groups-rings-fields

[agile burrow]

we're claiming that S is a p-Sylow subgroup of G

[formal ermine]

oh then it's a typo in my lecture script

[agile burrow]

maybe I'm not clear on what exactly you're proving. Is it just Sylow I?

[formal ermine]

yes

case 1 when p doesn't divide |Z(G)|

[agile burrow]

ok

[formal ermine]

but yeah I got the induction now

thx

[agile burrow]

np

[formal ermine]

I'm having a hard time when they just say "by induction" without specifying what the induction hypothesis, base case, and step is

[agile burrow]

fair enough, it's good to write those all out explicitly to see what exactly is being used

[formal ermine]

if $p$ divides $|Z(G)|$ then there exists a subgroup $H \subseteq Z(G)$ with order $p$ because $Z(G)$ is abelian

[coral shale]

the thinking is all in the step more often than not

[formal ermine]

why is $H$ cyclic

[coral shale]

order p

[agile burrow]

every group of prime order is cyclic

[formal ermine]

[agile burrow]

by like Cauchy or whatever

[formal ermine]

I keep forgetting these simples things

[agile burrow]

there are a lot of moving parts

[formal ermine]

lagrange

[agile burrow]

right, that's the one

though I suppose Cauchy works too

[formal ermine]

"this easily follows from euler's theorem"
"which one of them?"

ok another induction question

why does $G' \coloneqq G/H$ have a subgroup $S'$ of order $p^{m - 1}$ by induction over $|G|$

[south patrol]

Induction on power of p dividing the group order

basically you use existence of H so that you can quotient it out and get smth smaller than G

And then I guess next step will be correspondence to get back to G

[agile burrow]

explicitly, if |G| = p^m r and |H| = p, then |G/H| = p^{m-1} r

[south patrol]

but there is a cooler noninductive proof of sylow 1 lol

[agile burrow]

which one is that?

[formal ermine]

ah

[south patrol]

Which I have basically forgotten but I'll try to remember in a sec aha

[formal ermine]

what does this mean

like the $A \to B \to C$

[agile burrow]

composition

[chilly ocean]

the map is A -> C but you want to emphasize you're composing A -> B and B -> C

common notation

[formal ermine]

ah

why is $|G/\pi\inv(S')| = \frac{p^{m - 1}k}{p^{m - 1}}$

[agile burrow]

what are G' and S' here?

[formal ermine]

.

[agile burrow]

isomorphism theorems give that the quotient is isomorphic to (G/H) / (pi^-1(S')/H) = (G/H) / S which has order p^{m-1}k/p^{m-1}

at least on the level of cosets, since pi^{-1}(S') need not be normal

[south patrol]

ok gimme a sec lol

[formal ermine]

can you elaborate pls

[agile burrow]

Not really, it's the third isomorphism theorem

$\pi^{-1}(S')/H = \pi \circ \pi^{-1}(S') = S'$

[cloud walrus]

walter

[agile burrow]

alternatively, you could note that $|\pi^{-1}(S')| = |H| \cdot |S'| = p^m$

[cloud walrus]

walter

[formal ermine]

|G'/(pi^-1(S')/H)| = |G'/S'| = |G'|/|S'| = (|G|/|H|)/|S'| = (p^m k / p)/(p^(m - 1) = (p^(m - 1) k)/p^(m - 1)

ah

[formal ermine]

I don't understand the first equality here

[agile burrow]

the projection map is literally quotienting by H

[formal ermine]

projection map?

[agile burrow]

the map pi: G -> G/H

[formal ermine]

wdym

[agile burrow]

I don't know how to elaborate. The quotient map pi: G -> G/H is the map sending g -> gH, its coset

[formal ermine]

why is pi^-1(S')/H = pi(pi^-1(S'))

[agile burrow]

pi^{-1}(S') is the subgroup of G obtained by pulling back S' along this map. If you aren't sure this is a subgroup then you should verify that. But this is essentially by construction.

Let xH be in pi^{-1}(S')/H. Then x is in pi^{-1}(S'), so pi(x) = xH is in pi(pi^{-1}(S')). The other direction is equally straightfoward

[formal ermine]

ahhhhhhhhhhhh

thank you so much

[agile burrow]

no worries, sorry if i sounded frustrated at the end

[formal ermine]

lol dw

[agile burrow]

i still don't understand how you hate group actions though lol

[formal ermine]

I don't even know examples for group actions

[agile burrow]

.

[formal ermine]

we did them within one lecture and then we were supposed to fully know them

[agile burrow]

action of galois group

conjugation

multiplication within a group, automorphisms, linear transformations, the list is endless

[formal ermine]

but then it's just like a second group operation

[agile burrow]

or rather, the group operation is just another action

and anyways, conjugation happens on more than just the elements. You could also consider the action of conjugation on the lattice of subgroups

[formal ermine]

lattice of subgroups?

[agile burrow]

you know how there's a lattice of subfields

[formal ermine]

wut

[agile burrow]

you form a poset of subgroups of a given group ordered by inclusion

[formal ermine]

what's a poset of subgroups

[agile burrow]

partially ordered set of subgroups

it's not terribly important right now, my point is just that the group acts on the set of subgroups by conjugation in a way that preserves some structure

[formal ermine]

let $X$ be the set of all sylow p-subgroups and let $G$ act on $X$ by conjugation. since all sylow p-subgroups are conjugate there is only one orbit in $X$ wrt $G$. if $p \in X$ then $|X| = |Gp|$ which means that $|X|$ divides $|G|$

[formal ermine]

is that reasoning correct

[agile burrow]

Yeah, that seems right

[formal ermine]

epic

[south patrol]

Wielandt proof of Sylow + some extra intuition ig: Let $G$ be a group. Note that if $X$ is the set of non-empty subsets of $G$, then $G$ acts on $X$ by left mutliplication and if $S \in X$ then $S$ is a subgroup of $G$ iff $\mathrm{Stab}(S) = S$; this will motivate the following.

Now if $G$ is a finite group and $|G|=p^k m$ where $m \not \equiv 0 \mod p$, then we can restrict our attention to $\Omega$, the set of subsets of $G$ of size $p^k$ and search for stabilisers of size $p^k$. Note if $S \in \Omega$ then in fact $|\mathrm{Stab}(S)| \le p^k$, e.g. because $G = \bigcup_{T \in \mathrm{Orb}(S)} T$ and taking sizes shows $|G| \le |\mathrm{Orb}(S)| p^k$ so $|\mathrm{Orb}(S)| \ge m$.

Now Suppose, for contradiction, then, that $|\mathrm{Stab}(S)| < p^k$ for all $S \in \Omega$. Then $p^k$ doesn't divide $|\mathrm{Stab}(S)|$, so $p$ divides the size of each orbit. But this means that $|\Omega| \equiv 0 \mod p$. This is a contradiction, because $|\Omega| = {p^k m \choose p^k} = \frac{(p^k m)(p^k m -1) \dots (p^k(m-1) + 1)}{p^k \cdot \dots \cdot 1}$ and the same power of $p$ divides the numerator and denominator. Hence by this contradiction and the remark above, there's $S$ with $|\mathrm{Stab}(S)|=p^k$; $\mathrm{Stab}(S)$ is then the desired subgroup.

