groups-rings-fields

[rustic crown]

... 0 --> 0 --> M --> 0 --> 0 --> ...

[long nebula]

Okay that makes sense

[rustic crown]

and you wanna study this object

[long nebula]

Naturally

[rustic crown]

a nice thing you can associate to a chain complex is it's (co)homology

which is the quotient of ker/im

at each degree

[long nebula]

Hm

[rustic crown]

this is a nice invariant and you wanna study it

[long nebula]

Oh so that measures failure of exactness

[rustic crown]

yep, you could say that

[long nebula]

I see

(This emoji is so cute)

degree meaning like the i in M_i?

[rustic crown]

it's the entire reason i loved the server as soon as i joined

[rustic crown]

you know everything uwu

i don't have to say half the things :3

nice nice

okie so, to understand this weird object you make a weird definition

say you have two complex M* --> N*

[long nebula]

Okay

[rustic crown]

and a map between them

which means you get a ladder of commutative squares

you say it's a quasi-isomorphism if it induces isomorphism at homologies

[long nebula]

ooooh

[rustic crown]

so you're trying to understand which complexes have the same homological information

[long nebula]

ah okay

[rustic crown]

one way i think about homology is this

if you have a complex M*, then you can compute H_i(M*)

but you can also put these together to form a complex again

[long nebula]

ooooh

[rustic crown]

--> H1(M*) --> H0(M*) --> H-1(M*) -->

where the maps are just 0

[long nebula]

so H_i is the ker/im at each degree

[rustic crown]

yeep

[long nebula]

hmmmm why are the maps zero

think

[rustic crown]

i had some weird objects

i needed to connect them to get a complex

did the only obvious thing i could

[long nebula]

gotcha

[rustic crown]

okie notice this now

if in a complex N*

if all maps are 0s

then what is H_i(N*)

[long nebula]

it's just N*?

ker is everything and im is nothing

[rustic crown]

yea, at the ith degree, it's ker(N_i --> N_{-1})/im(N_{i+1} --> N_i)

which is just N_i becuase maps are 0

but notice what we just did

we had M*

we formed a complex H*(M*)

with all maps 0

[long nebula]

right

is H*(M*) the same as H_i(M*)

[rustic crown]

yea but the complex, i'm writing * when i want to mean the whole complex

[long nebula]

ohh

okay

[rustic crown]

and i when i want to talk about that particular object at the ith degree

[long nebula]

so H*(H*(M*) = H*(M*)

[rustic crown]

exactly!!!

[long nebula]

:o

[rustic crown]

so it's like homology projects all this information and think of it like a trivial complex

(from linear alg you probably recall that proejctions are idempotent p^2 = p)

[long nebula]

oooooh

yes

what do we call H*?

[rustic crown]

i haven't seen an actual word for it

[long nebula]

Haha okay

how would you pronounce H*(M*) then

[rustic crown]

homology complex of M?

idk :p

yee back to what i wanted to say

so say you had M* --> N*

and say it's a quasi-iso

which means at each H_i(M*) --> H_i(N*) this is an iso

and now notice what you defined as a free resolution

it was this weird thing

--> F2 --> F1 --> F0 --> 0 --> 0 --> ...

and you have the trivial complex of A

--> 0 --> 0 ---> A --> 0 --> 0 --> ...

the resolution also gave you a map F0 --> A

the exactness at then end, F1 --> F0 --> A --> 0

exactly tells you that this map F* --> A is a quasi-iso

so to understand the homological information of A

it's same as understanding homological information of F*

the sad part is the notion of quasi-isomorphism is very not uwu

[long nebula]

:OOOO

[rustic crown]

like the name probably suggests that this would be an equivalence relation

if M* --> N* is a quasi-iso

then you might expect that there is N*--> M*

but that not true 😭

[long nebula]

SAD

cool tho :o

[rustic crown]

but topology comes to rescue

there is a notion of homotopy equivalence of complexes!!

and this a slightly stronger relation than being a quasi-iso

but much more well-behaved

[long nebula]

what is the topology on complexes o.O

[rustic crown]

weird righttt

[long nebula]

very

:o

[rustic crown]

that's what @coral spindle was talking about

[long nebula]

that's so cool

[rustic crown]

F

[long nebula]

So are chain homotopies the same as homotopies in topology

[rustic crown]

i tagged by mistake

[coral spindle]

👁️ 👄 👁️

[rustic crown]

yee, you abstract out the algebra from homotopies

and you get the corresponding notion for complexes

[coral spindle]

There's no topology involved, that's kinda the neat part

[rustic crown]

anyway fun part

why do you care about projective/free resolutions specifically?

[coral spindle]

I think you could compare it to a (relative?) CW complex, but idk it's been too long

[long nebula]

Why!

[rustic crown]

because if you look at a quasi-isomorphism between two projective complexes (which only have non-zero objects on the left)

then this q-iso is automatically a homotopy equivalence

!!

so this makes a full circle

we started with free resolution

alked about q-iso

[coral spindle]

Comparison theorem is the keyword if you want to look this up

[rustic crown]

then homotopy

and then back to free resolutiosn

[long nebula]

ooooh

is there any time that you would care about a projective resolution vs caring about a free resolution

[coral spindle]

Well

Let's focus on modules here

free resolutions are easier to calculate in general

But when you look at homology, sometimes there's a gap that occurs, like there's a nonzero free module in your resolution that shouldn't be there

This happens when you calculate Ext modules, but let's not get into it

Basically for some rings, this 'gap' is reflecting the fact that there's a big difference between being projective and free

[long nebula]

Huh, interesting

[rustic crown]

okei so one thing is you don't need to stick to the category of R-mods to do all of these
you can replace R-mod with any abelian category, and there it's not so obvious what a free object is... but you can define projective objects in the same way

and these things naturally pop up

[coral spindle]

but if you look at projective resolutions, there are no gaps :) keywords: projective dimension, global dimension

[rustic crown]

for example the category of sheaves on topological spaces is abelian, but here it's not so obvioius to think of these as modules or something

[coral spindle]

^

it's also not necessarily even possible to construct a projective resolution

idk sheaves, but my understanding is that injective resolutions are the way to go with those

[rustic crown]

which is why you worry about injective objects and do the dual thing :3

[long nebula]

Ahhhh

Ty so much!!! :)))

[rustic crown]

i just gave you a brief overview of what you do in homological algebra at the start :3

hopefully you'll have a fun ride now

[long nebula]

That's so neat

Our alg prof says we will vote on what we do at the end of the semester

So we may choose homological algebra

But he also said that that's the driest option to teach in like two weeks lol

[rustic crown]

yea it can be very boring depending on how you do it

the way i said makes me the most happy

[toxic zephyr]

i'm lost as to how to show a subgroup is normal based on its order. the hint my prof gave is that |HK|=|H||K|/|H intersect K|, but i don't know how to use that

[tender wharf]

H cap K can be either 1 or 5

[oblique river]

well K is a subgroup of H here, so it's 5

I don't think you should apply the hint to the given H and K

here are some not very well formed ideas: you might have seen the proof that index 2 subgroups are always normal, that means K is normal in H. If K is not normal in G, then there must be another subgroup K' of G of order 5 (why?). Maybe looking at KK' or HK' could be fruitful

[rustic crown]

oooh

[tender wharf]

also ||since K is normal KK' is a subgroup||

pls ignore this statement is wrong

[rustic crown]

we can say that H has a unique subgroup of order 5

[oblique river]

not quite. K is normal in H, not in G, and K' won't be a subgroup of H.

