groups-rings-fields

[next obsidian]

This is classified by Ext groups

As those C’s are extensions of H by G (or maybe of G by H I forget lol)

[hidden haven]

of G by H

[next obsidian]

[fallow plume]

what exactly is it meant by "a matrix corresponding to a module"?

it seems like this is different than the homomorphism the matrix encodes, but I can't tell if this is just how I'm reading it or if it really is just the same thing

[hidden haven]

It seems that module corresponding to a matrix is the cokernel of the linear transformation that the matrix encodes

[fallow plume]

okay that makes sense logically. tho it loses me to why this is important/useful

[hidden haven]

You get a free presentation of the module

ie you are writing the module as a quotient of a free module by the image of a free module

It gives you structure theorems from matrix theorems

But also you get an exact sequence
R^m → R^n → M → 0
for any M

Which allows you to extend results about free modules to finitely presented modules by 5 lemma

[fallow plume]

do you have any tips for "getting more familiar" with sequences/cokernels?
i think right now they're probably the most unfamiliar/unintuitive ideas I've encountered so far

[hidden haven]

no clue, I just worked with them until it was good

[fallow plume]

pain

this class is gonna be the death of me lollll

[latent anvil]

Read ahead in an algebra book to the part where it talks about homological algebra

[fallow plume]

✨ finals week
and then next term we do Field/Galois theory and Idk if I'm ready to be on grades

[latent anvil]

The problems usually aren't so much hard as require you to understand the definitions

so it's good for learning those definitions, and not as scary as it sounds

I see the emote and I want to be clear that was a sincere suggestion

[fallow plume]

lmaooo I knew it was and it is a good suggestion

but also like lmfao, that's what our prof has been (trying) to do

[latent anvil]

lol

[hidden haven]

has been to do

[fallow plume]

we've talked about stuff like Krull Dimensions and shit and man it's just too much at once lmaooo

[latent anvil]

Oof

(not to minimize this but krull dimension are more a commutative algebra thing)

[fallow plume]

not to mention he kinda just gives proofs of results in other fields, but doesn't really talk about why they're important---obviously some I can see but it's hard because I only am coming from basic algebra backgrounds

[latent anvil]

(the module structure stuff is also just more advanced algebra, not homological)

[fallow plume]

i'm pretty sure that's what he does for a living

[latent anvil]

It's good stuff!

[next obsidian]

I am a fan

Some might even call me CHMISTER COMMUTATIVE ALGEBRA

(Like myself)

[fallow plume]

i was trying to figure out what my prof does and I can't help but laugh at this:

Email: my lastname at math dot fsu dot edu

LIKE BRUH LMAO

[next obsidian]

Is your prof Aluffi??

[fallow plume]

yea he's a visiting prof tho I think?

[next obsidian]

I don’t think so…

[hidden haven]

[next obsidian]

I think Aluffi is at FSU

[latent anvil]

YOOOO

That's poggers

[fallow plume]

well I definitely don't go to FSU

[next obsidian]

Or at least some Florida university

Also yeah

[hidden haven]

What

[latent anvil]

Don't shatter my dream

[next obsidian]

Ohhhhhh

I see

[fallow plume]

I'm on west coast

[next obsidian]

BRUH

[fallow plume]

best coast

[next obsidian]

That’s sick

My name is on his website

When he talks about errata

[fallow plume]

[latent anvil]

Damm

[fallow plume]

small world, that's pretty neat tho

[next obsidian]

Well

If you’re using chapter 0

[latent anvil]

I emailed him when I was 16

[fallow plume]

i'm actually curious who our new prof is gonna be 'cause he's apparently pretty new

[next obsidian]

I’d jump to chapter VIII

There’s good exercises there

[fallow plume]

that and "Notes from the UNderground"

[next obsidian]

Gotcha

[fallow plume]

mm okay, maybe I shall

[next obsidian]

That’s the new one

[latent anvil]

Is that the homological algebra chapter?

[next obsidian]

Yeah

[latent anvil]

bc do the bit in the ring theory chapter first

[next obsidian]

Introduced Ext and Tor

[latent anvil]

Do not skip all the way ahead

[next obsidian]

Ah yeah

[fallow plume]

our new prof is "Oswal"

no idea

[next obsidian]

I forgot that part exists too lol

[latent anvil]

In fact, just do that bit for what you need rn

[next obsidian]

Also idk, exactness is like

Maybe this isn’t a satisfactory answer but

Just have faith it’ll be useful

Because t will be

You just will find yourself caring more and more and more

[fallow plume]

yea ngl Idek what these are
maybe i will eventually but honestly rn I'm worried about next term before then

[next obsidian]

The easiest example is how you can characterize injectivity / surjectivity via exactness

Oh yeah don’t worry about that rn

[fallow plume]

esp since I think the anal(ytical) track does "Calculus on Manifolds"???

and I've barely even done multivar so

[next obsidian]

I was just telling Sham what was in Chapter VIII

[latent anvil]

Oh yeah I know what's in there

I was asking bc I wanted to make sure you were talking about that chapter

Bc I think Thomas would not have a good time if he looked at it

[next obsidian]

Yeah I forgot that

[fallow plume]

thank you for keeping my sanity in mind

[next obsidian]

There’s more basic stuff in the ring theory section

[latent anvil]

And that's what I was recommending to get good at cokernels

[next obsidian]

I just remember doing that over wpring break 2 years ago

Oh yeah

Makes sense

[latent anvil]

Bro do you remember aluffi linear algebra?

Followed by 1 day of d&f rep theory?

[fallow plume]

what do y'all think about when you guys think about cokernels out of curiosity?
lately i've been thinking about it in terms of like,,,, "structure outside the image" but idk. it seems like it's way more important than that and I'm struggling to see why it comes up as much as it does

[next obsidian]

I just view it kinda formally

[fallow plume]

this literally sounds so much like Aluffi

[next obsidian]

Like it measures failure of exactness

[hidden haven]

Structure outside the image is good lol

[latent anvil]

I think of it as a way to take quotients

[fallow plume]

our class is supposed to be module/ring theory.

first week or two was category theory
I do enjoy the fact i was introduced to it so early on but man

[latent anvil]

Blessed

[next obsidian]

So like

My real answer is

Sheafy

Or topological

So it’s really hard to like say how I think about it

Idk have you seen localization?

[fallow plume]

yea, that was like 2 sets ago

[next obsidian]

Okay so you know what M_p is for p a prime

And M_f where f is an element of A?

[fallow plume]

it's loosely in my head, yea

[next obsidian]

Okay so if you know what Spec A is as a topological space, p is a point, and going to M_f is like an open set

[latent anvil]

Lmfao

Jesus christ

[next obsidian]

So sometimes like

Uhhh

[latent anvil]

Chmonkey talk to someone who is new to algebra

[fallow plume]

this yea?

