [e_j in terms of S] ofc we can or else we have an issue

And any word in G can be written as a word in s

So we just kinda take G -> <S> -> G -> <S> sort of back and forth

And even if we expanded out s_j into T arbitrarily many times, it’s been reduced to a word in terms of S (pre-expansion) by reducing only the a a^-1 = e type pairs

Which I think should be an issue if T is not reducible by pairing

[rocky cloak]

I don't really see the issue, but maybe it could work

Maybe you sort of end up recreating the part of the MSE proof where you map to a finite group

[topaz solar]

It’s pretty much just using these rewrite rules and thereby reducing the T that shouldn’t be reducible

By virtue of being a nontrivial relation

Or uhh, maybe we could argue no such relation exists among the e_i then the f: G -> <S> from e_i = s_i would satisfy f(e_j) = T a nontrivial relation, but e_i don’t have this, so f (f^-1(T) e_j) = e and nontrivial kernel so not iso

[rocky cloak]

What is f^-1 ?

Just a choice of splitting?

[topaz solar]

Ye

Or more precisely what splitting, the one where we take some appropriate writing of T in terms of s_i

And then saying it’s that same corresponding indices order for the e_i in a s_i ~> e_i rewrite

Which is fine to consider if it’s an isomorphism

But that’s problematic since kernel

And this is probably a way more effective form of my argument (and might actually work if you don’t think my rewriting reduction one does Dr 2808 )

[rocky cloak]

I still don't follow

f^-1(T)e_j is in the kernel of f, why is that a problem?

[topaz solar]

That implies 1 of two things

It’s reducible by pairing (it’s not) or f is not an isomorphism

[rocky cloak]

But we want to prove that f is an isomorphism

[topaz solar]

And if no such relation exists, then we do have an isomorphism

[rocky cloak]

Sure, but are we any closer to showing that such a relation can't exist?

[topaz solar]

Idk I feel like the rewriting works but im also recently waking up

Pea brained as usual

[rocky cloak]

Idk, there's lots of complicated rewriting we're doing here. It's hard to keep it all in mind at the same time

[topaz solar]

It at the very least will get our s_j -> T expanded thing back into another thing without it by just using e_i = [chosen word in s_i] and s_i = the word in e_i and the pairings

So it won’t be any nontrivial relations

But it’s reducible to a trivial one?

Might be instead better to look at f: <S> -> G by s_i to e_i and showing this has trivial kernel idk

[topaz solar]

The fact that it’s a trivial relation in terms of e_i isn’t anything because lmao free basis, but we can then rewrite e_i = word in terms of S, so that gets it trivialized in terms of S

[rocky cloak]

How do you know such a map exists?

[topaz solar]

Well, obviously S -> G defined by s_i to e_i is a function

[rocky cloak]

If you had such a map we would be done

[solemn dew]

does anyone have an idea to make p-sylow groups from s_{p^k}?

[topaz solar]

and we know we can write e_i e_k = f(s_i s_k)

[solemn dew]

i know how to do this with s_4, but s_8 is just wrong

[rocky cloak]

Why do we know that, isn't that what we want to prove?

[topaz solar]

Well pairing things off isn’t an issue unless like s_i = s_k^-1

[rocky cloak]

Sure, but for more complicated expressions

[topaz solar]

I’m thinking like kinda just m a k i n g the thing, but this is obviously way inconvenient

[rocky cloak]

Wikipedia lists a construction
https://en.m.wikipedia.org/wiki/Symmetric_group#:~:text=In general%2C the Sylow p,base p expansion of n).

[solemn dew]

i'll check it thanks

[topaz solar]

Idea: taking S -> X a map into a group, if it makes sense maybe we could do something like e_i as words in S and hoping that works

@rocky cloak https://en.m.wikipedia.org/wiki/Nielsen_transformation

There’s these things which are pretty related

And using this to get S -> basis

[rocky cloak]

Seems maybe you consider S4xS4 inside S8, then take the sylow subgroups inside S4 and a permutation that swaps the two (like (15)(26)(37)(48))

[solemn dew]

sorry i am just mega confused right now

[rocky cloak]

Yeah, seems you can probably do it like this somehow then. But I don't know if I'll be able to wrap my head around it

[topaz solar]

I mean it seems related to my thing

Probably more careful though lmao

[rocky cloak]

Why is that?