[tender wharf]

oh right we don't have that K is normal in G

that makes sense

[oblique river]

ahhhh this is it! very clever

[rustic crown]

[rotund aurora]

zeta is 17th root of unity, sigma is the automorphism that sends zeta to zeta^3 (3 is a primitive root modulo 17)

[toxic zephyr]

honestly, i have no idea why there must be another subgroup of order 5

[rotund aurora]

so I wanted to compute the minimal poly of zeta+zeta^{-1} over the field below it (above in the picture)

The galois group of that field extension has order 2 and is given by sigma^4, where sigma sends zeta to zeta^3 (3 is a primitive root modulo 17)

and 3^4=-4, so that the Galois conjugate of zeta+zeta^{-1} is zeta^4+zeta^{-4}. Multiplying this should give something in the base field

but you get x^3+x^5+x^12+x^14

x=zeta

???

(thats the norm)

ok nvm me

so you just have to do more work

lmao

[oblique river]

G acts on its subgroups by conjugation. what does it mean for a subgroup to be normal, in that language?

[toxic zephyr]

if the way it acts doesn't change it? if you're talking about group actions, that's next chapter 😅

i'm not sure what G acting on a subgroup by conjugation means...

[oblique river]

nevermind, ignore that then

for any g in G, gKg^(-1) is a subgroup of G

what can you say if K is not normal

[toxic zephyr]

then gKg^(-1) is not equal to K?

and that conjugation does not change the order, right?

[oblique river]

yes

[toxic zephyr]

aha. so then conjugating a subgroup either gives you the original subgroup (if it's normal), or yields a different subgroup with the same order (if it's it not normal)?

[oblique river]

yes

[void cosmos]

why can we write every function like this? isnt this what he is trying to prove?

prop 3.29

[oblique river]

no

f is some given random function

g is some new function that he defines

he shows that f = g

[void cosmos]

he said let f be in the alternating functions k linear

he claims that f = sum(coefficints*alpha^i wedge ...)

isnt this saying that f is in the span of wedge(alpha^i)?

[oblique river]

yes, and then he proves that

[void cosmos]

how

[oblique river]

by showing that the function on the left and the function on the right are equal

the function on the right is a well-defined function in this same vector space

[void cosmos]

okaayy

so for each f

he assigned sum(f(e_j)*alpha wedge...))

and showed that they are equal as they agree on all basis elements ?

[oblique river]

that's right

[void cosmos]

cool

ty

sm

question tho

is this the exterior algebra?

[oblique river]

this is a piece of it

the exterior algebra would be the direct sum over all k

of \Lambda^k(v)

[void cosmos]

and with wedge product being the multiplication operatio nright?

[oblique river]

yeah

[south patrol]

I guess also it'd be the dual of the exterior algebra

[oblique river]

or, i guess more formally, would be the dual

yeah

well

[void cosmos]

so

[oblique river]

the dual of the exterior algebra would be bigger than this

[south patrol]

Well in each degree the dual at least

[void cosmos]

but its not the dual for me tho

[oblique river]

yeah

haha

[void cosmos]

cuz

its defined for me over A_k(V)

not over V

[oblique river]

okay that's fine

[void cosmos]

cuz idk what v wedge u would mean

ik what f wedge g would mean

for f and g being in A_k(V)

okay so the exterior algebra is the direct sum of these vec spaces

does it make sense to think about a basis

for

this

woudl it be just the direct sum of each basis for A_k(V)? as wev shown that they are the k wedge product of the dual basis?

[toxic zephyr]

so i used the hint to say that HK' must have an intersection with order>1. so then there is some non identity g^-1 kg in H. but i'm not sure that's the right path to go down searching for a contradiction.

[oblique river]

what is the order of the intersection, exactly?

[white oxide]

in general, is it true that given an isomorphism phi, from G to G', for any a in G phi(a^-1) = phi^-1(a)? would this hold by the homomorphism property?

[oblique river]

that statement just doesnt really make sense

phi(a^(-1)) is an element of G'

but phi^(-1) is a map from G' to G

so it doesnt make sense to say phi^(-1)(a)

do you mean phi^(a^(-1)) = [phi(a)]^(-1)?

that is true but doesn't require phi to be an isomorphism

[white oxide]

oh i meant $\phi(a^{-1}) = \phi^{-1}(a)$

oops

wait

[cloud walrus]

okeyokay

[oblique river]

that's what you said the first itme

again that doesnt make sense

[white oxide]

oh because a in G right

[tender wharf]

phi inverse is the inverse of phi which is a function

you probably meant phi(a inverse) = ( phi(a) )inverse

[white oxide]

oh yea i think i meant that

[tender wharf]

which is true

[white oxide]

like

$\phi(a^{-1}) = \phi(a)^{-1}$

[cloud walrus]

okeyokay

[white oxide]

is correct right

[tender wharf]

yes

you can check, since phi preserves the group operation

[oblique river]

i said that :c

it also doesnt require phi to be an isomorphism

[white oxide]

yeah i know

just wanted to make sure

[oblique river]

it's true for any homomorphism phi

[white oxide]

ok thank you

i'm struggling to see how this is true though

[toxic zephyr]

oh it has to be 5, right?

[tender wharf]

.

[white oxide]

ohh

i see what you m ean

okay thank you guys both

[void cosmos]

bad question ig hahaha

[oblique river]

the basis of a sum of vector spaces is the union of the bases of the individual spaces

it's not a bad question, it just doesnt have an interesting answer, and there's nothing special about this context

[void cosmos]

oh yeah

mb

forgot about that

ty ty

obv

[toxic zephyr]

so H has two subgroups of order 5. i saw det said earlier that H has a unique subgroup of order 5, which seems to be the contradiction. but i don't see why H can't have two subgroups of order 5?