[next obsidian]

Right

[latent anvil]

If you had talked to you at this point like that you would be annoyed at you

[next obsidian]

Okay so the cokernel measures failure of surjectivity

I know

I’m trying to explain in a nicer way

[fallow plume]

i appreciate the patience btw lmaoo

[next obsidian]

But like if you have M -> N -> C

Where C is the cokernel

[fallow plume]

as a warning I have like,,, no experience with topology

[next obsidian]

Maybe I just want to know that M -> N is surjective “local enough to p”

Think about it like maybe

In analysis

You want to know that for some point, you surject onto an entire open ball around it

The ball can be any ball

I don’t care how small it is

Algebraically this is like saying M_f -> N_f is surjective

Because that “at f” kind of models an open nbd

Oh lord…

I think I cannot explain this well

[hidden haven]

[fallow plume]

i don't blame you lmao

[next obsidian]

The point is it’s all controlled by C!

When C_f = 0

Your map is surjective on that open ball

But if you know you’re “surjective at a point”

Aka M_p -> N_p is surjective

Then you know C_p = 0

[fallow plume]

i def understand the whole C = 0 => surj., but I think I'm getting lost as to why there's localizations

[hidden haven]

Just keep in mind that it's dual to kernel because large kernel = very non injective and large cokernel = very non surjective

[next obsidian]

And you can lift that C_p = 0 at a point to an entire neighborhood

Yeah this is why like

I’d have to explain algebraic geometry

[hidden haven]

[next obsidian]

MonkaS

I guess what I can say

[fallow plume]

lmfao another day, hopefully I'll just get more comfortable with it in time

[next obsidian]

Is that it’s useful to be able to measure when something is surjective

Via an algebraic object

Because you can now apply algebra to the cokernel

[hidden haven]

Also this is good

[next obsidian]

To try and understand it via all the tools you have to understand modules

[hidden haven]

You can get any quotient of M by taking cokernel of some map into M

[fallow plume]

i think (?) i see what you mean here

and that actually seems pretty huge

[next obsidian]

Yeah like

This doesn’t exist in analysis

At least not what I know

The only way you have to see if something is surjective is like, on the level of sets

And like here’s a maybe good example if you’ve seen Nakayama’s lemma

Have you seen Nakayama’s lemma?

[fallow plume]

sounds familiar, but I'mma say probably not

[next obsidian]

Ah okay

[fallow plume]

i think he's mentioned it

[next obsidian]

It would say like if M = mM where M is finitely generated and m is the maximal ideal of a local ring

Then M = 0

The upshot here is like, if A is a local ring, and N is finitely generated, then M -> N is surjective iff M/mM -> N/mN is surjective

The proof uses the cokernel

Because you can apply Nakayama’s lemma to it

And this is super nice because M/mM and N/mN are vector spaces over A/m

So now it’s literal linear algebra

It’s super easy to know when maps of vector spaces are surjective

I am so sorry because I think none of this is going to make any sense

[fallow plume]

nah nah you're fine

parsing + someone was asking me smth.

makes sense, still confusing, but I can understand the purpose now :)

and that's mostly just the fact that I'm not getting into the details

[next obsidian]

Well the details are not too bad. If you accept Nakayama’s lemma as a given, then it goes like this.

The cokernel of M/mM -> N/mN is just C/mC. So if M/mM -> N/mN is surjective, C/mC = 0, aka C = mC. But Nakayama’s lemma says C = 0, so M -> N is surjective

The fact that the cokernel is just C/mC is not obvious, but it’s true

Probably follows from 3rd isomorphism theorem??

But here we are using algebra on the cokernel, and that’s what makes this proof work

So that’s why having the cokernel be an algebraic thing is so useful

[latent anvil]

In the case where you're looking at the cokernel of an associated matrix, it's definitely right to think of it as a way to take a quotient

You're defining M(A) = A^k/A^r, but where the embedding A^r -> A^k is the columns A, so you're modding out by the spanning set of the columns of A

Does that make sense?

[fallow plume]

I think I'm getting confused on the notation, but if I understand correctly, yea.

i think i might need a break tho for now tho haha

[toxic zephyr]

in part b should this say a bijection from A->Q?

[proud bear]

pretty sure yeah

how do you know this property can be expressed in first order logic and why doesn't first order expressibility matter? i thought the compactness theorem only applied to first order theories? (sorry if these are dumb questions. i really don't know much about this subject)

[hidden haven]

Ye you need first order expressibility. What I had in mind was that given any 2 elements, they either generate a cyclic group or they generate the entire group, but now I'm not sure how to say that they generate the entire group since that seems to require infinite disjunction

Ah yes

It's doable for abelian groups

I think

Hmm let me think after lunch

[proud bear]

[hidden haven]

Nvm got it already lol

Take the language of ℤ modules

So you have scalar multiplication by integers

Then the theory is gonna be the theory of ℤ modules and

For all x, y, either
There are z, a, b, c, d such that
z = ax + by
x = cz
y = dz

Or

Nvm this don't work

Because I'm quantifying over the unary functions

I guess it doesn't work

[proud bear]

[proud bear]

but this is just {0,+} \cup {all integers}?

[hidden haven]

Yes

[proud bear]

well thanks anyways. i will try to learn more about this in future. model theory seems to be bery cool

[gritty sparrow]

I think i have a proof that it can’t work: Assume that such a thing existed, now consider the language of Z modules along with a function symbol f. Put in the rules for this, along with axioms making f an injective but not surjective homomorphism. Then Z is a model which is infinite but all models are atmost countable since any model is cyclic. This contradicts lowenheim skolem

[hidden haven]

Bruh

[gritty sparrow]

I’m quite proud of this proof

[hidden haven]

Ye nice that makes sense

Though the wording at the end doesn't seem good

[gritty sparrow]

Wdym?

[hidden haven]

"any model is cyclic"

What does that mean?

But if all proper subgroups of a group are countable then the group must have an upper bound on cardinality so it works out

[gritty sparrow]

I don’t think this is entirely trivial. In any case what i meant is that any model for these axioms is in particular an abelian group (by forgetting about the f), and in fact is a cyclic abelian group since f makes it a proper subgroup of itself

[hidden haven]

It is, because if group has size ℵ_2 then there must be a subgroup of size ℵ_1

[hidden haven]

And ye that makes sense

[gritty sparrow]

How do you prove this, i don’t know how. It also seems like we need something that is (at least superficially) stronger, any group of size atleast aleph_2 has a subgroup of size aleph_1

[hidden haven]

Pick ℵ_1 many distinct elements, and take the subgroup generated by them

[gritty sparrow]

Oh right lol

[hidden haven]

No, LS says that all infinite cardinalities above cardinality of the language are hit, not just arbitrarily large ones

[gritty sparrow]

Right

[fallow plume]

did he also give you guys page long problems

i don't even know if I have the activation energy to read this

[next obsidian]

He wasn't our professor

We just used his textbook

[fallow plume]

oo gotcha

[tawdry crystal]

What is the method of decomposing an arbitrary element of S_n using the least number of 2-cycles?