[cloud walrus]

potato

[south patrol]

@agile burrow if u were interested lol

[agile burrow]

this is a neat proof

I've thought about that criterion for subgroups before but haven't thought about applying it that way

thanks for sharing

[south patrol]

Oh that was smth I like added in here for the intuition lol

Well like

Ye idk lol

I guess this is sort of proof by wishful thinking in a sense lol like if there's a subgroup then we can find it as a stabiliser

so we will examine stabilisers and see what happens

But yeah idk I think writing like this is interesting cause it shows that we reduce it tons like
S = Stab(S) reduced to |S| = |Stab(S)| reduced to |S| <= |Stab(S)|

hm

[formal ermine]

let $P$ be a sylow p-subgroup of $G$. let $X$ be the set of all sylow p-subgroups and let $P$ act on $X$ by conjugation. why is $|X| \equiv |X_P| \pmod{p}$

[south patrol]

Orbits partition X

[formal ermine]

but that's the stabiliser?

[wind steeple]

what is XP

[south patrol]

Isn't X_P the set of elements of X fixed by P

which isn't the stabiliser

i.e. the elements of X with orbits of size 1

[agile burrow]

if you have a non-trivial orbit, then it has size divisible by p by orbit-stabilizer

[wind steeple]

use class formula

[south patrol]

Beat me to it lol

[formal ermine]

wait so what is $X_P$

[south patrol]

i've just gvien two descriptions oop

but ye more eexplicitly like

$X_P = { x \in X \mid p\cdot x = x \text{ for all } p \in P} = {x \in X \mid \mathrm{Stab}(x) = P} = {x \in X \mid \mathrm{Orb}(x) = {x}}$

[cloud walrus]

potato

[south patrol]

Wait a minutet hough lol

[formal ermine]

what is $p \cdot x$

[south patrol]

Wouldn't you usually use the notation X^P not X_P or am I thick

[south patrol]

notation for group actions

which action would you normally use

[formal ermine]

ah

I've only seen it as $p.x$

[agile burrow]

yeah, X^P is the typical notation for fixed points

[south patrol]

Oh fair

I guess uh

X_P would be the invrse thing like

[formal ermine]

ok wait before we proceed

[agile burrow]

the set of orbits

[south patrol]

well

if G a group acting on X and s in X then G_s i guess would be the stabiliser of s

[formal ermine]

does someone have a simple proof for the congruence of p-sylow-subgroup size

[south patrol]

which is the kinda inverse

[formal ermine]

because the one I'm currently reading

is like really confusing

[south patrol]

G \cdot s would be orbit for me

[formal ermine]

and I couldn't find a better one

[south patrol]

What exactly do you mean by this

[agile burrow]

I'm just saying because when M is a G-module, I always see M^G for invariants and M_G for coinvariants lol

[south patrol]

or do you mean this

[formal ermine]

$s_p \equiv 1 \pmod{p}$ where $s_p$ is the amount of $p-$sylow subgroups

[south patrol]

yee

[south patrol]

Oh okay uhhh

Meh screw it i might just write up some spiel on sylow theorems since it's good recall for me lol

been a while

Okay right tbh the wikipedia article gives a nice proof

but if you are stuck on that maybe i can give some help

[karmic moat]

is universal property sufficient and necessary?

[tender wharf]

for?

[karmic moat]

just in general

e.g. if something meets the universal property for a tensor product then it's a tensor product, and vice versa?

[next obsidian]

Yeah

[vivid tiger]

universal properties are another way of looking at the thingy. looks to me like category theorists use universal properties as definitions of e.g. a product

[agile burrow]

I remember now, it's that if k is a field and G is finite, then kG modules are projective iff they're injective

[next obsidian]

Semisimple ring

This true for those

Actually it characterize them I think

[agile burrow]

that sounds right

[lethal dune]

for semisimple, any module over it is already injective and projective

not too sure about that being characterization tho

any arguments?

[wooden ember]

A ring is semi simple iff every ses of modules over it splits?

[lethal dune]

Yes

[wooden ember]

Cause you’re semi simple iff every module over you is semi simple, iff for each module over you sub modules are direct summands

And then it follows each module over R is projective and injective no?

[agile burrow]

oh yeah, you guys are right

[lethal dune]

Yes

[agile burrow]

wonder if there's a criterion for rings where modules are projective iff injective

[lethal dune]

But thats not the question I asked

[wooden ember]

Hmm fair enough

[next obsidian]

Is a summand of an injevtive injective?

I think so right?

[lethal dune]

If R is an ID then M proj and injective forces M=0 unless R is a field

[agile burrow]

this is right, yeah

[next obsidian]

Then this is true when you only restrict to fg modules iff A is self-injevtive

And if you want to restrict to Noeth rings where direct sums of injectives are always injevtive

It’s true even for non-fg

[agile burrow]

neat

[next obsidian]

Wait no

Uhhh

You get one direction

[lethal dune]

Didn’t get what’s happening tbh

[next obsidian]

If projective <=> injective then self injevtive

And self injevtive => (projective => injective)

With fg hypotheses if the ring is non-Noeth

But for the other direction I dunno

[wooden ember]

Maybe I’m being dumb but can’t you say that if every module is projective, then every ses splits since Ext^1(M,N)=0 for all M,N? And then you conclude?

Im probably misunderstanding your question @lethal dune cause you could also do simpler by just saying every projective module is a direct summand of a free module

Ahhh sorry I see you’re asking if projective <=> injective implies semi simple mb

[next obsidian]

Why use Ext for this, just lift id

[wooden ember]

Yeah yeah shut up

I don’t know what a simple argument is

[wooden ember]

Why do we need fg hypotheses here?

[next obsidian]

You know a projective is a summand of a free module

So you want to know the free module is injective

[wooden ember]

Oh infinite direct sum of injectives isn’t injective?

[next obsidian]

But if the free module is infinite rank it won’t be true

[wooden ember]

Yeah okay I see

[next obsidian]

Noetherian <==> every arbitrary direct sum of injectives is injective

[woeful flint]

This might be really obvious but I'm awful at group theory. Is it true that if $G^\prime \leq G$ where $G$ is abelian then if $N\leq G^\prime$ we have that $G^\prime/N \leq G/N$?

[cloud walrus]

Kraft Macaroni

[somber sleet]

Hey guys, if I look at the group Q, what is for every prime number q this intersection

in our case G=Q

[lethal dune]

Yeah

[lethal dune]

Q is divisible meaning nQ=Q for all n

[somber sleet]

but why though?

[lethal dune]

Try yo show x to nx is surjective in Q

[south patrol]

With one exception lol

[hidden haven]

Burn in hell

[somber sleet]

guys if i know that the set of all left cosets G\N is cyclic, how do the elements look like?

