groups-rings-fields

[simple mulch]

and 1 for t = s on both sides

I think I understood, thanks Todd!

[hot hill]

I'm confused about the explanation here: https://i.uguu.se/kGwfpAEV.png . In particular I don't understand the sentence "Therefore, a prime group should be in the form ${e, g^1, g^2, ..., g^{N-1}}$." If every member of the group has order either 1 or N, then say, $x_1^1 = e$ or $x_1^N = e$. So with $e = x_j^1$ for some j, how do we get that for say $x_l$, $x_l = g^i$ for some $i$?

[cloud walrus]

joesmith1042

[tribal moss]

If an element g has order N, that means that the elements 1, g, g², g³, ..., g^{N-1} are all different (because otherwise the order of g would be smaller). If N is also the size of the group, this means that the powers of g are actually the entire group.

[hot hill]

@tribal moss Oh, yeah, that makes sense.

@tribal moss So that implies that for any x_i in G, that x_i must be equal to one of the g^j.

[tribal moss]

Yes.

[hot hill]

Cool, thanks, really appreciate your help 😀

[chilly ocean]

yes I mean this but a group that isn't simple I definitely didn't mean that theres no other group that maps onto G. also since I sent that I realized my group was wrong 😔😔 not all quotients are isomorphic to G itself unless the subgroup is finitely generated in my particular case

[delicate orchid]

consider g^n = g^m for |g| > n > m. Then g^(n-m) = e, so |g| = n-m but |g| > n > m > n-m, which is a contradiction so g^n cannot equal g^m

just if you wanted a proof

[latent anvil]

@chilly ocean chmonkey and I proved that there's no group which is abelian and iso to all of its nontrivial quotients

He wanted me to relay the proof

[chilly ocean]

so have i 😔😔😔

[latent anvil]

oh nice, how did you do it?

[chilly ocean]

well the group obviously has to be infinitely generated

[latent anvil]

Yep

[chilly ocean]

and if all subgroupa are normal

causw it's abelian

you can quotient out by a cofinite set of generators

[latent anvil]

How do you know that there's such a set with nontrivial quotient?

[chilly ocean]

my original example is abelian, and the result holds for finitely generated normal subgroups

[chilly ocean]

oh so

I thought about that

[latent anvil]

Q/Z?

[chilly ocean]

yes 😔😔

[latent anvil]

That was my guess too :P

[chilly ocean]

I have a sort of wishy washy way around that

[latent anvil]

When chmonkey dmed me I immediately said Q/Z haha

And then he pointed out the subgroup <1/3, 1/5, 1/7,...>

[chilly ocean]

yup exactly

ok so for the question you askwd

take a minimal set of generators (don't ask me how to get one, idk 😳😳😬😬)

[latent anvil]

I'm suspicious already

[delicate orchid]

what if there is no minimal set

👻

[chilly ocean]

yeah like Q has no minimal set of generators 😔

[latent anvil]

generation isn't stable under intersection, I think, so you can't do a Zorn type thing

okay so I don't buy your proof haha

[chilly ocean]

ok what's yours 😭😭

[latent anvil]

Okay so it's only for the abelian case

Let G be a group which blah blah blah

First, G has to be either torsion free or entirely torsion

This is because the subgroup of all torsion elements must be the entire group or trivial

[delicate orchid]

🤨

[chilly ocean]

otherwise you can quotient by it?

[latent anvil]

If it's inbetween then the quotient can be distinguished from G, bc G has torsion and the quotient doesn't

Right

[chilly ocean]

and be left with a torsion free nontrivial group

[latent anvil]

Torsion just means finite order

Exactly wren

[delicate orchid]

yeah, I was getting it confused with module torsion

[latent anvil]

They're related!

[delicate orchid]

is this true?
I'm thinking about Z x Z/2Z here, but this might not satisfy the "blah blah blah"

[chilly ocean]

G is the sum of a torsion group and torsion free, right??

[latent anvil]

I never get the Indexing right but Tor_1(Z/nZ,G) = n torsion of G or something

[latent anvil]

I'm not sure if this holds in general

Outside of fg groups

[chilly ocean]

yeah I just realized this condition

[latent anvil]

This is for G a group isomorphic to each of its finite quotients

That's the blah blah blah

Okay so

For the torsion free case

[delicate orchid]

ah ok, then yes I agree

[latent anvil]

Take any nontrivial element g and look at G/<g^2>

g is not in <g^2> so it's nontrivial in the quotient and has order 2

[chilly ocean]

yup yup yup

[latent anvil]

So G must be entirely torsion

[chilly ocean]

that's what I was thinking too

[latent anvil]

Let n = min { |g| : g ≠ e }

n is prime because if n = ap and g^n = e then (g^a)^p = e, so |g^a| = p

[chilly ocean]

yup

[latent anvil]

I claim every nontrivial element of G has order n

Let H be the subgroup of e and all elements of order n

[chilly ocean]

oh I seeee

[latent anvil]

If x, y have order n then n(x+y) = 0, so the order of x+y divides n and then it actually is n by minimality/primality. So H is a subgroup

[chilly ocean]

H is a F_n vector space

[latent anvil]

Yep!

[chilly ocean]

and a subgroup

[latent anvil]

And H = G

[chilly ocean]

yup

[latent anvil]

So you take a codim 1 subspace and quotient by that

tada

[chilly ocean]

yessss

[latent anvil]

:^)

Chmonkey go drive

[next obsidian]

Yeah

[chilly ocean]

ok now for non wvelan groups

[latent anvil]

Still no idea about the nonabelian case

[chilly ocean]

abelian*

[latent anvil]

Even the torsion elements aren't a subgroup

[chilly ocean]

yeah 😬

[latent anvil]

So we immediately fail

[next obsidian]

I was gonna say tho structure thm fails in non fg

Q has no torsion, isn’t free

Anyway I drive

[latent anvil]

I think fg should still be doable

for nonabelian groups

[chilly ocean]

yeah for sur

[latent anvil]

here you can actually take a minimal set of generators

and then get a cyclic quotient

[chilly ocean]

exactly

[latent anvil]

Or you're cyclic and it's easy

[chilly ocean]

for non abelian infinitely generated

somehow I want something where all infinitely generated subgroups are G

that seems like too much to ask

or actually all I want is all cofinite sets of generators to be like that

[latent anvil]

I don't like thinking in terms of infinite sets of generators

Bc they can just be all the elements of the group

And there's no really a difference between G and G equipped with the set of all of its elements as generators

idk

I don't know that I believe this proof anymore

The step where you show G = H

Seems suspect now

[chilly ocean]

well H is a subgroup

G/H has elements of order other than n

unless G=H

[latent anvil]

how do you know that?

Oh sorry

[chilly ocean]

if H isn't the whole of G, then some element g has order other than n

[latent anvil]

I don't see why G only has elements of order n

Is what I'm saying

That's what the shorthand G = H meant to me, sorry

[chilly ocean]

you're in the abelian case still?

[latent anvil]

Yep

s

so

G/H is isomorphic to G

I think what this gives us is that every element of G has order divisible by n?