[fallow plume]

D&F has something about how to decompose into transpositions

[paper flint]

A fairly minimal way could be to decompose into disjoint cycles first, and then break each cycle into disjoint transpositions

[fallow plume]

i forgot exactly how it's suppose to be, otherwise I'd tell you, but basically you can break an n-cycle into n transpositions iirc?

[paper flint]

Yeah, I think that's true

[fallow plume]

pretty sure it's something like:
(12345) -> (54)(43)(32)(21)(15)
this probably isn't right, check it

[paper flint]

Right

[wooden ember]

yeah you can write (a_1 a_2 ...) as ...(a_1 a_3)(a_1 a_2)

[fallow plume]

ye that looks familiar

[wooden ember]

idk if this is minimal for every permutation after cancelling though 🤔

well not cancelling sorry since you take disjoint cycles

yeah no its probably minimal

[paper flint]

Yes, seems like it

[wooden ember]

cause you get n-1 transpositions for an n-cycle idk how you could get less

[hidden haven]

It seems so weird to think that aluffi is a prof, all this time I've been thinking he's a book 😵‍💫

[wooden ember]

https://math.stackexchange.com/questions/2235010/prove-or-disprove-the-minimum-number-of-transpositions-needed-to-decompose-si

[wooden ember]

its seems so weird to think moldi is a human, all this time I've been thinking he's a nerd

[hidden haven]

[fallow plume]

It's so weird to think that Moldi is a human, all this time I've been thinking he's a walking encyclopedia 😵‍💫

[lethal dune]

what a weird conversation

[hidden haven]

Nerds aren't humans and talking on discord implies ability to walk

[fallow plume]

is there always a way to do reduced row echelelon form for arbitrary general linear rings?

i'm assuming no since you're not guaranteed units in R

[lethal dune]

[fallow plume]

talking on discord implies ability to walk
I present a counterexample: myself

im too lazy to get up

[lethal dune]

thought you are Hawkins 2.0

[hidden haven]

[fallow plume]

yea I realized after typing that I should clarify

i wish i was half as ingenuitive as Hawking tho

im just some dumbass first year trying to get by

[hidden haven]

You're a first year?

[lethal dune]

I'm a dumbass 5th yr trying to get by

[tawdry crystal]

The big S should be the small s right?

[lethal dune]

[hidden haven]

Bruh moment

[fallow plume]

yea, just turned 20 basically

[hidden haven]

[tawdry crystal]

cuz n-S makes no sense since S stands for permutation group

[fallow plume]

i think my education is a bit strange 'cause Idk jack shit about multivar/analysis/etc but I know a bit more algebra than I think the typical first year but idfk

[wooden ember]

I think in their indices they use a big S but it’s pretty obvious what they mean yeah

[fallow plume]

algebra > analysis though, so worth it

[hidden haven]

✓

[hidden haven]

✓

[tawdry crystal]

capital small S haha

[wooden ember]

See I half agree with this

The theory I’ve seen in algebra is a lot more fun

[lethal dune]

strongly disagree

[wooden ember]

But I’ve done a lot more fun exercises in analysis than algebra, at least at uni

[lethal dune]

algebra cool but I like this [insert meme here]

[wooden ember]

I imagine analysis really kicks up a notch later down the line

[fallow plume]

i just really like the study of structure. stuff like the monster groups/moonshine/galois

[wooden ember]

PDEs seem fun

[south patrol]

Algebra is beauty, analysis is hoping all your <= signs give you smth small enough

[wooden ember]

So true

[hidden haven]

✓

[lethal dune]

buncho guys saying why alg is better in a algebra channel

[wooden ember]

Wtf someone is howling in my building

[fallow plume]

this reminds me of the Subway dude who contributed to Twin Primes

[lethal dune]

aau kavi haveli pe

[fallow plume]

don't call me out like that bro

[wooden ember]

Lmao

[paper flint]

Probably a seething analyst

[wooden ember]

Probably

Aight I should go see if they’re okay this isn’t normal

[fallow plume]

i should not fucking be laughing at this so much

[paper flint]

I should switch mz kezboard to English, Iäm getting funnz szmbols

[fallow plume]

i think the weirdest part is i'm not even sure if I wanna get into math

[south patrol]

German keyboard detected

[lethal dune]

[south patrol]

z and y swapped lol

[paper flint]

Yeah lmao

[paper flint]

That's okay

[lethal dune]

lol what

those Germans

[sharp sonnet]

manan acting ßuß

[fallow plume]

i think I would if we paid mathematicians even 1% of what some NFL players get

[paper flint]

[wooden ember]

Can’t find them

[paper flint]

Probably camping in Montana at this point, lamenting the industrial society and its consequences

[fallow plume]

casual reminder everyone that you or someone you know will probably make a significant contribution for the development of the next 50 years, but they're still not going to make anything close to a pop star/model/NFL player/actor/etc

✨

[paper flint]

Your reward is doing something you enjoy But yeah, no need to do math for its own sake if other priorities exist

[south patrol]

Manan acting süß more like

[sharp sonnet]

i will make a significant contribution?

[south patrol]

I will make a Significant contribution towards ruining my life

[wooden ember]

They obviously didn’t mean you

[paper flint]

Loch Fields Medal arc

[fallow plume]

oh yea totally agree---I'm not a CS major to clarify . just kinda sad

[sharp sonnet]

currently i am trying to make some algorithms better

but the (conjectured) runtime depends on GRH

[paper flint]

[sharp sonnet]

so i think my best course of action will be to prove GRH

[fallow plume]

GRH?

[sharp sonnet]

generalized riemann hypothesis

[fallow plume]

nvm

[wooden ember]

Generalized Riemann hypothesis

[fallow plume]

yea it just clicked

my first thought was "General Relativity Something"

[paper flint]

Disprove GRH with your algorithm

[sharp sonnet]

ye there is weird stuff

like the algorithms run in subexponential time iff GRH

[paper flint]

[fallow plume]

that sounds sick

[paper flint]

Sub-exponential is not necessarily polynomial time, right?

[sharp sonnet]

it wont be polynomial

well

that is conjecture as well

[fallow plume]

inb4 pvnp

[sharp sonnet]

it depends on the growth of another invariant that we dont know

[paper flint]

[fallow plume]

this sounds like some really amazing research tho

[sharp sonnet]

it ok

its really weird bcs the algorithms are heavily algebraic

but to analyze the runtime you need heavy analytic number theory

because they depend on asymptotics of certain invariants and GRH

[kind temple]

is there any like… empirical data for the run times which might sway u to think that GRH is true?

[fallow plume]

it's weird to me that you can guess at the runtime without knowing the runtime without GRH

[sharp sonnet]

ye there is

but you can just collect numerical data on GRH directly

[kind temple]

true

[sharp sonnet]

ofc this is no proof

[fallow plume]

i should sleep :/ 4 finals, an essay, and 2 absurd #groups-rings-fields sets due in 5 days

night y'all

[sharp sonnet]

damn

good luck

[fallow plume]

thanks, I'mma need it

[paper flint]

Goodluck!