[chilly ocean]

G/N, G\N is the set difference

[somber sleet]

this set here, sorry made a mistake

[tender wharf]

Elements of G or G/H

[chilly ocean]

If G/N is cyclic then we have some generator xN in G/N
and all elements are powers of xN, that is of the form (xN)^n = x^n N

[somber sleet]

oh that makes a lot of sense, thx a lot

I'm sorry if my question was kinda confusing 😦

[chilly ocean]

it was open to discussion

that's all

[tender wharf]

Wouldnt G/H just have 1 coset tho

Oh wait

Mb I was confused

If G/Z(G) is cyclic then G is abelian sorry

[chilly ocean]

yeah, that sounds about right

[tender wharf]

I evidently need sleep

[somber sleet]

I'm doing an exercises where N is a normal subgroup of G, which is contained in Z(G) (centre of G) and we also know that G/N is cyclic

[tender wharf]

Yeah then G is abelian

[somber sleet]

and I have to show that G is abelian

yeah exactly

but I'm kinda confused though

[tender wharf]

Where are you stuck

[chilly ocean]

every element of G is in some element of G/N

I think that should do it

[somber sleet]

I mean we knoe that G is abelian if Z(G)=G right? and Z(G) included in G is trivial

[tender wharf]

Yeah

[somber sleet]

on the other side i get kinda stuck

[chilly ocean]

I'd just take two arbitrary elements of G and show they commute

[somber sleet]

dont you have to do the whole construction first?

[chilly ocean]

what construction, I don't understand

[tender wharf]

If you take two arbitrary elements of G say x and y

Say G/N = <gN>

then xN = g^i N and yN = g^j N

[chilly ocean]

I'd just say, x in g^i N and y in g^j N

but that was precisely what I was saying

[tender wharf]

Then you can go from there

[somber sleet]

I completely understand the method you want to use, but the solution I'm lookin at are slightly different

[tender wharf]

In what sense

[chilly ocean]

so, are you trying to solve the exercise, or analyze the solution?

[somber sleet]

mostly analyze the solution, because I did it with your method and it worked well for me too

[chilly ocean]

ah okay. Then post it, lets analyze it

[somber sleet]

it's in german sadly

give me a sec to translate it

[chilly ocean]

it's okay

I'll understand

[somber sleet]

this is the german version

[chilly ocean]

I don't understand

[somber sleet]

I'm translating it

but i don't get it either haha

[elder wave]

what part do you not get

[chilly ocean]

yeah I just thought I wouldn't need to know that much German to understand it

[somber sleet]

the part where Cent_G(x)=G

[paper flint]

(Timo understands German)

[somber sleet]

why the hell

[somber sleet]

thx for trying though

[tender wharf]

g = x^i n for some n in N

so g is in the center of x

because N is a subgroup of Z(G)

(I dont know german sadly)

[elder wave]

das Erzeugnis x^n*N liegt im Zentralisator für alle n und man kann jedes g in G so darstellen

also ist G schon im Zentralisator

[chilly ocean]

ah. I gave up too early, the clues were all there

[somber sleet]

choose x in G, s.t. G/N is spanned by the coset xN. For each element g in G there exists a whole number n s.t. gN=x^n*N, which means that g is also contained in x^nN. From it we recognize that N and x span whole G. Because we also know that N (which is contained in Z(G)) and x are contained in Cent_G(x), and because it is a subgroup of G, we can say that Cent_G(x)=G. This is equivalent with saying that x is in Z(G), and because N is in Z(G) we say that Z(G) contains the spanned subgroup of x and N. Which means that G is contained in Z(G). That's why G is abelian.

I don't completely understand how the centralisator of x has to be G

[elder wave]

have you looked at the answers given so far

[somber sleet]

ist es darum?

[elder wave]

ja

[tender wharf]

because since n is in N subgroup Z(G) so n commutes with everything

And then g is arbitrary

im assuming thats what fur jedes element g means

[somber sleet]

yes yes

yes I get it now

thx a lot guys, sry for disturbing y'all

[elder wave]

[tender wharf]

Lol ur not disturbing at all

[elder wave]

that's what this channel is for

[tender wharf]

Ok you might have been disturbing a reddit scroll session

Which is good

[bitter epoch]

is Symmetric group of order 3 solvable?

[chilly ocean]

yes

[bitter epoch]

nice

thank you

makes sense, it's like 2 cyclic groups superimposed

[delicate orchid]

it's the semi-direct product of 2 cyclic groups yeah

[bitter epoch]

so is every symmetric group solvable?

I would think so

[delicate orchid]

no

S_5 and beyond are not solvable

[bitter epoch]

oh, that's weird

what operation exists in S_5 that isn't cyclic?

isn't that just like rotating a pentagon?

[delicate orchid]

odd way of thinking about solvability, and groups tbh

every "operation" is "cyclic" in that you can get a cyclic group from every element - I have no clue what you mean by this as it's trivial

[bitter epoch]

if it's cyclic isn't it abelian?

[delicate orchid]

The problem is that in the derived series you reach A_5 after the first step, which is equal to it's commutator subgroup

[bitter epoch]

and according to this youtube video, if a group is solvable it can be broken into abelian groups

[delicate orchid]

so it's not solvable

I'd go off of the actual definitions over wishy-washy youtube videos

[bitter epoch]

oh, yeah makes sense

[delicate orchid]

that is true though I think, a group is solvable <=> there is some chain of group extenstions from an abelian group that give you that group

[glossy crag]

If R\subset R' are Dedekind domains (without R' necessarily being integral over R), is P\cap R' guaranteed to be non-zero in R? I know it's non-zero if R'/R is integral, but does it happen without the assumption?

[formal ermine]

why does any element x in omega not fixed by H lie in an orbit of order |H|/|H_x|?

is it because uhh

orbit stabilizer

|Gx| = |G/G_x| = |G|/|G_x|

[agile burrow]

yes

[formal ermine]

but then why is it only for elements not fixed

wouldn't it also be true for fixed elements

[agile burrow]

the second part where they say |H|/|H_x| is a multiple of p relies on H_x not being all of H

[formal ermine]

ah because if it were fixed the stabilizer would be all of H

so $\Omega_0$ is the set of all elements whose stabilizer is the entirety of $H$?

[agile burrow]

yes

[formal ermine]

ahhhhh

[agile burrow]

that's what it means to be fixed under an action

[formal ermine]

yeah

why is |H|/|H_x| a multiple of p by assumption

[agile burrow]

H is a p-group so it has order p^n

H_x is a proper subgroup, what can you say about its order?

[formal ermine]

it's a divisior of p^n

so it's also a p group

[agile burrow]

right, namely it has order p^m for m < n

[formal ermine]

yeah

[agile burrow]

so |H| / |H_x| = ?

[formal ermine]

p^(n - m)

yeah and if x is fixed then H_x = H so |H|/|H_x| = 1

ok

then we use the fact that |X| = sum of all sizes of different orbits

because orbits are either identical or different

idk what you call the different thing in english

I forgor what it's called in german even

[agile burrow]

disjoint?

[formal ermine]

yes

disjoint

[agile burrow]

yeah, orbits partition the set

[formal ermine]

yeah

so $|\Omega| = \sum_i p^{r_i}$

[agile burrow]

you seem to have left out the fixed points, unless you want to count those as p^{0}

[formal ermine]

ah

right

,, |\Omega| = |\Omega_0| + \sum_i p^{r_i}

[formal ermine]

taking $\pmod{p}$ then gives us [ |\Omega| \equiv |\Omega_0| \pmod{p} ]

[agile burrow]

yes

[formal ermine]

right ok thanks

[agile burrow]

happy to help

[formal ermine]

now to the actual theorem

what does the highlighted part mean

like

aren't we letting P act on omega

instead of G

[agile burrow]

Yes, but P is a subgroup of G. This is saying that every element of P normalizes Q, so P is a subgroup of the normalizer of Q in G.