[void cosmos]

how

[white oxide]

would the number of automorphisms on $\mathbb{Z}_n$ be $n!$ ?

[void cosmos]

no

[cloud walrus]

okeyokay

[void cosmos]

@white oxide no

[white oxide]

oh wait

my reasoning was that permutations were automorphisms

but they're not

they're just functions right

[void cosmos]

automorphisms are isomoprhisms from the group to itself

[white oxide]

right

and the book defined permutations to be bijective functions on a set A

but that's not the same as an isomorphism

so that's where i got confused

[void cosmos]

yea u need to preserve the structure aswell

[white oxide]

right

thanks!

[void cosmos]

np

try to think of what Aut(Z_n) might be

[white oxide]

does that notation denote the set of all automorphisms on Z_n

[oblique river]

maybe you should think about that more then

that is indeed the contradiction

[rotund dragon]

yes

[white oxide]

thanks

[void cosmos]

mb didnt say that

[white oxide]

all good

so im doing an exercise that's asking me to find the number of automorphisms for different values of n on Z_n

and so i'm taking Z_2

the hint is asking me what must be the image of a generator under an automorphism

so clearly an automorphism preserves structure

so it maps to a generator

so that leads me to thinking that maybe it's the number of generators in a cyclic group G?

or am i totally lost here

or maybe it would be m! where m is the number of generators in a cyclic group G

[void cosmos]

keep going

[void cosmos]

no ur not

try it out

with different values of n

[white oxide]

so let's try Z_4 = {0, 1, 2, 3} where 1 and 3 are generators

so say we map 1 to 3 and 3 to 1

and every other element gets mapped to itself

so that should give an automorphism of Z_4 i believe

lol

[toxic zephyr]

I'm obviously missing something because i don't see it

[white oxide]

and the only other option is mapping the set Z_4 to itself, but i'm not sure if that constitutes an isomorphism

[void cosmos]

it does

[white oxide]

so it would be 2

well

let's see if i can find any other automorphisms

[void cosmos]

okay so

ur conjecture was

[void cosmos]

this right?

does this example work?

[white oxide]

well i'm not sure because i'm trying to find some other automorphism, if i can find one more then that would be 3 > 2 which is the amount of generators in Z_4

[void cosmos]

as you said

you have to map a generator to a generator

[white oxide]

oh right

so m = 2, and m! = 2

so so far that holds

i think

[void cosmos]

if what ur saying is true

[toxic zephyr]

doesn't C5xC5 have two subgroups of order 5?

[void cosmos]

then Z_n has n! generators

right?

[white oxide]

wait how

because Z_4 has 4 generators where n = 4

and in this case there are 2 generators

and 4 \neq 2

[void cosmos]

i meant this line

what did you mean by m!

!

[white oxide]

oh

should be automorphisms

m is the number of generators

in a cyclic group G

[void cosmos]

but this tho

[white oxide]

oh

this was my final line

of logic

sorry

[void cosmos]

try it for Z_5

[white oxide]

if this statement were to hold under my conjecture then Z_n would have n generators

which is not true

[white oxide]

ok

so here there are 4 generators

1, 2, 3, 4

since gcd(a, 5) = 1 for a = 1, 2, 3, 4

so we have to have a generator map to a generator...

so 0 has to map to itself

therefore it is the number of permutations of the set {1, 2, 3, 4}

so m = 4 because there are 4 generators

[void cosmos]

ur talking about the addition group here

right

[white oxide]

yeah

[void cosmos]

ok

okayy

[white oxide]

and therefore there are 4!

automorphisms

ig i would have to prove it by means of induction

[void cosmos]

so ur saying that there are 24 automorphisms

[white oxide]

yes

[void cosmos]

in Z_5

[white oxide]

yeah

[void cosmos]

if ur mapping a generator to a generator

or wait lets do this like

manually

so u like see it better

[white oxide]

wdym

[void cosmos]

lets look at a specific element in Aut(Z_n)

so in Z_n let x be a generator

or not x

think about 1

1 is always a generator right

[white oxide]

yeah

[void cosmos]

so phi(1) must be a generator of Z_n aswell

[white oxide]

right

OHH WAIT

[void cosmos]

but look here tho

what did u do here

explain this form e

for me*

[white oxide]

used the fact that if a is a generator of G, then a^r is a generator of G where gcd(r, n) = 1 where n is the order of G

[white oxide]

i forgot that you would have to account for the other permutations of the elements that are not generators

wait

is that right

yeah

so maybe it would be something like m!(n-m)! where n is the order of G and m is the number of generators of G?

well let's see with like

Z_4

nah

doesnt work

rip

[void cosmos]

np

but there is something

that you know

about generators in Z_n

that will make this easier for you

[white oxide]

did it have to do with

[white oxide]

this?

[void cosmos]

yea

[white oxide]

i mean

would it just be m! where m is the number of elements in G that are relatively prime to n where n is the order of G?

[void cosmos]

lets not try to count it just yet

lets just use what we already know

and build up from there

if we like obtain something

then this will lead this naturally to the correct answer

what must a generator , call it a, in Z_n satisify

[white oxide]

all elements g in Z_n can be written as a^n for some integer n right

[void cosmos]

yes thats what it means for a to be a generator

but you know that |a^k| = n/gcd(n,k) tho right

and if an element has order n then its a what

[white oxide]

then it's the identity correct

[white oxide]

how did you get this

[void cosmos]

its a well known proposition and u can prove it by yourself

just consider a^k^(n/gcd(n,k))

and argue why this is the least integer such that its the identity

does it make sense

[white oxide]

ye imma just return to this later my brain is not functioning rn

[void cosmos]

no ur doing great

lets take this for granted

do u know that if |a^k| is the same as the order of the group then its a generator?

if so then u can reach the conclusion that every generator must satisify gcd(n,k) = 1

so now we know the number of generators is the number of elements satisfying this

we gucci?

now the main problem with what your saying is

that you keep forgetting that u need structure preserving maps

ur trying to count how many permutations between the generators

u will be able to count how many automoprhisms

when u try to actually do it manually ie grab one and see where it maps something and so on

and u can do it generally for any n cuz for all n ,1 is a generator in Z_n

[median pawn]

could someone explain why the stated map from K[alpha] to sigma(K)[beta] an isomorphism?

thanks

[next obsidian]

If you understand why it’s defined, it is sort of clear

Surjectivity: beta is in the image, K maps iso morphically to sigma(K)

Injectivity: maybe the easiest way to see this is that they’re the same dimension over K (note sigma(K) is isomorphic to K). Follows because we added a single root of an irreducible polynomial of the same degree, so they’re of dimension that degree

Note that once you have the same rank and surjective, rank-nullity says it’s injevtive

Try to fill in any details that aren’t clear

[toxic zephyr]

can someone please explain why a group of order 10 can't have two distinct subgroups of order 5? wouldn't C5xC5 have that?