[fallow plume]

does anyone have reccs for videos/lectures on Ring/Module theory?
i don't think i can sleep yet so i figure I should just watch something and learn as I chill and hopefully pass out

i'm thinking Borcherds will have something

bless that man, can't believe he won the fucking Fields Medal and just decided "I should try out YouTube"

[spice whale]

Lagrange's theorem is so damn cool

[spice whale]

is this a good enough proof

it's kind of long, but the two cases are on opposite sides which should make it easier

[dull inlet]

Are there any good twitter accounts to follow for abstract algebra

[hidden haven]

[south patrol]

'since <x> is not a subgroup of <y> and vice versa, they must only share 1 in common' what is your justification for that

[delicate orchid]

Sylow theorems omggggg

[south patrol]

I do think the proof can be simplified considerably though and still just use a Lagrange-style argument

Ig the point is ||<x> is a proper subgroup of <x,y>|| and then that kills it if I'm not mistaken

[paper flint]

That's a slick argument, potato.

[coral shale]

saving paper? so clumped up

[wise igloo]

...

do you like

skip lines when you write?

tf

[latent anvil]

How do you express the property (in particular stuff about being/not being cyclic) in a first order way?

Chm told me to look at the model theory discussion earlier and this was the most recent one I could find

[broken stirrup]

Hi I have a quick question, If A is R-module and B is an ideal of A, doesn't it automatically imply B is a submodule of A?

[hidden haven]

Lol there was a discussion just today about how it doesn't work

[hidden haven]

Here

[latent anvil]

Ah okay lmao

[spice whale]

if there is more than one element in common between <x> and <y>, then x^a = y^b
i think you can do something with modular arithmetic at this point but I'm not sure what

[coral shale]

uh definitely. To space out my arguments and make it more readable

New thought new line was how I was taught

Not to write in paragraphs

[south patrol]

Sure but ye the point is ig you've not proven that so the proof seems incomplete as it is to me

[spice whale]

yeah

[wise igloo]

weird

[south patrol]

that was advice i was given too tbh, or at least going to a new line more often

[hidden haven]

Yes it is good advice

An assignment without spacing is the worst kind of assignment to grade

[coral shale]

When we define the group axioms... we write them as if the identity is unique, although that isn't necessarily true apriori - but is the 1st thing we prove right after.

For the inverse axiom, which would be preferred (assuming non-unique identity, apriori):

• There exists an identity e such that for all x in G, there exists y in G, such that xy = yx = e
• For all x in G, there exists y, e in G such that e is an identity and xy = yx = e

Written badly, but --- I am just asking about the order of the quantifiers, basically

$$(\exists e)(\forall x)(\exists y)...$$
or
$$(\forall x)(\exists e, y)...$$

[coral shale]

Which makes more 'sense' as a definition

[spice whale]

neither of these state that e must be the identity

[coral shale]

im just being lazy with notation

I'm asking about the quantifier order we 'want'

(within the '...' I can condition e to be an identity)

[delicate orchid]

top

[coral shale]

gut instinct? 👀

I think I agree

[spice whale]

well

[delicate orchid]

e doesn't depend on the choice of x

so put it before

imo

[spice whale]

yeah

[coral shale]

actually, if we define left/right identities, left/right inverses, the order will matter

for something other than a group (uh... right? or maybe I remembered wrong)

In those cases, they go with top?

🤔

[delicate orchid]

if inverses aren't two sided I just leave tbh

[spice whale]

this is why i prefer not using invertibility

but divisibility

which doesn't require identity

[coral shale]

I'm writing what is basically a set of notes

And don't want to lose rigor (like most stuff does - and just writes it like there is only 1 identity apriori)

[spice whale]

write it out as 2 axioms

[coral shale]

how 🤔

[latent anvil]

Shuri, I would write the inverse axiom like "exists e.(forall x.ex = xe = x) and (forall x.exists y.xy = e)"

[tribal moss]

It's not just a matter of the order of quantifiers. You also have a choice of whether to make "e" a constant in the language of groups or not. And similarly for the invertion function.

[latent anvil]

if I didn't want to think of e as a constant

[spice whale]

for all x, y there exists a,b such that x = ay = yb

then you define inverses as the special case where x = e

[coral shale]

Could you elaborate? 🤔

[tribal moss]

I think you need the same rule with multiplication from the right too, in the general case.

You can say either of

1. a group is a set with a binary operation, satisfying some operations (where some of them go \exists e
2. a group is a set together with a binary operation and a distinguished element e and a univary function ("inversion") such that bla bla bla

[spice whale]

just realised i could prove this with divisibility property of groups

[coral shale]

2. probably sounds like it would be easier to understand

And then right after we would have to prove such a thing is well defined?

or am I misthinking 🤔

[tribal moss]

The styles turn out to be equivalent for groups, for example for rings it makes a difference whether 1 is a constant in the language or not, because that controls whether the "natural" concept of ring homomorphism is required to preserve 1-ness.

[coral shale]

urghhh would you mind pointing out the ring case explicitly

I don't think I have fully understood

[spice whale]

monoids

[coral shale]

just monoid case would suffice ig?

[tribal moss]

Yeah, we get the same for monoids.

[tribal moss]

I don't immediately see a way this is right.

[coral shale]

(M, *, 1) is a monoid if

• • an associative binary operation on M
• 1 is an identity wrt * on M

So this is our definition (using method 2) yes?

Then for morphism...

I'm still 🤔 on how this is fundamentally different

[tribal moss]

Yes. Method 1 would just have: (M, *) is a monoid if * is associative and there exists a two-sided identity.

[coral shale]

f : M -> N is a homomorphism iff

(forall x, y in M) f(x*y) = f(x)#f(y)
f(1) = e```

We still define it like this for method 2?

[tribal moss]

Under method 1 the "natural" definition of homomorphism would just require preservation of the things that are in the signature -- that is f(1)=e would not be required. So if we have two monoids:
A: integers modulo 2 under multiplication
B: integers modulo 6 under multiplication
the map f: A -> B with f(0)=0 and f(1)=3 would count as a homomorphism.

We could still choose to define "homomorphism" to require that the image of an identity element must itself be an identity element in the codomain -- all I'm saying is that this would not be the obvious definition that could be derived mechanically from the language signature.

[spice whale]

why exactly does the proof of group identity preservation not work for monoids

[coral shale]

you use inverse to show it? in groups

That explicit example above shows why it might not be preserved

[latent anvil]

Also subring change

[tribal moss]

Because groups satisfy a cancellation law, so there can only be one identity for a given element.

[latent anvil]

*subrings

[spice whale]

ohhh

[hidden haven]

ℤ → ℤ multiplication by 2

[latent anvil]

Z × 0 is not a subring of Z × Z if you include 1 in the signature

[spice whale]

cancellation

[coral shale]

Well here I am thinking about all this when trying to write a basic set of notes but I find this so interesting

[latent anvil]

But it is if you just require existence

[spice whale]

what maths does to a mf

[coral shale]

I think I must prefer method 2, then

it makes so much more sense

when thought in this kinda context

[tribal moss]

They each have advantages and disadvantages, so it pays to be familiar with both.