[formal ermine]

ah

why does P = Q

I understand that P and Q are conjugate in N_G(Q)

and Q is normal in N_G(Q)

but how does that imply that P = Q

ah wait

P and Q are conjugate means that there exists a g in N_G(Q) such that P = gQg^-1

but because Q is normal P = gQg^-1 = Q

ok got it

[agile burrow]

yes

[formal ermine]

now to sylow 2

[next obsidian]

Walter

[elder wave]

Walter

[agile burrow]

chmonkey and timo

[formal ermine]

what is $[G:P]$

[agile burrow]

the index of P in G

[formal ermine]

ah

I've never seen that notation before

only in field theory

[agile burrow]

i find that surprising

it's pretty common in group theory

[formal ermine]

we defined the index as $|G/P|$

[delicate orchid]

which is true for finite groups

actually no, it's just true in general, sorry

[formal ermine]

what's the definition of the index for infinite groups

oh

[delicate orchid]

I was thinking of |G|/|P|

[formal ermine]

what's the usual definition of the index

[delicate orchid]

|G/P|

I did a whoopsie in my brain

[agile burrow]

my subgroups stay finite index

[barren sierra]

what is this font

[formal ermine]

why does $p$ not divide the index

[barren sierra]

[formal ermine]

comic sans

[barren sierra]

I hate it

[elder wave]

same

[agile burrow]

P is a maximal p-subgroup, right?

[formal ermine]

yes

ah

[agile burrow]

so if |G| = p^n k where k is not divisible by p, then |P| = p^n

[barren sierra]

Suppose p divided the index, implies not maximal

[delicate orchid]

because if |G| = p^mk then P \in Syl_p(G) has order p^m by definition

[formal ermine]

yeah

right

[barren sierra]

I love how we just said the same thing just with increasingly more explanation

[formal ermine]

why does $g\inv HgP = P$ imply that $g\inv Hg$ is a subgroup of $P$

[agile burrow]

it's the same thing with like xP = P implies x is in P

[formal ermine]

ah

iff

[delicate orchid]

just with the whole dang thang at once

[formal ermine]

ah ok

ye

finally done with the proofs for sylow

thanks walter and wew

[agile burrow]

happy to help

[delicate orchid]

one day I will use the sylow theorems

[agile burrow]

that day is not today

it's been a while, wew

hope you're well and enjoying the new year

[delicate orchid]

I'm ill and having a crisis, but I did get the first greggs of the new year at the airport so that's good!!

[agile burrow]

hope you feel better soon and sort out your crisis

[delicate orchid]

Oh yeah I do actually have a question for chat

does the artin-webberburn theorem actually construct the isomorphism between the algebra and the funny matrix algebras or do you have to go cry about it yourself - cause I can't find much information on the actual map itself beyond "yeah dey iso"

[agile burrow]

iirc tox and bacono asked some stuff about this a few months back for their rep theory course, could be worth asking them? i haven't thought about artin wedderburn in a while so i don't have an answer right away

[delicate orchid]

makes sense - I think I've found the map anyway it would just be nice to have a "canonical" way of doing it yk

oh and the map only works for the complex group-algebra, very sad!

[agile burrow]

hm

well the simple modules are irreps

[delicate orchid]

yeah - the main issue is I want the action induced by a fusion system to behave nicely through whatever isomorphism we come up with

cause then you can take orbit-sums and get wholesome F-stable characters

[agile burrow]

you lost me, but i'll try to think about it

https://math.stackexchange.com/questions/2845364/are-these-two-isomorphisms-of-the-group-algebra-with-a-matrix-algebra-distinct

this post seems somewhat promising

[delicate orchid]

I'll take a peep

[warm wyvern]

Omg, yellow timo

[elder wave]

[wooden ember]

I know it’s given by the Fourier transform for C[G] when G is finite

[wooden ember]

Oh and I see you’ve already found that, sadge

[hollow fjord]

why in finite fields the polynomials are solvable by radicals?

[delicate bloom]

this answer seems to lay it out pretty clearly I think https://math.stackexchange.com/a/2232489/741168

[warm wyvern]

why did aluffi draw the spectrum of Z like this? lol

he just throws this here so randomly, I don't get it

(this is the first time he's introducing prime ideals)

[south patrol]

Isnt he just ordering them by inclusion ig

But ye it is a bit intrresting lol

[chilly ocean]

drawings of spectrums are pretty famous though I don't know much about those
so I guess this is just an example where it looks a little less complicated

than usual

[formal ermine]

how do you read this

[chilly ocean]

[rustic crown]

that looks like Spec Z[x]

[rustic crown]

i don't understand the complete picture, but i can say a few things

[rotund aurora]

Show that if G is a finite subgroup of PGL_2(Q) then G occurs as a Galois group of Q.

Can anyone give me any hint or some directions to go? I'm not really thinking hard about this atm, but I'd like to solve this, and I have no idea where to start really. Maybe its not too difficult idk

this exercise comes after proving that PGL_2(F) iso Gal(F(x)/F)

but stating the exercise in terms of Aut(F(x)/F) or in terms of PGL_2(F) seems as challenging

[agile burrow]

i have no idea how to solve this, but fwiw the finite subgroups of PGL(2, Q) were classified by Dresden

which is probably not how you're supposed to approach it lol, but still interesting

[rotund aurora]

I think its a really nice problem

because

1. I have no idea why I should care
2. I have no idea where to start

xD

but its probably important that you get groups as Galois groups somehow

[agile burrow]

the isomorphism is via fractional linear transformations, right?

or like, (a b \ c d) sends f(x) to f((ax + b)/(cx + d))

[rotund aurora]

this is from Isaacs

[agile burrow]

that's neat

[rotund aurora]

I should reread that chapter in more detail tbh

because he actually proves S_n occurs as a galois group of Q

[rustic crown]

ig only way i see how to start is by hoping we can write PGL(2,Q) as semidirect product of N and G for some normal N

that way Gal(Q(x)^N/Q) = G

this is the first time i see someone looking at automorphisms of non-algebraic extensions lol

[rotund aurora]

like, the chapter is on transcendental extensions

[warm wyvern]

wdym by inclusion?

[tribal furnace]

subset relation ig

[warm wyvern]

oh, if some ideal is included in another then they're connected?

why would he do that?

[agile burrow]

there are some deeper ideas here that i'm not entirely sure about, but one explanation for this picture is that there's a topology we can put on the spectrum called the zariski topology

the picture is supposed to capture something about this topology. idrk much about this, but in this case my understanding is that (0) is a "generic point" (whatever that means). A way to make this explicit is that the closure of (0) is all of Spec Z. In this sense, (0) kind of touches every point of Spec Z or something like that

other people can probably elaborate but maybe this gives a very rough idea of why someone might draw a picture at all

[warm wyvern]

[rustic crown]

if p is prime in Spec R, then the closure of {p} in that topology is all the primes q that contain p, so if q is a bigger prime, then it's like super close to p in this topology which is why you also say q is a specialization of p or p is a generalization of q

[rotund aurora]

I'm gonna say something nonsensical but

doesnt every member of PGL_2(Q) satisfy either x^2-1=0 or x^3-1=0

?

wait no uhh idk. I dont think you can talk about addition in PGL_2(Q) ?

mmh my reasoning was that every thing in GL2(Q) satisfies a polynomial of deg 2

and since scalars get killed, the coefficients of the polynomial also get killed (they are either 1 or 0)

[rustic crown]

what about something like

[1 1]
[0 1]

[rotund aurora]

uhm yeah right.