[chilly ocean]

yeah I just realized elements can have order n²

[latent anvil]

right

that's the same mistake we made

so I don't actually see how to use this info about G = G/H but I did prove the divisibility thing in a completely different way

for a prime p, let T_p(G) = { g in G : p is coprime to |g| }. This is a subgroup

T_n(G) is not all of G

so either it's trivial (everything in G has order divisible by n) or G/T_n(G) is iso to G

ugggh I think I'm running into the same issue

[chilly ocean]

if G has elements whose orders are all powers of n

T_n(G) is trivial

[latent anvil]

right, I am trying to prove this

[chilly ocean]

oh I have a proof

[latent anvil]

oh sorry I wrote divisible by n when I meant "a power of n"

oh cool!

what's the proof?

I'm getting all twisted around

[chilly ocean]

if the order of g has a prime factor other than n

then let's say n^k is the largest power of n dividing |g|

then g^(|g|/n^k) has order coprime to n

both in G and in H

[latent anvil]

what is H?

[chilly ocean]

H is the subgroup of elements of order n

[latent anvil]

thinking

[latent anvil]

this element has order n^k, right?

[chilly ocean]

oh oops

I defined it wrong

[latent anvil]

g^(n^k)?

[chilly ocean]

yes hahaha

[latent anvil]

I don't see why this is an element of H

and otherwise I don't see where you can get a contradiction

[chilly ocean]

I had a contradiction earlier but I forgot what it was 😐

[latent anvil]

hmm

[chilly ocean]

are there such things as p-sylow subgroups of infinite grouos

[latent anvil]

I don't think so

[chilly ocean]

I have an idea

[latent anvil]

it really feels like n should divide the order of everything haha

[chilly ocean]

you defined T_p(G)

[latent anvil]

yep

[chilly ocean]

take the intersection of all such subgroups for p not equal n

[latent anvil]

sure

[chilly ocean]

aka just the group of elements whose orders are powers of n

[latent anvil]

this is stuff of order divisible by n, right?

right

oh sorry

you're right

just the powers

[chilly ocean]

yes

then G/H is either trivial or G

[latent anvil]

ah the quotient should have no n torsion

[chilly ocean]

if it's trivial, great

[chilly ocean]

exactly

[latent anvil]

cool

so

so G = H?

[chilly ocean]

yes

[latent anvil]

nice

I knew there was something going on haha

okay so

we know that G has to be a p group for some p

nice

[chilly ocean]

yup

[latent anvil]

what are the subgroups of Z_p?

the p adics

[chilly ocean]

I was just thinking that

[latent anvil]

haha

[chilly ocean]

cause Z_p has the property where Z_p/H=Z_p

[latent anvil]

oh but hm aren't there infinite order elements here?

like (1, 1, 1,...)?

[chilly ocean]

yeah

[latent anvil]

so it can't be this

maybe Q_p/Z_p lol

I don't want to think about that

[chilly ocean]

how are you notating elements of Z_p btw

through their normalized series cofficienfs

or

as elements of the product of pⁿ cyclic groups

[latent anvil]

second one

I'm thinking of the p adics as a limit

[chilly ocean]

ok yeah

Q_p/Z_p is a good candidate

its a p group after all

[latent anvil]

right the issue with Q/Z is you could select out a prime

[chilly ocean]

and hey this looks like my original guess 🤧🤧

[latent anvil]

hahaha

[chilly ocean]

does Z_p have elements of all orders

I'm forgetting

orders coprime to p

[latent anvil]

I don't really know sorry

[chilly ocean]

ok

ok now what are the subgroups of Q_p/Z_p 😳

[latent anvil]

lol

[chilly ocean]

in this case we have G/H=G

so that's good

[latent anvil]

is there some external characterization of Q_p/Z_p?

I want to just say "G/K still has property Nice so it's iso to Q_p/Z_p"

[chilly ocean]

hmmmm

I was thinking more just like

classifying all subgroups 😳😳😳

but that is probably pointless

[latent anvil]

seems hard

[chilly ocean]

yes it does

pointlessly hard

are there any p groups that have elements of every order tbag don't contain a subgroup iso to Q_p/Z_p

[tribal moss]

Under which operation? I should think it's torsion-free under both multiplication and addition.

[chilly ocean]

really??

I could have sworn it had torsion elements

[latent anvil]

I don't know very much about the p adics ¯_(ツ)_/¯

[chilly ocean]

neither do I 🙁

I know someone who's researching the p adics rn

should I ask him hahahaha

[tribal moss]

It has solutions to $\underbrace{x+\cdots+x}_n = 1$ for all $n$ coprime to $p$, but since $1$ is not the additive identity, that doesn't a torsion element make.

[cloud walrus]

Troposphere

[latent anvil]

troposphere, do you know if Q_p/Z_p has any nontrivial quotients that are not isomorphic to Q_p/Z_p?

we're trying to find an example of such a group

or prove there are none

[tribal moss]

Hmm, that challenges my spotty knowledge about p-adics too :-)
Additive groups, I suppose?

[latent anvil]

yep

[chilly ocean]

I can think of a group that has quotients other than Q_p/Z_p 😈😈😈

[latent anvil]

we know that if such a group is abelian, it must not be fg and all elements must have order p^k for some p

[tribal moss]

Hmm, everything in sight looks abelian.

[latent anvil]

Q/Z doesn't work because of a subgroup like <1/3, 1/5, ...>, but the hope was that by selecting out one prime ahead of time you could avoid this

[tribal moss]

Or are you looking for "groups where all quotients are isomorphic to the group itself" in general?

[latent anvil]

That one (but where the group isn't simple)

[tribal moss]

How about Z?

[latent anvil]

Z/2Z is not isomorphic to Z

[tribal moss]

Oh whoops

[latent anvil]

so Q/Z has the property that every quotient by a fg subgroup is isomorphic to Q/Z

but it's not quite right

[chilly ocean]

ok what about the subgroup of Q_p/Z_p generated by 1/p,1/p²,1/p³,....

what does the quotient of this subgroup give

[tribal moss]

That's the entire Q_p/Z_p, isn't it?

[chilly ocean]

wait is that subgroup just all of Q_p/Zp

yes

fuck

ok interesting that makes me think

[latent anvil]

huh

weird

so Q_p/Z_p is countable?

[chilly ocean]

let H be a subgroup of Qp/Zp

[chilly ocean]

yes

[latent anvil]

weird

[chilly ocean]

it's very comparable to the standard integers base p, but just like backwards

that's how you can enumerate the elements

[chilly ocean]

let m=max{k| 1/p^k in H}

[chilly ocean]

if m doesn't exist, meaning if the set is infinite, then H is all of Qp/Zp (I think)

[tribal moss]

Sounds convincing.

But if m does exist then H is generated by the negative powers of p up to -m.

[chilly ocean]

otherwise H is just the multiples of 1/p^m, is it not

[chilly ocean]

yeah that's what I'm thinking

[tribal moss]

In sum, every nontrivial quotient of Q_p/Z_p is isomorphic to the whole group.