[median pawn]

same rank?

[next obsidian]

Dimension, whatever

[median pawn]

perhaps we have shown that they have the same finite dimension over K

yes

[next obsidian]

For modules we say rank

[lethal dune]

If it did, that subgroup cannot be normal (Syllow). But it is normal as index 2

[next obsidian]

So I just said rank

[oblique river]

C5 x C5 doesn't have order 10...

[toxic zephyr]

im dumb

[next obsidian]

Also cuz Sylow

Idk why ryu sama said what they did. Sylow says the number of Sylow-5s is 1 mod 5

2 is not 1 mod 5

[void cosmos]

i think he meant that normal subgroups are conjugates

to each other

[next obsidian]

Yeah but why mention this

[toxic zephyr]

i checked and sylow's first mentioned three chapters ahead of this question... it seems like there's probably a different way to go about it

[next obsidian]

A different one literally says it can’t be 2

[oblique river]

K is normal in H because it has index 2

[next obsidian]

Sorry, I didn’t see there was other context for this problem, if I saw what you were doing I wouldn’t have suggested Sylow

[oblique river]

just do the same thing you have been doing

you have one tool here, use it haha

KK' is a subset of H; what can its order be

[lethal dune]

that too lol

[toxic zephyr]

doesn't it have to be 10?

[oblique river]

does 10 divide 25?

[toxic zephyr]

no, 5 then. but sorry why are we looking at 25 when |H|=10?

[oblique river]

the order of KK'

using the one formula that you have

about orders of products of subgroups lol

[toxic zephyr]

oh yeah |KK'|=|K||K'|/|K intersect K'|=25/|K intersect K'|

[median pawn]

what's the justification for the last "why?"

[median pawn]

followup to this

[oblique river]

but what did you show about the intersection of K and K' from earlier

[toxic zephyr]

|K intersect K'| has to be strictly less than 5 doesn't it? since we assumed K != K', the intersection must have an order smaller than the lowest order of the pair.

[oblique river]

it has to be 1

because it's a subgroup of K

or, it doesnt even matter, because as you said earlier KK' has to be all of H

which has order 10

so you have 10 = |KK'| = |K||K'| / |K cap K'|

which is impossible

im goign to bed now btw

gl

[toxic zephyr]

there we go now it makes sense.

[toxic zephyr]

thank you for all your help

[median pawn]

why does alpha + a*beta have less than n conjugates over K?

[formal ermine]

what are L and K here

finite fields?

[median pawn]

subfields of complex numbers with [L:K] < infty

so finite field extensions

well if it has n conjugates in total then K[alpha + abeta] = K[x]/(f) (where f is the minimal polynomial of alpha + abeta) is a degree n extension of K, and L is also a degree n extension of K

that'd mean K[alpha + a*beta] = L i suppose

makes sense?

[next obsidian]

of conjugates = degree of extension (in characteristic 0)

It’s less than [L:K] by assumption

[median pawn]

and two extensions of the same degree are isomorphic as K-vector spaces right?

or as fields?

[next obsidian]

As K-vector spaces

[median pawn]

later in the proof, how do you show that the denominator is non-zero?

[next obsidian]

If tau(beta) = sigma(beta) then we know
Sigma(alpha) + a sigma(beta) = sigma(alpha + abeta) = tau(alpha + abeta) = tau(alpha) + atau(beta)

Middle equality cuz they send alpha + abeta to the same element

Conclusion, sigma(alpha) = sigma(beta), but then sigma = tau because alpha and beta generate K[alpha,beta]

[median pawn]

thank you!!

[next obsidian]

Wait this should say sigma(alpha) = tau(alpha) lol

[pastel cliff]

is there a parallel between annihilators of a module and stabilizers of a group action

[next obsidian]

Chmonkey

If you think of a module as like a ring actuon (good analogy)

They’re kinda similar a little bit

[pastel cliff]

ive been kinda seeing that thoo

it's the only way it makes sense

the other analogy i see everywhere is a more general vector space

[pastel cliff]

does this mean there's some equivalent of the class equation for modules?

[median pawn]

why must alpha be sent to one of its conjugates over Q?

[median pawn]

perhaps this is a basic question

[pastel cliff]

pretty standard proof but can someone check this pls

asking bc i feel like i tripped up somewhere towards the end

[pastel cliff]

also still curious about this

bc if annihilators seem to be kinda dual to stabilizers

then torsion is similar to an orbit for a group action

also is there anything interesting in considering a "module action"

like say we have a module and think of it as a "ring acting on group", is there any use to then asking how that group acts on a set?

that could be very faulty logic feel free to sully me

,ti

[cloud walrus]

The current time for stμ₂dying is 02:34 AM (EST) on Mon, 06/02/2023.

[lethal dune]

any automorphism must send a root of an irred to another root of the same irred poly

[lethal dune]

looks fine just some typos also, don't write ψ (rx)=ψ (r) ψ (x)

[vast quiver]

here’s a relevant thought, maybe motivating modules as ring actions, and maybe a reason why we don’t bother giving a special definition for module actions.

I’m not sure what definition you’ve seen for group actions, but you can see them either as a function G x X -> X satisfying some axioms, or a group homomorphism G -> Perm(X), where Perm(X) is the permutation group on X. This second way is what I want to talk about.

This is the way I like to think about actions—you have some algebraic structure and some object with symmetries (eg a set with permutations), and you look at homs from the structure to the symmetries. Beyond actions on sets, you can also talk about linear actions on vector spaces, which are given by group homomorphisms
G -> GL(V),
where GL(V) are the invertible linear maps on V. So this is maybe more interesting than a group action on a set, because more structure is involved.

So you’ve seen group actions, linear group actions (also called representations), now you might be interested in other kinds of actions.

We want to respect structure, so for a “ring action”, you would want to do something like a homomorphism from R to some object, whose symmetries form a ring (so we get ring homs instead of just group homs). We get this situation if we try acting on an abelian group G, because the set of endomorphisms (ie group homs G->G) form a ring (with composition as multiplication). Therefore, you might be interested in ring homs
R -> End(G).
An exercise is to prove that these ring homs correspond exactly with R module structures on G.

Now, to talk about module actions, where M is an R-module, you would be interested in an object which has an R-module for its symmetries. There are objects which have this, but at this point I think we start shying away from the “action” terminology, and just write down homomorphisms from M to the symmetries on your object

[somber sleet]

hey guys, which isomorphismtheorem implies that the second equality holds?

is it this one?

can I also use the theorems for groups?