[rustic crown]

i was trying to come up with some nice examples of homomorphisms out of PGL_2(Q)

one is det

more precisely det : PGL_2(Q) --> Q*/(Q*^2) = infinite direct sum of Z/2Z

[warm wyvern]

where's when you need him?

[rustic crown]

[warm wyvern]

damn algebraic geometry

[agile burrow]

I need more group theory facts

I'm running low

Or maybe I'll just do another write up and talk about trace stuff this time

[chilly ocean]

all groups are abelian

[agile burrow]

Grow up

[chilly ocean]

[agile burrow]

I'm sorry, it was in jest 🫂

[warm wyvern]

aww

[chilly ocean]

[warm wyvern]

👴

walter be seeking blood

where do I learn about this zariski topowology?

do I have to wait till I do AG?

[coral spindle]

The holomorph is the terminal object in the full subcategory of split extensions of some fixed pair of groups

I learned that here

that was nice

[agile burrow]

Holomorph is a word I've heard before. I'll read about this. Thank you for your contribution

[rustic crown]

[coral spindle]

Trying to think of more facts I find cool...

Oh

here's something. I guess it's more topological in nature? Fuck it

[agile burrow]

Topology is fine too

[coral spindle]

Here is a very short proof that the fundamental group (with basepoint the identity) of a topological group is Abelian

A topological group is a group object in Top. Group objects in Grp are Abelian groups. The fundamental group is a functor, hence it preserves the diagrammatic property of being a group object. QED.

I like this proof quite a bit

[chilly ocean]

do you know Frattini subgroups

[agile burrow]

That's a good one.

[agile burrow]

I've heard of them before, haven't used them for anything though

[coral spindle]

Oh I guess you also need the fact that the fundamental group preserves the cartesian product. But this is well-known lol

[agile burrow]

Showing group objects in Grp are abelian uses Eckmann Hilton, right?

[chilly ocean]

woah. This is cool

[coral spindle]

Yeah, and in fact you can show that the fundamental group is Abelian directly using Eckmann–Hilton anyway lol

so it's kinda just pushing that fact forward

[agile burrow]

Right, that's the way I've seen it before so I was wondering if they were just the same under the hood

Still a cool way of phrasing it

[chilly ocean]

I thought it uses that x to x^-1 is a homomorphism

but maybe that's the same argument

[coral spindle]

However, it actually generalises a bit, as you can also argue that any topological monoid has Abelian fundamental group

[coral spindle]

Well maybe that's an alternative one haha

I think either one works tbh

[agile burrow]

I think you need Eckmann Hilton to argue that the underlying group operation agrees with the operation as a group object

From which you apply the fact that inversion ought to be a homomorphism to conclude that the group is abelian

But a priori if the operations were distinct then you wouldn't necessarily have the underlying group be abelian

I posted this maybe yesterday too in a different channel, but my current favorite group theory/topology fact is that if you have a finite K(G, 1) with nonzero Euler characteristic, then G has trivial center

[coral spindle]

I'm trying to come up with more interesting group theory facts

There's one I really like

but it is a bit complicated

Basically, if you have a linear algebraic group (think of some subgroup of GL_n that is defined by a polynomial equation in its entries) if you make some mild assumptions on it (connected reductive) you can actually get an explicit presentation of it in terms of the additive and multiplicative groups of the underlying field

This can sort of be seen as a generalisation of Gaussian elimination

[agile burrow]

Group presentations are great

[coral spindle]

Well this is a particularly interesting one, since it kinda relies on the field

so it's like, idk

an algebraic group representation is probably a good word for it

but it's somehow parametrised by the field in any case

[formal ermine]

are there any cool field/galois theory facts

[coral spindle]

Sure. A matrix in GL_n(F_p) is diagonalisable in the algebraic closure of F_p if and only if it's diagonalisable in GL_n(F_{p^n}) This follows easily from Galois theory.

[formal ermine]

"this easily follows from math"

[coral spindle]

Yes

[formal ermine]

so true

[coral spindle]

It really does.

This actually lets you do a cool thing in rep theory where you can characterise all representations of cyclic groups – even in really fucked up fields. I need to learn more about this though

[agile burrow]

That's a cool one

[warm wyvern]

I was ignored

[agile burrow]

I mean you can read the definition now if you want but it probably isn't helpful

But yeah, zariski topology is an algebraic geometry thing

[coral spindle]

I remembered something absolutely epic about this actually

Let's say you want to compute Sylow subgroups of SL_2(p) where p is some prime. I have been doing this recently.

SL_2(p) has order p(p-1)(p+1).

If you want to find the Sylow 2-subgroups, it's hard. If you want to find the Sylow p-subgroup, it's quite easy. What's interesting now is when you look at odd prime q subgroups. q must divide exactly one of p-1 or p+1.

It turns out if q divides p-1, you get diagonal matrices with entries being elements of order a power of q in F_p

However, this doesn't work for q dividing p+1. Uh-oh

But what you can do is really smart

You expand to F_{p^2}, and it turns out that you can diagonalise any Sylow q-subgroup of SL_2(p) when you expand to this

And lo and behold, now F_{p^2} has elements of order dividing p+1, since its multiplicative group of is order p^2-1 = (p+1)(p-1)

I think this is fucking awesome

Calculating this is a little tricky though

[agile burrow]

That's pretty dope

I love all these tricks for explicitly computing Sylow subgroups and stuff

Super neat

[coral spindle]

Those computational group theorists are so smart

really carrying the world of algebra on their backs tbh

[delicate orchid]

this idea of using F_{p^n} comes up when you try and find the conjugacy classes of SL_n as well

very cool idea

[coral spindle]

Oooh that's cool

[delicate orchid]

yeah so the idea is that you want to use distinct jordan normal forms to find conjugacy classes (there's some correspondence) - but this means you need eigenvalues, not all of which live in F_p

[coral spindle]

Right, but they do live in F_{p^n}! That's very smart.

[delicate orchid]

so you expand out to a larger field, which for SL_n turns out to be F_{p^n}

yeah it's sick

[coral spindle]

I'm honestly amazed you can do that at all. I have this idea that conjugacy classes of matrices is a hopeless problem

[delicate orchid]

it's actually pretty solved

but it is disgusting for large dimensions

[rustic crown]

aluffi does some affine algebraic geometry in chapter 7

[agile burrow]

I forget if he talks about zariski topology at all

[warm wyvern]

[rustic crown]

he does

[agile burrow]

But that section is all I know about algebraic geometry aside from the bits and pieces I've picked up elsewhere

[rustic crown]

but only for subsets of K^n

[warm wyvern]

I'm doing that after this chapter

[agile burrow]

Probably good as an introduction anyway

[warm wyvern]

why do we restrict PIDs to integral domains?

does assuming every ideal of a ring is a principle ideal imply the ring is an integral domain?

[formal ermine]

no

take e.g. K[x]/(x^2) where K is any field

such rings are called principial ideal rings

fun fact every pir is the direct product of quotients of pids

[south patrol]

Z/4Z too is a noice example

[warm wyvern]

I see

[rustic crown]

how to see the forward direction?