[coral shale]

ofc, method 1 and method 2 generally converge in most notes - not long after the definition in (1), you will prove uniqueness of identity and inverses, hence the well-definedness of a symbol to denote identity and the functional notation for inverses

[spice whale]

just use semigroups and you don't have to use either method

[coral shale]

Huh so if I use method 2

I will skip on proving identities are unique

🤔

[spice whale]

yes

although the uniqueness proof takes 2 seconds

[tribal moss]

On the other hand, you would have to prove that the identity is preserved each time you want to say something is a homomorphism.

[hidden haven]

No

[coral shale]

oh ok yh I wont skip

[hidden haven]

You might still want to show that there's no other element e' such that e'x = xe' = x for all x

[coral shale]

right.

[hidden haven]

Oh I thought you were asking if you can

[spice whale]

literally 1e = e = 1 so e = 1

[hidden haven]

Lol

[coral shale]

i didnt realise your point til just now

speaking of '

is it improper to use that to denote inverses

[spice whale]

yes

[hidden haven]

Perhaps

[coral shale]

like... what notational clash is there

[hidden haven]

I just use that on discord

[tribal moss]

I've only seen it widely used in the cubing literature.

[coral shale]

me too

^-1? hell no

[spice whale]

many things

depending on context

[hidden haven]

Set up a macro/snippet if you are doing latex

[coral shale]

ur like just casually plain texting on phone

5 buttons to do ^-1

D:

[hidden haven]

Also looks like shit

[delicate orchid]

5 buttons or as I like to call it 0.8 seconds

[coral shale]

a^-1b^-1 = (ba)^-1

[hidden haven]

If not actually superscript

[hidden haven]

What a gamer

[coral shale]

a'b' = (ba)' much easier to read plaintext

[delicate orchid]

' as a short hand isn't the worst thing in the world

[coral shale]

like what though. within group theory

[spice whale]

differentiable automorphisms

[coral shale]

you what

[hidden haven]

Let x and x' be arbitrary elements of the group

[coral shale]

😂 lemme look at that nvm i wont

yeah, you lose that, but I think i always preferred x1 x2

[hidden haven]

Lol I literally used this here

[spice whale]

yeah

[coral shale]

'as long as its clear from context anything goes '

[spice whale]

basically differentiable functions from a set to itself

[coral shale]

but at least, I dont think there is an immediate clash in gt

[hidden haven]

Lie groups

[coral shale]

g'^n will look dumb ig

g^n'

[spice whale]

g^-n

[hidden haven]

Let n and n' be arbitrary integers

[coral shale]

yh ok, now i think about it, lets keep ' to plaintext lel

[spice whale]

$g^{n} '$ vs $g^{n'}$

[cloud walrus]

all functions are alison
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate orchid]

I am on the verge of tears

[spice whale]

literally what is the error here texit

[hidden haven]

✓

[delicate orchid]

unrelated but moldi do you know of anything funny that happens when all of your eigenvectors are in the span of a single vector

[hidden haven]

That vector is then an eigenvector

Or there are no eigenvectors

[delicate orchid]

yeah, I already knew that though

just thought I'd double check

[hidden haven]

So your question is just

[delicate orchid]

(1, 0, ..., 0) is an eigenvector that spans the space

idk there might've been a funny kernel space!!!

[hidden haven]

"tell me a joke about an eigenvector"

[coral shale]

ok, let's not use '

$g$\rotatebox[origin=c]{180}{$g$}$=$\rotatebox[origin=c]{180}{$g$}$g=1$

[delicate orchid]

although this space is fixed as an FG-module there might be something here
https://cdn.discordapp.com/attachments/359052581022203914/940656085893799936/unknown.png

[coral shale]

$g^n$\rotatebox[origin=c]{180}{$g^n$}$=$\rotatebox[origin=c]{180}{$g^n$}$g^n=1$

[spice whale]

STOP

[delicate orchid]

no way do I just use first iso

POG

[spice whale]

yo first iso

best theorem

[hidden haven]

Shuri can't keep getting away with this

[spice whale]

they must be stopped

trap them in a monoid

no inverses to abuse there

[delicate orchid]

if we have an order 4 element can we write $g^3$ as \rotatebox[origin=c]{270}{$g$}

[cloud walrus]

Wew Lads Tbh (201 🍇) ✓

[delicate orchid]

new theorem chat

[coral shale]

are we just gonna dihedral everything now

[hidden haven]

This is not what I meant when I said they should make more graphical calculi

[spice whale]

this is wrong g³ would be upside down

[coral shale]

what about symmetrical letters

[hidden haven]

Self inverse

[coral shale]

oh so this is why 1^2 = 1, 0^2 = 0

[hidden haven]

It was right in front of us all along

[delicate orchid]

let $g$ be an element of order $k$, then we can denote $g^n$ as "\textbackslash rotatebox\lbrack origin=c\rbrack {360*k/n}{$g$}"

[cloud walrus]

Wew Lads Tbh (201 🍇) ✓

[delicate orchid]

yesss

mmm good

I should write all my latex such that when it complies it's still latex source code

[hidden haven]

I don't think that works

Bad tex and bad theorem

[delicate orchid]

I had to make up for the pog moment I had earlier

[coral shale]

What if we taught cayley graphs before the algebra.

I swear this could work

[hidden haven]

What if we taught category theory before counting

I swear it would be funny

[delicate orchid]

what if we taught

[spice whale]

have you read mathematics made difficult

[hidden haven]

Nope

[coral shale]

Its so good

[hidden haven]

I shall take a look

[coral shale]

#chill if u scroll down the pins, I think rok referenced a few screenshots

and a link

[hidden haven]

[coral shale]

that book was made for moldi

nami, not rok
#chill message

[spice whale]

the preface is already amazing

[hidden haven]

I was looking for rok for 5 mins

[frail zealot]

difficult math is the most fun tbh

[hidden haven]

Anyway I'll take a look tomorrow

[coral shale]

👋

https://tenor.com/view/tired-good-night-sweet-dreams-sleep-well-sleep-tight-gif-15812577

[frail zealot]

btw why does nlab feel the need to say shit like "this is rather trivial proof" to things that aren't trivial at all like sure it's trivial to someone who's been doing category theory for 30 years but for someone new and getting to that point is really hard

[tribal moss]

All the better to help you have fun, my dear.

difficult math is the most fun tbh

[frail zealot]

funny

[prisma ibex]

to the extent that the notes on the nlab are mostly written for experts, it's useful for experts to be clear about which proofs are trivial/definition chases and which proofs actually require really nontrivial ideas

there's some pages on nlab that are explicitly written for students and those are a lot easier to read and don't handwave proofs as much, even if they are kinda trivial

[frail zealot]

oooh i didn't know about this

[prisma ibex]

the main ones that come to mind are like

this really big series of notes on homological algebra and stable homotopy theory and chromatic homotopy theory

which was written for a semester/year course

that and the mathematical physics notes which were for a semester course

very readable and having all the definitions hyperlinked to other nlab pages is really insanely useful

[fallow plume]

should i ask questions about generalized modules/matrices here or #linear-algebra btw?