Cause we didn't go through the ones of rings in the lecture

[vast quiver]

just wanting to check, are you considering these things as groups or as rings? either way there is indeed an isomorphism theorem like the one you quoted

[vast quiver]

I don’t think there’ll be a class equation here. The problem is that while in group theory, your orbits are disjoint, in ring/module theory, your orbits very often overlap. This makes counting arguments not work very easily.

For example, the integers Z are a Z-module by left multiplication. Then, the “orbit” of 2 is none other than the ideal generated by 2, it’s 2Z. This definitely intersects the ideal generated by 1, which is Z. So we have distinct orbits that intersect nontrivially. It won’t always be inclusion either, for example k[x,y] as a module over itself has orbits (x) and (y) which intersect (eg at xy), but don’t contain each other

Modules have plenty of helpful structure though. If you are concerned about classifying modules, you might be interested in the structure theorem for finitely generated modules over PIDs. It’s a very nice theorem that generalizes the structure theorem for fin gen abelian groups.

[somber sleet]

I think I have to consider them as groups, cause we didn't do any isomorphism theorems for rings

just the one that has to do with ideals

[vast quiver]

hmm I’m not exactly sure how to help. Maybe I need more context to make sure I have the right goal in mind?

There are multiple versions of the isomorphism theorems (versions for groups, versions for rings without 1, etc), and definitely they are applicable, depending if you want to prove if this is just a group isomorphism or more strongly a ring isomorphism.

if you just need to prove group isomorphism, this theorem is applicable

but keep in mind that wikipedia has written this with “multiplication” as the group operation, instead of addition

so to apply the theorem, you should change all the multiplication you see for this theorem statement into addition

and you’ll get something very similar to the theorem statement you posted about rings

[somber sleet]

I can show you the solution I was looking at

so ggt is the gcd and the kgv is the lcm in this case

so they were trying to prove the equality given by using the isomorphism theorem

and I was trying to figure it out, because they use the isomorphism theorem for groups here

I assume that it is possible to use because all those rings are additive groups

so as you said, I assume that we use the isomorphism theorems by changing the multiplication to an addition

[rustic crown]

won't say that's changing multiplication to addition

more like forgetting multiplication existed

[woeful sage]

so im doing analysis rn

but I've heard of archimedian fields, so in the case of general fields, do you have something analogous to N? I was about to say "if we view R as a ring, then N would be some subring" but then remembered N is a semi-ring

[rustic crown]

you can talk about the submonoid generated by 1 inside any field

might as well look at the subring generated by 1 lol

i.e. image of the canonical map Z --> F, where F is your field

one equivalent definition of archimedean (for normed fields) is that the image of this map is unbounded.

[woeful sage]

[delicate orchid]

What we staring at boss

[woeful sage]

how's it going wew, long time no wew

[rustic crown]

wew as in view

🧠

[delicate orchid]

Initial object moment

[rustic crown]

[woeful sage]

every archimedian field is isomorphic to R innit?

I heard something like that

[rustic crown]

C

[woeful sage]

si?

[rustic crown]

is archimediean

[woeful sage]

oh

[rustic crown]

idk spelling

[woeful sage]

okay then that's not true

[delicate orchid]

Surely it would just be iso to some subfield of C, no?

[rustic crown]

ig R and C are the only two

[rotund aurora]

every archimedean field is isomorphic to a subfield of R

[rustic crown]

if you put extra stuff like completeness

[delicate orchid]

Actually yeah it can’t be C lol C isn’t ordered

[woeful sage]

does D&F have all this stuff on fields?

[rustic crown]

oh we talking about ordered fields? (i had normed fields in head)

[delicate orchid]

Don’t you need an ordering for a field to be Archimedean?

[rotund aurora]

yes you do

[rustic crown]

norm is enough

[woeful sage]

and the ordering needs to play nice with field axioms no?

[rustic crown]

you define it as Z being unbounded

[woeful sage]

i.e ordered field

[woeful sage]

i see

[rustic crown]

that's a nicer definition because the p-adics aren't ordered

[delicate orchid]

I can easily see the definition being expanded to fit a norm tbf

[rustic crown]

but they're still complete normed fields

and here Z is bounded

so they non-archimedean

[woeful sage]

ur a PhD student, im a lowly undergrad

[rotund aurora]

I think no, I would check Isaacs (chapter: Artin-Schreier theorem) and Clark's field theory notes (towards the end)

Lang does have a chapter

[woeful sage]

lang...

[woeful sage]

okay

[delicate orchid]

“nx > y -> ||nx|| > ||y||”

Thanks discord

[woeful sage]

\||y||

[delicate orchid]

Yeah if you hide ur numbers it’ll work for more fields

[woeful sage]

oh right ofc

[rustic crown]

i used to do \|\|x\|\|

[delicate orchid]

I simply refuse to work with non-finite spaces so I don’t write norms down ever

Easiest solution imo

[rustic crown]

[woeful sage]

okay im gonna go back to analysis thanks guys

[coral shale]

[echo granite]

Hi there, my math teacher told me that every object in math is a set. (for instance a function is a set of sets with two elements). So a class in math is also a set but with particular properties (like groups, structures, etc.) or am I wrong? And can someone explain me how classes resolve the Russel's paradox?

[formal ermine]

there was this one question on my exam which no one was able to figure out

let R be a commutative ring and I an ideal with R\I having only units

then I is the only maximal ideal

how do you do this

[formal ermine]

#foundations

[tender wharf]

[chilly radish]

Take any ideal that contains I

It's either I or strictly contains it

What happens if it strictly contains it

[delicate orchid]

Nvm I’ll let shin do it 😒

[coral shale]

is that really the best place to send them

[formal ermine]

better than here

[coral shale]

||isnt foundations a scary place no one sane ventures in||

deep in the depths of advanced

[chilly radish]

If you can figure that out, try and see how the same argument lets you show that no other maximal ideals exist

[formal ermine]

my brain is apple juice right now I'll think about it maybe later

but thanks

[chilly radish]

Alright

If you give up: ||Any ideal strictly containing I contains an element of R\I, hence has a unit, so it has to be all of R, hence I is maximal. Similarly, any other maximal ideal is neither contained in nor contains I, since they are both different maximal ideals, so it also contains an element of R\I, i.e. a unit, so it's all of R, so there are no other maximal ideals.||

[delicate orchid]

If ||M is an ideal not contained within I then M contains a unit thus is R, hence I is maximal and unique||

[chilly radish]

Sometimes it's better to break down a solution to better understand it

[delicate orchid]

^ cope

[south patrol]

Ok

[rotund aurora]

how this true

so the automorphisms are zeta --> zeta^g where (g,n)=1

[rotund aurora]

I actually solved this more than a year ago

but I knew I had to use arithmetical functions and transformations lul

[south patrol]

I'd be inclined to use properties of cyclotomic polynomials e.g. do you know what \Phi_{p^m r} is in terms of \Phi_{pr} where (p,r)=1 but I think this can be done most easily by chopping up into smaller extensions too, which is cute (using trans. of trace)

[rotund aurora]

I thought about using transitivity of trace, but didnt really try

[hot lake]

I would just say that 0 = 1+z+...+z^(n-1) = sum for d|n of the trace of z^d

[south patrol]

Okay that is a nice way to do it

lol

[delicate orchid]

one of my top 3 favourite properties of gauss sums right there

[rotund aurora]

that is some arithmetical convolution, right?