[solemn dew]

Alright I'll try to explain why I am so confused 😅

Suppose G is a group with o(G) = p^k * m, then we can create a subgroup of order p^k, right?

it is called a p-subgroup ?

[rocky cloak]

It's called a sylow p-subgroup yes

[solemn dew]

Alright, so if I want to create a 2-Subgroup from say S_4, it would have 2^3 = 8 elements, since o(s-4) = 1.2.3.4 = 2^3 . 3

[rocky cloak]

Yes

[solemn dew]

Yes

And in my book we have a subgroup of order 4 since the elements are 4

I am refering to this:

So they claim that the 2-Sylow subgroup is generated by (13)(24), but how is it that T can be the p-sylow subgroup?

[rocky cloak]

They are saying the group generated by (13)(24) and T, so this should have 2*4 elements

[topaz solar]

The way that’s written I say isn’t the best

[solemn dew]

but where is the identity element then?

[rocky cloak]

Where it is? What do you mean?

[solemn dew]

(13)(24) * {(12),(34), (12)(34),e} = {(

I am trying to figure out

How it is generated by (13)(24) and T

I just figured you have to multiply them

[topaz solar]

Well for one e is literally in T

[solemn dew]

But I am wrong hahah

[rocky cloak]

You just consider the smallesr subgroup that contain both

[topaz solar]

And yes you are wrong

[rocky cloak]

So all possible products of elements (including empty product)

And inverses

[topaz solar]

That empty product is exactly the identity, just in case

[solemn dew]

I have never seen something being generated by two sets

[topaz solar]

{x} u T

It’s one set

[rocky cloak]

It's exactly the same as being generated by some elements

It just mean generated by the elements in the set

[topaz solar]

Ye

[solemn dew]

Oh okay

[topaz solar]

If it’s generated by S, T then it’s generated by S u T

[solemn dew]

So for example if something is generated by (12) and (23) it should be {e, (12),(23),(12)(23),(23)(12)} ?

Is that S union T?

if S = (12), T = (23)?

[topaz solar]

S = {(12)}, T={(23)}

But don’t forget to consider, say, (12)(23)(12)

[solemn dew]

Alright haha

[topaz solar]

It’s arbitrary finite products, so that’s not just pairs

[solemn dew]

yes yes

But for a more complicated case:

T= {e, (12), (23)}, S = (45)
Then T u S would be:

{e, (45)e, (45)(12),(45)(12), (23)((45)), (45)(23), [ALL COMBINATIONS OF 45 AND T), (12), (23)}?

[void cosmos]

bad question

why does R being noetherina imply Mat_n(R) noetherian?

i know how ideals in Mat_n(R) look like

wait

let I_1 < I_2 ... be a chain in Mat_n(R) , each is of the form Mat_n(J_i) where J is an ideal of R , these J_is form a chain , so there exist a k such that J_k = J_n for al n>=k

so the Mat_n(of those) are equal and then the chain in Mat stablizies?

is this right?

[molten viper]

I seem to remember that n | |G| need not imply we can construct a subgroup of order n, is the sylow n-subgroup one of the cases where we can?

[topaz solar]

Sylow theorems time let’s goooo

[solemn dew]

n a prime number?

[void cosmos]

any1

[solemn dew]

@molten viper If p is a prime and p^m | o(G), then G has a subgroup of order p^m.

[wraith cargo]

Maybe prove that claim ABT the ideals of M_n

[void cosmos]

yea i can do that

[wraith cargo]

But yeah Ur reasoning works

[uncut girder]

Solomon Friedberg

[wraith cargo]

The first Sylow theorem shows you can (note that n has to be a prime)

[molten viper]

Wack

[void cosmos]

THANK YOU

[rocky cloak]

Noetherianess is usually framed in terms of left or right ideals, but you're looking at two-sided ideals.

One way to see that it's Noetherian is that it's finitely generated as an R-module, and ideals are in particular R-submodules.