[chilly ocean]

yes

I'm mostly convinced

[latent anvil]

so is Q_p/Z_p presented by <x_1, x_2, ... | x_1^p = e, x_{i+1}^p = x_i>?

This sounds convincing but I don't really feel like I have a handle on the group

[chilly ocean]

it's a weird group for sure

[latent anvil]

essentially like

there's some non-existent element and we're looking at all it's pth powers

sort of

[chilly ocean]

yes

sort of

[latent anvil]

start with an element with x^p = e and add in all the iterated pth roots of x

weird

but cool

[chilly ocean]

yeah sort of

Q_p is a really weird field and Q_p/Z_p is a really weird group

[tribal moss]

It's also the same as "the ordinary rationals whose denominator is a power of p" quotiented by Z.

[latent anvil]

this makes more sense lol

so Z[1/p]/Z

[tribal moss]

Yes.

[chilly ocean]

there's a result that I have been procrastinating on to read the proof of which is that if f(x) factors in F_p[x], it factors in Q_p[x] as well

[chilly ocean]

yeah oops that makes it much simpler hahaha I was thinking about it as some weird convoluted "reverse integers" this whole time

[latent anvil]

yeah haha I don't want to think about p adics if I don't have to

[chilly ocean]

so we ended up classifying all subgroups anyway

hopefully my p adics friend will be impressed 😈

[latent anvil]

Oh hey, this is an example of a group where all cofinite subsets of any generating set generate it

Bc otherwise your proof would go through wren

[chilly ocean]

yessss

[tribal moss]

Note that for p=2, Z[1/p] is exactly the "dyadic" rationals, which to help with the confusion is something different from "2-adic" rationals

[latent anvil]

right

[chilly ocean]

in fact all infinite sets generate it

not just cofinite

[latent anvil]

Makes sense to me

[chilly ocean]

what a good way to start my day

and all because of my false observation about Q/Z

[latent anvil]

it was fun to think about

But I was reading a book when Alex dmed me and now I get to back to reading the book bc this isn't going to keep occupying my thoughts

[chilly ocean]

what book

[barren sierra]

anyone have a good way to get intuition for group extensions and semi direct products

it just seems like

"yea ok we have this injection, this surjection, if this other function actually has an inverse it's split"

etc etc

but it all seems so arbitrary

how is something "extending" something else

[hidden haven]

Extending A by B is like taking the group A, then putting B in place of each element in a way that makes everything work

This is my intuition for why it's called an extension

As for why we care about exact sequences, there's a lot of stuff, mostly in homological algebra and its applications

Idk why about extensions of groups in particular because I've never worked with those

@barren sierra

🎷 looking a lot like ✓

[pastel cliff]

is there any practical use for knowing what a semi ring or semi group is

[barren sierra]

lol

[pastel cliff]

not that it's hard to remember but still

brain real estate limited

[barren sierra]

mood

[barren sierra]

hm

[hidden haven]

More practical than literally all of math, I've seen those used in automata theory

[barren sierra]

I'll think on this

semi groups show up in automata theory?

[pastel cliff]

this sign cant stop me cuz idk what that is :D

[barren sierra]

I'm taking an automata theory course rn I'll have to ask my prof

[hidden haven]

Monoids and semirings do, a lot

[barren sierra]

automata are extremely simple machines

[hidden haven]

I assume semigroups sometimes too

[barren sierra]

that in short are various versions of machines that have a state and take an input and give you a new state

they're really cool

[hidden haven]

Because the set of all strings is a monoid

Over any alphabet

[barren sierra]

yea

[pastel cliff]

ok good to know tho, ty moldi

[hidden haven]

And you can use this to do some shit

[barren sierra]

imma have to ask my prof lol

[hidden haven]

I haven't really encountered them much in math

But no harm knowing definitions, not like you have a finite amount of storage in your brain

Oh wait that's just me, sorry

[pastel cliff]

.

[delicate orchid]

semigroups
demisemigroups
hemidemisemigroups

I've yet to see a semigroup in the wild

[hidden haven]

Consider the semigroup of contracting endomorphisms of a metric space 😌

[delicate orchid]

no

actually yes I have

natural numbers under addition, classic

[hidden haven]

Are we allowing monoids in your semigroups

Because monoids are very important

[pastel cliff]

idk im just going through my notes to make sure i know all this mf terminology

it's a lot

[delicate orchid]

I know monoids are important but (N, +) is not a monoid

https://cdn.discordapp.com/attachments/342850939306246145/942439972206698596/unknown.png

[hidden haven]

It will very quickly feel like a lot

[pastel cliff]

it has 0

[delicate orchid]

not in this case

[pastel cliff]

then it's Z^+

[delicate orchid]

0 is both positive and negative and is thus in Z^+

[hidden haven]

Bro have you seen the monoid of endomorphisms of A other than identity?

[delicate orchid]

what's A moldi

[pastel cliff]

fuck you <3

[hidden haven]

Any object

In any category

[delicate orchid]

set

[hidden haven]

ok

[pastel cliff]

why u gotta ruin my logic

[delicate orchid]

ok wait

ok yeah I can see that being a monoid

cause you always got the identity map, and it's associative

[hidden haven]

Lol remove identity from any monoid and you're left with semigroup

[delicate orchid]

yeah ok but that seems like a big deal

[hidden haven]

Hence the other than identity

That's so arbitrary

[delicate orchid]

"the non-identical endomorphisms"
"identical to what?"
"what?

[hidden haven]

Are G and H isomorphic?

No, only G is

[delicate orchid]

ok I'm gonna be honest I don't get that one

[hidden haven]

smol brain go play f6

[pastel cliff]

yall weren't kidding the other day

[delicate orchid]

knife f5

[delicate orchid]

magma theory is incredibly important and widely studied

[hidden haven]

By geologists

[pastel cliff]

are you memeing or not

[delicate orchid]

well, most people call "magma theory" "algebra" for some reason

[pastel cliff]

cuz it's an algebra

[delicate orchid]

no... all algebras are magmas acting on magmas (acting on magmas)

[hidden haven]

[delicate orchid]

I did have an actual question when I clicked on this channel

I cannot remember it

[pastel cliff]

trolled

[delicate orchid]

lemme think

oh ok yes I've remembered

let me get the diagram

[upbeat swift]

Before you ask, could I ask a quick question? How do you prove that a set is a partition?

[delicate orchid]

a partition is a collection of subsets

[hidden haven]

Not sure what that means

[delicate orchid]

it's a partition if the union of them all is the entire set and they're disjoint

anyway I have a composition $i_1f = gi_2$ where $g$ is unique and $i_1$ and $i_2$ are some fixed inclusion maps - and are thus also unique, would it be correct to conclude that $f$ must be unique?