[delicate orchid]

module stuff goes in here

[fallow plume]

thanks!
Ig then my question is whether or not i'm on the right path here.
I'm trying to find the cokernel of a matrix, but I'm not sure if I'm calculating the image properly.
I can't reduce it into row echelon form since I'm working with a polynomial ring, so basically I just said "okay, let's just assign this coordinate arbitrarily, and examine what everything else has to be"
Is this reasonable? no giveaways though please

think I figured it out. I was apparently looking for the Smith-Normal form.

[chilly ocean]

So I dont know name of this but Given CRing A, D(A) is set of prime ideals p such that there exists a in A where p is minimal prime ideal containing (0:a)

Am asked to prove that for multiplicatively closed set S, D(S^-1A)=D(A) intersect Spec(S^-1A)

sorry but im slightly stuck

Im really only at definition of x in D(S^-1A) means there exists y/s in S^-1A such that x is minimal prime ideal of S^-1A containing (0:y/s). I want to show that this translates to there existing z in A such that x is minimal prime containing (0:z) and that x is a prime ideal of S^-1A which comes trivially from x being in D(S^-1A)

my intuition is that if x is minimal prime containing (0:y/s) then maybe x minimal prime containing (0:ys)

[maiden ocean]

Have you seen any theorems characterizing the prime ideals of S^-1A

@chilly ocean

[chilly ocean]

no i dont thik so

is it just how prime ideals transform with different ideal operations?

like if they commute with localization or not?

[maiden ocean]

You should have something like, "the prime ideals of S^-1 A" are in bijection with the prime ideals of A disjoint from S"

So you need to show that the minimal elements containing the annihilator of a among this set are exactly the minimal prime ideals of A, containing the annihilator, which are disjoint from S

[chilly ocean]

oh yeah

ty

[next obsidian]

Try #combinatorial-structures

[subtle ivy]

i think the point wasn't that this question isn't suited to this channel, but more that there might be more people familiar with what you're talking about over there

(i for one have encountered basically no graph theory so far in my algebra studies)

[magic root]

fair enough

[spice whale]

I've got that if a or c is equal to 0, the other must be 1 or -1, but I'm stuck on how to do the general case

[delicate orchid]

have you thought about the determinant of this matrix

[spice whale]

yes

that's how i got the special cases

because it nicely cancels out one of the terms

[delicate orchid]

ok lemme think

yeah I think this'll work

ok, so if a and c are non 0 we have the relation ad-bc = -1 or +1, focusing on this form for now: ad+(-b)c = 1 noting that d and b are completely arbitrary - does this expression remind you of any other lemma/theorem?

[spice whale]

ohh

they have to be coprime

[delicate orchid]

I believe so yeah

[spice whale]

wait how did you get the first bit tho

ad-bc = 1 or -1

[delicate orchid]

is this not a 2 by 2?

[spice whale]

yes

[delicate orchid]

formula of a determinant of a 2 by 2 then

[spice whale]

yes

i know

ad-bc

it has to divide a,b,c,d

[delicate orchid]

when you take the inverse you divide by the determinant

-1 and 1 are the only units in Z

[spice whale]

yeah but you have to prove that a,b,c,d can't have any other common factors
which I don't understand

how to do

[delicate orchid]

I'm starting to see why that might be a misguided assmuption

and I thought I was soooo clever

no wait wait wait you literally just

cannot divide

by any other integer

[spice whale]

i started replacing divisibility with multiplication by arbitrary constants but it went a bit haywire and i ended up with 6 more arbitraries

[delicate orchid]

it doesn't matter if the result is an integer 1/n is not in Z

[spice whale]

but

if you have eg
a,b,c,d divisible by 2
then couldn't you construct a matrix with determinant 2 or -2

which would still have an integer inverse

[delicate orchid]

I know what you're saying and I'm trying to remember if GL_n(R) is implicitly an R-module or you can pick and chose what the scalars are

[spice whale]

wait no you can't
because then ad-bc would be divisible by k² for k our common factor of a,b,c,d

that's actually pretty cool

[delicate orchid]

ohhh SHITTTT

nice spot

also the determinant just straight up maps into the group of integer units

I thought so

[spice whale]

lmao

[delicate orchid]

your explanation is a lot

nicer

yeah, nicer

[spice whale]

it explains why

and it generalises to any UFD i think

[delicate orchid]

I think so too

[latent anvil]

Well you can still construct an integer matrix whose determinant is divisible by 2

[spice whale]

nono

[delicate orchid]

yeah it just won't invert to an integer matrixz

[latent anvil]

I guess I'm confused where the bit about the common divisor came from

[delicate orchid]

(2 0)
(0 1) for example

[latent anvil]

imo this problem is about det(A^-1) = 1/det(A)

[delicate orchid]

hold on lemme scroll up

[latent anvil]

To be clear I agree with your proof wew lads

[delicate orchid]

yeah I'd just like to clear up my own confusion on this

[latent anvil]

I'm just not sure I get what this means

[delicate orchid]

if a_i = n_ip for some integer p, then a_1a_2-a_3a_4 = p^2(n_1n_2-n_3n_4)

[latent anvil]

yep

But why does that matter?

like we get that det(A) is divisible by p^2

But aren't you just using the fact that it's gotta be ±1 to get a contradiction from this?

[delicate orchid]

k is non-zero and, if we're assuming we can cancel (not multiply by an inverse but just omit) the factor of k from both the determinant and all the elements of the matrix we're still left with a 1/k

[latent anvil]

but det(kA) = k^2 det(A)?

Like this is the scaling you'd expect

[spice whale]

here's my simpler explanation

[delicate orchid]

I've never seen that identity

my lin alg is so spotty it's unreal

[latent anvil]

If you scale a column/row of a matrix by c you scale the determinant by c

[delicate orchid]

ok yeah now I follow

[spice whale]

basically, det² divides det, so det = ±1

[latent anvil]

So scaling every element of an n by n matrix scales the det. by c^n

How do you know that det divides the gcd of a, b, c, d at the start?

I agree that the gcd(a, b, c, d)^2 divides det(A)

[spice whale]

it must divide all of them because the inverse matrix must be an integer matrix

[latent anvil]

Oh, I see

[delicate orchid]

yeah so if we took GL_2(Z) as like, a Q-vector space, what we're saying would make sense

[latent anvil]

You're using the 2 by 2 inverse formula

Right?

[delicate orchid]

hence my comment here

[spice whale]

yep

[latent anvil]

Ahh okay I'm on board now

Nice!