[void cosmos]

wait this doesnt work if R is noncommutative right

[chilly ocean]

all rings are commutative

[solemn dew]

what do they mean by "number"?

[void cosmos]

and yea that classification of ideals in Mat_n is not true for like left ideals as u pointed out

last time

[chilly ocean]

how many there are

[solemn dew]

oh

every p-sylow subgroup is isomorphic to each other?

so then the numbers of p-sylow groups would be 5 or 5*5?

in this case ?

[wraith cargo]

Refer to third Sylow theorem

[solemn dew]

1+ kp = 1 + 8 * 3 = 25 divides 25, hence there are 25 3-Sylow subgroups?

[wraith cargo]

Well that's not the only option?

What if the subgroup is normal?

The number of 5-subgroups is more simple

[cinder onyx]

yo quick q

I'm confused in what sense a subset S can generate a normal subgroup N of G

like I understand how it can generate A subgroup by generating up to closure, but why normal

[topaz solar]

Well, surely it’s contained in a normal subgroup

Like, uhh

G

[cinder onyx]

• wth does this mean

[cinder onyx]

well yeah

but how is that subgroup 'generated' by the subset

[topaz solar]

So, you can also define <S> as the smallest subgroup containing S

Consider a similar definition by “smallest normal subgroup N containing S”

[cinder onyx]

ohhh and then you can define a subgroup N by including all conjugates

thereby generating a normal subgroup

[topaz solar]

Or, intersecting normal subgroups containing it

[cinder onyx]

true

okay that makes sense

[topaz solar]

So R script is the smallest normal subgroup containing R

[formal ermine]

worst formulation of the 3rd sylow theorem

[cinder onyx]

yeah that's a bit wank

just say like,
for |G| = p^e * m, then the number of Sylow p-subgroups divides m and is congruent to 1 mod p

surely

[cinder onyx]

ohhh nvm i get it now

[formal ermine]

thiz

[topaz solar]

This is essentially just saying the largest group possible on the generators where the relations all hold

[cinder onyx]

yeah it's saying that largest group possible is that quotient group

yh got it

[rocky cloak]

Curious question: if n = p^k m, is it true that for any t that divides m for which t = 1 mod p there exists a group of order n with t sylow p groups?

[delicate orchid]

C_{p^k} \rtimes C_m works for the t = m case I think

interesting question

[solemn dew]

@wraith cargo You're right, and k = 0 can also be true

But how do you know if a P-Sylow subgroup is normal?

[rocky cloak]

What is the action here, and are there more restrictions on m other than congruent to 1 mod p?

[cinder onyx]

p-sylow subgroups are conjugate with one another

if it is unique, naturally conjugation will just map to itself

i.e. it is normal

[solemn dew]

ouu ur right

[delicate orchid]

I'm picturing C_{p^k} as Z/p^kZ and C_m as some subgroup of (Z/p^kZ)^{\times}

[rocky cloak]

I see, that makes sense

Not obvious that there will be m sylow subgroups though...

[solemn dew]

but is this only true if the p-sylow group is unique? meaning there is only one p-sylow group?

[delicate orchid]

true, perhaps we would have to take some m' such that m' has m divisors and then just direct product them?

[cinder onyx]

yes

wait

hang on

[delicate orchid]

but at that point we're well outside the scope of what you're asking

[cinder onyx]

i think so?

[topaz solar]

If you had multiple and it’s abelian then uh I am idiot as per usual

They’d all be normal

[delicate orchid]

A sylow subgroup is normal if and only if it's unique

[cinder onyx]

ok yeah

[solemn dew]

alright then

[cinder onyx]

cuz they're conjugate with one another

[solemn dew]

yeah

[cinder onyx]

so if you have 2 p-Sylow subgroups H_1 and H_2, then there will be a g s.t. gH_1g^-1=H_2

[solemn dew]

hence not normal

[cinder onyx]

that's just the 2nd theorem

!!!

[delicate orchid]

sylow B-subgroup

[rocky cloak]

I found a paper stating that there are no finite groups with 22 sylow 3 subgroups (among some other examples)

[delicate orchid]

that's such a disgusting result I hate it

but in like

a good way

[cinder onyx]

ew

[rocky cloak]

And none with 1+3p sylow p groups when p>7

[cinder onyx]

why does group theory have so many vile results like that like bro

[delicate orchid]

ok that's a bit nicer

I imagine very little is known about the p = 2 case?