[cloud walrus]

GM Wew Lads Tbh

[hidden haven]

No

[delicate orchid]

it feels like it's either "yes duh" or "no you buffoon here's an obvious counter example" and nothing inbetween

[upbeat swift]

Thank you

[delicate orchid]

interesting

[hidden haven]

Wait I misinterpreted

I still believe the answer should be no

Or maybe not

Lul it's true I think

[delicate orchid]

yeah it's a tricky one

oh nice

[pastel cliff]

quick question sorry to interrupt you wew - for a set to be a semi group, does it multiplication specifically have to be associative or simply whatever its binary operator is

,w semigroup

[delicate orchid]

it has to be associative

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by nitezba#9923

[pastel cliff]

wolfram words it like it's specifically multiplication so idk

[delicate orchid]

what

[hidden haven]

So composition of injectives is injective, and then the first map in an injective composite is injective

[delicate orchid]

oh you're still in the mindset that multiplying = numbers, ok

[pastel cliff]

[hidden haven]

First meaning the one applied first

[pastel cliff]

so multiplying im gonna assume just means applying the binary operation then

[hidden haven]

Yes

[pastel cliff]

https://tenor.com/view/dumbfounded-shocked-im-shattered-gif-22235755

[delicate orchid]

we know g is injective btw

wait no we don't

[hidden haven]

Like when you multiply sets you get A × B their Cartesian multiple

[delicate orchid]

ignore me

[long obsidian]

Does it make sense to speak of the diagonalizability of an endomorphism? Similar to the diagonalizability of a square matrix

[hidden haven]

Oh wait g isn't given injective

Bruh

[delicate orchid]

yeah it's just some group homomorphism

it's more free group universal property shenanigans if you want context

[hidden haven]

Definitely if it's an endomorphism of some direct sum

Otherwise I'm not sure

Then I see no reason for f to be unique

What was I even proving bruh

Can't read

[delicate orchid]

it's alright moldi, I have another proof in mind (that doesn't require that composition) it's just a bit more spooky

[hidden haven]

Wait

f is unique

[long obsidian]

Hmmmm okay I think I understand about the free group stuff.

But when I think of square matrices I think of diagonalization as a factorization via eigenvalues.

So in the case of an endomorphism is a diagonalizable endomorphism a factoring into several endomorphism?

Not sure if that makes sense

[hidden haven]

Injective set maps are left cancellable

[delicate orchid]

oh yeah

[delicate orchid]

it was the first option

[hidden haven]

The free group stuff was unrelated I think lol

2 parallel conversations

[delicate orchid]

it was

[plucky flicker]

Is that true that a Lie algebra of dimension at most 2 solvable? If the dimension is 1 it is trivial I guess, becaus then L = Rad(L). But what is the dimension is 2?

[hidden haven]

But I was saying that end(some direct sum) has a notion of diagonalisability for direct sums of anything, be it vector spaces, abelian groups, modules or even exotic stuff like chain complexes and sheaves

[delicate orchid]

15625 convos in abstract alg at once

[hidden haven]

This is because you can write endomorphisms of a direct sum as a matrix

In the following sense

If you have A ⊕ B, then every endomorphism of this is uniquely determined by maps
f_AA, f_AB, f_BA, f_BB

Where f_CD is a map from C to D

Vector spaces are a special case of this

This extends to direct sums of more things too

So you get n × n matrices

And matrix multiplication still corresponds to composition

[hidden haven]

Every vector space is a direct sum of 1 dimensional spaces, and maps of 1 dimensional spaces are just scalars, so there the matrices just become numbers

But here our matrices will have functions as entries

@long obsidian

[hidden haven]

This is by the universal property of the product and the coproduct and finite direct sums are both in all additive categories, which includes all the above mentioned situations

Basically whenever the symbol ⊕ is used

[delicate orchid]

gonna retract my devastation

[tribal moss]

And diagonalization would a direct-sum decomposition where the cross-maps f_AB and f_BA vanish?

[hidden haven]

Yeah, so you need a notion of 0 morphisms

[tribal moss]

But how do you know you've decomposed into enough pieces to deserve the "diagonal" name?

Require that each summand has an abelian automorphism group?

[pastel cliff]

how do yall remember all this terminology is it just constant usage

[hidden haven]

I assumed that we are given a direct sum decomposition

[hidden haven]

Perhaps indecomposability

[tribal moss]

Hmm, that might work.

[hidden haven]

It is only daunting when you don't know what it means

[delicate orchid]

says the mf speaking an entire language of 1,000,000+ words

[pastel cliff]

two of them actually

[delicate orchid]

wow thanks for making my point twice as valid

[pastel cliff]

worth it for the smug

[tribal moss]

Add something like German which has denumerably many words.

[delicate orchid]

the set of german words is uncountably infinite

[hidden haven]

I remember my first year in ug constantly listening to my batchmates talk about tychonoff and hausdorff and thinking holy shit those sound like proper mathematician words

[pastel cliff]

half the things yall talk about in here give me that energy

but it's third year ug

[tribal moss]

Donau­dampfschifffahrts­elektrizitäten­hauptbetriebswerk­bauunterbeamten­gesellschaft

[pastel cliff]

though i really only started second yr

[hidden haven]

Ah damn that's one thing we didn't cover properly in our deqnastral course

[pastel cliff]

my case in point

[hidden haven]

[delicate orchid]

baby moldi

[hidden haven]

👶

[pastel cliff]

,w semiring

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by nitezba#9923

[sharp sonnet]

i think there is an error in this

[hidden haven]

I sometimes read semiring as present continuous tense of semire

[pastel cliff]

wait does a semi ring not have additive id then???

[delicate orchid]

no

[tribal moss]

Whereas actually it's the present contrantinuous tense.

[pastel cliff]

so uhhhh

that's awkward

[hidden haven]

Semiring is just rng right

[delicate orchid]

nah it's double monoid

I think it does have an additive identity...?

yes

it does

[pastel cliff]

wikipedia says additive identity

[tribal moss]

I think it's rig, actualy.

[pastel cliff]

but wolfram doesnt point it out

[hidden haven]

Ah yes

I keep confusing these

[delicate orchid]

oh come on "rig" is actually on the wikipedia semiring page

[hidden haven]

Is a semiring a monoid object in the category of abelian monoids

Then it would actually make sense to call it double monoid 😌

[tribal moss]

2-monoid, perhaps?

[delicate orchid]

hmm yes indeed

[hidden haven]

That could be easily confused with a 2-category with 1 object

Ngl half the time I say extra cat bs just to get a reaction out of you wew

It would be a tragic day when you learn cat theory

[delicate orchid]

I am naught but a play thing in this category hell

an abelian monoid is an interesting one though

hmmmm is the free abelian monoid nicely behaved like the free module and free abelian group are

I must answer this question

is it just N^r I feel like it is

yeah it is

thanks for your help #groups-rings-fields in cracking this case

[hidden haven]

Lul np

Monoids are very cool

There's a more general definition of a monoid than a set with operations

Under that, the usual monoid is just a monoidal set

A ring is exactly a monoidal abelian group

Similarly there are monoidal categories

[delicate orchid]

lemme guess, categories of objects with an operation that a monoid distributes over?