[formal ermine]

idk the proof is above my level

https://msp.org/pjm/1968/25-3/pjm-v25-n3-p11-s.pdf

[rustic crown]

direct sum

[formal ermine]

uhhhhhhh

[rustic crown]

i was thinking you're allowed to have infinite products

[formal ermine]

I think I read this on wikipedia

[warm wyvern]

you read wikipedia for math??

[formal ermine]

sometimes when I'm bored I just click myself through wikipedia

[celest cairn]

How do I find the minimal polynomial of $\sqrt{13}$ over $\mathbb{Z}_5$?
Is it $x^{2} - 3$?

[cloud walrus]

Sapphire Gaming

[celest cairn]

I’m actually only in precalc so this might be a stupid question.

[tribal moss]

Yeah, x²-3 sounds right.

Typing sqrt(13) is a bit of an odd way to express oneself in Z/5, but not really your fault, I suppose.

[formal ermine]

for what do you need that

[chilly ocean]

sapphire gaming spends all of their time here picking random fields and random elements and computing minimal polynomials

[celest cairn]

Yea basically.

Just doing stuff for fun.

[formal ermine]

is sqrt(13) even in Z5

cuz like x^2 = 3 mod 5 has no integer solutions

[lament dawn]

it is not, but that doesn't mean it doesnt have a min poly there

i has min poly x^2+1 in R

[tribal moss]

No -- if it were, its minimal polynomial would be linear.

[oblique river]

you say this as if the answer "no" means that the original question doesnt make sense

[formal ermine]

I forgor

don't we need to specify the extension field for minimal polynomial to make sense

[tribal moss]

The question should really be understood as: assuming we have an extension of Z5 where 13 has a square root; then what is the minimal polynomial of that square root over Z5?

[rotund aurora]

There is no notion of evennes modulo 5 tho?

[oblique river]

i mean, pick an algebraic closure of F_5

[lament dawn]

or just adjoin sqrt(13) to anything

[tribal moss]

Correct.

[oblique river]

there is clearly an element (well, two elements) in there which could reasonably be called "sqrt(13)"

[tribal moss]

But why "tho"?

[agile burrow]

Lol they didn't mean even in the sense of parity

[south patrol]

lol

[oblique river]

the answer to the question doesn't depend on which algebraic closure and which sqrt(13) you choose

[rotund aurora]

Brooo

[south patrol]

okay took me a while to understand this was what was meant

[rotund aurora]

Lol Im stupid

[tribal moss]

Oooooh.

[south patrol]

but ye i mean more generally like if a is in F and K an extension of F containing an element b with b^2 = a with b not in F then the min poly of b over F is just x^2 - a

[tribal moss]

Unless b is itself in F, of course.

[south patrol]

was just adding that lol

my apologies

[rotund aurora]

Why don't you try other algebra problems? There are many cool problems involving polynomials

[celest cairn]

Idk I just like field theory.

[rustic crown]

try finding the minimal polynomial of e + pi over Q

you'll be famous

[oblique river]

you should try learning some field theory then

like, some theory, not just the computations

[celest cairn]

Ok

[oblique river]

for example, you could try to learn how to understand statements like this

which you can use to answer not just the question "what is the minimal polynomial of sqrt(13) over F_5" also "what is the minimal polynomial of sqrt(13) over F_7" or any other field you choose

[formal ermine]

what's an intuition behind $R[[x]] \coloneqq (R^\bN, +, \cdot)$

[south patrol]

That is not really what I'd call a definition lol

[rustic crown]

cute font

[formal ermine]

I'm reading an analysis book and it defined that

[oblique river]

it's power series

[next obsidian]

Wtf is this comic sans ass font

[formal ermine]

ah

[oblique river]

if you try to define R[[x]] as "things of the form a + bx + cx^2 + ..." then you have to answer "what is x"

[cloud walrus]

potato

[formal ermine]

ring of sequences with elements in R

[rustic crown]

[south patrol]

which is?

[oblique river]

but this way you don't have to, you can just say "R[[x]] is the set of sequences of elements of R with pointwise addition and this funky ass multiplication"

[formal ermine]

wait

what's the difference between R[x] and R[[x]]

[oblique river]

potato just said

[south patrol]

what i said essentially

whether you allow for infinitely many non-zero coefficients or not

[formal ermine]

ah

[oblique river]

polynomials have a highest-degree term

[south patrol]

so for example 1 + x + x^2 + ... isn't in R[x]

[oblique river]

and power series don't necessarily

[formal ermine]

ah

ok thanks

[south patrol]

Note that like

[formal ermine]

what's a power series btw

[south patrol]

{1,x,x^2,...} form a basis of R[x] for example over R

a (formal) power series is an element of R[[x]]

in this context

[rustic crown]

formal

[south patrol]

true

the word formal is weird in math

[formal ermine]

what's the difference between a formal power series and a power series

[oblique river]

it just means like, "we are writing this down without regard to whether or not it actually makes sense"

[south patrol]

indeed i just find it weird lol cause it's like

idk formal power series make more sense arguably xd

[oblique river]

sort of nothing tbh. i think when people emphasize the word "formal" i think they are just emphasizing that like

we're not assuming this power series converges

or actually defines a function

[rustic crown]

it's emphasizing the "form" and not the "meaning"

[formal ermine]

how can a power series converge

[spice whale]

if you want that you need a topology

[oblique river]

have you taken calculus?

[rustic crown]

like 1 + x + x^2 + ... converges when |x|<1

[oblique river]

like did you learn about taylor series? and the radius of convergence of a power series?

[formal ermine]

so a power series converges when it converges everywhere?

[oblique river]

the point is that when we are talking about formal power series, we don't even care

it's just not a question we're bothering with

[rustic crown]

nah, the usual definition is that a convergent power series is something which converges in some open set, i.e. has positive radius of convergence

[south patrol]

i mean that would contradict what det said

[formal ermine]

ah

[next obsidian]

Power series in analysis means “converges somewhere which we care about”

Power series in algebra is just an abstract object which forms a ring in the way described above

[formal ermine]

right ok

thanks

[next obsidian]

To make clear we aren’t talking about the former ppl tend to put “formal” when doing the latter

[south patrol]

okay i'm glad i wasn't crazy for omitting the "formal" lol

topology is fake anyway

[spice whale]

true

[south patrol]

people just try to reduce it to algebra for that reason

[rustic crown]

but you can topologize R[[x]] where every power series would converge :p

[spice whale]

and we let them!

[next obsidian]

Shush your mouth

[south patrol]

how about we topologise it where no series converges

pain

[next obsidian]

Tfw 0 + 0x + 0x^2 + … doesn’t converge

[south patrol]

yes

[next obsidian]

Sadge

[south patrol]

jk it always will happen but uh

adugkjaybldgh1eb

[spice whale]

you can't get close enough to 0 it has covid

[south patrol]

who says that 0 is the same as 0 + 0x + 0x^2 + ...

intentionality ftw

after all, if you plug in x =infinity then you get either 0 or 0 + undefined + undefined + ...

[lament dawn]

is x-0 the minimal polymomial of 0 over F_7

[cloud walrus]

potato

[south patrol]

now i want macaroni

[rustic crown]

me too >.<

[south patrol]

Joe this is #groups-rings-fields not discussion

[tribal moss]

Random jokes go in #chill, please.

[agile burrow]

Here's something I've been thinking about. For nilpotent groups, there are two "natural" series to consider, namely the upper central series and the lower central series. Each are built inductively using the extremal properties that a central series ought to satisfy (I can expand on this if needed).