[hot lake]

yeah

oh and except for n=1 where it's 1 and not 0

[rotund aurora]

that was what I was talking about

so the same

[hot lake]

yeah

[novel parrot]

same thing as saying R/I is a field

so I is a maximal ideal

[oblique river]

They want to prove that I is the only maximal ideal

Not just that I is maximal

Another maximal ideal would have to contain an element not in I. What then?

[somber sleet]

how do you know which numbers are prime in a set like Z[i]

I know that in a factorial ring irreducible iff prime

and generally if R is an integral domain then every prime number is reducible

I understood that in an euclidean domain it's enough to prove the fact that if p=ab, then N(p)=N(a)*N(b), where N is the euclidian function, and if N(a) and N(b) are not in Z, then p has to be irreducible and so prime (euclidean ->factorial)

but what do I do, when I am not in an euclidean domain

do I have to go through the definition? Or is there another way

[novel parrot]

wait but there could be still multiple maximal ideals?

[formal ermine]

you cry

[rotund aurora]

If you know primes are maximal, you can try to compute the quotient, and see if you obtain a field

maximal ideals are always prime, so you can always try that I guess

[oblique river]

The ring Z has infinitely many maximal ideals.

[novel parrot]

then his question doesnt make sense?

[oblique river]

Yes it does…..

[novel parrot]

to show that I is the only maximal ideal?

Z is counter example?

[oblique river]

There is an additional condition…..

The ideal (2) in Z does not satisfy the property that “everything outside of the ideal is a unit”

3 is not in (2) and is not a unit

[novel parrot]

i dont see that in the original question

[oblique river]

It says

R\I has only units

[rotund aurora]

you should always get a domain if you are quotiening by a prime, no matter where you are

[novel parrot]

but R/I also contains 0

[oblique river]

The question says R\I, not R/I

There is no hypothesis given about R/I

[novel parrot]

oh

oops misread

as R/I

[pastel cliff]

im doing this wrong arent i

[delicate orchid]

it looks close

I would really use the fact that M has a basis at some point, makes it easier

[lethal dune]

don't think you need basis for this

[delicate orchid]

I can't explain what I want to without just giving it away

[lethal dune]

[oblique river]

[lethal dune]

||m ∈ T(M) then ∃ nonzero r s.t. rm=0 or r'rm=r'0=0 => m=0|| where basis

[oblique river]

It doesnt make it any easier to use a basis…

Smh dont give away the answer

[delicate orchid]

it's spoilered it's fine

[lethal dune]

spoilered

[oblique river]

Im sure the person asking for homework help will respect the spoiker

[delicate orchid]

||M free => M is torsion-free|| is what I was going for

[oblique river]

Im sure thats the intended solution

For this basic question about vector spaces

[delicate orchid]

yeah, actually, it could be - especially since you don't tend to learn about torsion at the same time vector spaces get introduced

[lethal dune]

why not Tor¹_k(V, _)=0

[oblique river]

No it absolutely is not lmao

[delicate orchid]

and what I cited is a pretty basic fact about torsion

it's like claiming using the fact that euclidean domains are PIDs is some outlandish specialist knowledge

[lethal dune]

at least not first course of module theory

[oblique river]

It is, if the problem is to “use the euclidean algorithm to prove that for any integers a, b there are integers x, y such that gcd(a,b) = ax + by”

[delicate orchid]

I just don't agree that this is a "basic question about vector spaces", sorry

[oblique river]

Youre right, its a basic question about modules

[delicate orchid]

yes, torsion specifically

[oblique river]

The question doesnt even use the phrase “vector space”

[delicate orchid]

and one of the first things I learnt about torsion was the inclusion ||Free > Projective > Torsion-Free|| (for integral domains, anyway)

[oblique river]

And i wouldnt be surprised if the facts “modules over a field are vector spaces” and “all vector spaces have a basis” have even been pointed out in class yet. (Well probably the first one has)

Ok you are clearly trolling if you say that one of the first things you learned about torsion involves projective modules lmao

[delicate orchid]

no it really did

this was the same course that introduced the tensor product via a universal property

[south patrol]

That is how it should be introduced right

[chilly ocean]

how else do you introduce it

explicit construction is unhandy

[delicate orchid]

I do agree with you looking back it was the best way to introduce it but when we didn't really know what a universal property was it was very confusing

[chilly ocean]

sounds like an introduction to category theory course in schools would do wonders

[delicate orchid]

I agree

but yeah, they then phrased projective and free modules using the same thing and my brain just kinda picked up on "le triangle" so it was fine from then on

[rotund aurora]

Let R be a ring, I, J ideals of R. In what cases do we have [R : IJ]=[R : I][R : J] ?

If I,J are coprime, then multiplicativity follows from CRT

[barren sierra]

I think so?

[charred bison]

somehow i cant find a proof of the impossibility of squaring the circle, can anyone point me to a textbook where i could reference it?

[void cosmos]

dummit and foote galois chapters

i got stuck at second line
how did he go from second to third
i understand it from there
this is problem 3.8

[chilly ocean]

#diff-geo-diff-top message

#diff-geo-diff-top message

[void cosmos]

so i had sum(sgn(phi) * phi(sum(a_j^1 * y^j) * sum(a_j^2y^j) * ..... *))

^ this is for problem 3.7

and i got stuck

so ur saying these sums can be written as one sum with this index notation

and this is the same as saying = sum(a_phi(j)^1*y^phi(j))

over all phi in S_n

[chilly ocean]

i am explaining how the sum over distinct i_1, ..., i_k from 1, ..., k is the same as a sum over elements of S_k in your first picture

[void cosmos]

yea so this works too right?

both of these problems use what u said so im trying to understand them both

[rustic crown]

[chilly ocean]

Let $G$ be a group. Call $H\subseteq G$ a $N$-embedded subgroup if for any normal subgroup $K\subseteq H$ of $H$ there is a normal subgroup $K_0\subseteq G$ of $G$ with $K = K_0\cap H$. What is known about $N$-embedded subgroups?