[hidden haven]

Which are categories like the category of abelian groups with the tensor product as its monoid "operation"

[hidden haven]

No, we are talking about a category that itself is a monoid

[delicate orchid]

Wait holy shit the tensor produc

omg of course it's a monoid

holyyy shitttt

[hidden haven]

Lol

But that also works with ordinary product, ordinary coproduct etc

There are lots of monoidal structures

And once you have a monoidal category you can define monoidal objects in that category

[delicate orchid]

I remember someone the other day talking about some construction with the tensor product that resulted in a group but I never made the further leap to consider that the tensor product in general is a monoid

I gotta get a ring out of this thing

[hidden haven]

A monoidal set is a monoidal object of (Set, ×). A monoidal abelian group (a ring) is a monoidal object of (Ab, ⊗)

[delicate orchid]

hmmm

are we viewing abelian groups as Z-modules here or something?

I've never heard of taking the tensor product of groups

[hidden haven]

And whenever you have a monoidal object, you immediately also get something called a simplicial object that you can do a lot of topology and homology on

[hidden haven]

Yes

[delicate orchid]

wait a minute is the tensor product of two abelian groups just an abelian group that satisfies the universal property?

[hidden haven]

Yes

[delicate orchid]

ah yeah but we don't know that it actually exists

I forget about that slight catch

[hidden haven]

You can talk about bilinear maps of abelian groups

[delicate orchid]

I'll just go with the Z-module view as I'm slightly more familiar with it

[hidden haven]

Ye whichever works

Same everything in both

Also, notice that rings are exactly ℤ algebras

So ℤ algebras are exactly rings which are exactly monoidal abelian groups which are exactly monoidal ℤ modules

R-algebras in general are exactly monoidal R-modules

[delicate orchid]

horrifying

[hidden haven]

It's such a great definition

You would never forget a condition to check whether something is an R-algebra if you remember this

[delicate orchid]

that would require remembering this

[hidden haven]

[delicate orchid]

I can kinda see it though

funny algbera over a ring is funny module over a ring with some multiplication on the module elements

[hidden haven]

Do you see why tensor product comes in

[delicate orchid]

lemme re-read

my only thought is you're defining the multiplication of the module elements to be their tensor but that doesn't make sense to me

[hidden haven]

So a monoidal set (ordinary monoid) is a set M, a map M × M → M (multiplication) and a map 1 → M (unit) where 1 is the singleton set, such that certain nice conditions are satisfied

An important point to note here is that all of these nice conditions can actually be stated using only commutative diagrams, no elements required

[delicate orchid]

I gathered

[inner acorn]

👀

[hidden haven]

Sasha feel free to post anything lol we'll move

[inner acorn]

xD I dun wanna interrupt the fun conversation flow

but if you insist

I'mma throw this here just to check I'm not being dumb
Counting the number of group homomorphisms from Zm₁ x Zm₂ to Zn₁ x Zn₂ I get gcd(m₁, n₁) gcd(m₂, n₁) gcd(m₁, n₂) gcd(m₂, n₂)

Using the fundamental theorem of presentations (or whatever it's called), given a presentation G = < X | R >, every function f: X --> H which satisfies the image of the relations R in H, extends to a unique group homomorphism.
The presentation Zm₁ x Zm₂ = < x, y | m₁x = m₂y = [x, y] = 0 >, means I just need to pick f(x) & f(y) so that m₁f(x) = m₂f(y) = 0 (the abelian requirement is automatically satisfied in Zn₁ x Zn₂).
Pretty sure this breaks down to 4 equations where I need to count g in Znⱼ where mᵢ g = 0 mod nⱼ, which is gcd(mᵢ, nⱼ), and so the total number of such functions is the product gcd(m₁, n₁) gcd(m₂, n₁) gcd(m₁, n₂) gcd(m₂, n₂).

[delicate orchid]

ah so that's why you were typing for 10 minutes

[inner acorn]

[delicate orchid]

I can't see anything wrong there

[hidden haven]

monoids

[delicate orchid]

assuming |Hom(Z_m1, Z_n1)| = gcd(m1, n1) is actually true because I forgor if it is

[inner acorn]

yeah, that's true

[delicate orchid]

I feel like a simple counting argument would work here rather than needing the universal property of the free group though

[inner acorn]

I'm just quoting the universal property, because it confirms I only need to count functions that preserve the order condition

the actual counting said functions is a different calculation entirely

because again it breaks down to counting x such that mx = 0 mod n with x in Zn
and if we write d = gcd(m, n)
then m = da & n = db with a & b coprime

dax = 0 mod db
ax = 0 mod b
b divides x
so counting said x gives b, 2b, 3b, ..., db = n
i.e. d = gcd(m, n) possibilities

[pastel cliff]

is there a more subtle understanding to what a homomorphism is than just the raw definition?

ok maybe asking that is dumb bc the definition is easy to remember

i mean like how i can remember a homeomorphism as a map between topological spaces with no tears

[inner acorn]

it's just a map between groups that preserves the group structure, i.e. f(ab) = f(a)f(b)

other than that

I don't really think of it in a more interesting way

[delicate orchid]

it gives you a copy of the group you're mapping from in the group you're mapping to

[inner acorn]

wouldn't that be an injective homo?

[pastel cliff]

^^

didnt think that was a requirement

[delicate orchid]

ok wise guy

[inner acorn]

[pastel cliff]

:(

[delicate orchid]

ok ok

so it gives you uhh

wait

ok the groups abelian ok? groups are just abelian now ok that's just how it works

[pastel cliff]

[delicate orchid]

it gives you a copy of a subgroup of the group in where ever you're mapping

[pastel cliff]

🍇

[delicate orchid]

thing of add is add of thing

[pastel cliff]

this shit makes my brain mush man

[delicate orchid]

don't look in the monoids thread

[chilly ocean]

do all uncountable groups have some countable quotient? this seems obvious but idk how to show it

[dense pumice]

does (R,+) even have one?

[chilly ocean]

oh wait yeah idk

[delicate orchid]

it doesn't

waittt a minuite

[chilly ocean]

R is a Q-vectpr space, so it has a basis and you can project onto a one dimensional space

[delicate orchid]

it has an uncountable basis as a Q-vector space

[chilly ocean]

yeah

[dense pumice]

I believe you can either have a minimum positive absolute value in which case the subgroup is basically Z, else 0 is a limit point so you should be able to get any real number as a combination?