[topaz solar]

What in the world kind of vile methodology can accomplish such unholy results

[delicate orchid]

combinatorics would be my guess

send the paper jagr

[cinder onyx]

the most ugliest mathematic

[rocky cloak]

https://www.sciencedirect.com/science/article/pii/0021869367900762

[delicate orchid]

using the tensor product for direct product of groups
so based

[topaz solar]

This is one of the most cursed results I’ve ever seen

[rocky cloak]

Seems like a lot of reductions through normal subgroups etc until you reduce down to simple groups

Then some black magic

[topaz solar]

Apparently any odd number of subgroups works though

[delicate orchid]

lets think about that

ok I've thought of examples for 1,3, and 5 so I believe it (C_2, S_3, SL(2,5))

[topaz solar]

And numbers which work play nicely with multiplication

[delicate orchid]

yeah that's the first thing the paper says and is kinda obvious

[topaz solar]

Still it’s a lot of things packed into this introduction

And many are cursed, like 1+3p, p >= 7

[delicate orchid]

oh I've read papers that are far worse than that

let me pull it up

[topaz solar]

I mean there’s more cursed looking results, but this is a very cursed result in how it all was

[delicate orchid]

this result comes to mind

what the fuck is going on here

it's like how every graph is planar unless it contains either K_5, B_3,3 but with some literally who group

[topaz solar]

I don’t know who would like

Think this would be true

[delicate orchid]

it's probably very similar to this result

"oh this doesn't always work? but what's the simplest case?"

[topaz solar]

Forget proving it, who designed this theorem statement

[delicate orchid]

"oh look, everything that doesn't work contains this case!"

[topaz solar]

I assume all the things there make sense and are well known (I don’t but I’m dumb), but like

It looks awful

[delicate orchid]

oh yeah the F_p(...) is garbage

actually no, even the J thing is absurd iirc

[wraith cargo]

Generally finding the number of Sylow p-subgroups is definitely not a trivial problem
My algebra prof gave us some really vile Sylow problems to do
Tho usually you can do some stuff with group actions to find out the number as well
Like we had a problem to show that an arbitrary group of order 36 isn't simple and the way to do it was to show that the action of the group on Syl_3(G) had non trivial kernel which then gives you enough info to show that the number of subgroups isn't 1

[topaz solar]

Oh no

Ok so yeah this is spaghetti

[wraith cargo]

I LOVE GROUP THEORY

[delicate orchid]

yeah J(P) is the group generated by all maximal abelian groups of P

so it's a little bit of a mess

[topaz solar]

But what is P

[solemn dew]

haven't done anything with group actions yet

[delicate orchid]

and it doesn't work for p = 2 holy shit this is such jank

[delicate orchid]

some p-group

[solemn dew]

but i am excited when i will be exploring that chapter!!!

[delicate orchid]

Sylow p-subgroup of some G specifically

[topaz solar]

I hate it thanks

[delicate orchid]

that result is immediately followed by this one

now this result is actually cool

[topaz solar]

What is F_P

[delicate orchid]

the fusion system of G on P

dw about it

[topaz solar]

Ok that’s still French

[delicate orchid]

do not call me french

[wraith cargo]

What paper is this???

[delicate orchid]

https://web.mat.bham.ac.uk/C.W.Parker/Fusion/fusion-intro.pdf

[topaz solar]

Idk how to interpret F, but this it looks useful?