Solvable groups have a natural descending series, namely the derived series, which is defined inductively starting from the whole group and terminating at the identity. To my knowledge, there isn't a canonical way to define an ascending series that starts from the identity and terminates at G iff G is solvable, but I haven't thought about this very much so maybe there's an obvious thing.

I guess the issue is that for nilpotent groups, when building the upper central series the condition is that $G_{i+1}/G_i \subseteq Z(G/G_i)$ so you have an obvious choice for how to maximize this. On the other hand, for solvable groups the fact that the first term only needs to be abelian doesn't yield a canonical ``maximal" abelian subgroup

[cloud walrus]

walter

[agile burrow]

In any case, I maintain that the correct way to think about solvable groups is as a group which is inductively built out of extensions by abelian groups.

[oblique river]

i'm not quite sure what your question is

but is there any doubt about your last statement?

[agile burrow]

I guess the definition of a derived series which terminates didn't really click with me

Or at least not the first time around

[oblique river]

idk i guess i just see that as a measure of solvability

like, the quotients in the derived series are exactly the abelian groups which are being used to build up the solvable group via extensions

[agile burrow]

Right, that makes sense

I guess these days I'm trying to rethink a lot of my arguments about nilpotent/solvable groups by phrasing them in terms of extension 1 -> A -> G -> G/A -> 1

There's a neat result by Hirsch, which says that if G is a nilpotent group acting linearly on a finite-dimensional vector space M such that H^0(G, M) = 0, then all the higher cohomology vanishes too

And the proof is essentially to show it for abelian groups and then apply Lyndon-Hochschild-Serre to the SES to proceed inductively

[oblique river]

that's nice

[agile burrow]

I don't really have a good grasp of spectral sequences yet. I wonder if there's some weaker result that holds for solvable groups, but I can't trace the same argument because the differentials are probably different if the extension isn't central lol

[oblique river]

they are magic

[agile burrow]

very much so

[oblique river]

do not stare directly at the spectral sequence

[rustic crown]

[south patrol]

How do spectral sequences relate to this?

Oh group cohomology

[ruby sundial]

what is a good reference for them?

hatcher’s?

[oblique river]

i learned them on the streets

not from a book

[ruby sundial]

yeah ive been looking for good resources but i get lost somewhere

i was trying to read through Serge Lang’s introduction

and it was messy and confusing

[hidden haven]

Highly recommend John Rognes's book

His examples are topological because he's a topologist

But the theory chapters don't have topology prereqs

Ravi Vakil also has this Venn diagram calculus thing that you can use to intuitively understand why they work

[ruby sundial]

oh cool

[hidden haven]

Look for his post on 3b1b's blog, it is called puzzling through exact sequences

It's a very informal treatment, and is not as general as the stuff Rognes proves, but you can formalize it with some effort and also adapt it to Rognes's treatment

[lethal dune]

what's the name of the book

[hidden haven]

Spectral sequences

[ruby sundial]

https://www.3blue1brown.com/blog/exact-sequence-picturebook

[lethal dune]

https://www.amazon.com/Spectral-Sequence-Topological-Mathematical-Monographs/dp/1470456745

is this the one?

[hidden haven]

Nope let me see if I can find it

Or I'll just forward my pdf lol

[lethal dune]

[hidden haven]

https://www.uio.no/studier/emner/matnat/math/MAT9580/v21/dokumenter/spseq.pdf

[south patrol]

Learning about spectral sequences through the adams one would be a baptism of fire lol

[hidden haven]

haha ASS

Still laugh when I see it written in all caps in serious books

[elder wave]

[hidden haven]

Another unfortunate piece of terminology from topology is of an A-null space given a set A of maps

Could have chosen any letter other than A

[next obsidian]

Ass(M)

[lethal dune]

where example

Idk what I was expecting

[hidden haven]

Every spectral sequence known to man is an example

His notes on the ASS have a chapter dedicated to examples of exact couples that come up (and then give rise to corresponding spectral sequences)

But the most notable example is that of a filtered chain complex

Probably just google "exact couple associated to a filtered chain complex" and you should find it

[lethal dune]

slowly making my way through

I'll get there eventually

[woeful flint]

Hi guys I have a bit of a difficult question whose answer I wasn't able to find on google. So I'm doing some research surrounding PQC and I'm coming across a set of numbers which is very obviously a multiplicative group. Precisely, this is the set of totally positive elements of the number field. To better understand these I'm trying to find a more familiar group which this set is isomorphic to, but I'm having a really hard time. I've tried many different ways without success. I was now hoping to use representation theory (which I have not studied) and so I was wondering if this is even possible i.e. can you recover the group structure of a group through a representation of said group?

[lethal dune]

if the representation is faithful then you just get an isomorphic copy of G, otherwise don't think

[woeful flint]

Ah ok

Ngl I don't really know enough about rep theory to know what that means but it's good to go into something knowing there is at least hope of it working

Thx

[lethal dune]

It means that G → GL(V) injective. Otherwise you can have trivial representation, say g.v=v for all v in V. That really doesn't say much, does it?

[woeful flint]

Yeah I get you

Complete shot in the dark but I'll see if it works

[warm wyvern]

to prove k[x] is PID if k is a field I just reasoned take any ideal (call it I) and choose a polynomial with the lowest degree (call it f(x))

now (f(x))\subset I

and the remainder of the division of any polynomial in I by f(x) is 0 (coz otherwise f(x) won't have the lowest degree) so I\subset (f(x))

but shouldn't I be careful when saying "choose a polynomial with the lowest degree"?

since not all sets have a minimum

[elder wave]

that's not an issue here

think about what the set you're worried about here is

[warm wyvern]

the set of degrees of polynomials in the ideal?

[elder wave]

yes

why would that not have a minimum

[warm wyvern]

I dunno

I haven't done a lot of set theory

it's subset N ig

[elder wave]

this doesn't need set theory

[elder wave]

yes

[warm wyvern]

ig you can just start counting from 0 and go till the degree of some arbitrary polynomial in I

[elfin furnace]

just well ordering right?

[warm wyvern]

and the first number you count that's in this set must be the lowest

[oblique river]

you can just say that every set of natural numbers has a minimum

you don't need to re-prove that fact in the context of degrees of polynomials

[elder wave]

yes

[warm wyvern]

I see

[elder wave]

rudin has ruined you

[warm wyvern]

ig I'm just being overly pedantic ¯_(ツ)_/¯

[warm wyvern]

btw, hello buncho

haven't seen you in a while

[oblique river]

hi

i havent been around in a while

[barren sierra]

What does $f^{[\beta]}$ mean? $f$ is a polynomial in $n$ variables.

[cloud walrus]

Spamakin🎷

[barren sierra]

from this: http://people.seas.harvard.edu/~madhusudan/MIT/ST15/scribe/lect21.pdf

[glossy crag]

I'm pretty sure it's just the coefficient of $f$ for the monomial $x^\beta$, like how you would write $f(x)=\sum_{i=0}^nf_ix^i$ for a one-variable polynomial.

[cloud walrus]

Ocean Man

[barren sierra]

ahhh

ok that makes sense

[glossy crag]

If zeta is a complex primitive n-th root of unity, why are the two highlighted expressions equal? i,j are the numbers from 0 to n-1.