[spice whale]

the misc problems in artin are cool

[latent anvil]

It's not, because the sum of two invertible matrices might not be invertible

And you can't multiply by 0

[delicate orchid]

oh yes

[coral shale]

I + -I gg

[latent anvil]

You can think of GL_n(R) as a group that contains the group of units of R

So relatedly, M_n(R) is a ring containing the ring R

M_n(R) meaning n by n matrices with entries in R

[delicate orchid]

however if we take the extenstion of scalars $GL_2(\bZ) \bigotimes_{\bZ} \bQ$ :troll:...

[latent anvil]

And then we're saying the units of the subring are units of the bigger ring

[cloud walrus]

Wew Lads Tbh (201 🍇) ✓

[spice whale]

[1 0] + [-1 0]
[0 1] [0 -1]
one example

[delicate orchid]

I'm crying in the club again

[gilded gull]

Is it normal for aa to feel much more difficult than previous classes

Usually there aren't many concepts that take me a while to conceptualize but at this point it feels like everything is just being shoved down my throat and there are so many theorems that build that it feels overwhelming

[latent anvil]

A lot of people feel this way, it's okay

All the abstract definitions and theorems about objects that don't feel tangible can make it hard to get a grip on what's going on in the class

The best remedy is doing a lot of examples so that when you see a theorem, you can think eg "oh okay how does this work for the group S_3?"

[gilded gull]

o

that seems helpful

[spice whale]

in other news: artin is committing crimes against algebra

[gilded gull]

idek what to say to that lmfao

[delicate orchid]

interesting question

[spice whale]

he's saying semigroups must have identity

[delicate orchid]

cyclic semigroups if you will

[spice whale]

nono

cyclic monoids

[delicate orchid]

hold on a fuckin moment

[spice whale]

he's blatantly using the wrong terminology

[delicate orchid]

you got a gosh darn point

when was this book written

[spice whale]

1991
he has literally no excuse

the second edition was 2011

[delicate orchid]

lol

[spice whale]

they had a chance to edit it

[delicate orchid]

anyway "cyclic" monoids are interesting to think about

[spice whale]

ye

I'll do that q in a min

currently classifying all groups of order 6

[delicate orchid]

wholesome family friendly fun

[spice whale]

not when every element is order 1 or 2

[delicate orchid]

huh?

[delicate orchid]

alison what did you mean by this

[spice whale]

nevermind

it's easy

[delicate orchid]

I mean I think I know all groups of order 6 just by exposure lol

do you even have to do sylow memes

[latent anvil]

Here's a fun problem: classify all groups where every element has order 1 or 2

[spice whale]

no

[delicate orchid]

C_2^n :troll:

[latent anvil]

I don't really know either of your background so I cannot guarantee you have the tools to do this

[spice whale]

he hasn't introduced sylow yet

[latent anvil]

😉

I don't think you need sylow

[delicate orchid]

I'm pretty sure if a group is nothing but elements of order 2 it has to be abelian

[spice whale]

yeah

[delicate orchid]

ergo, (C_2)^n

[spice whale]

i am beginning to believe that there are only two groups of order 6

[latent anvil]

Why is this?

[spice whale]

C6 and S3

[delicate orchid]

I am beginning to agree with you

[delicate orchid]

I cannot remember I saw this result like a few years ago

[latent anvil]

lol

[delicate orchid]

it's coset related

[latent anvil]

then that does not count as a proof

There is a direct proof by looking at elements

[delicate orchid]

what if I cite the result

[spice whale]

coset memes

[delicate orchid]

ab
=aba^2
=abab^2a = (ab)(ab)(ba) = (ab)^2(ba) = ba

ok there we go

that took some thinking

[upper pivot]

you can use that a=a^(-1) so ab=(ab)^(-1) = b^(-1) a^(-1) = ba

in general if a -> a^-1 is a homomorphism (in this case its the identity so its a homo) then its abelian

[latent anvil]

This is the proof I was thinking of

[spice whale]

also isn't this just finite cyclic groups and the free monoid on one generator

[latent anvil]

although without saying the word homomorphism

[spice whale]

that was surprisingly unexciting

[latent anvil]

Cyclic is a very strong property

[spice whale]

yo category memes

[latent anvil]

So @delicate orchid how do you go from cyclic to C_2^n?

[delicate orchid]

I mean

[latent anvil]

Er

Abelian

Sorry

[delicate orchid]

yeah

I was gonna say

structure theorem

[latent anvil]

That works

But it doesn't generalize to infinite groups :)

[delicate orchid]

when it doubt, structure theorem

[upper pivot]

group doesnt need to be finitely

yeah

[latent anvil]

I didn't say classify all finite G

[delicate orchid]

WOWWW I see how it is

[spice whale]

klein four group is definitely cyclic trust me bro

[delicate orchid]

C_2^\infty...

[latent anvil]

Not wrong

But why?

[delicate orchid]

I mean it's

[upper pivot]

C_2^infty doesnt make sense

[delicate orchid]

ok as the limit of some sequence of objects it might not be wrong but that's not how I'd write it

[delicate orchid]

yeah

[spice whale]

this is just the nimber group lol

[upper pivot]

different cardinalities give non-isomorphic groups

[latent anvil]

But it's not a good idea

[spice whale]

i take it back it's not the nimber group

[latent anvil]

you just need to figure out why these "C_2^infty"'s are the only possibilities

[spice whale]

it's close tho

[delicate orchid]

ok here's what I'm thinking

additive group of the closure of F_2

[latent anvil]

What do you mean by closure?

[delicate orchid]

sorry

completion

[latent anvil]

Algebraic closure?

I'm not sure what you mean by completion

Thinking about F2 is the way to go though

[delicate orchid]

I wanna generate a ring with of sequences with elements in F_2, completion seemed like the natural choice but now I'm realising that the fact that it's a field runs into some difficulties with actually formalising this in terms of a completion

F_2[[X]] is the other option that strikes me as possibly being a thing

I don't know how that would generalise to uncountable abelian groups though

[latent anvil]

Yeah, how are you relating this ring to a group?

[delicate orchid]

well the only group in there is the additive group

could take the group of units under multiplication I suppose but I'm pretty sure that's just F_2 so that gets us nowhere

[latent anvil]

Ah I see

[wraith obsidian]

what are you trying to do in general?

[latent anvil]

Classify groups where g^2 = 1

I gave it as a challenge

[delicate orchid]

it certainly is a thinker

[latent anvil]

The additive groups of F2[x] and F2[[x]] are both good examples

These are the direct sum and direct product of countably many copies of F2

Respectively

[delicate orchid]

logical

[wraith obsidian]

ah, good challenge

[delicate orchid]

now for uncountable I believe I will just cope and say it's a topology problem now

[latent anvil]

you would be incorrect

[delicate orchid]

hence why it's a cope

[latent anvil]

And F2[[x]] is uncountable!

[delicate orchid]

is it?

[latent anvil]

It's got the same cardinality as R

What's an element of the power series ring?

It's basically a sequence of 1s and 0s, right?