[wraith cargo]

Wew you always know how to make a drool OwO

[delicate orchid]

it's the only known (and is probably the only) family of "exotic" fusion systems over a 2-group

[topaz solar]

Interesting

[delicate orchid]

called the solomon fusion systems after Dr. Fusion Systems

[topaz solar]

Interesting stuff but studying finite groups sounds so vile

Infinite groups too dw

[delicate orchid]

yeah it's all vile

[topaz solar]

But finite ones have all these wacky factorization things

[delicate orchid]

wdym

[topaz solar]

Like prime numbers and counting subgroups and all

Probably have some related infinite cardinality cursed results in infinite groups but that’s gonna have nasty consequences

[delicate orchid]

oh right yeah

I don't really know what's going on in the infiniosphere but there's probably some awful nonsense up there too

[cinder onyx]

oh my

being able to kinda comprehend these papers makes me feel so smort holy shit

😭

[topaz solar]

Well it’s hard to have subgroups such that |subgroup| < |group| lmao

[delicate orchid]

ofc it's the fuckin finite simple groups

there's something to those guys

[topaz solar]

But in the vein of these sylow results uhh you’d possibly have some cursed cofinality adjacent results

[topaz solar]

Not so simple consequences of the finite simple group revolution

[delicate orchid]

wow they actually wrote proofs for this stuff back in the day?

[delicate orchid]

I stg the finite simple groups are like, the law of small numbers or something

but for groups

there's a pervasive weirdness "around" 0 that permeates through all of mathematics

[topaz solar]

I feel like that’s less “things near 0 weird”

More “big things made from small things” type of a lot of math

[cinder onyx]

and I guess kinda there's a lot of possible values of big

not a lotta values of small

[topaz solar]

Also they’re easier to see what they look like

Compared to like

[cinder onyx]

yeah that's true

[delicate orchid]

True the monster is pretty big

[cinder onyx]

fuckin 1 kajillion dimensional group representation

[topaz solar]

Giant products of the monster group and wacky things

[delicate orchid]

I can give you a 1 dimensional group rep of the monster

[topaz solar]

0

[delicate orchid]

1

[untold turret]

i want to take a break from commutative algebra and start a bit of algebraic geometry, can i do that without being 100% comfortable in commutative algebra? in terms of algebra progress i'm up to the proof of the nullstellensatz using noether normalization

[next obsidian]

Have you done dimension theory for fg integral k-algebras?

[untold turret]

not yet, would you suggest I do that first?

[next obsidian]

If yes, then go learn (or just learn the statement of) Krull’s Hauptidealsatz

Yes

Do those then start AG

[untold turret]

noted, thanks for the recommendation!

[dim bane]

I’m too small brain to understand this joke, 0 can’t be in the image of a representation so that’s clearly not the joke right

[next obsidian]

If you have Noether Normalization and the Cohen Seidenberg theorems (going up / going down) this follows easily

[untold turret]

yeah i've done integral extensions too

[next obsidian]

I guess also learn going up and going down if you don’t know them

This is needed to justify a lot of dimension theory in classical AG

The Hauptidealsatz is literally the statement “a hypersurface is cutout by one equation” but algebraic

Once you have this you can do a good bit of AG, and then you’ll probably appreciate commutative algebra more once you see t being used

And if you hit a point where you algebraic knowledge is lacking, you’ll be much more motivated to learn commutative algebra

[untold turret]

fantastic, thanks a lot for the recommendations!

[next obsidian]

Are you using AM for comm alg?

[untold turret]

yep, a mix of atiyah-macdonald and steps in commutative algebra by sharp

[next obsidian]

Got it. I don’t know AM as well, but look for a section which says dim A = tr. deg Frac(A)

And learn that stuff, it should also prove stuff like dim A/p + dim A_p = dim A

For integral domains fg over a field

And then I believe the last section of the book does the Hauptidealsatz via the Samuel function / Hilbert function

[untold turret]

noted, i've done transcendence degrees too, nice to see dimension stuff seems to use a lot of previous material

[wraith cargo]

Idk if other books do this but I know hartshorne cites results that are more obscure than what's in atiyah macdonald

[next obsidian]

You can just blackbox them for the most part

And they show up either deep in the book, or in chapter 1

You can learn almost all of chapter II with less than all of AM

[wraith cargo]

Ig true
I know hartshorne used some obscure result thats in Matsumura in ch II
Tho he lists all the CA results he uses

[next obsidian]

Okay true

I mean they aren’t too obscure, but they will be for your first pass I guess

II.8 has stuff about differentials, but those proofs are actually easy

[wraith cargo]

The differential appendix in Eisenbud nearly ended me lol

[next obsidian]

And then he uses Japaneseness of fields when he uses that the integral closure of a fg k-algebra is finite