[delicate orchid]

I don't think the fact they're roots of unity comes into play here, we're just substituting k = j-i

and then doing some commuting of products

[glossy crag]

lol you're right, that was silly

[chilly ocean]

I'd denote that by Z[A], where Z(f) is the zero set of f

[ruby sundial]

ty for resource

i only made a correspondence before with kernel injecting into X and then surjecting onto image

[hidden haven]

My bad, A is a set of spaces, not maps

A space Z is A-null if the space of maps from every space in A to Z is contractible

I think

[hidden haven]

Yeah it's very cool, especially the idea of denoting quotients on one side of the Venn diagram

One thing that I think makes it easier though is not actually gluing things together

Like for a map X → Y, Vakil glues its coimage and image together to get a single big diagram

When there aren't many maps involved, this works well

But I've had to draw infinitely many maps in this notation before lmao

So I prefer to keep X and Y separate, draw the filtrations on them, and just remember which regions are supposed to be glued together

[white yoke]

Can anyone help with this question? I tried asking in the help channel but it seems like no one can help there 😦

[formal ermine]

(those questions are way too advanced for the help channels fwiw)

[solar glacier]

Good text just saying Hatcher is better for alg top

[long nebula]

this is just currying right?

the notation is killing me

[chilly ocean]

this is painful to look at

but yes

[long nebula]

I agree

[wooden ember]

Just to make sure I’m not being dumb here, instead of writing “for different primes p_i” and later “if we assume that p_i does not divide p_j” can’t we just straight off the bat say “for non associated primes p_i”?

[next obsidian]

I think so

[wooden ember]

Okay I’m very much unfamiliar with trans finite induction so I just need an indicator to know if I’m not doing something illegal here

I was reading this post https://math.stackexchange.com/questions/162945/submodule-of-free-module-over-a-p-i-d-is-free-even-when-the-module-is-not-finit

Now in the first comment they say that to complete the proof you should use trans finite induction

So my “attempt” at showing the u_i generate U is to just say that for an element x of U I can take the biggest i such that p_i(x) is non zero, so that x is in U_i. Then subtract some multiple of u_i to get an element in U_j for some j<i. I suppose by trnsfinite induction that U_j is generated by the set of u_k’s for all j less than i to conclude

This feels too broken

Like intuitively I don’t understand why this works

It makes me feel like axiom of choice (well ordering) + transfinite induction just allows you to apply induction to any set you want

Which feels like a big no no

I guess this is just another weird consequence of choice though

[hot lake]

yeah a well-ordering on a set allows you to do induction on that set

what's broken is that for anything larger then N, you don't have any explicit order that's a well-ordering

[warm wyvern]

you didn't use spivak on manifolds?

cringe.

[pallid oracle]

Is Z/2 supposed to be Z/2Z or the half-integers?

(Usually it denotes the half-integers but it looks like you're assuming it's Z/2Z)

[wind steeple]

Z/2 is Z/2Z

[formal ermine]

I've never seen anyone refer to Z/n as 1/n * Z

[coral shale]

youre going to have to make it explicit if u want it to mean that

otherwise Z/n := Z/(n) is the norm

,,\frac{1}{n}\bZ \coloneqq \left{\frac kn : k\in\bZ\right}

[coral shale]

however this is probably understood

[pallid oracle]

I see, my bad, must have confused it with (1/2)Z.

In that case, I think it might help to consider that the map from R to the product R1xR2 needs to be injective according to A.1.
If you're sending 0 to 0x0, where are you sending (+1) + (-1)? ||I would try changing f2 to sums instead of substractions, (a+e) + ...||

[wooden ember]

This really hurts my belief of AC 😭

[warm wyvern]

skill issue

[rotund aurora]

imagine not accepting AC

[chilly ocean]

without AC it's way worse than with it

maybe your issue lies in mathematics itself

[wooden ember]

Honestly probably

It’s just weird how in some situations you’re completely fine with using it and don’t see a problem and in others it feels like the most contradictory thing

[white yoke]

Z/2Z

[white yoke]

Yeah if it was the half integers I would write (1/2)Z

[white yoke]

Ah I see

What is your reasoning behind using sums and not subtractions?

[white yoke]

Oh I just realised, that is a big brain move

Thanks!

[left estuary]

Is this a correct way to describe an ideal of a ring?

[coral spindle]

There is no real description of what it means to be an ideal in that diagram

you've only described a subset

[left estuary]

Well an ideal is a special subset

[coral spindle]

It is, but it's not particularly helpful to see it that way.

[left estuary]

So I wanted to try to describe it using that one subset analogy

[formal ermine]

you could describe a subring that way

but the important property of an ideal is lost

[left estuary]

What could I do to fix that

Any ideas?

[formal ermine]

...don't use that diagram?

[coral spindle]

I have no idea what you're trying to achieve here

[left estuary]

Trying to connect 2 ideas together to better understand ideals

[coral spindle]

OK

The use you will immediately see in an ideal is defining quotient rings. Go look at that and it might become clearer

You can derive the definition of an ideal entirely from the requirement that it should provide quotients

[left estuary]

ah yeah that wasn’t clear in the definition of an ideal

[coral spindle]

This is one fruitful way of seeing ideals

[left estuary]

the words “special subset” seemed a bit vague

[coral spindle]

You don't say

It is in fact entirely vague what that means.

[left estuary]

Also I understand better with graphics and stuff like that do you happen to know any?

[coral spindle]

I don't think graphics are helpful for rings.

[chilly ocean]

algebraic geometry btfo

[coral spindle]

yeah we've all seen drawings of spec Z[x]

When you learn about quotient groups or quotients in general, I've seen some helpful diagrams though

[chilly ocean]

yeah they're silly

"""generic point"""

[delicate orchid]

simply visualise each ideal as the image of a closed set in the zariski topology 😌

[left estuary]

Like as in ℤ/nℤ ?

[coral spindle]

Yes

[left estuary]

Arnt those like cyclic groups or something like that

[formal ermine]

yes

$\bZ_k \coloneqq \bZ/k\bZ$ is the cyclic group of order $k$

[coral spindle]

If you forget the ring structure, yes in particular Z/kZ is the cyclic group of order k

[left estuary]

So that’s an ideal?

[coral spindle]

The quotient notation is the same for groups and rings

What's an ideal? A quotient group?

No

I'm confused. Have you actually read the definition of an ideal?

[left estuary]

yes

[left estuary]

^

[coral spindle]

So you haven't read the definition of an ideal.

No wonder you're confused

What book are you working from?

[left estuary]

Not a book but articles/documents

[elder wave]

use a book

[left estuary]

I don’t have access to the books

[coral spindle]

OK, find a book on ring theory and read it

[left estuary]

I don’t think I do

[coral spindle]

You do; often they're free pdfs online

we have a channel to discuss finding books: #book-recommendations

[left estuary]

I get long term that will help but rn that just seems like a tedious way just to get one definition

[formal ermine]

you've been refusing to read a book for the past months

[coral spindle]

Oh wait you're that guy ok lmao

[left estuary]

No I just don’t know how to find them

I googled the names and couldn’t find them

[coral spindle]

Get a book. It's not just one definition, it's the whole field that you're learning.

[warm wyvern]

tf are you gonna do with the definition of an ideal alone? lmao

[formal ermine]

first google search result