[chilly ocean]

are you talking about subgroups of R?

they're all either cyclic or dense

[dense pumice]

yea

[chilly ocean]

but I'm talking about quotients of R

[dense pumice]

oh can dense be countable quotient?

maybe

[delicate orchid]

to get a countable quotient you'll need to have an uncountable subgroup I believe

I'm not sure though my set theory is very rusty

[chilly ocean]

yeah you definitely do

[dense pumice]

yea that was my thought

[delicate orchid]

yeah

I'm pretty sure the set of algebraic reals is the largest subfield of R

and it's countable

[chilly ocean]

no it's not

you could pick any transcendental number and throw it in

and get a larger field

[delicate orchid]

Hmm

At what point do we just get R

[dense pumice]

this does seem feasible but its very jank

[sharp sonnet]

when you added continuum many

[desert dome]

Hi, I am working on this problem. I proved the first half, but had a hard time figuring out the later half. I assume a^m = 1 mod p for contradiction, so a^n - a^m = 0 mod p. Then p | a^m or p | (a^{n-m} - 1). p|a^m is not possible because if so, p|a^n. I dunno how to show p|(a^{n-m} - 1) is not possible and how to use (p,n) = 1. Any hint would be appreciated

[delicate orchid]

sounds more like a #elementary-number-theory problem to me

[desert dome]

yeah, I feel the same way. I'll ask in that channel

[prisma shuttle]

can someone help explain what is the difference between a direct sum and a direct product

i googled it and found some stuff but it was pretty confusing

for example is $\mathbb{R}^{\oplus2}$ a direct sum or a direct product of $\mathbb{R}$ and $\mathbb{R}$?

[cloud walrus]

JustKeepRunning

[delicate orchid]

I don't think the direct product is a thing

[prisma shuttle]

wait are u sure

cuz i've heard ppl using it

[delicate orchid]

the Cartesian product is a thing

maybe they're synonyms I'll google it

[dense pumice]

direct product is a thing

do u know qm by any chance

[delicate orchid]

yeah they're synonyms

[prisma shuttle]

wut is qm

[delicate orchid]

whoopsie

[delicate orchid]

it's the direct sum then

[prisma shuttle]

i don't think so

[delicate orchid]

\oplus is the direct sum

[prisma shuttle]

then wuts direct product

[delicate orchid]

\times

[prisma shuttle]

no like i don't mean the symbol

i mean like literally wut is it

also i've seen ppl use the \oplus symbol and say the "direct product of..." so that means the notation is often misused ig

[dense pumice]

quantum mechanics

[delicate orchid]

yeah they're the same for finite groups

[prisma shuttle]

no idk qm

[dense pumice]

I ask because I only understood linalg after learning quantum

[prisma shuttle]

no

only for abelian

according to d&f at least

[delicate orchid]

you can't direct sum non-abelian groups

I was kind of implying the "abelian"

but you're right I should be more explicit

[dense pumice]

theres some categorical definition I dont know

[proud bear]

so for finite direct products and direct sums, they are basically the same. but they're different in the infinite case. in an infinite direct sum R(+)R(+)..., all but finitely many of the terms are 0. so the element (1,1,1,1...) isn't an element of this direct sum. but in the direct product, you don't have this restriction

[dense pumice]

but for vector spaces, the sum is like adding vectors from 1 vector space to vectors from another and the product is a new vector space generated by taking products of vectors from one space and the other

[delicate orchid]

(fun fact it's the same direct sum for vspaces and abelian groups)

[dense pumice]

wait shoot, direct product is different from tensor product isnt it

ignore what I said

[delicate orchid]

yes it is very different

the tensor product is a quotient of the direct product

so they're related at least!?!

[proud bear]

i thought it was the quotient of some free module

[delicate orchid]

it might be?

every construction I've seen uses the direct product though

[proud bear]

yeah they take the free module on the set MxN and quotient by some massive submodule

[delicate orchid]

hmm

[proud bear]

i guess yeah, they're related

[delicate orchid]

I think you might be right

[upbeat swift]

If I have an equivalence relation on R×R, what does it look like geometrically? Say I have (x,y) ~ (u,v) iff ax^2 + by^2 = au^2 + bv^2 (where a,b > 0)

[thorn delta]

This is a special case of a more general equivalence relation: x ~ y if f(x) = f(y). The equivalence classes are the fibers of points in the image of f. i.e. level sets of f

[lavish nexus]

X, V, V* all unitary, Dx is unitary and diagonal

why is the second step true

[dense pumice]

the second equality?

expand the exponential and then notice both are the same due to V being unitary

[lavish nexus]

oof

ty

[lavish nexus]

is this saying sl(2) = su(2)+i*su(2)

[dense pumice]

it says they're isomorphic so I think that means there's a map between the basis that sends one set of structure constants to the other

or something

[prisma shuttle]

no i think (s)he's asking what the symbols on both side of the isomorphic symbol mean

[dense pumice]

tensor product with C, so like pretend the coefficients can take values in C and everything else is the same

[chilly ocean]

convoluted way to say it

but it just means the matrix groups are seen as modules over C

or ig ur right

but its a little much if u never seen them before

[dense pumice]

yea ignore the first 4 words

focus on the "coefficients are in C and everything else is the same"

[lavish nexus]

I see thank you

[chilly ocean]

if M is a maximal normal subgroup of G, is G/M always cyclic

[next obsidian]

Is G finite?

But I think so

If it wasn’t then there should be some x not in M such that <M,x> is not all of G

[chilly ocean]

no G is infinite

oh also is the frattini subgroup of infinite groups always nilpotent

[hidden haven]

G/M is simple I think

And not all simple groups are finite

[thorn delta]

nor cyclic

[hidden haven]

It's simple by the correspondence theorem

[chilly ocean]

that's good enough for me! and how do you show this

[hidden haven]

Oh ye I meant cyclic lol

[chilly ocean]

wait so it's cyclic

or simple

[hidden haven]

Simple, and not all simple are cyclic

[chilly ocean]

oh ok

how does lattice iso give that?

wait

I see nvm

[hidden haven]

Nice

[chilly ocean]

ok what about the frattini subgroup

is it always nilpotent

please say yes 🤞🤞🤞

[hidden haven]

Apparently it is if it is finite

I just googled it lol

[chilly ocean]

what about infinite 😢

if google specified finite, then probably not

[hidden haven]

Probably not since all the statements were mentioning finite

[chilly ocean]

yeahhhh

shit

[next obsidian]

The frattini subgroup can be empty I think

Or no it can be everything

Frattini is interscetion of maximal subgroups right?

Is Q nilpotent?

The frattini subgroup of Q is Q

[chilly ocean]

yes by definition since Q doesn't have any maximal subgroups

so if it's empty it's defined to be everything

[next obsidian]

Yah or you can phrase it in terms of uhhh

Non-essential generators

[chilly ocean]

yes

shamrock gave me a little nugget about the frattini subgroup now I'm struggling with something about it

[next obsidian]

Oof

[chilly ocean]

I just accidentally proved that the normalizer of a subgroup is always normal in the larger group, but this isn't true in general and idk where I went wrong 😢😢

let H be a subgroup of G. G acts on left cosets of H by left multiplication. the kernel of this action is N(H). how does this not imply that N(H) is normal

nvm I see my errors

[fallow plume]

Am I mistaken or is this incorrect? Isn't this a K-module morphism regardless of commutativity?

[open pilot]

That’s exactly what my first impression was. Either way Wikipedia’s treatment of generalized linear algebra is roundabout and confusing.