But just black box that

And Chapter I has the Cohen Structure Theorem you should also blackbox lol

I mean, you should learn it, it’s awesome

[wraith cargo]

That's a good one
I know Eisenbud covers it relatively early on

[next obsidian]

But not on your first pass

Huh

It’s at the end of Matsumura CA

But i guess it doesn’t have to be

[wraith cargo]

He does it like ch. 7

[next obsidian]

There’s a lot of ways to get the theorem

[wraith cargo]

And his book has 21 chapters

[next obsidian]

I like a very general way, so I think it’s very very advanced

[wraith cargo]

I mean Eisenbud proves it in the context of completions

[next obsidian]

How does he establish the existence of a coefficient field?

And does he handle the mixed characteristic case then?

[wraith cargo]

I haven't read the whole chapter yet I'm still in the weeds of it
He just mentions the theorem at the start of the chapter and talks Abt how the theorem also gives rise to Witt vectors in certain cases but I think he mentions that he'll just assume existence of the coefficient field in the proof? (I don't have the book on me I'd need to read that passage again to see exactly what he says)

[next obsidian]

Oh lmfao

Establishing the coefficient field is the hard part

Once you have that the argument is simple, I can sketch it rn

When you have a coefficient field, which is a k < A such that the projection map A -> A/m = k is an isomorphism, you get that A = m + k as a module

Let m have generators x1,…,xn, and take a in A

[wraith cargo]

Oh no he proves it lol

[next obsidian]

Write A = a_11x_1 + … + a_1nx_n + b_0 where b_i in k, a_ij in A

Oh okay

Well, the proof is almost done, iteratively keep writing each a_ij in this way

Then you get a power series in the x_i with coefficients in k

Completeness says it converges, giving your surjection k[[X1,…,Xn]] -> A

[wraith cargo]

I like how short Eisenbuds proof is lol

[next obsidian]

It’s an iso when A is regular because if it had a kernel then the x_i have a dependence contradicting regularity

How does he establish the coefficient field?

[formal ermine]

I didn't know you're a racist

[next obsidian]

Bruh

Don’t even get me started on this dude

Calling that name racist fucking pisses me off, as a half-Japanese person

I have trauma from a twitter thread about this shit

[wraith cargo]

He proves this

[next obsidian]

Lmfao, so he deals with differential bases oog

[wraith cargo]

NGL this ch of Eisenbud is actually one of the hardest things I've ever read
Completions are so fucked

[wraith cargo]

Yeah
He sorta assume you know what you're doing with them?
Which was like whaaaa for me

[next obsidian]

Yeah it’s tough field theory

Matsumura combines that with generalities on formal smoothness

[wraith cargo]

Every CA book falls into two categories
(1) Nice reads, covers enough to be useful
(2) They will make you feel like worthless scum and you'll know so much after you've read them that youll be a walking encyclopedia of CA knowledge

[next obsidian]

Me after reading Matsumura

[wraith cargo]

I took a break from Eisenbud to read some homological algebra

To understand his last few chapters that assume knowhow of that

[formal ermine]

I have yet to see (1)

[elder wave]

Gathmann comm alg

I've recommended this a lot of times

[next obsidian]

Undergraduate commutative algebra

By Miles Reid

[rocky gorge]

is it really enough to have that gs = tg condition at the end? doesn't that only get you gS is contained in Sg?

[vast quiver]

you're correct that it isn't enough, there's an example here which shows that the weaker condition can get you a bigger subgroup.
https://math.stackexchange.com/questions/401592/normal-subgroup-if-conjugate-subgroup-is-subset

for $G=\mathrm{GL}_2(\mathbb Q)$, you have a subgroup
$$H=\left{\begin{bmatrix}1 & n\ 0 & 1\end{bmatrix} \mid n\in \mathbb Z \right}$$
which when conjugated by
$$g=\begin{bmatrix}2 & 0 \ 0 & 1\end{bmatrix}$$
you get the strict subgroup
$$gHg^{-1} = \left{\begin{bmatrix}1 & 2n\ 0 & 1\end{bmatrix} \mid n\in \mathbb Z \right},$$
so you have $gHg^{-1} \subseteq H$, but not $gHg^{-1} = H$. We should then say that $g\not\in N_G(H)$, we need there to be actual equality to say that $g$ normalizes $H$.