[cloud walrus]

Blitz

[chilly ocean]

@coral spindle ?

[coral spindle]

Looking now!

[chilly ocean]

btw I made the name up

[coral spindle]

So in particular, H is normal, right?

[chilly ocean]

yeah

this came up because I was wondering when subgroups of quotients are the same as quotients of subgroups

so in general the two shouldn't be the same

but I think that a quotient of an N-embedded subgroup is a subgroup of a quotient

[coral spindle]

Yeah that makes sense to me

I get this horrible feeling it's going to be central

[hot lake]

is there a subgroup of S3 that's not N-embedded

[coral spindle]

OK, so if we have a normal subgroup K of H, then G acts on K/H = (K' n H)/H. I'd love to apply a theorem here...

I'm wondering if we can follow around the action of G on K/H by following G/K'

I'm writing K' for K_0 because I'm just writing text

[chilly ocean]

besides the ones that aren't normal?

[hot lake]

well

if I pick H = {1, (12)}

it has 2 normal subgroups

{1}

which is the intersection of {1} with H and {1} is normal in G

and H

[chilly ocean]

I think I might have wrote the definition in an unclear way

[hot lake]

which is the intersection of G with H and G is normal in G

ah

[chilly ocean]

Let $G$ be a group. Call $H\subseteq G$ a $N$-embedded subgroup if for any normal subgroup $K\subseteq H$ of $H$ there is a normal subgroup $K_0\subseteq G$ of $G$ with $K = K_0\cap H$. What is known about $N$-embedded subgroups?

[cloud walrus]

Blitz

[chilly ocean]

this is what I meant

[hot lake]

I'm not sure what's the difference

[chilly ocean]

the subgroups K have to be normal in H as opposed to being normal in G

[coral spindle]

A normal subgroup of a normal subgroup is not necessarily going to be normal

[hot lake]

yeah ?

do you see any other normal subgroup of my H

that I didn't describe

[coral spindle]

OK

S_3 is a bad example

because it's too small

It seemed to me (and evidently to Blitz) that there was some confusion about the definition, not about H

[hot lake]

idk I'm just showing a counter example to your claim that H has to be normal in G ?

[coral spindle]

Ah I see what you're saying

[chilly ocean]

oh!

[coral spindle]

Yeah good point, I did get that wrong

So ok, we could continue with this definition? I think there are interesting things to say about it

(probably)

I think there might be some minimal subgroup G' of G such that H still has the property in G'

That might be useful

Lmao that's just K, nvm

I'm trying to get to a position where the map (normal subgroups of G) -> (normal subgroups of H) is not only surjective but injective

[cloud walrus]

Blitz

[chilly ocean]

that way we would have that subgroups of quotients are the same as quotients of N-embedded subgroups

K+H_0 should be KH_0, sorry

[chilly ocean]

Also I suggest modifications of this concept to characteristic and fully characteristic subgroups

Maybe call those C-embedded and FC-embedded

Maybe studying the latter two could be related to extension properties of homomorphisms

For instance if we can extend endomorphisms in H then it should be FC-embedded

And similarly if we can extend automorphisms then it should be C-embedded

Maybe discord is too small for thoughts like this

[sonic coral]

Can anyone give me a reminder on how to show a function is surjective. I know I should take an element out of the pre-image and show that there exists an element in the image that maps to it. Im just not sure how to do it for this problem.

[pastel cliff]

does this look right?

did i ask about this before

i mightve

[rustic crown]

using the words pre-image and image isn't nice... use domain and co-domain

image by definition is all the elements you get from something in the domain

[sonic coral]

ive heard it called both pairs but was never sure of the difference, thanks

[pastel cliff]

surjective means that any element in the domain has a preimage

you should be able to check that directly

[sonic coral]

yes like for all elements in the co-domain, i can find an element that maps to it

[pastel cliff]

bump to this tho

[sonic coral]

oh lol do i just pick g to be e

[rustic crown]

what are f_i and g_i?

oh co

i didn't read co

mb

[pastel cliff]

yeah

[rustic crown]

yee looks uwu

[pastel cliff]

what if i finish proofs with UWU instead of the lil square

[rustic crown]

[pastel cliff]

what would my 70 year old professor thing of that

[rustic crown]

[pastel cliff]

[chilly radish]

I used to have a smiley face instead of a halmos on my psets

[rustic crown]

one of my graders at a math program used to draw cats after every nice proof i did

and i once drew a in my analysis midterm

[pastel cliff]

do yk how to do this in latex by any chance

[sonic coral]

i saw somewhere of a girl using smiley faces as her independent variable in calculus

[rustic crown]

because the prof missed writing something like "take X to be non-empty"

i wrote "what if X = empty "

[chilly radish]

I don't like this because it's just confusing to follow when grading

[rustic crown]

true

[coral spindle]

I often just put a smile instead of a halmos without thinking, just bc I'm happy

:)

[chilly radish]

Grader and I shared cat pics through psets and feedback. I posted memes and cat pics on my psets and on the last pset he went "my turn" and put a bunch of pics of his own cat

[coral spindle]

I also put a smile when someone gets full marks and I'm grading

[chilly radish]

Had a guy do this to me but with Nordic runes instead

[coral spindle]

That kinda hurts

[rustic crown]

i drew eevee :3

[coral spindle]

Haha cute

[chilly radish]

He said I was the first person to comment on the runes

[coral spindle]

LOL

Someone start writing hanzi for unknowns

2何 - 1 = 2, 何 = ???

OK this is getting discussion-y

I'm still thinking about the N-embedded subgroups thing

I'll post here if I come up with anything

[chilly radish]

I'll just say that you most likely need some package, idr which

[sonic coral]

am i doing the surjective parts right? injective is much easier for me

[chilly ocean]

Why index 13

[agile burrow]

I think you might be a bit confused. For surjectivity, you want to show that every element in Hg has a preimage in H; that is, for every y in Hg, there exists an x in H such that f(x) = y. As you have it written, you're applying f to h_3 which doesn't even lie in H so it doesn't make sense to apply f to it

To show surjectivity, you might want to review the definition of Hg. What do elements in Hg look like?

[sonic coral]

isnt it just the right coset

so hg

[agile burrow]

Right, elements in Hg have the form hg

Now can you tell me an element x in H such that f(x) = hg?