[hidden haven]

They're only defining algebras over commutative rings

[rustic crown]

the stress is over 'K-module homomorphism' and not 'if K is a commutative ring'

[hidden haven]

But also doesn't algebra over a ring only make sense if the ring is commutative

Since you require the structure map to be into the center

Why was that again

[rustic crown]

if you want the scalars to interact in certain way, then yea

[hidden haven]

Probably follows from monoidal R-module directly

[rustic crown]

probably something like (ax)*(by) = (ab)xy?

[hidden haven]

One reason I remember is that you want multiplication by a fixed scalar to be an endomorphism

Ah yes it also follows from monoidal R-module definition trivially lmao

Or like what follows is that the image of R has to be in the center

Not that R itself has to be commutative

Because we have the unit multiplication laws on both sides

[rustic crown]

for that we would require like M to be a (R, R) bimodule something right?

[hidden haven]

oh ye R-mod is not a monoidal category under tensor

So I guess yeah

[hidden haven]

So basically algebras over non commutative rings are more complicated in their definition and that's probably why they are not talking about them

Det did you see the monoid thread

That's what I spent 2 hours on last night

Talking about how monoids are amazing

[rustic crown]

i saw the existence

[hidden haven]

Did you know that monads are monoids in the endofunctor category where the monoidal bifunctor is functor composition

[rustic crown]

i've heard that phrase

[hidden haven]

Lol nice

I have started liking monoids and monads so much just because of

[rustic crown]

[fallow plume]

I agree but wouldn't it still be a K-Module homomorphism even if it wasn't a commutative ring?

[hidden haven]

Ye that is true

[hidden haven]

But also this lol

[fallow plume]

That feels like it should be editted and I'm too lazy rn so rip. Another time maybe

Also you guys got nice colors (I think that changed?)

If new role congrats but I think y'all already had it so looking good instead

[hidden haven]

Yes, you leave out piss to dry, its color becomes stronger

[fallow plume]

Uhg I wanna ask one more question but unfortunately I don't think i can ask it due to stupid honor code policy even though it's more of a clarification 😭
Imma just ere on the side of caution

We've moved on from category theory and homology (for now) and now our ring/module theory prof is trying to give us a taste into algebraic geometry lolll

I already hate the Rees Algebra

[open pilot]

Algeo is addicting

Don’t blame him

[fallow plume]

Apparently he's an algeo-ist primarily so I am rather curious

This course has been a lot of fun but also a massive pain in the ass

Is there a good intuition for an algebraic set at my level? He briefly talked about it but the audio was having issues in the recording

My understanding is we're letting polynomials trace out curves on spaces and their solution sets to 0 have something to do with algebraic sets but Idk much else

[rustic crown]

if you have a set of polynomials in k[x1, ..., xn] then their set of common zeros is called an algebraic set

idk what more to say about intuition about them

[open pilot]

Can you explain what the difference is between that and a variety?

[rustic crown]

my AG isn't very good, but i think classically they mean the same thing

modern AG defines variety as something like reduced separated finite type scheme over a field?

[fallow plume]

So literally just a set of shared 0s (basically) ? A lot simpler than I thought

[sharp sonnet]

classically you also consider varieties living in projective space

its also hard to write down a variety (in modern language) that cannot be embedded in some projective space

so the classical language is already pretty powerful

[waxen hedge]

usually you add some hypothesis like «the ideal of definition is prime»

or you patch together these sets of shared 0's

[rustic crown]

maybe one thing is this, if F was that set of polynomials, then the set of common zeros of F and the ideal generated by F is the same thing, but as k[x1, ..., xn] is noetherian, that ideal is finitely generated, so, that algebraic set would be shared 0s of finitely many polynomials

[fallow plume]

Nullstenstaz yea?

Or however it's spelled

[waxen hedge]

no not here

the nullstellensatz tells you that the system has a solution in some finite extension of the base field

[fallow plume]

Mm yea I got a lot more to cover, but getting there

Still got several more years for better or for worse (spoiler alert, probably the latter )

[waxen hedge]

(you need some hypothesis, your ideal should be proper)

[fallow plume]

Nice moon pfp btw

[open pilot]

I always assumed that to be true before reaching Nullstellensatz given the existence of field extensions, but I suppose there is no reason to without it

[waxen hedge]

field extensions works well for polynomials with one variable

for >= 2 variables it's way harder

[open pilot]

Could you elaborate?

[fallow plume]

I need to get some sleep so good night everyone, thanks as always 🧡

[waxen hedge]

For a one variable polynomial P , you know that in some extensions it has a root, you can split it in a finite extension, if the polynomial is separable you can study the field of decomposition through Galois theory

I don't know if there is Galois theory of >= 2 variables's polynomials

In order to show that polynomial in >= 2 variables has a root (in some extension, assuming it's not invertible), you need to use the Nullstellensatz, which is way harder to prove that looking at a quotient k[X]/P(X) if P is irreducible

algebraic geometry can be described as the study of systems of equations of polynomials in multiple variables

depending of the base ring/field you are considering (the integers, the rationnals, the reals, the complex numbers, the p-adics, a number field, a function field, …) you get different theories, with specific methods

solving $x^n+y^n=z^n$ over the integers does not raise the same questions/problems than solving it over the reals numbers (where it's trivial)

[cloud walrus]

Adrien

[open pilot]

You’re right, it quickly branches into many other subjects. I wasn’t even thinking of Fermat’s Last Theorem

[waxen hedge]

I can't give a proper background/overview of the subjet since I'm far from being an expert, but I want to emphasize that the problems gets usually way harder when you generalise them, and for some of them you may be close to research's questions !

[waxen hedge]

depending of the context you get algebraic number theory, real algebraic geometry, complex geometry, p-adic geometry, …

concerning Galois theory, it's far from being understood nowadays

the Langland's program studies the representations of the absolute Galois group of the rationals, and relate them to automorphic forms
it's still highly conjectural, and has been a field of research for 50 years

on the other hand, you can take a look at analytic number theory
here are some questions you could have about a systems of equations : «Does the system has solutions ?» «If it does, how many are they ?» «Can we give a formula ?» «Can we have asymptotic results» ? «How are they distributed ?»  …

basically, you try to give quantitative statements

[upbeat juniper]

random question, but is there a nice description of endomorphisms of F_2 up to inner automorphisms?

I know for instance that the outer automorphism group is GL_2(Z)

[hot lake]

F2 ?

[upbeat juniper]

the free group on 2 generators

[hot lake]

the noncommutative one ?

cuz GL2(Z) is the group of outer automorphisms of Z²

[upbeat juniper]

yeah it turns out to be the outer automorphism group of F_2 as well

[hot lake]

so I'm a little puzzled

oh really

[upbeat juniper]

it only works for n=2

[hot lake]

isn't the group of inner automorphisms isomorphic to F²

[upbeat juniper]

yes it is

[hot lake]

ah is the outer group the quotient of the whole automorphism group by the inner ones

[upbeat juniper]

yeah

[hot lake]

if you take an automorphism f and make a matrix M(f) by counting the occurences of a and b in f(a) and f(b) you get something in GL2 and it is not changed by composition with a inner automorphism

so maybe it's that ?