[cloud walrus]

thejoesully

[vast quiver]

oh lol is this on nlab?

[rocky gorge]

ah yep, that's a good counterexample

yeah

was pulling my hair out trying to check it lol. i found some other resources online that have similarly flawed defs

[vast quiver]

yeah I was trying to prove it for a bit too

it seems like one of those true things

then I gave up and stack exchanged it

[rocky gorge]

i have a vague memory of something similar being actually true for checking normality too, which made me think it might betrue here

like can't you say H is normal in G if gHg' is a subset of H or something?

[vast quiver]

yep that's right

if you know for all g, then you can plug in g^{-1} to get the reverse inclusion

[rocky gorge]

ah right

man i've forgotten a bunchof algebra lol. for some reason it's one of those topics that never sticks for me

[vast quiver]

there's just too much math to remember

[sly rain]

What makes his notes so good?

[vast quiver]

I like Gathmann’s notes too, I like that they have good geometric motivation and are very gentle.

I think they're maybe a bit lacking on exercises?

[vast quiver]

for example the "going-up" and "going-down" theorems are motivated geometrically by what they say about finite maps between varieties/schemes

that being said it looks like atiyah-macdonald still has this, it's just reserved for exercises. If you are well-motivated or have a course to help give structure then Atiyah Macdonald might be slightly more enriching? I found it a little too tough to start out with

[solemn dew]

how do you draw the conclusion that if at least 1 5-Sylow subgroup or 3-Sylow subgroup must be normal, then every 3-Sylow and 5-Sylow Subgroup must be normal?

Refering to b:

Is it since o(G) = 30 = 5.2.3, then there must be 3 Sylow subgroups, hence the product must be closed in G. But if not every 3-Sylow or 5-Sylow group is then it is not closed, hence closed => every 3 and 5 Sylow subgroup must be normal in G?

[wooden ember]

Note that if a sylow p-subgroup is normal, it is unique

Since all sylow p-subgroups are conjugate

Just to clear up the « there must be 3 » misconception

[solemn dew]

yes, but in the case where the 3-Sylow group isn't unique, then it can have at most 2 subgroups. then the elements in that group would be at least 2.3 + 5.1 < 30, so it works?

[wraith cargo]

Do a counting argument for a)

[wooden ember]

That’s the idea yeah, if it’s not unique G will have too many elements

[solemn dew]

done that

[wraith cargo]

Show that there isn't enough space for them to not be normal

[wooden ember]

Not sure I got this though

Oh wait sorry you’re doing b)

Mb

[wraith cargo]

Ur number of possible 3 Sylow subgroups is wrong

[solemn dew]

i calculated that if it is not normal, then the 3-Sylow subgroups would have to be 2. Then I calculated that 5 had to be normal, and the elements is less than 30, which is not a contradiction

[wooden ember]

I was thinking like irony incarnate

Yeah you either have 1 or 10 sylow 3 subgroups and 1 or 6 5 subgroups

2 is not = 1 mod 3

[solemn dew]

you are right

m ybad

oh okay then it works

since the elements would be less than 30 if they are normal, and if not we would get a contradiction

well thanks

do you guys have a good paper on group actions? my book doesn't really cover that

[wooden ember]

Could you elaborate your argument ?

Haven’t done sylow stuff in a bit

[solemn dew]

Sure

I just look at the cases when they are not normal:

If 3-Sylow subgroup is not normal, and 5-Sylow subgroup is normal:

3 . 10 + 1 . 5 > 30, hence contradiction

If 3-Sylow subgroup isnormal, and 5-Sylow subgroup is not normal:

3 . 1 + 5 . 5 > 30, hence contradiction

If 3-Sylow subgroup is not normal, and 5-Sylow subgroup is not normal:

3 . 10 + 5 . 5 > 30, hence contradiction

where o(G) = 30

[wooden ember]

You don’t have 3*10

That would be counting the identity multiple times

You can only guarantee 25 elements in the first scenario

Or ig 26 including order 2

Second scenario also doesn’t work

[solemn dew]

1 + 3*k divides 30, then k = 0 or k = 3, so there must be 1 or 10 3-Sylow subgroups

or am i thinking wrong?