[solar glacier]

Random thought but are there any infinite non Abelian groups idk why I can’t think of any

Unles u add some weird structure to R ?

Wait matrix groups

Right?

[agile burrow]

There are plenty - matrix groups, free groups, infinite dihedral and symmetric groups, etc.

[sonic coral]

would it be e

[agile burrow]

Not e

It's kind of staring you in the face in the way the problem is written

[sonic coral]

im looking for an element in H such that when i apply f to it i get hg?

[agile burrow]

Yes

[sonic coral]

g then

[agile burrow]

Not quite

[sonic coral]

its inverse, third trys the charm

[agile burrow]

It's h

The map f sends h to hg

[sonic coral]

yeah i dont get it

[agile burrow]

Ok so let's walk through how f is defined

I've fixed an element g in the group and I have a subgroup H, right?

[sonic coral]

yes

[agile burrow]

So now I have two sets, namely H and Hg, where Hg is the right coset

[sonic coral]

yep

[agile burrow]

My map f: H -> Hg is defined by taking an element h in H and mapping it to hg in Hg, so f(h) = hg

[sonic coral]

f sends H to Hg

yeah

[agile burrow]

So now you've managed to show that f is injective, which is very good

For surjectivity, I want to show that every element in Hg is in the image of f, right?

[sonic coral]

yeah thats the definition

[agile burrow]

If I have an arbitrary element y in Hg, then by the definition of right cosets, y = hg for some h in H, right?

[sonic coral]

yes

[agile burrow]

But then f(h) = hg = y, so y is in the image of f

Does that make any more sense?

[sonic coral]

not particularly, when i was doing this stuff with functions over the real numbers i was letting an element in the co-domain be arbitrary and then "undoing" the function to show if it was surjective or not

but that isnt going to work here

[agile burrow]

You can still "undo" the function here, if my guess as to what that means is correct

In particular, the fact that f is a bijection means that it has an inverse Hg -> H

Can you guess what that might be?

[sonic coral]

well i dont know its a bijection yet but yeah

all bijective functions have an inverse such that f^-1(f(x))=x

or the other way works too

[agile burrow]

right, so in particular one way you could show that f is a bijection is by constructing its inverse

so maybe a different approach here is for you to tell me what the inverse of f is in this case

[sonic coral]

maybe g^-1 h^-1

wait

[agile burrow]

not quite, remember that the inverse is a map Hg -> H. In particular, the image needs to lie in H

[sonic coral]

so like f(hg) = something just in h?

[agile burrow]

yeah

well maybe don't use f for the inverse function since we've already defined f: H -> Hg by f(h) = hg

[sonic coral]

yeah sure

im just trying to thing of what to compose in order to get h back

f^-1: Hg -> H by f^-1(hg)=g^-1 maybe

[agile burrow]

the issue is that g^-1 doesn't necessarily lie in H, g was just an arbitrary element in the group

[sonic coral]

is this way any easier than just guess and check

[agile burrow]

again, it's kind of just staring you in the face

[pastel cliff]

it's staring

[sonic coral]

h then like last time lol

[agile burrow]

yeah, exactly

the inverse maps hg to h

sorry if that wasn't particularly enlightening or illuminating, I'm not sure how to explain this further

[sonic coral]

theres not much depth to it so youre fine, i think im just not understanding the definitions properly

[agile burrow]

It could be worth reviewing definitions, yeah

[sonic coral]

so like f^-1: Hg -> H by f^-1(hg)=h implies that f(f^-1(hg))=f(h)=hg right

[agile burrow]

yeah, exactly. Just explicitly writing out the composition does the job

maybe another way to view f^-1 is as right-multiplication by g^-1, since hg * g^-1 = h

[agile burrow]

Recall that G acts on the set of 3-Sylow subgroups by conjugation and that this action is transitive (hence there is one orbit). Furthermore, the stabilizer of a sylow subgroup P under this action is N_G(P), the normalizer of P. Then by orbit-stabilizer, the |G|/|N_G(P)| is equal to the number of p-Sylow subgroups, but the former is just the index of N_G(P)

[pastel cliff]

@rustic crown can we talk about that problem from yesterday/earlier

showing that epimorphisms in Vect(R) are surjective

[rustic crown]

oh okie

[pastel cliff]

so suppose we have an epimorphism f in Vect(R)

R being reals

the defining characteristic of that epimorphism is how it interacts with two other maps g1 and g2 right

in order to bring this to the level of surjectivity then, i need to pick arbitrary g1 and g2 and show the g1 \circ f = g2 \circ f => g1 = g2 thing right

^^this is probably wrong but it's the first thing that comes to mind

i think we said the arbitrary part is wrong

bc in the example of Set we pick specific g1 and g2

but im not sure how using a specific map shows surjectivity for all epimorphisms

[rustic crown]

did you remember what we did in Set*

(pointed sets)

i said if X --> Y is a morphism, then we wanna find maps such that on the image they're the same

so idea was in one case we will kill everything

and in other we kill only the image

so we looked at the object Z = Y/(im f)

[pastel cliff]

is Set* anything in particular or just category of sets

[rustic crown]

can you try to do this for vector space?

[pastel cliff]

idk what the * is

[rustic crown]

a chosen base point

so that i don't have to deal with empty sets lmao

cuz this argument not very nice with that

objects are sets X with a distinguished element x

and a map (X, x) --> (Y, y) is a function X --> Y which sends x --> y

[pastel cliff]

this is the proof our prof gave us

[rustic crown]

this is a slightly different idea which would work exactly the same in Set*, Ab, R-mod, k-vect etc, cause all these are pointed :3

(more precisely i'm proving, surjective if and only if coker = 0, if you know what cokernel means)

[pastel cliff]

i cant get used to the term coker

or cok

coke addicts and genitals invading my notes

[warm wyvern]

This looks like Aluffi's typesetting

[pastel cliff]

naw it's prof notees

[pastel cliff]

but ok im gonna ignore this then

[rustic crown]

you could try to give a proof like that, but then would require you to choose basis and that's not uwu (when we don't need to)

[pastel cliff]

i think this is what im stuck on

that is we wanna find g1, g2 : Y -> Z right

[rustic crown]

yep

so first question is what to pick Z

[pastel cliff]

we wanna pick a vector space Z such that there are two maps from Y - > Z that "agree" with our given map X -> Y

[rustic crown]

yee

"if you don't know what to do then kill"
-Sun Tzu

[pastel cliff]

well what's stopping me from just being like

R^2

because X and Y stay arbitrary if im not mistaken

[rustic crown]

(btw if you like this idea more, we can do this too. that's what you were thinking yesterday as well)

[chilly ocean]

too flashy