[upbeat juniper]

the hard part to show is that it is left unchanged only by the inner automorphisms

in general the inner automorphism group is contained in the kernel of the map Aut(F_n) -> GL_n(Z)

[coral shale]

For the first time today, I saw a polynomial ring with possibly an uncountable number of variables being used in a proof in Galois Theory.

I ran back through my notes and all the results seem to be proven for single variable polynomials. Uhhh should every proof be the same if I try to extend it to the infinite case? Stuff like proving it is a euclidean domain, etc.

[upbeat juniper]

algebraic closure?

[lavish nexus]

is this a standard result I can find somewhere

[hidden haven]

Idk any resources but I spent like 3 hours proving this with Saketh the day before an exam once

The way we did this was by basically picking a value for f(1) then halving the input repeatedly and then arguing by continuity that after some number of halvings this value has to start choosing only the "principal" value as its square root and the point where it stabilizes gives you what it will be in the exponential form

something like that

[tribal moss]

Hmm, even with countably many variables, the polynomial ring stops being Noetherian, for example.

[quaint tree]

Even with two variables you lose PID. I can't remember about UFD off the top of my head though. You don't even need to get to the infinite case to get totally different behaviour from the one-variable case.

[coral shale]

Uhhhh ok

[hidden haven]

If D is a UFD, so is D[x]

[coral shale]

let me show

[hidden haven]

And I believe that the union of an ascending chain of UFDs should be a UFD

This would show UFD for all polynomials rings

big if true

[coral shale]

Proof goes onto 2nd page but im referring just to the 1st part

I thought we needed PID but maybe not uh (nvm just realised that dumb lol)

[tribal moss]

A particular element of the polynomial ring can only mention finitely many variables, so if polynomial rings with finitely many variable admit unique factorizations, then so does every polynomial ring.

[coral shale]

So reading up, what do I know about this poly ring with infinite var
Other than its a ring (im convinced about that at least)

[hidden haven]

I don't think you need anything in this proof

All you do is adjoin elements t_i which will act as roots for q_i(x)

And then quotient by (q_i(t_i))

but this need not be a field

So you take a maximal ideal containing all q_i(t_i)

Only issue is showing that this is a proper ideal

I don't think you need much for that

[coral shale]

I see

proper ideal is what the thing sets out to do

at the end there, i follow that barely

thanks for clearing that up

[hidden haven]

ah right

[coral shale]

I thought we needed other properties and stuff

but the proof aside

it is a ufd

but not pid

?

[coral shale]

From this?

[hidden haven]

Yeah UFD follows from transfinite induction + adjoining 1 variable doesn't break UFD + unioning ascending chains of UFDS doesn't break UFD

PID breaks as soon as you adjoin 2 variables

[coral shale]

That doesn't sound simple 👀

havent seen transfinite

haha

[hidden haven]

Do you see that it works for finitely many variables?

That's just normal induction

[coral shale]

yh

[hidden haven]

Now transfinite induction is just that but now our variables are numbered by a general well ordered set, rather than the natural numbers

Well ordered set because they are numbered by some set, and you can assume that that indexing set has a well ordering because every set can be well ordered

And then we are inducting by using the fact that every element of a well ordered set is either a successor or a limit

Oh you can view this as zorn's lemma lol

[coral shale]

We take zorn as an axiom essentially at the start of this course

But thanks for explaining will look into that too 👀

[hidden haven]

Ye it's not math if you don't assume zorn

Given K[infinitely many variables], you can take the poset of K[any subset of the variables]

Well the poset of K[any subset of the variables] that are UFDs

Apply zorn on this to get a maximal element, since the union of any ascending chain of UFDs is a UFD this works

And then prove that the maximal such UFD has to be the entire ring because if not, you can adjoin a missing variable and make it a larger UFD

[pastel cliff]

symmetric groups arent about symmetries in the middle school sense right

it's taking the word symmetric as just like... "the same"...?

[hidden haven]

Yeah, here symmetry doesn't mean symmetry of a geometric shape, it means symmetry of any object

In symmetric groups, you talk about symmetries of a set

[pastel cliff]

but if it's like the dihedral group then it technically does mean that kind of symmetry ig...?

or am i mixing things

[hidden haven]

Yes

[pastel cliff]

oh ok

[hidden haven]

You can talk about symmetries of any object, and they always form a group

[tribal moss]

This argument ought to work even if the set of variables is not well-orderable.

[pastel cliff]

then the problem is i only know an example of a symmetry, like how i only know open sets in R until topology :D

[hidden haven]

Ye I used well ordering to induct

That fact is fine

[hidden haven]

Ye you will learn about automorphisms soon if you haven't already

[pastel cliff]

starting week 4 rn

[hidden haven]

Did you learn about isomorphisms

[pastel cliff]

he defined them but it seems we're going back to it in two weeks

[hidden haven]

It is an invertible homomorphism

And you intuitively think of it as a relabelling of elements

[pastel cliff]

yeah i came across it yesterday as bijective homomorphism so same thing

and it actually made sense!

[hidden haven]

ie if you have a group G and another group H, then an isomorphism G → H is a way of identifying corresponding elements of G and H so that these groups are essentially the same

And only differ in the names of their elements

The analagous thing for sets is bijections

ie bijections are isomorphisms of sets

[pastel cliff]

:O

[hidden haven]

Now you can talk about isomorphisms from an object to itself

That is a way of relabelling elements of that object so that the object is still essentially the same

[pastel cliff]

✓

[hidden haven]

Isomorphisms from an object to itself are called automorphims

ie symmetries of that object

And for any object, the set of symmetries on that object forms a group

[pastel cliff]

so then we're probably gonna do that without him actually saying the word automorphism lol

[hidden haven]

You have seen a specific instance of that

[pastel cliff]

the dihedral group

[hidden haven]

I think he will after you see isomorphisms

Ye and the symmetric group as an example of Aut(S)

Aut(S) meaning group of automorphisms of a set S

[pastel cliff]

[hidden haven]

ie bijections S → S

And now you can do Aut(G) for a group

or Aut(R) for a ring R

and all the other stuff

[pastel cliff]

rings are week ten

but thank you moldi this actually helps

lecture time

[hidden haven]

All of them are groups because symmetries can be composed to give symmetries and composition of functions is associative and identity is symmetric and symmetries are invertible

That's all I have rn

[pastel cliff]

i'll be back to bother later dw <3

[hidden haven]

Good thing I will not be home for the next 4 days 😌

[hidden haven]

Anyway what does nitezba mean

[pastel cliff]

what does ✓ mean

[hidden haven]

That I have accepted tinky dinky as the supreme overlord

[pastel cliff]

[hidden haven]

You should try it

[pastel cliff]

"meg" "kanga gang" "✓"

[delicate orchid]

much better 😌

[hidden haven]

Ye idk what all those other weird things are, ✓ is just tinky cult, the least weird

[pastel cliff]

what is tinky lol

tokidoki?

[hidden haven]

yes that is what some call tinky dinky