[wooden ember]

10 sylow subgroups doesn’t mean 30 elements

Since they all share the identity

[solemn dew]

oh right

[wooden ember]

No I don’t think part b will just be a counting argument

[solemn dew]

damn

[wooden ember]

Okay I’ve got one case

Let P be a sylow 3 subgroup and Q a sylow 5 subgroup and suppose Q is normal

Then PQ is iso to Z/15Z so that P is normal in PQ and thus Q < N_G(P). Thus |G:N_G(P)|<=2 => n_3 = 1

As a matter of fact this also covers the other case

[solemn dew]

not sure how you found out it is isomorphis to Z/15Z

but you proof works i guess

[wooden ember]

All groups of order 15 are cyclic

It’s of the form pq with p not dividiving q-1 (p<q)

ie reapply sylow

[solemn dew]

alright alright

i'll try to remember that haha

but speaking of subgroups of order 15

look at (c)

hahah

[wooden ember]

You can already do that with a) anyways lol

It’s just PQ again

[solemn dew]

hahah okay

well thank you brother

[wooden ember]

Aight I originally came to ask a question so here it is: I’m asked to prove that any integral domain for which every ideal is a finite product of prime ideals must be dedekind but it always feels im missing something in my attempts

Just want to make sure that the statement is correct and that uniqueness was not forgotten

[rocky cloak]

A Dedekind domain is an integral domain in which every nonzero proper ideal factors into a product of prime ideals. It can be shown that such a factorization is then necessarily unique
-Wikipedia

[wooden ember]

Aight okay I’ll keep giving it a go then

It’s the whole primes possibly being contained in one another I can’t get rid of

[rocky cloak]

So I guess you're trying to prove the uniqueness part?

[wooden ember]

No not necessarily

Trying to get something of dimension 1 and then localizing should do the trick

[rocky cloak]

So what's your definition of Dedekind domain?

[wooden ember]

This is an exercise with a bunch of equivalent defs I gotta prove. The only things I’ve proven so far are
1.Noetherian domain which is locally a DVR
2.Noetherian domain which is integrally closer and of dimension 1
3.Domain in which all prime ideals are invertible

It feels like proving the last one would be easiest

[rocky cloak]

Okay, so you're trying to prove that factorization of ideals implies one of these?

[wooden ember]

Yeah

[wooden ember]

I think I’ve got something for this

Nvm something goes wrong

[rocky cloak]

I managed to find a proof online, but it's actually pretty tricky.

It goes by showing that every invertible prime is maximal, and that any principal ideal can be factored into invertible primes.

Then you can prove that any prime is maximal, and from there it's pretty straight forward.

[wooden ember]

Aight I’ll try to follow that sketch thanks a bunch

[hidden wing]

How would I go about proving coz(z) = cos(x) cosh(y) - i sin(x) sinh(y)?

[solemn dew]

guys if a subgroup is normal, is it in the center?

[wraith cargo]

I am dummy but how does gamma generate Gamma here??
Gamma is the image of the group of units of an ultrametric, locally compact non discrete field under mod_k

[south patrol]

No, most strikingly note the whole group is a normal subgroup of itself

But if you want smth less trivial, S_3 has trivial centre whilst A_3 is normal

[delicate orchid]

the converse is true though, centric => normal

[south patrol]

Oh wew I had a reps q lol

And it's about symmetric groups

[delicate orchid]

fine

[solemn dew]

⚔️

[south patrol]

Lol it's like

Say $E$ is a complex vector space, $\Sigma_k$ acts on $E^{\otimes k}$ in the obvious fashion. Then $E^{\otimes k} \simeq \bigoplus_{\pi} \Big(V_\pi \otimes \mathrm{Hom}{\Sigma_k}(V\pi, E^{\otimes k})\Big)$ as $\Sigma_k$-representations

[solemn dew]

I don't see why this is wrong though

[cloud walrus]

potato

[south patrol]

Where {V_pi} is the set of all irreps

Apparently this is v standard, do you know why it holds?