[delicate orchid]

it's an "infinite polynomial" which is a countable sum I thought

[latent anvil]

True

But you get to make countably many independent choices from a finite set

[delicate orchid]

oh I see you diagonal argument it

[latent anvil]

Well I'm thinking something a little more direct

But you could do the diagonal argument

I was going to say that if you have a power series f = Σ a_i X^i, let S_f = { i : a_i = 1 }

Does this definition make sense?

[delicate orchid]

yeah

[latent anvil]

Then f -> S_f gives a bijection between F2[[x]] and the powerset of N

[delicate orchid]

just the set of indexes where the coeff of the power series is non-zero

ohhh

[latent anvil]

Also you could think about the binary number 0.a1 a2 a3...

This would give a surjection onto [0,1]

Either way, same cardinality as R

[delicate orchid]

continuum hypothesis

[latent anvil]

R and 2^N have the same cardinality no matter what

That's not what the continuum hypothesis says

CH is about whether there's anything inbetween this cardinality and the cardinality of N

[delicate orchid]

I shall endeavour to find such a cheeky set

[rapid bramble]

${\text{cheeky}}$

[cloud walrus]

Mosh

[rapid bramble]

there you go

[delicate orchid]

oh yeah it's 2^aleph_null = aleph_1 not 2^|N| = |R|

[delicate orchid]

the continuum hypothesis has been disproven

[latent anvil]

Very common misconception

[delicate orchid]

yeah I just kind of have aleph_1 = |R| stapled into my head as a fact even though it's technically not

[wraith obsidian]

I've never done that before actually, confused this w/ another exercise.
My first line of thought is that ||(ab)²=1 implies that ab=ba, hence it must be abelian||
shouldn't this basically tell us by ||a zorn argument|| that we can write it as ||C₂^X for some set X||?

[latent anvil]

Yes

Wait

Explain what you mean by ^?

Is that a direct sum or a direct product?

[wraith obsidian]

||direct product||, but now that I think about it that might be the wrong thing

[latent anvil]

Yeah, it is

[wraith obsidian]

b/c what I'm doing in ||the zorn argument is taking a colimit||

[latent anvil]

Yeah, that's the direct sum

||it's easier to think of the Zorn argument as choosing an F2 basis||

[rapid bramble]

For the converse, getting the fact there exists r in R st ra=1 is sufficient right?

[wraith obsidian]

yes, that's how I'm thinking about it.
Because ||for me it's basically an inductive enlargement of something that we know to be free, and extend to something bigger that we know to be free||
One might say in my eyes ||every zorn argument is kinda transfinite induction||

[wraith obsidian]

yep

[rapid bramble]

go back to whatever you're talking about lol

[latent anvil]

for sure, I like thinking of Zorn like that

I recently proved a generalization of Zorn and I had to redo the whole transfinite induction thing

[wraith obsidian]

one might even go as far as to say every other conceptualization is a ||cardinal sin||

[fallow plume]

me too prof, me too

[latent anvil]

lol based

[fallow plume]

meanwhile on the set he's making us do God knows what

[wraith obsidian]

that being?

[fallow plume]

i mean obviously a lot of y'all will probs understand where this is going

but i have not a clue. and unfortunately i can't really receive help

[wraith obsidian]

hate to take you back to reality, but I have absolutely no clue what Grassmannians have to do with young diagrams. But that's probably more #point-set-topology related

[fallow plume]

oh boy

all of this is supposed to be module/ring theory so I'm honestly just lost what field we're even in anymore

[latent anvil]

My Twitter username is @grassmannian

[patent crescent]

Is there a difference between external and internal semidirect products?

Or is it just the same product with different ways to get there

[hidden haven]

Yes, once you take the external semidirect product, the resulting group is the internal semidirect product of the embeddings of the 2 groups you started with. Internal semidirect products are used to decompose a group you already have into simpler groups, and external semidirect product is a way to construct larger groups from smaller ones with some control on the properties

[patent crescent]

Oh I see, is that where the names external and internal come from as well?

[hidden haven]

Yes

Internal to a group, or just abstractly taking a semidirect product without being inside a group

[chilly ocean]

In proving that the only homomorphism between two groups of coprime order is the trivial homomorphism, I was thinking that the image of that homomorphism must always be a subgroup of the group, so the function will always be mapping a domain to a codomain of coprime order—this means that there will be an element left over in the domain which won't allow us to create a parallel algebraic structure between the two groups. First, I'm not sure if my intuition is correct, and second, I'm not sure how to rigorously prove this

[wraith obsidian]

I'm drunk but it seems you're overcomplicating

Think of image and Kernel, and what constraints they have with regards to their size

Or: why can there not be a nontrivial hom from C_9 to C_20?

[chilly ocean]

G/ker(phi) must be isomorphic to image(phi), but G/ker(phi) and image(phi) must be coprime via Lagrange, so the only homomorphism between the two is the trivial homomorphism?

[next obsidian]

The size of im phi divides |G|

But also divides |H| by Lagrange

Yuh-oh, so the image has size 1

[chilly ocean]

yeh

[oblique leaf]

Why is phi surjective?

[coral shale]

K1K2 is the smallest field containing them both

$\left{\frac{f(x,y)}{g(x,y)}\right}$

[coral shale]

I think we can express all elements of this form

where
x is from K1

y is from K2

hmmmmm

It's not clear to me how we can decompose something like $x+y$ into a product

[coral shale]

This must be vital https://i.imgur.com/k0BFY6o.png

https://i.imgur.com/PX56SkB.png

[oblique leaf]

I am sorry but i still dont really see why surjectivity is guaranteed

[coral shale]

Neither do I - I didn't realise this defn I am not familiar with is vital

[oblique leaf]

now you mention it, we were given the product definition as a hint

so i can in fact use it

i just dont know how

[coral shale]

$x+y$

[coral shale]

We need to show why something like this

can be decomposed into a product
utilizing that defn

[oblique leaf]

yea i had the same concern

[coral shale]

thinking back to my vague knowledge of tensors.......

if {xi} and {yj} is a basis

isnt

{xiyj} a basis

(looping over i and j)

[oblique leaf]

yes

ok i suppose xiyj spans K1K2

[coral shale]

yh is this enough? (Need to wrap my head around this myself)

[oblique leaf]

i mean this map is going to take each xi\otimes yj to xiyj, meaning that if its image is a field then it is surjective

is the image a field?

[coral shale]

closure under addition is necessary, so this brings us back to x + y

[oblique leaf]

what do you mean?

[coral shale]

(x, 1) -> x
(1, y) -> y

If the image is a field, then x + y needs to be in it.

[oblique leaf]

the image is a field

[coral shale]

ah

ofc

[oblique leaf]

so it is solved right?

[coral shale]

i guess (but im not 100% sure on this new stuff to me)

[coral shale]

im thinking about this rn

[oblique leaf]

yeah same, i am in a grad class rn but I took the undergrad version of algebra 1 so i havent actually seen anything about fields before lol